Chapter 12: Thermodynamics

Introduction

Thermodynamics Overview

• Study of heat, temperature, and energy conversion.
• Macroscopic science: deals with bulk systems, not molecular details.
• State variables: pressure (P), volume (V), temperature (T), internal energy (U), entropy (S).
• Focus on internal state changes, not motion of the system as a whole.
• Example: Kinetic energy of a bullet converts to heat on impact, raising temperature.
• Temperature relates to internal disordered motion energy, not bulk motion.

Solved Examples

• Ex 1: Calculate heat to raise 2 kg water by 10°C, s = 4200 J/kg°C.
  • Q = msΔT = 2 × 4200 × 10 = 84000 J
• Ex 2: Find molar specific heat if heat capacity H = 60 J/°C and molar mass μ = 20 g/mol.
  • C = H/μ = 60/20 = 3 J/mol°C

Ex 1

• A bullet of mass 0.01 kg moving at 300 m/s stops in wood. Calculate kinetic energy converted to heat.
  • Formula: KE = 1/2mv²
  • Calculation: KE = 0.5 × 0.01 × 300² = 450 J

Ex 1

• A bullet of mass 0.01 kg moving at 300 m/s stops in wood. Calculate kinetic energy converted to heat.
  • Formula: KE = 1/2mv²
  • Calculation: KE = 0.5 × 0.01 × 300² = 450 J

Practice Set

• Q1: Thermodynamics studies
  • A) Molecular motion
  • B) Bulk system heat and energy
  • C) Particle forces
  • D) Electric currents
• Q2: Which is NOT a thermodynamic state variable?
  • A) Pressure
  • B) Volume
  • C) Velocity of system
  • D) Temperature
• Q3: When a bullet stops in wood, kinetic energy converts to
  • A) Mechanical energy
  • B) Heat energy
  • C) Potential energy
  • D) Electrical energy

Answer Key

• Q1: B - Thermodynamics deals with heat and energy in bulk systems.
• Q2: C - Velocity of the system is not a thermodynamic state variable.
• Q3: B - Kinetic energy converts into heat energy raising temperature.

Memory Aids

• Thermodynamics = Heat + Energy conversion in bulk systems.
• State variables: P, V, T, U, S.
• Temperature = internal motion energy, not bulk motion.

Thermodynamics Overview

• Study of heat, temperature, and energy conversion.
• Macroscopic science: deals with bulk systems, not molecular details.
• State variables: pressure (P), volume (V), temperature (T), internal energy (U), entropy (S).
• Focus on internal state changes, not motion of the system as a whole.
• Example: Kinetic energy of a bullet converts to heat on impact, raising temperature.
• Temperature relates to internal disordered motion energy, not bulk motion.

Solved Examples

Ex 1

• A bullet of mass 0.01 kg moving at 300 m/s stops in wood. Calculate kinetic energy converted to heat.
  • Formula: KE = 1/2mv²
  • Calculation: KE = 0.5 × 0.01 × 300² = 450 J

Practice Set

• Q1: Thermodynamics studies
  • A) Molecular motion
  • B) Bulk system heat and energy
  • C) Particle forces
  • D) Electric currents
• Q2: Which is NOT a thermodynamic state variable?
  • A) Pressure
  • B) Volume
  • C) Velocity of system
  • D) Temperature
• Q3: When a bullet stops in wood, kinetic energy converts to
  • A) Mechanical energy
  • B) Heat energy
  • C) Potential energy
  • D) Electrical energy

Answer Key

• Q1: B - Thermodynamics deals with heat and energy in bulk systems.
• Q2: C - Velocity of the system is not a thermodynamic state variable.
• Q3: B - Kinetic energy converts into heat energy raising temperature.

Memory Aids

• Thermodynamics = Heat + Energy conversion in bulk systems.
• State variables: P, V, T, U, S.
• Temperature = internal motion energy, not bulk motion.

Thermodynamics Overview

• Study of heat, temperature, and energy conversion.
• Macroscopic science: deals with bulk systems, not molecular details.
• State variables: pressure (P), volume (V), temperature (T), internal energy (U), entropy (S).
• Focus on internal state changes, not motion of the system as a whole.
• Example: Kinetic energy of a bullet converts to heat on impact, raising temperature.
• Temperature relates to internal disordered motion energy, not bulk motion.

Solved Examples

• Ex 1: Calculate change in internal energy
  • Use first law: ΔU = Q - W
  • Given heat added (Q) and work done by system (W)
  • Calculate ΔU in Joules.

Ex 1

• A bullet of mass 0.01 kg moving at 300 m/s stops in wood. Calculate kinetic energy converted to heat.
  • Formula: KE = 1/2mv²
  • Calculation: KE = 0.5 × 0.01 × 300² = 450 J

Ex 1

• A bullet of mass 0.01 kg moving at 300 m/s stops in wood. Calculate kinetic energy converted to heat.
  • Formula: KE = 1/2mv²
  • Calculation: KE = 0.5 × 0.01 × 300² = 450 J

Practice Set

• Q1: Thermodynamics studies
  • A) Molecular motion
  • B) Bulk system heat and energy
  • C) Particle forces
  • D) Electric currents
• Q2: Which is NOT a thermodynamic state variable?
  • A) Pressure
  • B) Volume
  • C) Velocity of system
  • D) Temperature
• Q3: When a bullet stops in wood, kinetic energy converts to
  • A) Mechanical energy
  • B) Heat energy
  • C) Potential energy
  • D) Electrical energy

Answer Key

• Q1: B - Thermodynamics deals with heat and energy in bulk systems.
• Q2: C - Velocity of the system is not a thermodynamic state variable.
• Q3: B - Kinetic energy converts into heat energy raising temperature.

Memory Aids

• Thermodynamics = Heat + Energy conversion in bulk systems.
• State variables: P, V, T, U, S.
• Temperature = internal motion energy, not bulk motion.

Thermodynamics Overview

• Study of heat, temperature, and energy conversion.
• Macroscopic science: deals with bulk systems, not molecular details.
• State variables: pressure (P), volume (V), temperature (T), internal energy (U), entropy (S).
• Focus on internal state changes, not motion of the system as a whole.
• Example: Kinetic energy of a bullet converts to heat on impact, raising temperature.
• Temperature relates to internal disordered motion energy, not bulk motion.

Solved Examples

Ex 1

• A bullet of mass 0.01 kg moving at 300 m/s stops in wood. Calculate kinetic energy converted to heat.
  • Formula: \( KE = \frac{1}{2}mv^2 \)
  • Calculation: \( KE = 0.5 \times 0.01 \times 300^2 = 450 \text{ J} \)

Practice Set

• Q1: Thermodynamics studies
  • A) Molecular motion
  • B) Bulk system heat and energy
  • C) Particle forces
  • D) Electric currents
• Q2: Which is NOT a thermodynamic state variable?
  • A) Pressure
  • B) Volume
  • C) Velocity of system
  • D) Temperature
• Q3: When a bullet stops in wood, kinetic energy converts to
  • A) Mechanical energy
  • B) Heat energy
  • C) Potential energy
  • D) Electrical energy

Answer Key

• Q1: B - Thermodynamics deals with heat and energy in bulk systems.
• Q2: C - Velocity of the system is not a thermodynamic state variable.
• Q3: B - Kinetic energy converts into heat energy raising temperature.

Memory Aids

• Thermodynamics = Heat + Energy conversion in bulk systems.
• State variables: P, V, T, U, S.
• Temperature = internal motion energy, not bulk motion.

Thermal Equilibrium

Definition and Conditions

• Thermodynamic equilibrium: Macroscopic variables (P, T, V, mass, composition) constant over time.
• Thermal equilibrium: No net heat flow between systems in thermal contact.

Solved Examples

Ex 1

• Two bodies at temperatures 300 K and 300 K are brought into contact.
  • Heat flow? None, since temperatures are equal.

Practice Set

• Q1: Thermodynamic equilibrium means
  • A) Variables change with time
  • B) Variables constant with time
  • C) Heat flows continuously
  • D) System moves as a whole
• Q2: Two systems in thermal equilibrium have
  • A) Different temperatures
  • B) No net heat flow
  • C) Different pressures
  • D) Different volumes
• Q3: If heat flows between two bodies, they are
  • A) In thermal equilibrium
  • B) Not in thermal equilibrium
  • C) At same temperature
  • D) Isolated systems

Answer Key

• Q1: B - Equilibrium means variables do not change with time.
• Q2: B - Thermal equilibrium means no net heat flow.
• Q3: B - Heat flow indicates systems are not in thermal equilibrium.

Memory Aids

• Equilibrium = steady state, no change in variables.
• Thermal equilibrium = equal temperature, no heat flow.

Definition and Conditions

• Thermodynamic equilibrium: Macroscopic variables (P, T, V, mass, composition) constant over time.
• Thermal equilibrium: No net heat flow between systems in thermal contact.

Solved Examples

Ex 1

• Two bodies at temperatures 300 K and 300 K are brought into contact.
  • Heat flow? None, since temperatures are equal.

Practice Set

• Q1: Thermodynamic equilibrium means
  • A) Variables change with time
  • B) Variables constant with time
  • C) Heat flows continuously
  • D) System moves as a whole
• Q2: Two systems in thermal equilibrium have
  • A) Different temperatures
  • B) No net heat flow
  • C) Different pressures
  • D) Different volumes
• Q3: If heat flows between two bodies, they are
  • A) In thermal equilibrium
  • B) Not in thermal equilibrium
  • C) At same temperature
  • D) Isolated systems

Answer Key

• Q1: B - Equilibrium means variables do not change with time.
• Q2: B - Thermal equilibrium means no net heat flow.
• Q3: B - Heat flow indicates systems are not in thermal equilibrium.

Memory Aids

• Equilibrium = steady state, no change in variables.
• Thermal equilibrium = equal temperature, no heat flow.

Zeroth Law of Thermodynamics

Fundamental Concept

• States: If two systems (P and Q) are each in thermal equilibrium with a third system (R), then P and Q are in thermal equilibrium with each other.
• Establishes the concept of temperature as a measurable and transitive property.
• Thermal equilibrium means no net heat flow between systems.

Thermal Equilibrium Setup

• System consists of three regions: P, Q, and R.
• Initially, P and Q reach thermal equilibrium with R through conducting walls.
• Replacing the adiabatic wall between P and Q with a conducting wall shows P and Q remain in thermal equilibrium.
• Adiabatic wall: no heat transfer; Conducting wall: allows heat transfer.

Solved Examples

• Ex 1: If system P is in thermal equilibrium with R at 300 K, and Q is also in thermal equilibrium with R, what is the temperature of Q?
  • Since P and Q are in thermal equilibrium with R, by zeroth law, P and Q are in thermal equilibrium.
  • Therefore, temperature of Q = temperature of P = 300 K.

Practice Set

• Q1: What does the zeroth law of thermodynamics establish?
  • A) Conservation of energy
  • B) Concept of temperature
  • C) Entropy increase
  • D) Heat capacity
• Q2: If system P is in thermal equilibrium with R at 350 K, and Q is also in thermal equilibrium with R, what is the temperature of Q?
  • A) 300 K
  • B) 350 K
  • C) 400 K
  • D) Cannot be determined
• Q3: In the given system, if the wall between P and Q is adiabatic and both are in equilibrium with R, what happens when the wall is replaced by a conducting wall?
  • A) P and Q will exchange heat and reach different temperatures
  • B) P and Q remain in thermal equilibrium
  • C) Heat flows from R to P and Q
  • D) System loses thermal equilibrium

Answer Key

• Q1: B - Zeroth law defines temperature as a transitive property through thermal equilibrium.
• Q2: B - By zeroth law, Q must have the same temperature as P and R, i.e., 350 K.
• Q3: B - Conducting wall allows heat transfer, confirming P and Q remain in thermal equilibrium.

Memory Aids

• "Zeroth law links temperatures via a third system."
• Thermal equilibrium is transitive: if A = C and B = C, then A = B.
• Adiabatic wall = no heat transfer; Conducting wall = heat transfer allowed.

Fundamental Concept

• States: If two systems (P and Q) are each in thermal equilibrium with a third system (R), then P and Q are in thermal equilibrium with each other.
• Establishes the concept of temperature as a measurable and transitive property.
• Thermal equilibrium means no net heat flow between systems.

Thermal Equilibrium Setup

• System consists of three regions: P, Q, and R.
• Initially, P and Q reach thermal equilibrium with R through conducting walls.
• Replacing the adiabatic wall between P and Q with a conducting wall shows P and Q remain in thermal equilibrium.
• Adiabatic wall: no heat transfer; Conducting wall: allows heat transfer.

Solved Examples

• Ex 1: If system P is in thermal equilibrium with R at 300 K, and Q is also in thermal equilibrium with R, what is the temperature of Q?
  • Since P and Q are in thermal equilibrium with R, by zeroth law, P and Q are in thermal equilibrium.
  • Therefore, temperature of Q = temperature of P = 300 K.

Practice Set

• Q1: What does the zeroth law of thermodynamics establish?
  • A) Conservation of energy
  • B) Concept of temperature
  • C) Entropy increase
  • D) Heat capacity
• Q2: If system P is in thermal equilibrium with R at 350 K, and Q is also in thermal equilibrium with R, what is the temperature of Q?
  • A) 300 K
  • B) 350 K
  • C) 400 K
  • D) Cannot be determined
• Q3: In the given system, if the wall between P and Q is adiabatic and both are in equilibrium with R, what happens when the wall is replaced by a conducting wall?
  • A) P and Q will exchange heat and reach different temperatures
  • B) P and Q remain in thermal equilibrium
  • C) Heat flows from R to P and Q
  • D) System loses thermal equilibrium

Answer Key

• Q1: B - Zeroth law defines temperature as a transitive property through thermal equilibrium.
• Q2: B - By zeroth law, Q must have the same temperature as P and R, i.e., 350 K.
• Q3: B - Conducting wall allows heat transfer, confirming P and Q remain in thermal equilibrium.

Memory Aids

• "Zeroth law links temperatures via a third system."
• Thermal equilibrium is transitive: if A = C and B = C, then A = B.
• Adiabatic wall = no heat transfer; Conducting wall = heat transfer allowed.

Heat

Definition and Transfer

• Heat is a form of energy transferred due to temperature difference.
• Flows from higher to lower temperature regions.
• SI unit: Joule (J).

Solved Examples

• Ex 1: Calculate heat transferred when temperature difference exists.
  • Use formula: Q = m c ΔT
  • Given mass (m), specific heat (c), and temperature change (ΔT)
  • Calculate Q in Joules.

Solved Examples

• Ex 1: Calculate heat transferred when temperature difference exists.
  • Use formula: Q = m c ΔT
  • Given mass (m), specific heat (c), and temperature change (ΔT)
  • Calculate Q in Joules.

Solved Examples

• Ex 1: Calculate heat transferred when temperature difference exists.
  • Use formula: Q = m c ΔT
  • Given mass (m), specific heat (c), and temperature change (ΔT)
  • Calculate Q in Joules.

Practice Set

• Q1: Heat transfer occurs due to
  • A) Pressure difference
  • B) Temperature difference
  • C) Volume difference
  • D) Mass difference
• Q2: Heat is measured in
  • A) Newton
  • B) Joule
  • C) Watt
  • D) Pascal
• Q3: Heat flows from
  • A) Low to high temperature
  • B) High to low temperature
  • C) Equal temperatures
  • D) None of these

Answer Key

• Q1: B - Heat transfer is due to temperature difference.
• Q2: B - Heat is energy, unit is Joule.
• Q3: B - Heat flows from high to low temperature.

Memory Aids

• Heat = energy transfer by temperature difference.
• Unit Joule (J) like work and energy.
• Heat flows from hot to cold.

Definition and Transfer

• Heat is a form of energy transferred due to temperature difference.
• Flows from higher to lower temperature regions.
• SI unit: Joule (J).

Solved Examples

• Ex 1: Calculate heat transferred when temperature difference exists.
  • Use formula: Q = m c \(\Delta T\)
  • Given mass (m), specific heat (c), and temperature change (\(\Delta T\))
  • Calculate Q in Joules.

Practice Set

• Q1: Heat transfer occurs due to
  • A) Pressure difference
  • B) Temperature difference
  • C) Volume difference
  • D) Mass difference
• Q2: Heat is measured in
  • A) Newton
  • B) Joule
  • C) Watt
  • D) Pascal
• Q3: Heat flows from
  • A) Low to high temperature
  • B) High to low temperature
  • C) Equal temperatures
  • D) None of these

Answer Key

• Q1: B - Heat transfer is due to temperature difference.
• Q2: B - Heat is energy, unit is Joule.
• Q3: B - Heat flows from high to low temperature.

Memory Aids

• Heat = energy transfer by temperature difference.
• Unit Joule (J) like work and energy.
• Heat flows from hot to cold.

Internal Energy

Definition and Properties

• Sum of kinetic and potential energies of molecules inside a system.
• Excludes overall kinetic energy of the system as a whole.
• For ideal gas, depends only on temperature.
• For real gas, depends on temperature and volume.
• Includes molecular motions: translational, rotational, vibrational.

Solved Examples

• Ex 1: Internal energy change for ideal gas
  • Use ΔU = n C_V ΔT
  • Given moles (n), molar heat capacity at constant volume (C_V), and temperature change (ΔT)
  • Calculate ΔU in Joules.

Practice Set

• Q1: Internal energy excludes
  • A) Molecular kinetic energy
  • B) Molecular potential energy
  • C) Overall kinetic energy of system
  • D) Vibrational energy
• Q2: For ideal gas, internal energy depends on
  • A) Volume only
  • B) Temperature only
  • C) Pressure only
  • D) Temperature and volume
• Q3: Internal energy includes
  • A) Translational motion energy
  • B) Rotational motion energy
  • C) Vibrational motion energy
  • D) All of the above

Answer Key

• Q1: C - Overall kinetic energy of system is excluded.
• Q2: B - Ideal gas internal energy depends only on temperature.
• Q3: D - Includes all molecular motion energies.

Memory Aids

• Internal energy = molecular kinetic + potential energies.
• Ideal gas: depends only on temperature.
• Exclude system's bulk kinetic energy.

Definition and Properties

• Sum of kinetic and potential energies of molecules inside a system.
• Excludes overall kinetic energy of the system as a whole.
• For ideal gas, depends only on temperature.
• For real gas, depends on temperature and volume.
• Includes molecular motions: translational, rotational, vibrational.

Solved Examples

• Ex 1: Internal energy change for ideal gas
  • Use \( \Delta U = n C_V \Delta T \)
  • Given moles (n), molar heat capacity at constant volume (\(C_V\)), and temperature change (\(\Delta T\))
  • Calculate \(\Delta U\) in Joules.

Practice Set

• Q1: Internal energy excludes
  • A) Molecular kinetic energy
  • B) Molecular potential energy
  • C) Overall kinetic energy of system
  • D) Vibrational energy
• Q2: For ideal gas, internal energy depends on
  • A) Volume only
  • B) Temperature only
  • C) Pressure only
  • D) Temperature and volume
• Q3: Internal energy includes
  • A) Translational motion energy
  • B) Rotational motion energy
  • C) Vibrational motion energy
  • D) All of the above

Answer Key

• Q1: C - Overall kinetic energy of system is excluded.
• Q2: B - Ideal gas internal energy depends only on temperature.
• Q3: D - Includes all molecular motion energies.

Memory Aids

• Internal energy = molecular kinetic + potential energies.
• Ideal gas: depends only on temperature.
• Exclude system's bulk kinetic energy.

Work

Definition and Modes

• Work is energy transfer by mechanical means.
• Examples: moving piston, raising/lowering weights.
• SI unit: Joule (J).

Solved Examples

• Ex 1: Work done by gas expanding
  • Use formula: W = P ΔV
  • Given pressure (P) and volume change (ΔV)
  • Calculate work in Joules.

Practice Set

• Q1: Work is energy transfer by
  • A) Heat only
  • B) Mechanical means
  • C) Radiation
  • D) Chemical reaction
• Q2: Work done by gas is given by
  • A) P ΔV
  • B) m c ΔT
  • C) F d
  • D) m g h
• Q3: Unit of work is
  • A) Watt
  • B) Newton
  • C) Joule
  • D) Pascal

Answer Key

• Q1: B - Work is mechanical energy transfer.
• Q2: A - Work done by gas = pressure × volume change.
• Q3: C - Work unit is Joule.

Memory Aids

• Work = energy transfer by mechanical action.
• Formula: W = P ΔV for gases.
• Unit Joule same as energy and heat.

Definition and Modes

• Work is energy transfer by mechanical means.
• Examples: moving piston, raising/lowering weights.
• SI unit: Joule (J).

Solved Examples

• Ex 1: Work done by gas expanding
  • Use formula: \( W = P \Delta V \)
  • Given pressure (P) and volume change (\(\Delta V\))
  • Calculate work in Joules.

Practice Set

• Q1: Work is energy transfer by
  • A) Heat only
  • B) Mechanical means
  • C) Radiation
  • D) Chemical reaction
• Q2: Work done by gas is given by
  • A) \( P \Delta V \)
  • B) \( m c \Delta T \)
  • C) \( F d \)
  • D) \( m g h \)
• Q3: Unit of work is
  • A) Watt
  • B) Newton
  • C) Joule
  • D) Pascal

Answer Key

• Q1: B - Work is mechanical energy transfer.
• Q2: A - Work done by gas = pressure × volume change.
• Q3: C - Work unit is Joule.

Memory Aids

• Work = energy transfer by mechanical action.
• Formula: \( W = P \Delta V \) for gases.
• Unit Joule same as energy and heat.

Heat and Work Energy Transfer

Key Concept

• Heat and work are distinct modes of energy transfer.
• Both cause change in internal energy of a system.
• Heat: transfer due to temperature difference.
• Work: transfer by mechanical means.

Solved Examples

• Ex 1: Calculate change in internal energy
  • Use first law: ΔU = Q - W
  • Given heat added (Q) and work done by system (W)
  • Calculate ΔU in Joules.

Practice Set

• Q1: Heat and work are
  • A) Same mode of energy transfer
  • B) Distinct modes of energy transfer
  • C) Forms of potential energy
  • D) Forms of kinetic energy
• Q2: Change in internal energy depends on
  • A) Heat only
  • B) Work only
  • C) Heat and work
  • D) Neither heat nor work
• Q3: First law of thermodynamics is
  • A) ΔU = Q + W
  • B) ΔU = Q - W
  • C) ΔU = W - Q
  • D) ΔU = 0

Answer Key

• Q1: B - Heat and work are distinct energy transfer modes.
• Q2: C - Internal energy change depends on both heat and work.
• Q3: B - First law: ΔU = Q - W.

Memory Aids

• Heat and work change internal energy.
• First law: ΔU = Q - W.
• Heat by temperature difference, work by mechanical means.

Key Concept

• Heat and work are distinct modes of energy transfer.
• Both cause change in internal energy of a system.
• Heat: transfer due to temperature difference.
• Work: transfer by mechanical means.

Solved Examples

• Ex 1: Calculate change in internal energy
  • Use first law: \( \Delta U = Q - W \)
  • Given heat added (Q) and work done by system (W)
  • Calculate \( \Delta U \) in Joules.

Practice Set

• Q1: Heat and work are
  • A) Same mode of energy transfer
  • B) Distinct modes of energy transfer
  • C) Forms of potential energy
  • D) Forms of kinetic energy
• Q2: Change in internal energy depends on
  • A) Heat only
  • B) Work only
  • C) Heat and work
  • D) Neither heat nor work
• Q3: First law of thermodynamics is
  • A) \( \Delta U = Q + W \)
  • B) \( \Delta U = Q - W \)
  • C) \( \Delta U = W - Q \)
  • D) \( \Delta U = 0 \)

Answer Key

• Q1: B - Heat and work are distinct energy transfer modes.
• Q2: C - Internal energy change depends on both heat and work.
• Q3: B - First law: \( \Delta U = Q - W \).

Memory Aids

• Heat and work change internal energy.
• First law: \( \Delta U = Q - W \).
• Heat by temperature difference, work by mechanical means.

First Law of Thermodynamics

Concept and Formula

• Energy conservation applied to thermodynamic systems.
• Energy supplied as heat (ΔQ) partly increases internal energy (ΔU) and partly does work (ΔW).
• Mathematical form: ΔQ = ΔU + ΔW.
• Rearranged: ΔQ - ΔW = ΔU.
• ΔQ and ΔW depend on the path; ΔQ - ΔW is path independent.
• For ΔU = 0, ΔQ = ΔW.
• Work done at constant pressure: ΔW = PΔV so ΔQ = ΔU + PΔV.

Solved Examples

• Ex 1: Heat supplied ΔQ = 500 J, work done ΔW = 200 J. Find ΔU.
  • Use ΔU = ΔQ - ΔW = 500 - 200 = 300 J.
• Ex 2: Gas expands at constant pressure P = 2 × 10⁵ Pa, volume change ΔV = 0.001 m³, ΔU = 0. Find heat supplied ΔQ.
  • Work done ΔW = PΔV = 2 × 10⁵ × 0.001 = 200 J.
  • Since ΔU = 0, ΔQ = ΔW = 200 J.

Solved Examples

• Ex 1: Heat supplied ΔQ = 500 J, work done ΔW = 200 J. Find ΔU.
  • Use ΔU = ΔQ - ΔW = 500 - 200 = 300 J.
• Ex 2: Gas expands at constant pressure P = 2 × 10⁵ Pa, volume change ΔV = 0.001 m³, ΔU = 0. Find heat supplied ΔQ.
  • Work done ΔW = PΔV = 2 × 10⁵ × 0.001 = 200 J.
  • Since ΔU = 0, ΔQ = ΔW = 200 J.

Practice Set

• Q1: If ΔQ = 400 J and ΔU = 150 J, work done by the system is:
  • A) 550 J
  • B) 250 J
  • C) 150 J
  • D) 400 J
• Q2: For an isothermal process, ΔU = 0. If the gas does 300 J work, heat supplied is:
  • A) 0 J
  • B) 300 J
  • C) -300 J
  • D) 600 J
• Q3: A gas absorbs 500 J heat, internal energy increases by 200 J. Work done by the gas is:
  • A) 700 J
  • B) 300 J
  • C) -300 J
  • D) 200 J

Answer Key

• Q1: B) 250 J — Using ΔW = ΔQ - ΔU = 400 - 150 = 250 J.
• Q2: B) 300 J — For isothermal, ΔU=0, so ΔQ = ΔW = 300 J.
• Q3: B) 300 J — Work done = ΔQ - ΔU = 500 - 200 = 300 J.

Memory Aids

• First law: Heat in = Increase in internal energy + Work done.
• Formula: ΔQ = ΔU + ΔW.
• Work done by gas (expansion) is positive; on gas (compression) is negative.
• Heat absorbed positive; heat released negative.

Concept and Formula

• Energy conservation applied to thermodynamic systems.
• Energy supplied as heat (\(\Delta Q\)) partly increases internal energy (\(\Delta U\)) and partly does work (\(\Delta W\)).
• Mathematical form: \(\Delta Q = \Delta U + \Delta W\).
• Rearranged: \(\Delta Q - \Delta W = \Delta U\).
• \(\Delta Q\) and \(\Delta W\) depend on the path; \(\Delta Q - \Delta W\) is path independent.
• For \(\Delta U = 0\), \(\Delta Q = \Delta W\).
• Work done at constant pressure: \(\Delta W = P\Delta V\) so \(\Delta Q = \Delta U + P\Delta V\).

Solved Examples

• Ex 1: Heat supplied \(\Delta Q = 500\,J\), work done \(\Delta W = 200\,J\). Find \(\Delta U\).
  • Use \(\Delta U = \Delta Q - \Delta W = 500 - 200 = 300\,J\).
• Ex 2: Gas expands at constant pressure \(P = 2 \times 10^5\,Pa\), volume change \(\Delta V = 0.001\,m^3\), \(\Delta U = 0\). Find heat supplied \(\Delta Q\).
  • Work done \(\Delta W = P\Delta V = 2 \times 10^5 \times 0.001 = 200\,J\).
  • Since \(\Delta U = 0\), \(\Delta Q = \Delta W = 200\,J\).

Practice Set

• Q1: If \(\Delta Q = 400\,J\) and \(\Delta U = 150\,J\), work done by the system is:
  • A) 550 J
  • B) 250 J
  • C) 150 J
  • D) 400 J
• Q2: For an isothermal process, \(\Delta U = 0\). If the gas does 300 J work, heat supplied is:
  • A) 0 J
  • B) 300 J
  • C) -300 J
  • D) 600 J
• Q3: A gas absorbs 500 J heat, internal energy increases by 200 J. Work done by the gas is:
  • A) 700 J
  • B) 300 J
  • C) -300 J
  • D) 200 J

Answer Key

• Q1: B) 250 J — Using \(\Delta W = \Delta Q - \Delta U = 400 - 150 = 250\,J\).
• Q2: B) 300 J — For isothermal, \(\Delta U=0\), so \(\Delta Q = \Delta W = 300\,J\).
• Q3: B) 300 J — Work done \(= \Delta Q - \Delta U = 500 - 200 = 300\,J\).

Memory Aids

• First law: Heat in = Increase in internal energy + Work done.
• Formula: \(\Delta Q = \Delta U + \Delta W\).
• Work done by gas (expansion) is positive; on gas (compression) is negative.
• Heat absorbed positive; heat released negative.

Sign Conventions Used in Thermodynamics

Key Points

• Heat absorbed by system: positive ΔQ > 0.
• Heat rejected by system: negative ΔQ < 0.
• Internal energy increases with temperature rise: ΔU > 0.
• Internal energy decreases with temperature fall: ΔU < 0.
• Work done by gas during expansion: positive ΔW > 0.
• Work done on gas during compression: negative ΔW < 0.

Solved Examples

• Ex 1: Gas absorbs 100 J heat and expands doing 60 J work. Find ΔU.
  • ΔU = ΔQ - ΔW = 100 - 60 = 40 J (positive).
• Ex 2: Gas compressed with 50 J work done on it, releasing 20 J heat. Find ΔU.
  • Work done by system ΔW = -50 J, heat supplied ΔQ = -20 J.
  • ΔU = ΔQ - ΔW = -20 - (-50) = 30 J (positive).

Solved Examples

• Ex 1: Gas absorbs 100 J heat and expands doing 60 J work. Find ΔU.
  • ΔU = ΔQ - ΔW = 100 - 60 = 40 J (positive).
• Ex 2: Gas compressed with 50 J work done on it, releasing 20 J heat. Find ΔU.
  • Work done by system ΔW = -50 J, heat supplied ΔQ = -20 J.
  • ΔU = ΔQ - ΔW = -20 - (-50) = 30 J (positive).

Solved Examples

• Ex 1: Gas absorbs 100 J heat and expands doing 60 J work. Find ΔU.
  • ΔU = ΔQ - ΔW = 100 - 60 = 40 J (positive).
• Ex 2: Gas compressed with 50 J work done on it, releasing 20 J heat. Find ΔU.
  • Work done by system ΔW = -50 J, heat supplied ΔQ = -20 J.
  • ΔU = ΔQ - ΔW = -20 - (-50) = 30 J (positive).

Practice Set

• Q1: Heat rejected by system is:
  • A) Positive
  • B) Negative
  • C) Zero
  • D) Undefined
• Q2: When gas expands, work done is:
  • A) Negative
  • B) Positive
  • C) Zero
  • D) Depends on temperature
• Q3: If temperature falls, internal energy change is:
  • A) Positive
  • B) Negative
  • C) Zero
  • D) Cannot say

Answer Key

• Q1: B) Negative — Heat rejected means heat leaves system, negative sign.
• Q2: B) Positive — Expansion means work done by system, positive.
• Q3: B) Negative — Temperature fall decreases internal energy.

Memory Aids

• Heat in (+), heat out (-).
• Expansion work (+), compression work (-).
• Temp rise (+ΔU), temp fall (-ΔU).

Key Points

• Heat absorbed by system: positive \(\Delta Q > 0\).
• Heat rejected by system: negative \(\Delta Q < 0\).
• Internal energy increases with temperature rise: \(\Delta U > 0\).
• Internal energy decreases with temperature fall: \(\Delta U < 0\).
• Work done by gas during expansion: positive \(\Delta W > 0\).
• Work done on gas during compression: negative \(\Delta W < 0\).

Solved Examples

• Ex 1: Gas absorbs 100 J heat and expands doing 60 J work. Find \(\Delta U\).
  • \(\Delta U = \Delta Q - \Delta W = 100 - 60 = 40\,J\) (positive).
• Ex 2: Gas compressed with 50 J work done on it, releasing 20 J heat. Find \(\Delta U\).
  • Work done by system \(\Delta W = -50\,J\), heat supplied \(\Delta Q = -20\,J\).
  • \(\Delta U = \Delta Q - \Delta W = -20 - (-50) = 30\,J\) (positive).

Practice Set

• Q1: Heat rejected by system is:
  • A) Positive
  • B) Negative
  • C) Zero
  • D) Undefined
• Q2: When gas expands, work done is:
  • A) Negative
  • B) Positive
  • C) Zero
  • D) Depends on temperature
• Q3: If temperature falls, internal energy change is:
  • A) Positive
  • B) Negative
  • C) Zero
  • D) Cannot say

Answer Key

• Q1: B) Negative — Heat rejected means heat leaves system, negative sign.
• Q2: B) Positive — Expansion means work done by system, positive.
• Q3: B) Negative — Temperature fall decreases internal energy.

Memory Aids

• Heat in (+), heat out (-).
• Expansion work (+), compression work (-).
• Temp rise (+\(\Delta U\)), temp fall (-\(\Delta U\)).

Specific Heat Capacity

Definition and Formula

• Heat absorbed or evolved per unit mass for 1°C temperature change.
• Formula: Q = msΔT = ms(T_f - T_i)
• Specific heat capacity: s = Q/(mΔT)

Molar Specific Heat for Solid

• Heat required to raise 1 mole temperature by 1°C.
• Formula: C = H/μ = (1/μ) ΔQ/ΔT, where H = heat capacity.
• For N atoms vibrating, average energy per oscillator in 3D: 3k_B T.
• Total energy per mole: U = 3k_B T × N_A = 3RT.
• Molar specific heat: C = ΔQ/ΔT = ΔU/ΔT = 3R (valid at room temperature, except carbon).

Specific Heat Capacity of Water

• Varies slightly with temperature.
• Graph shows U-shaped curve: decreases till ~40°C, then increases.
• One calorie = heat to raise 1 g water from 14.5°C to 15.5°C.

Solved Examples

• Ex 1: Calculate heat to raise 2 kg water by 10°C, s = 4200 J/kg°C.
  • Q = msΔT = 2 × 4200 × 10 = 84000 J
• Ex 2: Find molar specific heat if heat capacity H = 60 J/°C and molar mass μ = 20 g/mol.
  • C = H/μ = 60/20 = 3 J/mol°C

Molar Specific Heat for Solid

• Heat required to raise 1 mole temperature by 1°C.
• Formula: C = H/μ = (1/μ) ΔQ/ΔT, where H = heat capacity.
• For N atoms vibrating, average energy per oscillator in 3D: 3k_B T.
• Total energy per mole: U = 3k_B T × N_A = 3RT.
• Molar specific heat: C = ΔQ/ΔT = ΔU/ΔT = 3R (valid at room temperature, except carbon).

Specific Heat Capacity of Water

• Varies slightly with temperature.
• Graph shows U-shaped curve: decreases till ~40°C, then increases.
• One calorie = heat to raise 1 g water from 14.5°C to 15.5°C.

Solved Examples

• Ex 1: Calculate heat to raise 2 kg water by 10°C, s = 4200 J/kg°C.
  • Q = msΔT = 2 × 4200 × 10 = 84000 J
• Ex 2: Find molar specific heat if heat capacity H = 60 J/°C and molar mass μ = 20 g/mol.
  • C = H/μ = 60/20 = 3 J/mol°C

Molar Specific Heat for Solid

• Heat required to raise 1 mole temperature by 1°C.
• Formula: C = H/μ = (1/μ) ΔQ/ΔT, where H = heat capacity.
• For N atoms vibrating, average energy per oscillator in 3D: 3k_B T.
• Total energy per mole: U = 3k_B T × N_A = 3RT.
• Molar specific heat: C = ΔQ/ΔT = ΔU/ΔT = 3R (valid at room temperature, except carbon).

Specific Heat Capacity of Water

• Varies slightly with temperature.
• Graph shows U-shaped curve: decreases till ~40°C, then increases.
• One calorie = heat to raise 1 g water from 14.5°C to 15.5°C.

Solved Examples

• Ex 1: Calculate heat to raise 2 kg water by 10°C, s = 4200 J/kg°C.
  • Q = msΔT = 2 × 4200 × 10 = 84000 J
• Ex 2: Find molar specific heat if heat capacity H = 60 J/°C and molar mass μ = 20 g/mol.
  • C = H/μ = 60/20 = 3 J/mol°C

Practice Set

• Q1: Specific heat capacity is the heat required to raise the temperature of
  • A) 1 mole of substance by 1°C
  • B) 1 gram of substance by 1°C
  • C) 1 kg of substance by 1°C
  • D) 1 liter of substance by 1°C
• Q2: Molar specific heat of a solid is
  • A) Heat to raise 1 g by 1°C
  • B) Heat to raise 1 mole by 1°C
  • C) Heat to raise 1 kg by 1°C
  • D) Heat to raise 1 atom by 1°C
• Q3: Total energy of 1 mole solid vibrating atoms at temperature T is
  • A) 3k_B T
  • B) 3RT
  • C) k_B T
  • D) RT

Answer Key

• Q1: B - Specific heat capacity is per gram per degree Celsius.
• Q2: B - Molar specific heat is heat for 1 mole per degree.
• Q3: B - Total energy for 1 mole is 3RT.

Memory Aids

• Specific heat: Q = msΔT
• Molar heat: C = H/μ
• Energy per mole: U = 3RT
• Water's specific heat varies with temperature (U-shaped curve).

Definition and Formula

• Heat absorbed or evolved per unit mass for 1°C temperature change.
• Formula: Q = ms\Delta T = ms(T_f - T_i)
• Specific heat capacity: s = \frac{Q}{m\Delta T}

Molar Specific Heat for Solid

• Heat required to raise 1 mole temperature by 1°C.
• Formula: C = \frac{H}{\mu} = \frac{1}{\mu} \frac{\Delta Q}{\Delta T}, where H = heat capacity.
• For N atoms vibrating, average energy per oscillator in 3D: 3k_B T.
• Total energy per mole: U = 3k_B T \times N_A = 3RT.
• Molar specific heat: C = \frac{\Delta Q}{\Delta T} = \frac{\Delta U}{\Delta T} = 3R (valid at room temperature, except carbon).

Specific Heat Capacity of Water

• Varies slightly with temperature.
• Graph shows U-shaped curve: decreases till ~40°C, then increases.
• One calorie = heat to raise 1 g water from 14.5°C to 15.5°C.

Solved Examples

• Ex 1: Calculate heat to raise 2 kg water by 10°C, s = 4200 J/kg°C.
  • Q = ms\Delta T = 2 \times 4200 \times 10 = 84000 J
• Ex 2: Find molar specific heat if heat capacity H = 60 J/°C and molar mass \(\mu = 20\) g/mol.
  • C = H/\mu = 60/20 = 3 J/mol°C

Practice Set

• Q1: Specific heat capacity is the heat required to raise the temperature of
  • A) 1 mole of substance by 1°C
  • B) 1 gram of substance by 1°C
  • C) 1 kg of substance by 1°C
  • D) 1 liter of substance by 1°C
• Q2: Molar specific heat of a solid is
  • A) Heat to raise 1 g by 1°C
  • B) Heat to raise 1 mole by 1°C
  • C) Heat to raise 1 kg by 1°C
  • D) Heat to raise 1 atom by 1°C
• Q3: Total energy of 1 mole solid vibrating atoms at temperature T is
  • A) 3k_B T
  • B) 3RT
  • C) k_B T
  • D) RT

Answer Key

• Q1: B - Specific heat capacity is per gram per degree Celsius.
• Q2: B - Molar specific heat is heat for 1 mole per degree.
• Q3: B - Total energy for 1 mole is 3RT.

Memory Aids

• Specific heat: Q = ms\Delta T
• Molar heat: C = H/\mu
• Energy per mole: U = 3RT
• Water's specific heat varies with temperature (U-shaped curve).

Relation Specific Heat Constant Volume Pressure

First Law and Definitions

• Heat absorbed: ΔQ = ΔU_int + PΔV
• At constant volume (ΔV=0): C_v = (ΔQ/ΔT)_v = (ΔU_int/ΔT)
• Internal energy depends only on temperature for ideal gas.

Heat at Constant Pressure

• At constant pressure: C_p = (ΔQ/ΔT)_p = C_v + P(ΔV/ΔT)_p
• Difference: C_p - C_v = P(ΔV/ΔT)_p
• Using ideal gas law for 1 mole: P(ΔV/ΔT)_p = R
• Mayer's relation: C_p - C_v = R

Solved Examples

• Ex 1: For 1 mole ideal gas, if C_v = 12.5 J/mol°C, find C_p.
  • C_p = C_v + R = 12.5 + 8.31 = 20.81 J/mol°C
• Ex 2: Calculate C_p - C_v for ideal gas.
  • C_p - C_v = R = 8.31 J/mol°C

Heat at Constant Pressure

• At constant pressure: C_p = (ΔQ/ΔT)_p = C_v + P(ΔV/ΔT)_p
• Difference: C_p - C_v = P(ΔV/ΔT)_p
• Using ideal gas law for 1 mole: P(ΔV/ΔT)_p = R
• Mayer's relation: C_p - C_v = R

Solved Examples

• Ex 1: For 1 mole ideal gas, if C_v = 12.5 J/mol°C, find C_p.
  • C_p = C_v + R = 12.5 + 8.31 = 20.81 J/mol°C
• Ex 2: Calculate C_p - C_v for ideal gas.
  • C_p - C_v = R = 8.31 J/mol°C

Practice Set

• Q1: Mayer's relation is
  • A) C_p + C_v = R
  • B) C_p - C_v = R
  • C) C_v - C_p = R
  • D) C_p × C_v = R
• Q2: At constant volume, heat capacity is
  • A) C_p
  • B) C_v
  • C) R
  • D) 0
• Q3: For 1 mole ideal gas, P(ΔV/ΔT)_p equals
  • A) 0
  • B) R
  • C) C_p
  • D) C_v

Answer Key

• Q1: B - Mayer's relation states C_p - C_v = R.
• Q2: B - At constant volume, heat capacity is C_v.
• Q3: B - From ideal gas law, P(ΔV/ΔT)_p = R.

Memory Aids

• Mayer's relation: C_p - C_v = R
• At constant volume: C_v = (ΔU/ΔT)
• At constant pressure: C_p = C_v + R

First Law and Definitions

• Heat absorbed: ΔQ = ΔU_int + PΔV
• At constant volume (ΔV=0): C_v = (ΔQ/ΔT)_v = (ΔU_int/ΔT)
• Internal energy depends only on temperature for ideal gas.

Heat at Constant Pressure

• At constant pressure: C_p = (ΔQ/ΔT)_p = C_v + P(ΔV/ΔT)_p
• Difference: C_p - C_v = P(ΔV/ΔT)_p
• Using ideal gas law for 1 mole: P(ΔV/ΔT)_p = R
• Mayer's relation: C_p - C_v = R

Solved Examples

• Ex 1: For 1 mole ideal gas, if C_v = 12.5 J/mol°C, find C_p.
  • C_p = C_v + R = 12.5 + 8.31 = 20.81 J/mol°C
• Ex 2: Calculate C_p - C_v for ideal gas.
  • C_p - C_v = R = 8.31 J/mol°C

Practice Set

• Q1: Mayer's relation is
  • A) C_p + C_v = R
  • B) C_p - C_v = R
  • C) C_v - C_p = R
  • D) C_p × C_v = R
• Q2: At constant volume, heat capacity is
  • A) C_p
  • B) C_v
  • C) R
  • D) 0
• Q3: For 1 mole ideal gas, P(ΔV/ΔT)_p equals
  • A) 0
  • B) R
  • C) C_p
  • D) C_v

Answer Key

• Q1: B - Mayer's relation states C_p - C_v = R.
• Q2: B - At constant volume, heat capacity is C_v.
• Q3: B - From ideal gas law, P(ΔV/ΔT)_p = R.

Memory Aids

• Mayer's relation: C_p - C_v = R
• At constant volume: C_v = (ΔU/ΔT)
• At constant pressure: C_p = C_v + R

First Law and Definitions

• Heat absorbed: ΔQ = ΔU_int + PΔV
• At constant volume (ΔV=0): C_v = (ΔQ/ΔT)_v = (ΔU_int/ΔT)
• Internal energy depends only on temperature for ideal gas.

Heat at Constant Pressure

• At constant pressure: C_p = (ΔQ/ΔT)_p = C_v + P(ΔV/ΔT)_p
• Difference: C_p - C_v = P(ΔV/ΔT)_p
• Using ideal gas law for 1 mole: P(ΔV/ΔT)_p = R
• Mayer's relation: C_p - C_v = R

Solved Examples

• Ex 1: For 1 mole ideal gas, if C_v = 12.5 J/mol°C, find C_p.
  • C_p = C_v + R = 12.5 + 8.31 = 20.81 J/mol°C
• Ex 2: Calculate C_p - C_v for ideal gas.
  • C_p - C_v = R = 8.31 J/mol°C

Heat at Constant Pressure

• At constant pressure: C_p = (ΔQ/ΔT)_p = C_v + P(ΔV/ΔT)_p
• Difference: C_p - C_v = P(ΔV/ΔT)_p
• Using ideal gas law for 1 mole: P(ΔV/ΔT)_p = R
• Mayer's relation: C_p - C_v = R

Solved Examples

• Ex 1: For 1 mole ideal gas, if C_v = 12.5 J/mol°C, find C_p.
  • C_p = C_v + R = 12.5 + 8.31 = 20.81 J/mol°C
• Ex 2: Calculate C_p - C_v for ideal gas.
  • C_p - C_v = R = 8.31 J/mol°C

Practice Set

• Q1: Mayer's relation is
  • A) C_p + C_v = R
  • B) C_p - C_v = R
  • C) C_v - C_p = R
  • D) C_p × C_v = R
• Q2: At constant volume, heat capacity is
  • A) C_p
  • B) C_v
  • C) R
  • D) 0
• Q3: For 1 mole ideal gas, P(ΔV/ΔT)_p equals
  • A) 0
  • B) R
  • C) C_p
  • D) C_v

Answer Key

• Q1: B - Mayer's relation states C_p - C_v = R.
• Q2: B - At constant volume, heat capacity is C_v.
• Q3: B - From ideal gas law, P(ΔV/ΔT)_p = R.

Memory Aids

• Mayer's relation: C_p - C_v = R
• At constant volume: C_v = (ΔU/ΔT)
• At constant pressure: C_p = C_v + R

First Law and Definitions

• Heat absorbed: \Delta Q = \Delta U_{int} + P\Delta V
• At constant volume (\(\Delta V=0\)): C_v = \left(\frac{\Delta Q}{\Delta T}\right)_v = \left(\frac{\Delta U_{int}}{\Delta T}\right)
• Internal energy depends only on temperature for ideal gas.

Heat at Constant Pressure

• At constant pressure: C_p = \left(\frac{\Delta Q}{\Delta T}\right)_p = C_v + P\left(\frac{\Delta V}{\Delta T}\right)_p
• Difference: C_p - C_v = P\left(\frac{\Delta V}{\Delta T}\right)_p
• Using ideal gas law for 1 mole: P\left(\frac{\Delta V}{\Delta T}\right)_p = R
• Mayer's relation: C_p - C_v = R

Solved Examples

• Ex 1: For 1 mole ideal gas, if C_v = 12.5 J/mol°C, find C_p.
  • C_p = C_v + R = 12.5 + 8.31 = 20.81 J/mol°C
• Ex 2: Calculate C_p - C_v for ideal gas.
  • C_p - C_v = R = 8.31 J/mol°C

Practice Set

• Q1: Mayer's relation is
  • A) C_p + C_v = R
  • B) C_p - C_v = R
  • C) C_v - C_p = R
  • D) C_p \times C_v = R
• Q2: At constant volume, heat capacity is
  • A) C_p
  • B) C_v
  • C) R
  • D) 0
• Q3: For 1 mole ideal gas, P(\Delta V/\Delta T)_p equals
  • A) 0
  • B) R
  • C) C_p
  • D) C_v

Answer Key

• Q1: B - Mayer's relation states C_p - C_v = R.
• Q2: B - At constant volume, heat capacity is C_v.
• Q3: B - From ideal gas law, P(\Delta V/\Delta T)_p = R.

Memory Aids

• Mayer's relation: C_p - C_v = R
• At constant volume: C_v = (\Delta U/\Delta T)
• At constant pressure: C_p = C_v + R

Thermodynamic State Variables

Definition and Types

• Thermodynamic state variables describe equilibrium states of a system.
• Two types: Extensive and Intensive.
• Extensive variables depend on system size (e.g., volume V, internal energy U, mass μ).
• Intensive variables do not depend on system size (e.g., pressure P, temperature T).
• Product of an intensive and an extensive variable is extensive (e.g., P × V is extensive).

Solved Examples

• Ex 1: Identify if pressure and volume are intensive or extensive.
  • Pressure (P) is intensive.
  • Volume (V) is extensive.

Solved Examples

• Ex 1: Identify if pressure and volume are intensive or extensive.
  • Pressure (P) is intensive.
  • Volume (V) is extensive.

Solved Examples

• Ex 1: Identify if pressure and volume are intensive or extensive.
  • Pressure (P) is intensive.
  • Volume (V) is extensive.

Practice Set

• Q1: Which of the following is an intensive variable?
  • A. Volume
  • B. Mass
  • C. Temperature
  • D. Internal energy
• Q2: The product of an intensive and an extensive variable is:
  • A. Intensive
  • B. Extensive
  • C. Neither
  • D. Depends on variables
• Q3: Which of the following is an extensive variable?
  • A. Pressure
  • B. Temperature
  • C. Volume
  • D. Density

Answer Key

• Q1: C - Temperature is intensive; it does not depend on system size.
• Q2: B - Product of intensive and extensive variables is extensive.
• Q3: C - Volume is extensive; it depends on system size.

Memory Aids

• Intensive: "I" for Independent of size (P, T).
• Extensive: "E" for Extends with size (V, U, μ).
• Product rule: Intensive × Extensive = Extensive.

Definition and Types

• Thermodynamic state variables describe equilibrium states of a system.
• Two types: Extensive and Intensive.
• Extensive variables depend on system size (e.g., volume V, internal energy U, mass μ).
• Intensive variables do not depend on system size (e.g., pressure P, temperature T).
• Product of an intensive and an extensive variable is extensive (e.g., P × V is extensive).

Solved Examples

• Ex 1: Identify if pressure and volume are intensive or extensive.
  • Pressure (P) is intensive.
  • Volume (V) is extensive.

Practice Set

• Q1: Which of the following is an intensive variable?
  • A. Volume
  • B. Mass
  • C. Temperature
  • D. Internal energy
• Q2: The product of an intensive and an extensive variable is:
  • A. Intensive
  • B. Extensive
  • C. Neither
  • D. Depends on variables
• Q3: Which of the following is an extensive variable?
  • A. Pressure
  • B. Temperature
  • C. Volume
  • D. Density

Answer Key

• Q1: C - Temperature is intensive; it does not depend on system size.
• Q2: B - Product of intensive and extensive variables is extensive.
• Q3: C - Volume is extensive; it depends on system size.

Memory Aids

• Intensive: "I" for Independent of size (P, T).
• Extensive: "E" for Extends with size (V, U, μ).
• Product rule: Intensive × Extensive = Extensive.

Definition and Types

• Thermodynamic state variables describe equilibrium states of a system.
• Two types: Extensive and Intensive.
• Extensive variables depend on system size (e.g., volume V, internal energy U, mass μ).
• Intensive variables do not depend on system size (e.g., pressure P, temperature T).
• Product of an intensive and an extensive variable is extensive (e.g., P × V is extensive).

Solved Examples

• Ex 1: Identify if pressure and volume are intensive or extensive.
  • Pressure (P) is intensive.
  • Volume (V) is extensive.

Solved Examples

• Ex 1: Identify if pressure and volume are intensive or extensive.
  • Pressure (P) is intensive.
  • Volume (V) is extensive.

Solved Examples

• Ex 1: Identify if pressure and volume are intensive or extensive.
  • Pressure (P) is intensive.
  • Volume (V) is extensive.

Practice Set

• Q1: Which of the following is an intensive variable?
  • A. Volume
  • B. Mass
  • C. Temperature
  • D. Internal energy
• Q2: The product of an intensive and an extensive variable is:
  • A. Intensive
  • B. Extensive
  • C. Neither
  • D. Depends on variables
• Q3: Which of the following is an extensive variable?
  • A. Pressure
  • B. Temperature
  • C. Volume
  • D. Density

Answer Key

• Q1: C - Temperature is intensive; it does not depend on system size.
• Q2: B - Product of intensive and extensive variables is extensive.
• Q3: C - Volume is extensive; it depends on system size.

Memory Aids

• Intensive: "I" for Independent of size (P, T).
• Extensive: "E" for Extends with size (V, U, μ).
• Product rule: Intensive × Extensive = Extensive.

Definition and Types

• Thermodynamic state variables describe equilibrium states of a system.
• Two types: Extensive and Intensive.
• Extensive variables depend on system size (e.g., volume V, internal energy U, mass μ).
• Intensive variables do not depend on system size (e.g., pressure P, temperature T).
• Product of an intensive and an extensive variable is extensive (e.g., P × V is extensive).

Solved Examples

• Ex 1: Identify if pressure and volume are intensive or extensive.
  • Pressure (P) is intensive.
  • Volume (V) is extensive.

Practice Set

• Q1: Which of the following is an intensive variable?
  • A. Volume
  • B. Mass
  • C. Temperature
  • D. Internal energy
• Q2: The product of an intensive and an extensive variable is:
  • A. Intensive
  • B. Extensive
  • C. Neither
  • D. Depends on variables
• Q3: Which of the following is an extensive variable?
  • A. Pressure
  • B. Temperature
  • C. Volume
  • D. Density

Answer Key

• Q1: C - Temperature is intensive; it does not depend on system size.
• Q2: B - Product of intensive and extensive variables is extensive.
• Q3: C - Volume is extensive; it depends on system size.

Memory Aids

• Intensive: "I" for Independent of size (P, T).
• Extensive: "E" for Extends with size (V, U, μ).
• Product rule: Intensive × Extensive = Extensive.

Equation of State

Ideal Gas Equation

• Relates thermodynamic state variables: pressure (P), volume (V), temperature (T), and amount of substance (μ).
• Ideal gas equation: PV = μRT
• R = universal gas constant (8.314 J/mol·K)
• Valid for ideal gases under equilibrium.

Solved Examples

• Ex 1: Calculate efficiency of heat engine absorbing 800 J and rejecting 500 J.
  • Work done W = 800 - 500 = 300 J
  • Efficiency η = W/Q₁ = 300/800 = 0.375 = 37.5%
• Ex 2: Calculate COP of refrigerator absorbing 200 J from cold reservoir and work input 50 J.
  • COP = Q/W = 200/50 = 4

Practice Set

• Q1: For 1 mol ideal gas at 273 K and 22.4 L, pressure is:
  • A. 1 atm
  • B. 2 atm
  • C. 0.5 atm
  • D. 3 atm
• Q2: If volume doubles at constant temperature and moles, pressure:
  • A. Doubles
  • B. Halves
  • C. Remains same
  • D. Quadruples
• Q3: Calculate volume of 3 mol gas at 300 K and 2 atm.
  • A. 36.9 L
  • B. 22.4 L
  • C. 44.8 L
  • D. 11.2 L

Answer Key

• Q1: A - Standard molar volume at STP is 22.4 L at 1 atm.
• Q2: B - P inversely proportional to V at constant T and μ.
• Q3: A - V = μRT/P = (3 × 0.0821 × 300)/2 = 36.9 L (using R=0.0821 L·atm/mol·K)

Memory Aids

• Efficiency η = 1 - T₂/T₁
• Maximum efficiency engine is ideal Carnot engine.

Ideal Gas Equation

• Relates thermodynamic state variables: pressure (P), volume (V), temperature (T), and amount of substance (μ).
• Ideal gas equation: PV = μRT
• R = universal gas constant (8.314 J/mol·K)
• Valid for ideal gases under equilibrium.

Solved Examples

• Ex 1: Calculate pressure of 2 mol ideal gas at 300 K in 10 L container.
  • Given: μ=2 mol, T=300 K, V=10 L = 0.01 m³, R=8.314 J/mol·K
  • Use PV = μRT → P = μRT/V
  • P = (2 × 8.314 × 300) / 0.01 = 498840 Pa = 498.84 kPa

Practice Set

• Q1: For 1 mol ideal gas at 273 K and 22.4 L, pressure is:
  • A. 1 atm
  • B. 2 atm
  • C. 0.5 atm
  • D. 3 atm
• Q2: If volume doubles at constant temperature and moles, pressure:
  • A. Doubles
  • B. Halves
  • C. Remains same
  • D. Quadruples
• Q3: Calculate volume of 3 mol gas at 300 K and 2 atm.
  • A. 36.9 L
  • B. 22.4 L
  • C. 44.8 L
  • D. 11.2 L

Answer Key

• Q1: A - Standard molar volume at STP is 22.4 L at 1 atm.
• Q2: B - P inversely proportional to V at constant T and μ.
• Q3: A - V = μRT/P = (3 × 0.0821 × 300)/2 = 36.9 L (using R=0.0821 L·atm/mol·K)

Memory Aids

• Ideal gas: PV = μRT
• At STP: 1 mol gas occupies 22.4 L
• Pressure inversely ∝ volume at constant T and μ

Ideal Gas Equation

• Relates thermodynamic state variables: pressure (P), volume (V), temperature (T), and amount of substance (μ).
• Ideal gas equation: PV = μRT
• R = universal gas constant (8.314 J/mol·K)
• Valid for ideal gases under equilibrium.

Solved Examples

• Ex 1: Calculate efficiency of heat engine absorbing 800 J and rejecting 500 J.
  • Work done W = 800 - 500 = 300 J
  • Efficiency η = W/Q₁ = 300/800 = 0.375 = 37.5%
• Ex 2: Calculate COP of refrigerator absorbing 200 J from cold reservoir and work input 50 J.
  • COP = Q/W = 200/50 = 4

Practice Set

• Q1: For 1 mol ideal gas at 273 K and 22.4 L, pressure is:
  • A. 1 atm
  • B. 2 atm
  • C. 0.5 atm
  • D. 3 atm
• Q2: If volume doubles at constant temperature and moles, pressure:
  • A. Doubles
  • B. Halves
  • C. Remains same
  • D. Quadruples
• Q3: Calculate volume of 3 mol gas at 300 K and 2 atm.
  • A. 36.9 L
  • B. 22.4 L
  • C. 44.8 L
  • D. 11.2 L

Answer Key

• Q1: A - Standard molar volume at STP is 22.4 L at 1 atm.
• Q2: B - P inversely proportional to V at constant T and μ.
• Q3: A - V = μRT/P = (3 × 0.0821 × 300)/2 = 36.9 L (using R=0.0821 L·atm/mol·K)

Memory Aids

• Efficiency η = 1 - T₂/T₁
• Maximum efficiency engine is ideal Carnot engine.

Ideal Gas Equation

• Relates thermodynamic state variables: pressure (P), volume (V), temperature (T), and amount of substance (μ).
• Ideal gas equation: PV = μRT
• R = universal gas constant (8.314 J/mol·K)
• Valid for ideal gases under equilibrium.

Solved Examples

• Ex 1: Calculate pressure of 2 mol ideal gas at 300 K in 10 L container.
  • Given: μ=2 mol, T=300 K, V=10 L = 0.01 m³, R=8.314 J/mol·K
  • Use PV = μRT → P = μRT/V
  • P = (2 × 8.314 × 300) / 0.01 = 498840 Pa = 498.84 kPa

Practice Set

• Q1: For 1 mol ideal gas at 273 K and 22.4 L, pressure is:
  • A. 1 atm
  • B. 2 atm
  • C. 0.5 atm
  • D. 3 atm
• Q2: If volume doubles at constant temperature and moles, pressure:
  • A. Doubles
  • B. Halves
  • C. Remains same
  • D. Quadruples
• Q3: Calculate volume of 3 mol gas at 300 K and 2 atm.
  • A. 36.9 L
  • B. 22.4 L
  • C. 44.8 L
  • D. 11.2 L

Answer Key

• Q1: A - Standard molar volume at STP is 22.4 L at 1 atm.
• Q2: B - P inversely proportional to V at constant T and μ.
• Q3: A - V = μRT/P = (3 × 0.0821 × 300)/2 = 36.9 L (using R=0.0821 L·atm/mol·K)

Memory Aids

• Ideal gas: PV = μRT
• At STP: 1 mol gas occupies 22.4 L
• Pressure inversely ∝ volume at constant T and μ

Ideal Gas Equation

• Relates thermodynamic state variables: pressure (P), volume (V), temperature (T), and amount of substance (μ).
• Ideal gas equation: PV = μRT
• R = universal gas constant (8.314 J/mol·K)
• Valid for ideal gases under equilibrium.

Solved Examples

• Ex 1: Calculate efficiency of heat engine absorbing 800 J and rejecting 500 J.
  • Work done W = 800 - 500 = 300 J
  • Efficiency η = W/Q₁ = 300/800 = 0.375 = 37.5%
• Ex 2: Calculate COP of refrigerator absorbing 200 J from cold reservoir and work input 50 J.
  • COP = Q/W = 200/50 = 4

Practice Set

• Q1: For 1 mol ideal gas at 273 K and 22.4 L, pressure is:
  • A. 1 atm
  • B. 2 atm
  • C. 0.5 atm
  • D. 3 atm
• Q2: If volume doubles at constant temperature and moles, pressure:
  • A. Doubles
  • B. Halves
  • C. Remains same
  • D. Quadruples
• Q3: Calculate volume of 3 mol gas at 300 K and 2 atm.
  • A. 36.9 L
  • B. 22.4 L
  • C. 44.8 L
  • D. 11.2 L

Answer Key

• Q1: A - Standard molar volume at STP is 22.4 L at 1 atm.
• Q2: B - P inversely proportional to V at constant T and μ.
• Q3: A - V = μRT/P = (3 × 0.0821 × 300)/2 = 36.9 L (using R=0.0821 L·atm/mol·K)

Memory Aids

• Efficiency η = 1 - T₂/T₁
• Maximum efficiency engine is ideal Carnot engine.

Ideal Gas Equation

• Relates thermodynamic state variables: pressure (P), volume (V), temperature (T), and amount of substance (μ).
• Ideal gas equation: PV = μRT
• R = universal gas constant (8.314 J/mol·K)
• Valid for ideal gases under equilibrium.

Solved Examples

• Ex 1: Calculate pressure of 2 mol ideal gas at 300 K in 10 L container.
  • Given: μ=2 mol, T=300 K, V=10 L = 0.01 m³, R=8.314 J/mol·K
  • Use PV = μRT → P = μRT/V
  • P = (2 × 8.314 × 300) / 0.01 = 498840 Pa = 498.84 kPa

Practice Set

• Q1: For 1 mol ideal gas at 273 K and 22.4 L, pressure is:
  • A. 1 atm
  • B. 2 atm
  • C. 0.5 atm
  • D. 3 atm
• Q2: If volume doubles at constant temperature and moles, pressure:
  • A. Doubles
  • B. Halves
  • C. Remains same
  • D. Quadruples
• Q3: Calculate volume of 3 mol gas at 300 K and 2 atm.
  • A. 36.9 L
  • B. 22.4 L
  • C. 44.8 L
  • D. 11.2 L

Answer Key

• Q1: A - Standard molar volume at STP is 22.4 L at 1 atm.
• Q2: B - P inversely proportional to V at constant T and μ.
• Q3: A - V = μRT/P = (3 × 0.0821 × 300)/2 = 36.9 L (using R=0.0821 L·atm/mol·K)

Memory Aids

• Ideal gas: PV = μRT
• At STP: 1 mol gas occupies 22.4 L
• Pressure inversely ∝ volume at constant T and μ

Ideal Gas Equation

• Relates thermodynamic state variables: pressure (P), volume (V), temperature (T), and amount of substance (μ).
• Ideal gas equation: PV = μRT
• R = universal gas constant (8.314 J/mol·K)
• Valid for ideal gases under equilibrium.

Solved Examples

• Ex 1: Calculate efficiency of heat engine absorbing 800 J and rejecting 500 J.
  • Work done W = 800 - 500 = 300 J
  • Efficiency η = W/Q₁ = 300/800 = 0.375 = 37.5%
• Ex 2: Calculate COP of refrigerator absorbing 200 J from cold reservoir and work input 50 J.
  • COP = Q/W = 200/50 = 4

Practice Set

• Q1: For 1 mol ideal gas at 273 K and 22.4 L, pressure is:
  • A. 1 atm
  • B. 2 atm
  • C. 0.5 atm
  • D. 3 atm
• Q2: If volume doubles at constant temperature and moles, pressure:
  • A. Doubles
  • B. Halves
  • C. Remains same
  • D. Quadruples
• Q3: Calculate volume of 3 mol gas at 300 K and 2 atm.
  • A. 36.9 L
  • B. 22.4 L
  • C. 44.8 L
  • D. 11.2 L

Answer Key

• Q1: A - Standard molar volume at STP is 22.4 L at 1 atm.
• Q2: B - P inversely proportional to V at constant T and μ.
• Q3: A - V = μRT/P = (3 × 0.0821 × 300)/2 = 36.9 L (using R=0.0821 L·atm/mol·K)

Memory Aids

• Efficiency η = 1 - T₂/T₁
• Maximum efficiency engine is ideal Carnot engine.

Ideal Gas Equation

• Relates thermodynamic state variables: pressure (P), volume (V), temperature (T), and amount of substance (μ).
• Ideal gas equation: PV = μRT
• R = universal gas constant (8.314 J/mol·K)
• Valid for ideal gases under equilibrium.

Solved Examples

• Ex 1: Calculate pressure of 2 mol ideal gas at 300 K in 10 L container.
  • Given: μ=2 mol, T=300 K, V=10 L = 0.01 m³, R=8.314 J/mol·K
  • Use PV = μRT → P = μRT/V
  • P = (2 × 8.314 × 300) / 0.01 = 498840 Pa = 498.84 kPa

Practice Set

• Q1: For 1 mol ideal gas at 273 K and 22.4 L, pressure is:
  • A. 1 atm
  • B. 2 atm
  • C. 0.5 atm
  • D. 3 atm
• Q2: If volume doubles at constant temperature and moles, pressure:
  • A. Doubles
  • B. Halves
  • C. Remains same
  • D. Quadruples
• Q3: Calculate volume of 3 mol gas at 300 K and 2 atm.
  • A. 36.9 L
  • B. 22.4 L
  • C. 44.8 L
  • D. 11.2 L

Answer Key

• Q1: A - Standard molar volume at STP is 22.4 L at 1 atm.
• Q2: B - P inversely proportional to V at constant T and μ.
• Q3: A - V = μRT/P = (3 × 0.0821 × 300)/2 = 36.9 L (using R=0.0821 L·atm/mol·K)

Memory Aids

• Ideal gas: PV = μRT
• At STP: 1 mol gas occupies 22.4 L
• Pressure inversely ∝ volume at constant T and μ

Thermodynamic Processes

Key Concepts

• Follows equation: PV^x = constant, where x is polytropic index
• Special cases: x=0 (isobaric), x=1 (isothermal), x=γ (adiabatic)
• Other relations: TV^x-1 = constant, P^1-x T^x = constant
• Work done: W = nR(T₁ - T₂)/(x-1) = (P₁V₁ - P₂V₂)/(x-1) for x ≠ 1
• Molar heat capacity: C = C_v - R/(x-1) = R/(γ - 1) - R/(x-1)

Solved Examples

• Ex 1: For a cyclic process, find ΔU if initial and final internal energies are equal.
  Answer: ΔU = 0
• Ex 2: Identify process where temperature remains constant.
  Answer: Isothermal process

Practice Set

• Q1: What is the change in internal energy in a cyclic process?
  A. Positive
  B. Negative
  C. Zero
  D. Depends on process
  
• Q2: Which process keeps volume constant?
  A. Isobaric
  B. Isochoric
  C. Adiabatic
  D. Isothermal
  
• Q3: In a quasi static process, the system is always in:
  A. Non-equilibrium state
  B. Equilibrium state
  C. Rapid change
  D. Isolated state
  

Answer Key

• Q1: C - ΔU = 0 in cyclic process as system returns to initial state.
• Q2: B - Isochoric process keeps volume constant.
• Q3: B - Quasi static process maintains equilibrium at every stage.

Memory Aids

• Cycle → ΔU = 0, T constant
• Quasi static = slow equilibrium
• Isobaric = P constant, Isochoric = V constant

Key Concepts

• Quasi static process: Infinitely slow process maintaining equilibrium at every stage (P, T, V change slowly).
• Cyclic process: Series of changes returning system to initial state; ΔU = 0, ΔT = 0 (temperature constant).
• Types of processes: Isobaric (P constant), Isochoric (V constant), Adiabatic (Q = 0), Isothermal (T constant).

Solved Examples

• Ex 1: For a cyclic process, find ΔU if initial and final internal energies are equal.
  Answer: ΔU = 0
• Ex 2: Identify process where temperature remains constant.
  Answer: Isothermal process

Practice Set

• Q1: What is the change in internal energy in a cyclic process?
  A. Positive
  B. Negative
  C. Zero
  D. Depends on process
  
• Q2: Which process keeps volume constant?
  A. Isobaric
  B. Isochoric
  C. Adiabatic
  D. Isothermal
  
• Q3: In a quasi static process, the system is always in:
  A. Non-equilibrium state
  B. Equilibrium state
  C. Rapid change
  D. Isolated state
  

Answer Key

• Q1: C - ΔU = 0 in cyclic process as system returns to initial state.
• Q2: B - Isochoric process keeps volume constant.
• Q3: B - Quasi static process maintains equilibrium at every stage.

Memory Aids

• Cycle → ΔU = 0, T constant
• Quasi static = slow equilibrium
• Isobaric = P constant, Isochoric = V constant

Key Concepts

• Follows equation: PV^x = constant, where x is polytropic index
• Special cases: x=0 (isobaric), x=1 (isothermal), x=γ (adiabatic)
• Other relations: TV^x-1 = constant, P^1-x T^x = constant
• Work done: W = nR(T₁ - T₂)/(x-1) = (P₁V₁ - P₂V₂)/(x-1) for x ≠ 1
• Molar heat capacity: C = C_v - R/(x-1) = R/(γ - 1) - R/(x-1)

Solved Examples

• Ex 1: For a cyclic process, find ΔU if initial and final internal energies are equal.
  Answer: ΔU = 0
• Ex 2: Identify process where temperature remains constant.
  Answer: Isothermal process

Practice Set

• Q1: What is the change in internal energy in a cyclic process?
  A. Positive
  B. Negative
  C. Zero
  D. Depends on process
  
• Q2: Which process keeps volume constant?
  A. Isobaric
  B. Isochoric
  C. Adiabatic
  D. Isothermal
  
• Q3: In a quasi static process, the system is always in:
  A. Non-equilibrium state
  B. Equilibrium state
  C. Rapid change
  D. Isolated state
  

Answer Key

• Q1: C - ΔU = 0 in cyclic process as system returns to initial state.
• Q2: B - Isochoric process keeps volume constant.
• Q3: B - Quasi static process maintains equilibrium at every stage.

Memory Aids

• Cycle → ΔU = 0, T constant
• Quasi static = slow equilibrium
• Isobaric = P constant, Isochoric = V constant

Key Concepts

• Quasi static process: Infinitely slow process maintaining equilibrium at every stage (P, T, V change slowly).
• Cyclic process: Series of changes returning system to initial state; ΔU = 0, ΔT = 0 (temperature constant).
• Types of processes: Isobaric (P constant), Isochoric (V constant), Adiabatic (Q = 0), Isothermal (T constant).

Solved Examples

• Ex 1: For a cyclic process, find ΔU if initial and final internal energies are equal.
  Answer: ΔU = 0
• Ex 2: Identify process where temperature remains constant.
  Answer: Isothermal process

Practice Set

• Q1: What is the change in internal energy in a cyclic process?
  A. Positive
  B. Negative
  C. Zero
  D. Depends on process
  
• Q2: Which process keeps volume constant?
  A. Isobaric
  B. Isochoric
  C. Adiabatic
  D. Isothermal
  
• Q3: In a quasi static process, the system is always in:
  A. Non-equilibrium state
  B. Equilibrium state
  C. Rapid change
  D. Isolated state
  

Answer Key

• Q1: C - ΔU = 0 in cyclic process as system returns to initial state.
• Q2: B - Isochoric process keeps volume constant.
• Q3: B - Quasi static process maintains equilibrium at every stage.

Memory Aids

• Cycle → ΔU = 0, T constant
• Quasi static = slow equilibrium
• Isobaric = P constant, Isochoric = V constant

Isothermal Process

Key Concepts

• Temperature remains constant: ΔT = 0
• First law: ΔU = 0, so ΔQ = ΔW (heat supplied = work done)
• Equation: PV = constant (Boyle’s law)
• Indicator diagram: P-V curve is rectangular hyperbola (isotherm)
• Slope of curve: (dP/dV) = -P/V
• Work done: W = nRT ln(V_f/V_i) = nRT ln(P_i/P_f)
• Specific heat: Infinite (C = Q/(mΔT) = ∞)

Solved Examples

• Ex 1: Calculate work done when 1 mole of ideal gas expands isothermally from 10 L to 20 L at 300 K.
  Use W = nRT ln(V_f/V_i).
  Answer: W = 1 × 8.314 × 300 × ln 2 ≈ 1728 J

Practice Set

• Q1: In isothermal process, what is the change in internal energy?
  A. Positive
  B. Negative
  C. Zero
  D. Depends on volume
  
• Q2: Work done in isothermal expansion is:
  A. Zero
  B. Equal to heat absorbed
  C. Negative
  D. Equal to change in internal energy
  
• Q3: Which law is obeyed in isothermal process?
  A. Charles’s law
  B. Boyle’s law
  C. Gay-Lussac’s law
  D. Avogadro’s law
  

Answer Key

• Q1: C - ΔU = 0 as temperature is constant.
• Q2: B - Work done equals heat absorbed (ΔQ = ΔW).
• Q3: B - Boyle’s law (PV = constant) applies.

Memory Aids

• Isothermal = T constant
• Work = heat absorbed
• PV = constant (Boyle’s law)

Key Concepts

• Temperature remains constant: ΔT = 0
• First law: ΔU = 0, so ΔQ = ΔW (heat supplied = work done)
• Equation: PV = constant (Boyle’s law)
• Indicator diagram: P-V curve is rectangular hyperbola (isotherm)
• Slope of curve: (dP/dV) = -P/V
• Work done: W = nRT ln(V_f/V_i) = nRT ln(P_i/P_f)
• Specific heat: Infinite (C = Q/(mΔT) = ∞)

Solved Examples

• Ex 1: Calculate work done when 1 mole of ideal gas expands isothermally from 10 L to 20 L at 300 K.
  Use W = nRT ln(V_f/V_i).
  Answer: W = 1 × 8.314 × 300 × ln 2 ≈ 1728 J

Practice Set

• Q1: In isothermal process, what is the change in internal energy?
  A. Positive
  B. Negative
  C. Zero
  D. Depends on volume
  
• Q2: Work done in isothermal expansion is:
  A. Zero
  B. Equal to heat absorbed
  C. Negative
  D. Equal to change in internal energy
  
• Q3: Which law is obeyed in isothermal process?
  A. Charles’s law
  B. Boyle’s law
  C. Gay-Lussac’s law
  D. Avogadro’s law
  

Answer Key

• Q1: C - ΔU = 0 as temperature is constant.
• Q2: B - Work done equals heat absorbed (ΔQ = ΔW).
• Q3: B - Boyle’s law (PV = constant) applies.

Memory Aids

• Isothermal = T constant
• Work = heat absorbed
• PV = constant (Boyle’s law)

Key Concepts

• Temperature remains constant: ΔT = 0
• First law: ΔU = 0, so ΔQ = ΔW (heat supplied = work done)
• Equation: PV = constant (Boyle’s law)
• Indicator diagram: P-V curve is rectangular hyperbola (isotherm)
• Slope of curve: (dP/dV) = -P/V
• Work done: W = nRT ln(V_f/V_i) = nRT ln(P_i/P_f)
• Specific heat: Infinite (C = Q/(mΔT) = ∞)

Solved Examples

• Ex 1: Calculate work done when 1 mole of ideal gas expands isothermally from 10 L to 20 L at 300 K.
  Use W = nRT ln(V_f/V_i).
  Answer: W = 1 × 8.314 × 300 × ln 2 ≈ 1728 J

Practice Set

• Q1: In isothermal process, what is the change in internal energy?
  A. Positive
  B. Negative
  C. Zero
  D. Depends on volume
  
• Q2: Work done in isothermal expansion is:
  A. Zero
  B. Equal to heat absorbed
  C. Negative
  D. Equal to change in internal energy
  
• Q3: Which law is obeyed in isothermal process?
  A. Charles’s law
  B. Boyle’s law
  C. Gay-Lussac’s law
  D. Avogadro’s law
  

Answer Key

• Q1: C - ΔU = 0 as temperature is constant.
• Q2: B - Work done equals heat absorbed (ΔQ = ΔW).
• Q3: B - Boyle’s law (PV = constant) applies.

Memory Aids

• Isothermal = T constant
• Work = heat absorbed
• PV = constant (Boyle’s law)

Key Concepts

• Temperature remains constant: ΔT = 0
• First law: ΔU = 0, so ΔQ = ΔW (heat supplied = work done)
• Equation: PV = constant (Boyle’s law)
• Indicator diagram: P-V curve is rectangular hyperbola (isotherm)
• Slope of curve: \( \frac{dP}{dV} = -\frac{P}{V} \)
• Work done: \( W = nRT \ln \frac{V_f}{V_i} = nRT \ln \frac{P_i}{P_f} \)
• Specific heat: Infinite (\( C = \frac{Q}{m\Delta T} = \infty \))

Solved Examples

• Ex 1: Calculate work done when 1 mole of ideal gas expands isothermally from 10 L to 20 L at 300 K.
  Use \( W = nRT \ln \frac{V_f}{V_i} \).
  Answer: \( W = 1 \times 8.314 \times 300 \times \ln 2 \approx 1728 \text{ J} \)

Practice Set

• Q1: In isothermal process, what is the change in internal energy?
  A. Positive
  B. Negative
  C. Zero
  D. Depends on volume
  
• Q2: Work done in isothermal expansion is:
  A. Zero
  B. Equal to heat absorbed
  C. Negative
  D. Equal to change in internal energy
  
• Q3: Which law is obeyed in isothermal process?
  A. Charles’s law
  B. Boyle’s law
  C. Gay-Lussac’s law
  D. Avogadro’s law
  

Answer Key

• Q1: C - ΔU = 0 as temperature is constant.
• Q2: B - Work done equals heat absorbed (ΔQ = ΔW).
• Q3: B - Boyle’s law (PV = constant) applies.

Memory Aids

• Isothermal = T constant
• Work = heat absorbed
• PV = constant (Boyle’s law)

Adiabatic Process

Key Concepts

• No heat exchange: ΔQ = 0
• First law: ΔU = -ΔW
• Temperature changes with work done (expansion cools, compression heats)
• Equation: PV^γ = constant, γ = C_p/C_v
• Slope of curve: (dP/dV) = -γP/V (steeper than isothermal)
• Specific heat: Zero (C = 0)
• Work done: W = nR(T₁ - T₂)/(γ - 1) = (P₁V₁ - P₂V₂)/(γ - 1)

Solved Examples

• Ex 1: Calculate work done during adiabatic compression of 1 mole of gas from 2 L to 1 L with initial pressure 1 atm, γ = 1.4.
  Use W = (P₁V₁ - P₂V₂)/(γ - 1).
  Answer: Calculate P₂ using P₁V₁^γ = P₂V₂^γ, then find W.

Practice Set

• Q1: In adiabatic process, heat exchanged is:
  A. Positive
  B. Negative
  C. Zero
  D. Depends on volume
  
• Q2: Which equation represents adiabatic process?
  A. PV = constant
  B. PV^γ = constant
  C. V/T = constant
  D. P/T = constant
  
• Q3: Specific heat in adiabatic process is:
  A. Infinite
  B. Zero
  C. Equal to C_p
  D. Equal to C_v
  

Answer Key

• Q1: C - No heat exchange (ΔQ = 0).
• Q2: B - Poisson’s law: PV^γ = constant.
• Q3: B - Specific heat is zero as no heat is absorbed or released.

Memory Aids

• Adiabatic = No heat exchange
• Work changes internal energy
• PV^γ = constant (Poisson’s law)

Key Concepts

• No heat exchange: ΔQ = 0
• First law: ΔU = -ΔW
• Temperature changes with work done (expansion cools, compression heats)
• Equation: PV^γ = constant, γ = C_p/C_v
• Slope of curve: (dP/dV) = -γP/V (steeper than isothermal)
• Specific heat: Zero (C = 0)
• Work done: W = nR(T₁ - T₂)/(γ - 1) = (P₁V₁ - P₂V₂)/(γ - 1)

Solved Examples

• Ex 1: Calculate work done during adiabatic compression of 1 mole of gas from 2 L to 1 L with initial pressure 1 atm, γ = 1.4.
  Use W = (P₁V₁ - P₂V₂)/(γ - 1).
  Answer: Calculate P₂ using P₁V₁^γ = P₂V₂^γ, then find W.

Practice Set

• Q1: In adiabatic process, heat exchanged is:
  A. Positive
  B. Negative
  C. Zero
  D. Depends on volume
  
• Q2: Which equation represents adiabatic process?
  A. PV = constant
  B. PV^γ = constant
  C. V/T = constant
  D. P/T = constant
  
• Q3: Specific heat in adiabatic process is:
  A. Infinite
  B. Zero
  C. Equal to C_p
  D. Equal to C_v
  

Answer Key

• Q1: C - No heat exchange (ΔQ = 0).
• Q2: B - Poisson’s law: PV^γ = constant.
• Q3: B - Specific heat is zero as no heat is absorbed or released.

Memory Aids

• Adiabatic = No heat exchange
• Work changes internal energy
• PV^γ = constant (Poisson’s law)

Key Concepts

• No heat exchange: ΔQ = 0
• First law: ΔU = -ΔW
• Temperature changes with work done (expansion cools, compression heats)
• Equation: PV^γ = constant, γ = C_p/C_v
• Slope of curve: (dP/dV) = -γP/V (steeper than isothermal)
• Specific heat: Zero (C = 0)
• Work done: W = nR(T₁ - T₂)/(γ - 1) = (P₁V₁ - P₂V₂)/(γ - 1)

Solved Examples

• Ex 1: Calculate work done during adiabatic compression of 1 mole of gas from 2 L to 1 L with initial pressure 1 atm, γ = 1.4.
  Use W = (P₁V₁ - P₂V₂)/(γ - 1).
  Answer: Calculate P₂ using P₁V₁^γ = P₂V₂^γ, then find W.

Practice Set

• Q1: In adiabatic process, heat exchanged is:
  A. Positive
  B. Negative
  C. Zero
  D. Depends on volume
  
• Q2: Which equation represents adiabatic process?
  A. PV = constant
  B. PV^γ = constant
  C. V/T = constant
  D. P/T = constant
  
• Q3: Specific heat in adiabatic process is:
  A. Infinite
  B. Zero
  C. Equal to C_p
  D. Equal to C_v
  

Answer Key

• Q1: C - No heat exchange (ΔQ = 0).
• Q2: B - Poisson’s law: PV^γ = constant.
• Q3: B - Specific heat is zero as no heat is absorbed or released.

Memory Aids

• Adiabatic = No heat exchange
• Work changes internal energy
• PV^γ = constant (Poisson’s law)

Key Concepts

• No heat exchange: ΔQ = 0
• First law: ΔU = -ΔW
• Temperature changes with work done (expansion cools, compression heats)
• Equation: \( PV^\gamma = \text{constant} \), \( \gamma = \frac{C_p}{C_v} \)
• Slope of curve: \( \frac{dP}{dV} = -\gamma \frac{P}{V} \) (steeper than isothermal)
• Specific heat: Zero (\( C = 0 \))
• Work done: \( W = \frac{nR(T_1 - T_2)}{\gamma - 1} = \frac{P_1V_1 - P_2V_2}{\gamma - 1} \)

Solved Examples

• Ex 1: Calculate work done during adiabatic compression of 1 mole of gas from 2 L to 1 L with initial pressure 1 atm, \( \gamma = 1.4 \).
  Use \( W = \frac{P_1V_1 - P_2V_2}{\gamma - 1} \).
  Answer: Calculate \( P_2 \) using \( P_1V_1^\gamma = P_2V_2^\gamma \), then find W.

Practice Set

• Q1: In adiabatic process, heat exchanged is:
  A. Positive
  B. Negative
  C. Zero
  D. Depends on volume
  
• Q2: Which equation represents adiabatic process?
  A. PV = constant
  B. PV^\gamma = constant
  C. V/T = constant
  D. P/T = constant
  
• Q3: Specific heat in adiabatic process is:
  A. Infinite
  B. Zero
  C. Equal to C_p
  D. Equal to C_v
  

Answer Key

• Q1: C - No heat exchange (ΔQ = 0).
• Q2: B - Poisson’s law: \( PV^\gamma = \text{constant} \).
• Q3: B - Specific heat is zero as no heat is absorbed or released.

Memory Aids

• Adiabatic = No heat exchange
• Work changes internal energy
• PV^γ = constant (Poisson’s law)

Isobaric Process

Key Concepts

• Pressure remains constant
• In expansion: ΔU > 0, ΔW > 0, ΔQ > 0
• In compression: ΔU < 0, ΔW < 0, ΔQ < 0
• Specific heat: C_p = (f/2 + 1) R
• Work done: ΔW = P (V_f - V_i) = μ R (T_f - T_i)

Solved Examples

• Ex 1: Calculate work done when 1 mole gas expands at constant pressure 2 atm from 5 L to 10 L.
  Use ΔW = P ΔV.
  Answer: ΔW = 2 × (10 - 5) = 10 atm·L = 1013 J

Practice Set

• Q1: In isobaric process, pressure is:
  A. Constant
  B. Increasing
  C. Decreasing
  D. Variable
  
• Q2: Work done in isobaric expansion is:
  A. Zero
  B. Positive
  C. Negative
  D. Depends on temperature
  
• Q3: Specific heat at constant pressure is:
  A. C_v
  B. C_p
  C. Zero
  D. Infinite
  

Answer Key

• Q1: A - Pressure remains constant in isobaric process.
• Q2: B - Work done is positive during expansion.
• Q3: B - Specific heat at constant pressure is C_p.

Memory Aids

• Isobaric = P constant
• Work = PΔV
• Heat changes internal energy and does work

Key Concepts

• Pressure remains constant
• In expansion: ΔU > 0, ΔW > 0, ΔQ > 0
• In compression: ΔU < 0, ΔW < 0, ΔQ < 0
• Specific heat: C_p = (f/2 + 1) R
• Work done: ΔW = P (V_f - V_i) = μ R (T_f - T_i)

Solved Examples

• Ex 1: Calculate work done when 1 mole gas expands at constant pressure 2 atm from 5 L to 10 L.
  Use ΔW = P ΔV.
  Answer: ΔW = 2 × (10 - 5) = 10 atm·L = 1013 J

Practice Set

• Q1: In isobaric process, pressure is:
  A. Constant
  B. Increasing
  C. Decreasing
  D. Variable
  
• Q2: Work done in isobaric expansion is:
  A. Zero
  B. Positive
  C. Negative
  D. Depends on temperature
  
• Q3: Specific heat at constant pressure is:
  A. C_v
  B. C_p
  C. Zero
  D. Infinite
  

Answer Key

• Q1: A - Pressure remains constant in isobaric process.
• Q2: B - Work done is positive during expansion.
• Q3: B - Specific heat at constant pressure is C_p.

Memory Aids

• Isobaric = P constant
• Work = PΔV
• Heat changes internal energy and does work

Key Concepts

• Pressure remains constant
• In expansion: ΔU > 0, ΔW > 0, ΔQ > 0
• In compression: ΔU < 0, ΔW < 0, ΔQ < 0
• Specific heat: C_p = (f/2 + 1) R
• Work done: ΔW = P (V_f - V_i) = μ R (T_f - T_i)

Solved Examples

• Ex 1: Calculate work done when 1 mole gas expands at constant pressure 2 atm from 5 L to 10 L.
  Use ΔW = P ΔV.
  Answer: ΔW = 2 × (10 - 5) = 10 atm·L = 1013 J

Practice Set

• Q1: In isobaric process, pressure is:
  A. Constant
  B. Increasing
  C. Decreasing
  D. Variable
  
• Q2: Work done in isobaric expansion is:
  A. Zero
  B. Positive
  C. Negative
  D. Depends on temperature
  
• Q3: Specific heat at constant pressure is:
  A. C_v
  B. C_p
  C. Zero
  D. Infinite
  

Answer Key

• Q1: A - Pressure remains constant in isobaric process.
• Q2: B - Work done is positive during expansion.
• Q3: B - Specific heat at constant pressure is C_p.

Memory Aids

• Isobaric = P constant
• Work = PΔV
• Heat changes internal energy and does work

Key Concepts

• Pressure remains constant
• In expansion: ΔU > 0, ΔW > 0, ΔQ > 0
• In compression: ΔU < 0, ΔW < 0, ΔQ < 0
• Specific heat: \( C_p = \left( \frac{f}{2} + 1 \right) R \)
• Work done: \( \Delta W = P (V_f - V_i) = \mu R (T_f - T_i) \)

Solved Examples

• Ex 1: Calculate work done when 1 mole gas expands at constant pressure 2 atm from 5 L to 10 L.
  Use \( \Delta W = P \Delta V \).
  Answer: \( \Delta W = 2 \times (10 - 5) = 10 \text{ atm·L} = 1013 \text{ J} \)

Practice Set

• Q1: In isobaric process, pressure is:
  A. Constant
  B. Increasing
  C. Decreasing
  D. Variable
  
• Q2: Work done in isobaric expansion is:
  A. Zero
  B. Positive
  C. Negative
  D. Depends on temperature
  
• Q3: Specific heat at constant pressure is:
  A. \( C_v \)
  B. \( C_p \)
  C. Zero
  D. Infinite
  

Answer Key

• Q1: A - Pressure remains constant in isobaric process.
• Q2: B - Work done is positive during expansion.
• Q3: B - Specific heat at constant pressure is \( C_p \).

Memory Aids

• Isobaric = P constant
• Work = PΔV
• Heat changes internal energy and does work

Isochoric Process

Key Concepts

• Volume remains constant
• Slope of indicator diagram: (dP/dV) = ∞
• Isometric heating: Pressure and temperature increase, ΔQ > 0, ΔU > 0
• Isometric cooling: Pressure and temperature decrease, ΔQ < 0, ΔU < 0
• Specific heat: C_v = f/2 R
• Work done: ΔW = 0 (since volume constant)

Solved Examples

• Ex 1: Calculate work done in isochoric heating.
  Answer: ΔW = 0

Practice Set

• Q1: In isochoric process, volume is:
  A. Constant
  B. Increasing
  C. Decreasing
  D. Variable
  
• Q2: Work done in isochoric process is:
  A. Positive
  B. Negative
  C. Zero
  D. Depends on pressure
  
• Q3: Specific heat at constant volume is:
  A. C_p
  B. C_v
  C. Zero
  D. Infinite
  

Answer Key

• Q1: A - Volume remains constant in isochoric process.
• Q2: C - Work done is zero as volume does not change.
• Q3: B - Specific heat at constant volume is C_v.

Memory Aids

• Isochoric = V constant
• Work done = 0
• Heat changes internal energy only

Key Concepts

• Volume remains constant
• Slope of indicator diagram: ∞
• Isometric heating: Pressure and temperature increase, ΔQ > 0, ΔU > 0
• Isometric cooling: Pressure and temperature decrease, ΔQ < 0, ΔU < 0
• Specific heat: C_v = f/2 R
• Work done: ΔW = 0 (since volume constant)

Solved Examples

• Ex 1: Calculate work done in isochoric heating.
  Answer: ΔW = 0

Practice Set

• Q1: In isochoric process, volume is:
  A. Constant
  B. Increasing
  C. Decreasing
  D. Variable
  
• Q2: Work done in isochoric process is:
  A. Positive
  B. Negative
  C. Zero
  D. Depends on pressure
  
• Q3: Specific heat at constant volume is:
  A. C_p
  B. C_v
  C. Zero
  D. Infinite
  

Answer Key

• Q1: A - Volume remains constant in isochoric process.
• Q2: C - Work done is zero as volume does not change.
• Q3: B - Specific heat at constant volume is C_v.

Memory Aids

• Isochoric = V constant
• Work done = 0
• Heat changes internal energy only

Key Concepts

• Volume remains constant
• Slope of indicator diagram: (dP/dV) = ∞
• Isometric heating: Pressure and temperature increase, ΔQ > 0, ΔU > 0
• Isometric cooling: Pressure and temperature decrease, ΔQ < 0, ΔU < 0
• Specific heat: C_v = f/2 R
• Work done: ΔW = 0 (since volume constant)

Solved Examples

• Ex 1: Calculate work done in isochoric heating.
  Answer: ΔW = 0

Practice Set

• Q1: In isochoric process, volume is:
  A. Constant
  B. Increasing
  C. Decreasing
  D. Variable
  
• Q2: Work done in isochoric process is:
  A. Positive
  B. Negative
  C. Zero
  D. Depends on pressure
  
• Q3: Specific heat at constant volume is:
  A. C_p
  B. C_v
  C. Zero
  D. Infinite
  

Answer Key

• Q1: A - Volume remains constant in isochoric process.
• Q2: C - Work done is zero as volume does not change.
• Q3: B - Specific heat at constant volume is C_v.

Memory Aids

• Isochoric = V constant
• Work done = 0
• Heat changes internal energy only

Key Concepts

• Volume remains constant
• Slope of indicator diagram: \( \frac{dP}{dV} = \infty \)
• Isometric heating: Pressure and temperature increase, ΔQ > 0, ΔU > 0
• Isometric cooling: Pressure and temperature decrease, ΔQ < 0, ΔU < 0
• Specific heat: \( C_v = \frac{f}{2} R \)
• Work done: \( \Delta W = 0 \) (since volume constant)

Solved Examples

• Ex 1: Calculate work done in isochoric heating.
  Answer: \( \Delta W = 0 \)

Practice Set

• Q1: In isochoric process, volume is:
  A. Constant
  B. Increasing
  C. Decreasing
  D. Variable
  
• Q2: Work done in isochoric process is:
  A. Positive
  B. Negative
  C. Zero
  D. Depends on pressure
  
• Q3: Specific heat at constant volume is:
  A. \( C_p \)
  B. \( C_v \)
  C. Zero
  D. Infinite
  

Answer Key

• Q1: A - Volume remains constant in isochoric process.
• Q2: C - Work done is zero as volume does not change.
• Q3: B - Specific heat at constant volume is \( C_v \).

Memory Aids

• Isochoric = V constant
• Work done = 0
• Heat changes internal energy only

Polytropic Process

Key Concepts

• Follows equation: PV^x = constant, where x is polytropic index
• Special cases: x=0 (isobaric), x=1 (isothermal), x=γ (adiabatic)
• Other relations: TV^x-1 = constant, P^1-x T^x = constant
• Work done: W = nR(T₁ - T₂)/(x-1) = (P₁V₁ - P₂V₂)/(x-1) for x ≠ 1
• Molar heat capacity: C = C_v - R/(x-1) = R/(γ - 1) - R/(x-1)

Solved Examples

• Ex 1: Identify the process type for x=1.
  Answer: Isothermal process
• Ex 2: Calculate work done for polytropic process with given P₁, V₁, P₂, V₂, x.
  Use W = (P₁V₁ - P₂V₂)/(x-1).

Practice Set

• Q1: For x=0, the process is:
  A. Isothermal
  B. Isobaric
  C. Adiabatic
  D. Isochoric
  
• Q2: Work done in polytropic process is:
  A. nR(T₁ - T₂)/(x-1)
  B. Zero
  C. nRT ln V_f/V_i
  D. P(V_f - V_i)
  
• Q3: Molar heat capacity for polytropic process is:
  A. C_v + R/(x-1)
  B. C_v - R/(x-1)
  C. C_p
  D. Zero
  

Answer Key

• Q1: B - For x=0, P=constant (isobaric).
• Q2: A - Work done formula for polytropic process.
• Q3: B - Molar heat capacity formula for polytropic process.

Memory Aids

• Polytropic: PV^x = constant
• x=0: Isobaric, x=1: Isothermal, x=γ: Adiabatic
• Work and heat capacity depend on x

Key Concepts

• Follows equation: PV^x = constant, where x is polytropic index
• Special cases: x=0 (isobaric), x=1 (isothermal), x=γ (adiabatic)
• Other relations: TV^x-1 = constant, P^1-x T^x = constant
• Work done: W = nR(T₁ - T₂)/(x-1) = (P₁V₁ - P₂V₂)/(x-1) for x ≠ 1
• Molar heat capacity: C = C_v - R/(x-1) = R/(γ - 1) - R/(x-1)

Solved Examples

• Ex 1: Identify the process type for x=1.
  Answer: Isothermal process
• Ex 2: Calculate work done for polytropic process with given P₁, V₁, P₂, V₂, x.
  Use W = (P₁V₁ - P₂V₂)/(x-1).

Practice Set

• Q1: For x=0, the process is:
  A. Isothermal
  B. Isobaric
  C. Adiabatic
  D. Isochoric
  
• Q2: Work done in polytropic process is:
  A. nR(T₁ - T₂)/(x-1)
  B. Zero
  C. nRT ln V_f/V_i
  D. P(V_f - V_i)
  
• Q3: Molar heat capacity for polytropic process is:
  A. C_v + R/(x-1)
  B. C_v - R/(x-1)
  C. C_p
  D. Zero
  

Answer Key

• Q1: B - For x=0, P=constant (isobaric).
• Q2: A - Work done formula for polytropic process.
• Q3: B - Molar heat capacity formula for polytropic process.

Memory Aids

• Polytropic: PV^x = constant
• x=0: Isobaric, x=1: Isothermal, x=γ: Adiabatic
• Work and heat capacity depend on x

Key Concepts

• Follows equation: PV^x = constant, where x is polytropic index
• Special cases: x=0 (isobaric), x=1 (isothermal), x=γ (adiabatic)
• Other relations: TV^x-1 = constant, P^1-x T^x = constant
• Work done: W = nR(T₁ - T₂)/(x-1) = (P₁V₁ - P₂V₂)/(x-1) for x ≠ 1
• Molar heat capacity: C = C_v - R/(x-1) = R/(γ - 1) - R/(x-1)

Solved Examples

• Ex 1: Identify the process type for x=1.
  Answer: Isothermal process
• Ex 2: Calculate work done for polytropic process with given P₁, V₁, P₂, V₂, x.
  Use W = (P₁V₁ - P₂V₂)/(x-1).

Practice Set

• Q1: For x=0, the process is:
  A. Isothermal
  B. Isobaric
  C. Adiabatic
  D. Isochoric
  
• Q2: Work done in polytropic process is:
  A. nR(T₁ - T₂)/(x-1)
  B. Zero
  C. nRT ln V_f/V_i
  D. P(V_f - V_i)
  
• Q3: Molar heat capacity for polytropic process is:
  A. C_v + R/(x-1)
  B. C_v - R/(x-1)
  C. C_p
  D. Zero
  

Answer Key

• Q1: B - For x=0, P=constant (isobaric).
• Q2: A - Work done formula for polytropic process.
• Q3: B - Molar heat capacity formula for polytropic process.

Memory Aids

• Polytropic: PV^x = constant
• x=0: Isobaric, x=1: Isothermal, x=γ: Adiabatic
• Work and heat capacity depend on x

Key Concepts

• Follows equation: \( PV^x = \text{constant} \), where \( x \) is polytropic index
• Special cases: \( x=0 \) (isobaric), \( x=1 \) (isothermal), \( x=\gamma \) (adiabatic)
• Other relations: \( TV^{x-1} = \text{constant} \), \( P^{1-x} T^x = \text{constant} \)
• Work done: \( W = \frac{nR(T_1 - T_2)}{x-1} = \frac{P_1V_1 - P_2V_2}{x-1} \) for \( x \neq 1 \)
• Molar heat capacity: \( C = C_v - \frac{R}{x-1} = \frac{R}{\gamma - 1} - \frac{R}{x-1} \)

Solved Examples

• Ex 1: Identify the process type for \( x=1 \).
  Answer: Isothermal process
• Ex 2: Calculate work done for polytropic process with given \( P_1, V_1, P_2, V_2, x \).
  Use \( W = \frac{P_1V_1 - P_2V_2}{x-1} \).

Practice Set

• Q1: For \( x=0 \), the process is:
  A. Isothermal
  B. Isobaric
  C. Adiabatic
  D. Isochoric
  
• Q2: Work done in polytropic process is:
  A. \( \frac{nR(T_1 - T_2)}{x-1} \)
  B. Zero
  C. \( nRT \ln \frac{V_f}{V_i} \)
  D. \( P(V_f - V_i) \)
  
• Q3: Molar heat capacity for polytropic process is:
  A. \( C_v + \frac{R}{x-1} \)
  B. \( C_v - \frac{R}{x-1} \)
  C. \( C_p \)
  D. Zero
  

Answer Key

• Q1: B - For x=0, P=constant (isobaric).
• Q2: A - Work done formula for polytropic process.
• Q3: B - Molar heat capacity formula for polytropic process.

Memory Aids

• Polytropic: PV^x = constant
• x=0: Isobaric, x=1: Isothermal, x=γ: Adiabatic
• Work and heat capacity depend on x

Second Law of Thermodynamics

Key Points

• A reversible process can be reversed by infinitesimal changes in external conditions.
• Pressure and temperature of the working substance must closely match surroundings at all times.
• Processes must be carried out infinitely slowly (quasi-static).
• Engine parts must be frictionless.
• No energy loss by conduction, convection, or radiation.
• Ideal case; real processes always have some energy loss.

Solved Examples

• Ex 1: Calculate maximum efficiency of a heat engine operating between 500 K and 300 K.
  • Use η_max = 1 - T₂/T₁ = 1 - 300/500 = 0.4 = 40%

Practice Set

• Q1: Which of the following is impossible according to the second law?
  • A. Heat engine with 100% efficiency
  • B. Refrigerator with finite COP
  • C. Heat transfer from hot to cold spontaneously
  • D. Work done by expanding gas
• Q2: The second law implies that
  • A. Energy can be created
  • B. Heat engines can convert all heat to work
  • C. Some heat must be rejected to sink
  • D. Refrigerators can have infinite COP
• Q3: A heat engine absorbs 600 J from a hot reservoir and rejects 400 J to a cold reservoir. Its efficiency is
  • A. 33.3%
  • B. 40%
  • C. 66.7%
  • D. 50%

Answer Key

• Q1: A - 100% efficiency violates the second law.
• Q2: C - Some heat must be rejected; full conversion is impossible.
• Q3: A - Efficiency = W/Q₁ = (600-400)/600 = 200/600 = 33.3%

Memory Aids

• Efficiency η < 1
• Heat flows hot to cold naturally
• Work output < heat input

Key Points

• The second law restricts processes allowed by the first law.
• No heat engine can have efficiency η = 1 (100% efficiency).
• No refrigerator can have an infinite coefficient of performance (COP).

Solved Examples

• Ex 1: Calculate maximum efficiency of a heat engine operating between 500 K and 300 K.
  • Use η_max = 1 - T₂/T₁ = 1 - 300/500 = 0.4 = 40%

Practice Set

• Q1: Which of the following is impossible according to the second law?
  • A. Heat engine with 100% efficiency
  • B. Refrigerator with finite COP
  • C. Heat transfer from hot to cold spontaneously
  • D. Work done by expanding gas
• Q2: The second law implies that
  • A. Energy can be created
  • B. Heat engines can convert all heat to work
  • C. Some heat must be rejected to sink
  • D. Refrigerators can have infinite COP
• Q3: A heat engine absorbs 600 J from a hot reservoir and rejects 400 J to a cold reservoir. Its efficiency is
  • A. 33.3%
  • B. 40%
  • C. 66.7%
  • D. 50%

Answer Key

• Q1: A - 100% efficiency violates the second law.
• Q2: C - Some heat must be rejected; full conversion is impossible.
• Q3: A - Efficiency = W/Q₁ = (600-400)/600 = 200/600 = 33.3%

Memory Aids

• Efficiency η < 1
• Heat flows hot to cold naturally
• Work output < heat input

Key Points

• A reversible process can be reversed by infinitesimal changes in external conditions.
• Pressure and temperature of the working substance must closely match surroundings at all times.
• Processes must be carried out infinitely slowly (quasi-static).
• Engine parts must be frictionless.
• No energy loss by conduction, convection, or radiation.
• Ideal case; real processes always have some energy loss.

Solved Examples

• Ex 1: Calculate maximum efficiency of a heat engine operating between 500 K and 300 K.
  • Use η_max = 1 - T₂/T₁ = 1 - 300/500 = 0.4 = 40%

Practice Set

• Q1: Which of the following is impossible according to the second law?
  • A. Heat engine with 100% efficiency
  • B. Refrigerator with finite COP
  • C. Heat transfer from hot to cold spontaneously
  • D. Work done by expanding gas
• Q2: The second law implies that
  • A. Energy can be created
  • B. Heat engines can convert all heat to work
  • C. Some heat must be rejected to sink
  • D. Refrigerators can have infinite COP
• Q3: A heat engine absorbs 600 J from a hot reservoir and rejects 400 J to a cold reservoir. Its efficiency is
  • A. 33.3%
  • B. 40%
  • C. 66.7%
  • D. 50%

Answer Key

• Q1: A - 100% efficiency violates the second law.
• Q2: C - Some heat must be rejected; full conversion is impossible.
• Q3: A - Efficiency = W/Q₁ = (600-400)/600 = 200/600 = 33.3%

Memory Aids

• Efficiency η < 1
• Heat flows hot to cold naturally
• Work output < heat input

Key Points

• The second law restricts processes allowed by the first law.
• No heat engine can have efficiency \( \eta = 1 \) (100% efficiency).
• No refrigerator can have an infinite coefficient of performance (COP).

Solved Examples

• Ex 1: Calculate maximum efficiency of a heat engine operating between 500 K and 300 K.
  • Use \( \eta_{max} = 1 - \frac{T_2}{T_1} = 1 - \frac{300}{500} = 0.4 = 40\% \)

Practice Set

• Q1: Which of the following is impossible according to the second law?
  • A. Heat engine with 100% efficiency
  • B. Refrigerator with finite COP
  • C. Heat transfer from hot to cold spontaneously
  • D. Work done by expanding gas
• Q2: The second law implies that
  • A. Energy can be created
  • B. Heat engines can convert all heat to work
  • C. Some heat must be rejected to sink
  • D. Refrigerators can have infinite COP
• Q3: A heat engine absorbs 600 J from a hot reservoir and rejects 400 J to a cold reservoir. Its efficiency is
  • A. 33.3%
  • B. 40%
  • C. 66.7%
  • D. 50%

Answer Key

• Q1: A - 100% efficiency violates the second law.
• Q2: C - Some heat must be rejected; full conversion is impossible.
• Q3: A - Efficiency = \( \frac{W}{Q_1} = \frac{600-400}{600} = \frac{200}{600} = 33.3\% \)

Memory Aids

• Efficiency \( \eta < 1 \)
• Heat flows hot to cold naturally
• Work output < heat input

Kelvin-Planck Statement

Key Points

• No process can convert all absorbed heat from a reservoir into work alone.
• Heat engine must reject some heat to a cooler sink.
• Mathematically, η = W/Q₁ < 1.

Solved Examples

• Ex 1: A heat engine absorbs 1000 J heat and does 700 J work. Is it violating Kelvin-Planck?
  • Efficiency = 700/1000 = 0.7 < 1 → No violation.

Practice Set

• Q1: Kelvin-Planck statement forbids
  • A. 100% efficient heat engine
  • B. Refrigerator operation
  • C. Heat transfer from cold to hot
  • D. Work done by gas expansion
• Q2: If a heat engine absorbs 500 J and does 500 J work, what happens?
  • A. Allowed by Kelvin-Planck
  • B. Violates Kelvin-Planck
  • C. Allowed by Clausius
  • D. Violates first law
• Q3: Which is true for Kelvin-Planck?
  • A. η = 1 possible
  • B. Some heat must be rejected
  • C. Heat flows cold to hot
  • D. Work input required

Answer Key

• Q1: A - 100% efficient heat engine is impossible.
• Q2: B - Complete conversion of heat to work is forbidden.
• Q3: B - Heat engine must reject some heat.

Memory Aids

• "No 100% heat to work"
• Heat engine rejects heat
• Efficiency always < 1

Key Points

• No process can convert all absorbed heat from a reservoir into work alone.
• Heat engine must reject some heat to a cooler sink.
• Mathematically, η = W/Q₁ < 1.

Solved Examples

• Ex 1: A heat engine absorbs 1000 J heat and does 700 J work. Is it violating Kelvin-Planck?
  • Efficiency = 700/1000 = 0.7 < 1 → No violation.

Practice Set

• Q1: Kelvin-Planck statement forbids
  • A. 100% efficient heat engine
  • B. Refrigerator operation
  • C. Heat transfer from cold to hot
  • D. Work done by gas expansion
• Q2: If a heat engine absorbs 500 J and does 500 J work, what happens?
  • A. Allowed by Kelvin-Planck
  • B. Violates Kelvin-Planck
  • C. Allowed by Clausius
  • D. Violates first law
• Q3: Which is true for Kelvin-Planck?
  • A. η = 1 possible
  • B. Some heat must be rejected
  • C. Heat flows cold to hot
  • D. Work input required

Answer Key

• Q1: A - 100% efficient heat engine is impossible.
• Q2: B - Complete conversion of heat to work is forbidden.
• Q3: B - Heat engine must reject some heat.

Memory Aids

• "No 100% heat to work"
• Heat engine rejects heat
• Efficiency always < 1

Key Points

• No process can convert all absorbed heat from a reservoir into work alone.
• Heat engine must reject some heat to a cooler sink.
• Mathematically, η = W/Q₁ < 1.

Solved Examples

• Ex 1: A heat engine absorbs 1000 J heat and does 700 J work. Is it violating Kelvin-Planck?
  • Efficiency = 700/1000 = 0.7 < 1 → No violation.

Practice Set

• Q1: Kelvin-Planck statement forbids
  • A. 100% efficient heat engine
  • B. Refrigerator operation
  • C. Heat transfer from cold to hot
  • D. Work done by gas expansion
• Q2: If a heat engine absorbs 500 J and does 500 J work, what happens?
  • A. Allowed by Kelvin-Planck
  • B. Violates Kelvin-Planck
  • C. Allowed by Clausius
  • D. Violates first law
• Q3: Which is true for Kelvin-Planck?
  • A. η = 1 possible
  • B. Some heat must be rejected
  • C. Heat flows cold to hot
  • D. Work input required

Answer Key

• Q1: A - 100% efficient heat engine is impossible.
• Q2: B - Complete conversion of heat to work is forbidden.
• Q3: B - Heat engine must reject some heat.

Memory Aids

• "No 100% heat to work"
• Heat engine rejects heat
• Efficiency always < 1

Key Points

• No process can convert all absorbed heat from a reservoir into work alone.
• Heat engine must reject some heat to a cooler sink.
• Mathematically, \( \eta = \frac{W}{Q_1} < 1 \).

Solved Examples

• Ex 1: A heat engine absorbs 1000 J heat and does 700 J work. Is it violating Kelvin-Planck?
  • Efficiency = \( \frac{700}{1000} = 0.7 < 1 \) → No violation.

Practice Set

• Q1: Kelvin-Planck statement forbids
  • A. 100% efficient heat engine
  • B. Refrigerator operation
  • C. Heat transfer from cold to hot
  • D. Work done by gas expansion
• Q2: If a heat engine absorbs 500 J and does 500 J work, what happens?
  • A. Allowed by Kelvin-Planck
  • B. Violates Kelvin-Planck
  • C. Allowed by Clausius
  • D. Violates first law
• Q3: Which is true for Kelvin-Planck?
  • A. \( \eta = 1 \) possible
  • B. Some heat must be rejected
  • C. Heat flows cold to hot
  • D. Work input required

Answer Key

• Q1: A - 100% efficient heat engine is impossible.
• Q2: B - Complete conversion of heat to work is forbidden.
• Q3: B - Heat engine must reject some heat.

Memory Aids

• "No 100% heat to work"
• Heat engine rejects heat
• Efficiency always < 1

Clausius Statement

Key Points

• Heat cannot spontaneously flow from a colder to a hotter body.
• Refrigerators require work input to transfer heat from cold to hot.
• Heat transfer from cold to hot without work violates this law.

Solved Examples

• Ex 1: Heat Q = 200 J moves from 270 K to 300 K with 50 J work input. Is this allowed?
  • Yes, work input enables heat transfer from cold to hot.

Practice Set

• Q1: Clausius statement forbids
  • A. Heat flow hot to cold
  • B. Heat flow cold to hot without work
  • C. Work done on gas
  • D. Heat engine operation
• Q2: Which device requires work to transfer heat from cold to hot?
  • A. Heat engine
  • B. Refrigerator
  • C. Thermal conductor
  • D. Adiabatic system
• Q3: Heat flows spontaneously from 280 K to 300 K. What is true?
  • A. Violates Clausius
  • B. Allowed naturally
  • C. Requires work
  • D. Impossible

Answer Key

• Q1: B - Heat cannot flow cold to hot without work.
• Q2: B - Refrigerator moves heat from cold to hot using work.
• Q3: B - Heat flows hot to cold naturally.

Memory Aids

• "Heat flows hot to cold"
• Refrigerator needs work
• Cold to hot needs energy

Key Points

• Heat cannot spontaneously flow from a colder to a hotter body.
• Refrigerators require work input to transfer heat from cold to hot.
• Heat transfer from cold to hot without work violates this law.

Solved Examples

• Ex 1: Heat Q = 200 J moves from 270 K to 300 K with 50 J work input. Is this allowed?
  • Yes, work input enables heat transfer from cold to hot.

Practice Set

• Q1: Clausius statement forbids
  • A. Heat flow hot to cold
  • B. Heat flow cold to hot without work
  • C. Work done on gas
  • D. Heat engine operation
• Q2: Which device requires work to transfer heat from cold to hot?
  • A. Heat engine
  • B. Refrigerator
  • C. Thermal conductor
  • D. Adiabatic system
• Q3: Heat flows spontaneously from 280 K to 300 K. What is true?
  • A. Violates Clausius
  • B. Allowed naturally
  • C. Requires work
  • D. Impossible

Answer Key

• Q1: B - Heat cannot flow cold to hot without work.
• Q2: B - Refrigerator moves heat from cold to hot using work.
• Q3: B - Heat flows hot to cold naturally.

Memory Aids

• "Heat flows hot to cold"
• Refrigerator needs work
• Cold to hot needs energy

Key Points

• Heat cannot spontaneously flow from a colder to a hotter body.
• Refrigerators require work input to transfer heat from cold to hot.
• Heat transfer from cold to hot without work violates this law.

Solved Examples

• Ex 1: Heat Q = 200 J moves from 270 K to 300 K with 50 J work input. Is this allowed?
  • Yes, work input enables heat transfer from cold to hot.

Practice Set

• Q1: Clausius statement forbids
  • A. Heat flow hot to cold
  • B. Heat flow cold to hot without work
  • C. Work done on gas
  • D. Heat engine operation
• Q2: Which device requires work to transfer heat from cold to hot?
  • A. Heat engine
  • B. Refrigerator
  • C. Thermal conductor
  • D. Adiabatic system
• Q3: Heat flows spontaneously from 280 K to 300 K. What is true?
  • A. Violates Clausius
  • B. Allowed naturally
  • C. Requires work
  • D. Impossible

Answer Key

• Q1: B - Heat cannot flow cold to hot without work.
• Q2: B - Refrigerator moves heat from cold to hot using work.
• Q3: B - Heat flows hot to cold naturally.

Memory Aids

• "Heat flows hot to cold"
• Refrigerator needs work
• Cold to hot needs energy

Key Points

• Heat cannot spontaneously flow from a colder to a hotter body.
• Refrigerators require work input to transfer heat from cold to hot.
• Heat transfer from cold to hot without work violates this law.

Solved Examples

• Ex 1: Heat \( Q = 200 \) J moves from 270 K to 300 K with 50 J work input. Is this allowed?
  • Yes, work input enables heat transfer from cold to hot.

Practice Set

• Q1: Clausius statement forbids
  • A. Heat flow hot to cold
  • B. Heat flow cold to hot without work
  • C. Work done on gas
  • D. Heat engine operation
• Q2: Which device requires work to transfer heat from cold to hot?
  • A. Heat engine
  • B. Refrigerator
  • C. Thermal conductor
  • D. Adiabatic system
• Q3: Heat flows spontaneously from 280 K to 300 K. What is true?
  • A. Violates Clausius
  • B. Allowed naturally
  • C. Requires work
  • D. Impossible

Answer Key

• Q1: B - Heat cannot flow cold to hot without work.
• Q2: B - Refrigerator moves heat from cold to hot using work.
• Q3: B - Heat flows hot to cold naturally.

Memory Aids

• "Heat flows hot to cold"
• Refrigerator needs work
• Cold to hot needs energy

Heat Engines and Refrigerators

Key Points

• Heat Engine (a):
  • Heat Q₁ absorbed from hot source at temperature T₁.
  • Work W done by the engine.
  • Heat Q₂ rejected to cold sink at temperature T₂.
• Refrigerator or Heat Pump (b):
  • Heat Q absorbed from cold source at temperature T₂ < T₁.
  • Work done on system to transfer heat to hot reservoir at T₁.

Solved Examples

• Ex 1: Calculate efficiency of heat engine absorbing 800 J and rejecting 500 J.
  • Work done W = 800 - 500 = 300 J
  • Efficiency η = W/Q₁ = 300/800 = 0.375 = 37.5%
• Ex 2: Calculate COP of refrigerator absorbing 200 J from cold reservoir and work input 50 J.
  • COP = Q/W = 200/50 = 4

Practice Set

• Q1: In a heat engine, heat rejected is 400 J and heat absorbed is 1000 J. Efficiency is
  • A. 40%
  • B. 60%
  • C. 50%
  • D. 30%
• Q2: A refrigerator absorbs 300 J heat from cold reservoir and work done is 100 J. COP is
  • A. 3
  • B. 4
  • C. 2
  • D. 1
• Q3: Which is true for refrigerators?
  • A. They convert heat fully to work
  • B. They transfer heat from cold to hot using work
  • C. They operate without work input
  • D. They violate second law

Answer Key

• Q1: B - Carnot engine is an ideal engine with maximum efficiency.
• Q2: B - Efficiency depends only on temperatures of source and sink.
• Q3: B - ( η = 1 - 400/500 = 0.2 = 20% )

Memory Aids

• Isothermal compression: ( Q = W = RT ln V₃/V₄ )
• Work done on gas is negative of expansion work.

Key Points

• Heat Engine (a):
  • Heat \( Q_1 \) absorbed from hot source at temperature \( T_1 \).
  • Work \( W \) done by the engine.
  • Heat \( Q_2 \) rejected to cold sink at temperature \( T_2 \).
• Refrigerator or Heat Pump (b):
  • Heat \( Q \) absorbed from cold source at temperature \( T_2 < T_1 \).
  • Work done on system to transfer heat to hot reservoir at \( T_1 \).

Solved Examples

• Ex 1: Calculate efficiency of heat engine absorbing 800 J and rejecting 500 J.
  • Work done \( W = 800 - 500 = 300 \) J
  • Efficiency \( \eta = \frac{W}{Q_1} = \frac{300}{800} = 0.375 = 37.5\% \)
• Ex 2: Calculate COP of refrigerator absorbing 200 J from cold reservoir and work input 50 J.
  • COP \( = \frac{Q}{W} = \frac{200}{50} = 4 \)

Practice Set

• Q1: In a heat engine, heat rejected is 400 J and heat absorbed is 1000 J. Efficiency is
  • A. 40%
  • B. 60%
  • C. 50%
  • D. 30%
• Q2: A refrigerator absorbs 300 J heat from cold reservoir and work done is 100 J. COP is
  • A. 3
  • B. 4
  • C. 2
  • D. 1
• Q3: Which is true for refrigerators?
  • A. They convert heat fully to work
  • B. They transfer heat from cold to hot using work
  • C. They operate without work input
  • D. They violate second law

Answer Key

• Q1: A - Efficiency = \( \frac{1000-400}{1000} = 0.6 = 60\% \)
• Q2: A - COP = \( \frac{300}{100} = 3 \)
• Q3: B - Refrigerators use work to transfer heat from cold to hot.

Memory Aids

• Heat engine: \( \eta = 1 - \frac{Q_2}{Q_1} \)
• Refrigerator: COP = \( \frac{Q}{W} \)
• Work input needed for cold to hot heat transfer

Reversible and Irreversible Processes

Reversible process

• A reversible process can be reversed by infinitesimal changes in external conditions.
• Pressure and temperature of the working substance must closely match surroundings at all times.
• Processes must be carried out infinitely slowly (quasi-static).
• Engine parts must be frictionless.
• No energy loss by conduction, convection, or radiation.
• Ideal case; real processes always have some energy loss.

Solved Examples

• Ex 1: A gas expands slowly and reversibly at temperature 300 K. What is the condition for reversibility?
  Answer: Pressure and temperature of gas ≈ surroundings; process is quasi-static and frictionless.

Practice Set

• Q1: Which condition is NOT required for a reversible process?
  A. Infinitely slow process
  B. Frictionless parts
  C. Large temperature difference with surroundings
  D. No energy loss
  
• Q2: A reversible process is characterized by:
  A. Rapid changes
  B. Finite temperature difference
  C. Infinitesimal changes in external conditions
  D. Presence of friction
  
• Q3: Which of the following is true for a reversible heat engine?
  A. Operates with friction
  B. Has energy loss by radiation
  C. Pressure of working substance equals surroundings
  D. Operates rapidly
  

Answer Key

• Q1: C - Large temperature difference causes irreversibility.
• Q2: C - Reversible processes involve infinitesimal changes.
• Q3: C - Pressure equality is essential for reversibility.

Memory Aids

• Reversible = Infinitesimal changes + No friction + No energy loss.
• Slow and steady = reversible.

Reversible process

• A reversible process can be reversed by infinitesimal changes in external conditions.
• Pressure and temperature of the working substance must closely match surroundings at all times.
• Processes must be carried out infinitely slowly (quasi-static).
• Engine parts must be frictionless.
• No energy loss by conduction, convection, or radiation.
• Ideal case; real processes always have some energy loss.

Solved Examples

• Ex 1: A gas expands slowly and reversibly at temperature 300 K. What is the condition for reversibility?
  Answer: Pressure and temperature of gas ≈ surroundings; process is quasi-static and frictionless.

Practice Set

• Q1: Which condition is NOT required for a reversible process?
  A. Infinitely slow process
  B. Frictionless parts
  C. Large temperature difference with surroundings
  D. No energy loss
  
• Q2: A reversible process is characterized by:
  A. Rapid changes
  B. Finite temperature difference
  C. Infinitesimal changes in external conditions
  D. Presence of friction
  
• Q3: Which of the following is true for a reversible heat engine?
  A. Operates with friction
  B. Has energy loss by radiation
  C. Pressure of working substance equals surroundings
  D. Operates rapidly
  

Answer Key

• Q1: C - Large temperature difference causes irreversibility.
• Q2: C - Reversible processes involve infinitesimal changes.
• Q3: C - Pressure equality is essential for reversibility.

Memory Aids

• Reversible = Infinitesimal changes + No friction + No energy loss.
• Slow and steady = reversible.

Irreversible Process

Irreversible process

• All natural processes are irreversible.
• Thermodynamic equilibrium conditions (mechanical, thermal, chemical) are not met.
• Processes are not quasi-static (not infinitely slow).
• Dissipative effects like friction, viscosity, electric resistance are present.
• Example: Spontaneous expansion of gas into vacuum.

Solved Examples

• Ex 1: Gas expands spontaneously into evacuated space. Is this reversible?
  Answer: No, due to dissipative effects and non-equilibrium.

Practice Set

• Q1: Which is a cause of irreversibility?
  A. Friction
  B. Infinitesimal changes
  C. No energy loss
  D. Quasi-static process
  
• Q2: Natural processes are generally:
  A. Reversible
  B. Irreversible
  C. Frictionless
  D. Infinitely slow
  
• Q3: Spontaneous expansion of gas into vacuum is:
  A. Reversible
  B. Irreversible
  C. Frictionless
  D. Quasi-static
  

Answer Key

• Q1: A - Friction causes energy dissipation and irreversibility.
• Q2: B - Natural processes are irreversible.
• Q3: B - Spontaneous expansion is irreversible.

Memory Aids

• Irreversible = Friction + Dissipation + Non-equilibrium.
• Natural processes ≠ reversible.

Irreversible process

• All natural processes are irreversible.
• Thermodynamic equilibrium conditions (mechanical, thermal, chemical) are not met.
• Processes are not quasi-static (not infinitely slow).
• Dissipative effects like friction, viscosity, electric resistance are present.
• Example: Spontaneous expansion of gas into vacuum.

Solved Examples

• Ex 1: Gas expands spontaneously into evacuated space. Is this reversible?
  Answer: No, due to dissipative effects and non-equilibrium.

Practice Set

• Q1: Which is a cause of irreversibility?
  A. Friction
  B. Infinitesimal changes
  C. No energy loss
  D. Quasi-static process
  
• Q2: Natural processes are generally:
  A. Reversible
  B. Irreversible
  C. Frictionless
  D. Infinitely slow
  
• Q3: Spontaneous expansion of gas into vacuum is:
  A. Reversible
  B. Irreversible
  C. Frictionless
  D. Quasi-static
  

Answer Key

• Q1: A - Friction causes energy dissipation and irreversibility.
• Q2: B - Natural processes are irreversible.
• Q3: B - Spontaneous expansion is irreversible.

Memory Aids

• Irreversible = Friction + Dissipation + Non-equilibrium.
• Natural processes ≠ reversible.

Carnot Engine

Definition and Characteristics

• Theoretical ideal heat engine with no practical imperfections.
• Has maximum possible efficiency.
• Cannot be realized in practice.

Solved Examples

• Ex 1: Calculate efficiency if T1 = 600 K and T2 = 300 K.
  • Use formula ( η = 1 - T₂/T₁ = 1 - 300/600 = 0.5 = 50% )

Practice Set

• Q1: What is the nature of Carnot engine?
  • A) Practical engine
  • B) Ideal engine with maximum efficiency
  • C) Engine with losses
  • D) Real engine
• Q2: Carnot engine efficiency depends on?
  • A) Working substance
  • B) Temperature of source and sink
  • C) Pressure only
  • D) Volume only
• Q3: If T1 = 500 K and T2 = 400 K, efficiency is?
  • A) 0.2
  • B) 0.1
  • C) 0.3
  • D) 0.4

Answer Key

• Q1: B - Carnot engine is an ideal engine with maximum efficiency.
• Q2: B - Efficiency depends only on temperatures of source and sink.
• Q3: B - ( η = 1 - 400/500 = 0.2 = 20% )

Memory Aids

• Efficiency ( η = 1 - T₂/T₁ )
• Maximum efficiency engine is ideal Carnot engine.

Definition and Characteristics

• Theoretical ideal heat engine with no practical imperfections.
• Has maximum possible efficiency.
• Cannot be realized in practice.

Solved Examples

• Ex 1: Calculate efficiency if T1 = 600 K and T2 = 300 K.
  • Use formula η = 1 - T₂/T₁ = 1 - 300/600 = 0.5 = 50%

Practice Set

• Q1: What is the nature of Carnot engine?
  • A) Practical engine
  • B) Ideal engine with maximum efficiency
  • C) Engine with losses
  • D) Real engine
• Q2: Carnot engine efficiency depends on?
  • A) Working substance
  • B) Temperature of source and sink
  • C) Pressure only
  • D) Volume only
• Q3: If T1 = 500 K and T2 = 400 K, efficiency is?
  • A) 0.2
  • B) 0.1
  • C) 0.3
  • D) 0.4

Answer Key

• Q1: B - Carnot engine is an ideal engine with maximum efficiency.
• Q2: B - Efficiency depends only on temperatures of source and sink.
• Q3: B - ( η = 1 - 400/500 = 0.2 = 20% )

Memory Aids

• Efficiency η = 1 - T₂/T₁
• Maximum efficiency engine is ideal Carnot engine.

Definition and Characteristics

• Theoretical ideal heat engine with no practical imperfections.
• Has maximum possible efficiency.
• Cannot be realized in practice.

Solved Examples

• Ex 1: Calculate efficiency if T1 = 600 K and T2 = 300 K.
  • Use formula ( η = 1 - T₂/T₁ = 1 - 300/600 = 0.5 = 50% )

Practice Set

• Q1: What is the nature of Carnot engine?
  • A) Practical engine
  • B) Ideal engine with maximum efficiency
  • C) Engine with losses
  • D) Real engine
• Q2: Carnot engine efficiency depends on?
  • A) Working substance
  • B) Temperature of source and sink
  • C) Pressure only
  • D) Volume only
• Q3: If T1 = 500 K and T2 = 400 K, efficiency is?
  • A) 0.2
  • B) 0.1
  • C) 0.3
  • D) 0.4

Answer Key

• Q1: B - Carnot engine is an ideal engine with maximum efficiency.
• Q2: B - Efficiency depends only on temperatures of source and sink.
• Q3: B - ( η = 1 - 400/500 = 0.2 = 20% )

Memory Aids

• Efficiency ( η = 1 - T₂/T₁ )
• Maximum efficiency engine is ideal Carnot engine.

Definition and Characteristics

• Theoretical ideal heat engine with no practical imperfections.
• Has maximum possible efficiency.
• Cannot be realized in practice.

Solved Examples

• Ex 1: Calculate efficiency if T1 = 600 K and T2 = 300 K.
  • Use formula \( \eta = 1 - \frac{T_2}{T_1} = 1 - \frac{300}{600} = 0.5 = 50\% \)

Practice Set

• Q1: What is the nature of Carnot engine?
  • A) Practical engine
  • B) Ideal engine with maximum efficiency
  • C) Engine with losses
  • D) Real engine
• Q2: Carnot engine efficiency depends on?
  • A) Working substance
  • B) Temperature of source and sink
  • C) Pressure only
  • D) Volume only
• Q3: If T1 = 500 K and T2 = 400 K, efficiency is?
  • A) 0.2
  • B) 0.1
  • C) 0.3
  • D) 0.4

Answer Key

• Q1: B - Carnot engine is an ideal engine with maximum efficiency.
• Q2: B - Efficiency depends only on temperatures of source and sink.
• Q3: B - \( \eta = 1 - \frac{400}{500} = 0.2 = 20\% \)

Memory Aids

• Efficiency \( \eta = 1 - \frac{T_2}{T_1} \)
• Maximum efficiency engine is ideal Carnot engine.

Carnot Cycle

Cycle Description

• Four quasi-static processes: two isothermal and two adiabatic.
• States: A, B, C, D with specific P, V, T.
• Path: A → B (isothermal expansion), B → C (adiabatic expansion), C → D (isothermal compression), D → A (adiabatic compression).
• Work done = area enclosed in P-V diagram.

Solved Examples

• Ex 1: Identify processes in Carnot cycle from P-V diagram.
  • AB and CD: Isothermal (constant T)
  • BC and DA: Adiabatic (no heat exchange)

Practice Set

• Q1: Which process in Carnot cycle occurs at constant temperature?
  • A) AB and BC
  • B) AB and CD
  • C) BC and DA
  • D) CD and DA
• Q2: During which process is no heat exchanged?
  • A) AB
  • B) BC
  • C) CD
  • D) AB and CD
• Q3: The net work done in Carnot cycle is represented by?
  • A) Area under AB
  • B) Area under BC
  • C) Area enclosed by ABCD
  • D) Area under CD

Answer Key

• Q1: B - AB and CD are isothermal processes.
• Q2: B - BC and DA are adiabatic with no heat exchange.
• Q3: C - Net work done equals area enclosed by ABCD.

Memory Aids

• Isothermal: constant T, heat exchange occurs.
• Adiabatic: no heat exchange, temperature changes.
• Cycle path: A-B-C-D-A.

Cycle Description

• Four quasi-static processes: two isothermal and two adiabatic.
• States: A, B, C, D with specific P, V, T.
• Path: A → B (isothermal expansion), B → C (adiabatic expansion), C → D (isothermal compression), D → A (adiabatic compression).
• Work done = area enclosed in P-V diagram.

Solved Examples

• Ex 1: Identify processes in Carnot cycle from P-V diagram.
  • AB and CD: Isothermal (constant T)
  • BC and DA: Adiabatic (no heat exchange)

Practice Set

• Q1: Which process in Carnot cycle occurs at constant temperature?
  • A) AB and BC
  • B) AB and CD
  • C) BC and DA
  • D) CD and DA
• Q2: During which process is no heat exchanged?
  • A) AB
  • B) BC
  • C) CD
  • D) AB and CD
• Q3: The net work done in Carnot cycle is represented by?
  • A) Area under AB
  • B) Area under BC
  • C) Area enclosed by ABCD
  • D) Area under CD

Answer Key

• Q1: B - AB and CD are isothermal processes.
• Q2: B - BC and DA are adiabatic with no heat exchange.
• Q3: C - Net work done equals area enclosed by ABCD.

Memory Aids

• Isothermal: constant T, heat exchange occurs.
• Adiabatic: no heat exchange, temperature changes.
• Cycle path: A-B-C-D-A.

Isothermal Expansion

Process Details

• Gas compresses at constant temperature ( T₂ ).
• Heat ( Q₂ ) is rejected to sink.
• Work done on gas ( W₃ = Q₂ = RT₂ ln V₃/V₄ ).
• Negative work indicates work done on the system.

Solved Examples

• Ex 1: Calculate work done if ( R=8.31 J/mol.K, T₁=300 K, V₂=4V₁ ).
  • ( W₁ = 8.31 × 300 × ln 4 = 8.31 × 300 × 1.386 = 3453.5 J )

Practice Set

• Q1: During isothermal expansion, temperature is?
  • A) Increasing
  • B) Constant
  • C) Decreasing
  • D) Variable
• Q2: Work done in isothermal expansion is?
  • A) Zero
  • B) ( RT₁ ln V₂/V₁ )
  • C) ( P ΔV )
  • D) ( C_V ΔT )
• Q3: Heat absorbed during isothermal expansion is?
  • A) Zero
  • B) Equal to work done
  • C) Greater than work done
  • D) Less than work done

Answer Key

• Q1: B - Temperature remains constant.
• Q2: B - Work done formula for isothermal expansion.
• Q3: B - Heat absorbed equals work done in isothermal process.

Memory Aids

• Isothermal: ( Q = W = RT ln V₂/V₁ )
• Temperature constant during expansion.

Process Details

• Gas expands at constant temperature ( T₁ ) absorbing heat ( Q₁ ) from source.
• Work done by gas ( W₁ = Q₁ = RT₁ ln V₂/V₁ ).
• Heat supplied maintains temperature during expansion.

Solved Examples

• Ex 1: Calculate work done if ( R=8.31 J/mol.K, T₁=300 K, V₂=4V₁ ).
  • ( W₁ = 8.31 × 300 × ln 4 = 8.31 × 300 × 1.386 = 3453.5 J )

Practice Set

• Q1: During isothermal expansion, temperature is?
  • A) Increasing
  • B) Constant
  • C) Decreasing
  • D) Variable
• Q2: Work done in isothermal expansion is?
  • A) Zero
  • B) ( RT₁ ln V₂/V₁ )
  • C) ( P ΔV )
  • D) ( C_V ΔT )
• Q3: Heat absorbed during isothermal expansion is?
  • A) Zero
  • B) Equal to work done
  • C) Greater than work done
  • D) Less than work done

Answer Key

• Q1: B - Temperature remains constant.
• Q2: B - Work done formula for isothermal expansion.
• Q3: B - Heat absorbed equals work done in isothermal process.

Memory Aids

• Isothermal: ( Q = W = RT ln V₂/V₁ )
• Temperature constant during expansion.

Process Details

• Gas expands at constant temperature T₁ absorbing heat Q₁ from source.
• Work done by gas W₁ = Q₁ = RT₁ ln V₂/V₁.
• Heat supplied maintains temperature during expansion.

Solved Examples

• Ex 1: Calculate work done if R=8.31 J/mol.K, T₁=300 K, V₂=4V₁.
  • W₁ = 8.31 × 300 × ln 4 = 8.31 × 300 × 1.386 = 3453.5 J

Practice Set

• Q1: During isothermal expansion, temperature is?
  • A) Increasing
  • B) Constant
  • C) Decreasing
  • D) Variable
• Q2: Work done in isothermal expansion is?
  • A) Zero
  • B) RT₁ ln V₂/V₁
  • C) P ΔV
  • D) C_V ΔT
• Q3: Heat absorbed during isothermal expansion is?
  • A) Zero
  • B) Equal to work done
  • C) Greater than work done
  • D) Less than work done

Answer Key

• Q1: B - Temperature remains constant.
• Q2: B - Work done formula for isothermal expansion.
• Q3: B - Heat absorbed equals work done in isothermal process.

Memory Aids

• Isothermal: Q = W = RT ln V₂/V₁
• Temperature constant during expansion.

Process Details

• Gas expands at constant temperature ( T₁ ) absorbing heat ( Q₁ ) from source.
• Work done by gas ( W₁ = Q₁ = RT₁ ln V₂/V₁ ).
• Heat supplied maintains temperature during expansion.

Solved Examples

• Ex 1: Calculate work done if ( R=8.31 J/mol.K, T₁=300 K, V₂=4V₁ ).
  • ( W₁ = 8.31 × 300 × ln 4 = 8.31 × 300 × 1.386 = 3453.5 J )

Practice Set

• Q1: During isothermal expansion, temperature is?
  • A) Increasing
  • B) Constant
  • C) Decreasing
  • D) Variable
• Q2: Work done in isothermal expansion is?
  • A) Zero
  • B) ( RT₁ ln V₂/V₁ )
  • C) ( P ΔV )
  • D) ( C_V ΔT )
• Q3: Heat absorbed during isothermal expansion is?
  • A) Zero
  • B) Equal to work done
  • C) Greater than work done
  • D) Less than work done

Answer Key

• Q1: B - Temperature remains constant.
• Q2: B - Work done formula for isothermal expansion.
• Q3: B - Heat absorbed equals work done in isothermal process.

Memory Aids

• Isothermal: ( Q = W = RT ln V₂/V₁ )
• Temperature constant during expansion.

Process Details

• Gas expands at constant temperature ( T₁ ) absorbing heat ( Q₁ ) from source.
• Work done by gas ( W₁ = Q₁ = RT₁ ln V₂/V₁ ).
• Heat supplied maintains temperature during expansion.

Solved Examples

• Ex 1: Calculate work done if ( R=8.31 J/mol.K, T₁=300 K, V₂=4V₁ ).
  • ( W₁ = 8.31 × 300 × ln 4 = 8.31 × 300 × 1.386 = 3453.5 J )

Practice Set

• Q1: During isothermal expansion, temperature is?
  • A) Increasing
  • B) Constant
  • C) Decreasing
  • D) Variable
• Q2: Work done in isothermal expansion is?
  • A) Zero
  • B) ( RT₁ ln V₂/V₁ )
  • C) ( P ΔV )
  • D) ( C_V ΔT )
• Q3: Heat absorbed during isothermal expansion is?
  • A) Zero
  • B) Equal to work done
  • C) Greater than work done
  • D) Less than work done

Answer Key

• Q1: B - Temperature remains constant.
• Q2: B - Work done formula for isothermal expansion.
• Q3: B - Heat absorbed equals work done in isothermal process.

Memory Aids

• Isothermal: ( Q = W = RT ln V₂/V₁ )
• Temperature constant during expansion.

Process Details

• Gas expands at constant temperature \( T_1 \) absorbing heat \( Q_1 \) from source.
• Work done by gas \( W_1 = Q_1 = RT_1 \ln \frac{V_2}{V_1} \).
• Heat supplied maintains temperature during expansion.

Solved Examples

• Ex 1: Calculate work done if \( R=8.31 \mathrm{J/mol.K}, T_1=300 K, V_2=4V_1 \).
  • \( W_1 = 8.31 \times 300 \times \ln 4 = 8.31 \times 300 \times 1.386 = 3453.5 \mathrm{J} \)

Practice Set

• Q1: During isothermal expansion, temperature is?
  • A) Increasing
  • B) Constant
  • C) Decreasing
  • D) Variable
• Q2: Work done in isothermal expansion is?
  • A) Zero
  • B) \( RT_1 \ln \frac{V_2}{V_1} \)
  • C) \( P \Delta V \)
  • D) \( C_V \Delta T \)
• Q3: Heat absorbed during isothermal expansion is?
  • A) Zero
  • B) Equal to work done
  • C) Greater than work done
  • D) Less than work done

Answer Key

• Q1: B - Temperature remains constant.
• Q2: B - Work done formula for isothermal expansion.
• Q3: B - Heat absorbed equals work done in isothermal process.

Memory Aids

• Isothermal: \( Q = W = RT \ln \frac{V_2}{V_1} \)
• Temperature constant during expansion.

Adiabatic Expansion

Process Details

• No heat exchange (Q=0).
• Temperature falls from ( T₁ ) to ( T₂ ).
• Work done ( W₂ = R(T₁ - T₂)/(γ - 1) ).
• Follows ( PV^γ = constant ).

Solved Examples

• Ex 1: Calculate work done if ( R=8.31, T₁=400 K, T₂=300 K, γ=1.4 ).
  • ( W₂ = 8.31(400-300)/(1.4-1) = 8.31 × 100/0.4 = 2077.5 J )

Practice Set

• Q1: Heat exchange in adiabatic expansion is?
  • A) Positive
  • B) Negative
  • C) Zero
  • D) Variable
• Q2: Work done in adiabatic expansion is?
  • A) Zero
  • B) ( R(T₁ - T₂)/(γ - 1) )
  • C) ( RT ln V₂/V₁ )
  • D) ( C_P ΔT )
• Q3: Temperature change in adiabatic expansion is?
  • A) Increases
  • B) Constant
  • C) Decreases
  • D) No change

Answer Key

• Q1: C - No heat exchange in adiabatic process.
• Q2: B - Work done formula for adiabatic expansion.
• Q3: C - Temperature decreases during expansion.

Memory Aids

• Adiabatic: ( Q=0 ), ( W= R(T₁ - T₂)/(γ - 1) )
• Temperature falls during expansion.

Process Details

• No heat exchange (Q=0).
• Temperature falls from ( T₁ ) to ( T₂ ).
• Work done ( W₂ = R(T₁ - T₂)/(γ - 1) ).
• Follows ( PV^γ = constant ).

Solved Examples

• Ex 1: Calculate work done if ( R=8.31, T₁=400 K, T₂=300 K, γ=1.4 ).
  • ( W₂ = 8.31(400-300)/(1.4-1) = 8.31 × 100/0.4 = 2077.5 J )

Practice Set

• Q1: Heat exchange in adiabatic expansion is?
  • A) Positive
  • B) Negative
  • C) Zero
  • D) Variable
• Q2: Work done in adiabatic expansion is?
  • A) Zero
  • B) ( R(T₁ - T₂)/(γ - 1) )
  • C) ( RT ln V₂/V₁ )
  • D) ( C_P ΔT )
• Q3: Temperature change in adiabatic expansion is?
  • A) Increases
  • B) Constant
  • C) Decreases
  • D) No change

Answer Key

• Q1: C - No heat exchange in adiabatic process.
• Q2: B - Work done formula for adiabatic expansion.
• Q3: C - Temperature decreases during expansion.

Memory Aids

• Adiabatic: ( Q=0 ), ( W= R(T₁ - T₂)/(γ - 1) )
• Temperature falls during expansion.

Process Details

• No heat exchange (Q=0).
• Temperature falls from T₁ to T₂.
• Work done W₂ = R(T₁ - T₂)/(γ - 1).
• Follows PV^γ = constant.

Solved Examples

• Ex 1: Calculate work done if R=8.31, T₁=400 K, T₂=300 K, γ=1.4.
  • W₂ = (8.31(400-300))/(1.4-1) = (8.31 × 100)/(0.4) = 2077.5 J

Practice Set

• Q1: Heat exchange in adiabatic expansion is?
  • A) Positive
  • B) Negative
  • C) Zero
  • D) Variable
• Q2: Work done in adiabatic expansion is?
  • A) Zero
  • B) R(T₁ - T₂)/(γ - 1)
  • C) RT ln V₂/V₁
  • D) C_P ΔT
• Q3: Temperature change in adiabatic expansion is?
  • A) Increases
  • B) Constant
  • C) Decreases
  • D) No change

Answer Key

• Q1: C - No heat exchange in adiabatic process.
• Q2: B - Work done formula for adiabatic expansion.
• Q3: C - Temperature decreases during expansion.

Memory Aids

• Adiabatic: Q=0, W= R(T₁ - T₂)/(γ - 1)
• Temperature falls during expansion.

Process Details

• No heat exchange (Q=0).
• Temperature falls from ( T₁ ) to ( T₂ ).
• Work done ( W₂ = R(T₁ - T₂)/(γ - 1) ).
• Follows ( PV^γ = constant ).

Solved Examples

• Ex 1: Calculate work done if ( R=8.31, T₁=400 K, T₂=300 K, γ=1.4 ).
  • ( W₂ = 8.31(400-300)/(1.4-1) = 8.31 × 100/0.4 = 2077.5 J )

Practice Set

• Q1: Heat exchange in adiabatic expansion is?
  • A) Positive
  • B) Negative
  • C) Zero
  • D) Variable
• Q2: Work done in adiabatic expansion is?
  • A) Zero
  • B) ( R(T₁ - T₂)/(γ - 1) )
  • C) ( RT ln V₂/V₁ )
  • D) ( C_P ΔT )
• Q3: Temperature change in adiabatic expansion is?
  • A) Increases
  • B) Constant
  • C) Decreases
  • D) No change

Answer Key

• Q1: C - No heat exchange in adiabatic process.
• Q2: B - Work done formula for adiabatic expansion.
• Q3: C - Temperature decreases during expansion.

Memory Aids

• Adiabatic: ( Q=0 ), ( W= R(T₁ - T₂)/(γ - 1) )
• Temperature falls during expansion.

Process Details

• No heat exchange (Q=0).
• Temperature falls from ( T₁ ) to ( T₂ ).
• Work done ( W₂ = R(T₁ - T₂)/(γ - 1) ).
• Follows ( PV^γ = constant ).

Solved Examples

• Ex 1: Calculate work done if ( R=8.31, T₁=400 K, T₂=300 K, γ=1.4 ).
  • ( W₂ = 8.31(400-300)/(1.4-1) = 8.31 × 100/0.4 = 2077.5 J )

Practice Set

• Q1: Heat exchange in adiabatic expansion is?
  • A) Positive
  • B) Negative
  • C) Zero
  • D) Variable
• Q2: Work done in adiabatic expansion is?
  • A) Zero
  • B) ( R(T₁ - T₂)/(γ - 1) )
  • C) ( RT ln V₂/V₁ )
  • D) ( C_P ΔT )
• Q3: Temperature change in adiabatic expansion is?
  • A) Increases
  • B) Constant
  • C) Decreases
  • D) No change

Answer Key

• Q1: C - No heat exchange in adiabatic process.
• Q2: B - Work done formula for adiabatic expansion.
• Q3: C - Temperature decreases during expansion.

Memory Aids

• Adiabatic: ( Q=0 ), ( W= R(T₁ - T₂)/(γ - 1) )
• Temperature falls during expansion.

Process Details

• No heat exchange (Q=0).
• Temperature falls from \( T_1 \) to \( T_2 \).
• Work done \( W_2 = \frac{R(T_1 - T_2)}{\gamma - 1} \).
• Follows \( PV^\gamma = \text{constant} \).

Solved Examples

• Ex 1: Calculate work done if \( R=8.31, T_1=400 K, T_2=300 K, \gamma=1.4 \).
  • \( W_2 = \frac{8.31(400-300)}{1.4-1} = \frac{8.31 \times 100}{0.4} = 2077.5 \mathrm{J} \)

Practice Set

• Q1: Heat exchange in adiabatic expansion is?
  • A) Positive
  • B) Negative
  • C) Zero
  • D) Variable
• Q2: Work done in adiabatic expansion is?
  • A) Zero
  • B) \( \frac{R(T_1 - T_2)}{\gamma - 1} \)
  • C) \( RT \ln \frac{V_2}{V_1} \)
  • D) \( C_P \Delta T \)
• Q3: Temperature change in adiabatic expansion is?
  • A) Increases
  • B) Constant
  • C) Decreases
  • D) No change

Answer Key

• Q1: C - No heat exchange in adiabatic process.
• Q2: B - Work done formula for adiabatic expansion.
• Q3: C - Temperature decreases during expansion.

Memory Aids

• Adiabatic: \( Q=0 \), \( W= \frac{R(T_1 - T_2)}{\gamma - 1} \)
• Temperature falls during expansion.

Isothermal Compression

Process Details

• Gas compresses at constant temperature ( T₂ ).
• Heat ( Q₂ ) is rejected to sink.
• Work done on gas ( W₃ = Q₂ = RT₂ ln V₃/V₄ ).
• Negative work indicates work done on the system.

Solved Examples

• Ex 1: Calculate heat rejected if ( R=8.31, T₂=300 K, V₃=4V₄ ).
  • ( Q₂ = 8.31 × 300 × ln 4 = 3453.5 J )

Practice Set

• Q1: During isothermal compression, temperature is?
  • A) Increasing
  • B) Constant
  • C) Decreasing
  • D) Variable
• Q2: Work done in isothermal compression is?
  • A) ( -RT₂ ln V₄/V₃ )
  • B) Zero
  • C) ( RT₂ ln V₃/V₄ )
  • D) ( C_V ΔT )
• Q3: Heat rejected during isothermal compression is?
  • A) Zero
  • B) Equal to work done on gas
  • C) Greater than work done
  • D) Less than work done

Answer Key

• Q1: B - Temperature remains constant.
• Q2: A - Work done is negative of expansion formula.
• Q3: B - Heat rejected equals work done on gas.

Memory Aids

• Isothermal compression: ( Q = W = RT ln V₃/V₄ )
• Work done on gas is negative of expansion work.

Process Details

• Gas compresses at constant temperature ( T₂ ).
• Heat ( Q₂ ) is rejected to sink.
• Work done on gas ( W₃ = Q₂ = RT₂ ln V₃/V₄ ).
• Negative work indicates work done on the system.

Solved Examples

• Ex 1: Calculate heat rejected if ( R=8.31, T₂=300 K, V₃=4V₄ ).
  • ( Q₂ = 8.31 × 300 × ln 4 = 3453.5 J )

Practice Set

• Q1: During isothermal compression, temperature is?
  • A) Increasing
  • B) Constant
  • C) Decreasing
  • D) Variable
• Q2: Work done in isothermal compression is?
  • A) ( -RT₂ ln V₄/V₃ )
  • B) Zero
  • C) ( RT₂ ln V₃/V₄ )
  • D) ( C_V ΔT )
• Q3: Heat rejected during isothermal compression is?
  • A) Zero
  • B) Equal to work done on gas
  • C) Greater than work done
  • D) Less than work done

Answer Key

• Q1: B - Temperature remains constant.
• Q2: A - Work done is negative of expansion formula.
• Q3: B - Heat rejected equals work done on gas.

Memory Aids

• Isothermal compression: ( Q = W = RT ln V₃/V₄ )
• Work done on gas is negative of expansion work.

Process Details

• Gas compresses at constant temperature T₂.
• Heat Q₂ is rejected to sink.
• Work done on gas W₃ = Q₂ = RT₂ ln V₃/V₄.
• Negative work indicates work done on the system.

Solved Examples

• Ex 1: Calculate heat rejected if R=8.31, T₂=300 K, V₃=4V₄.
  • Q₂ = 8.31 × 300 × ln 4 = 3453.5 J

Practice Set

• Q1: During isothermal compression, temperature is?
  • A) Increasing
  • B) Constant
  • C) Decreasing
  • D) Variable
• Q2: Work done in isothermal compression is?
  • A) -RT₂ ln V₄/V₃
  • B) Zero
  • C) RT₂ ln V₃/V₄
  • D) C_V ΔT
• Q3: Heat rejected during isothermal compression is?
  • A) Zero
  • B) Equal to work done on gas
  • C) Greater than work done
  • D) Less than work done

Answer Key

• Q1: B - Temperature remains constant.
• Q2: A - Work done is negative of expansion formula.
• Q3: B - Heat rejected equals work done on gas.

Memory Aids

• Isothermal compression: Q = W = RT ln V₃/V₄
• Work done on gas is negative of expansion work.

Process Details

• Gas compresses at constant temperature ( T₂ ).
• Heat ( Q₂ ) is rejected to sink.
• Work done on gas ( W₃ = Q₂ = RT₂ ln V₃/V₄ ).
• Negative work indicates work done on the system.

Solved Examples

• Ex 1: Calculate heat rejected if ( R=8.31, T₂=300 K, V₃=4V₄ ).
  • ( Q₂ = 8.31 × 300 × ln 4 = 3453.5 J )

Practice Set

• Q1: During isothermal compression, temperature is?
  • A) Increasing
  • B) Constant
  • C) Decreasing
  • D) Variable
• Q2: Work done in isothermal compression is?
  • A) ( -RT₂ ln V₄/V₃ )
  • B) Zero
  • C) ( RT₂ ln V₃/V₄ )
  • D) ( C_V ΔT )
• Q3: Heat rejected during isothermal compression is?
  • A) Zero
  • B) Equal to work done on gas
  • C) Greater than work done
  • D) Less than work done

Answer Key

• Q1: B - Temperature remains constant.
• Q2: A - Work done is negative of expansion formula.
• Q3: B - Heat rejected equals work done on gas.

Memory Aids

• Isothermal compression: ( Q = W = RT ln V₃/V₄ )
• Work done on gas is negative of expansion work.

Process Details

• Gas compresses at constant temperature ( T₂ ).
• Heat ( Q₂ ) is rejected to sink.
• Work done on gas ( W₃ = Q₂ = RT₂ ln V₃/V₄ ).
• Negative work indicates work done on the system.

Solved Examples

• Ex 1: Calculate heat rejected if ( R=8.31, T₂=300 K, V₃=4V₄ ).
  • ( Q₂ = 8.31 × 300 × ln 4 = 3453.5 J )

Practice Set

• Q1: During isothermal compression, temperature is?
  • A) Increasing
  • B) Constant
  • C) Decreasing
  • D) Variable
• Q2: Work done in isothermal compression is?
  • A) ( -RT₂ ln V₄/V₃ )
  • B) Zero
  • C) ( RT₂ ln V₃/V₄ )
  • D) ( C_V ΔT )
• Q3: Heat rejected during isothermal compression is?
  • A) Zero
  • B) Equal to work done on gas
  • C) Greater than work done
  • D) Less than work done

Answer Key

• Q1: B - Temperature remains constant.
• Q2: A - Work done is negative of expansion formula.
• Q3: B - Heat rejected equals work done on gas.

Memory Aids

• Isothermal compression: ( Q = W = RT ln V₃/V₄ )
• Work done on gas is negative of expansion work.

Process Details

• Gas compresses at constant temperature \( T_2 \).
• Heat \( Q_2 \) is rejected to sink.
• Work done on gas \( W_3 = Q_2 = RT_2 \ln \frac{V_3}{V_4} \).
• Negative work indicates work done on the system.

Solved Examples

• Ex 1: Calculate heat rejected if \( R=8.31, T_2=300 K, V_3=4V_4 \).
  • \( Q_2 = 8.31 \times 300 \times \ln 4 = 3453.5 \mathrm{J} \)

Practice Set

• Q1: During isothermal compression, temperature is?
  • A) Increasing
  • B) Constant
  • C) Decreasing
  • D) Variable
• Q2: Work done in isothermal compression is?
  • A) \( -RT_2 \ln \frac{V_4}{V_3} \)
  • B) Zero
  • C) \( RT_2 \ln \frac{V_3}{V_4} \)
  • D) \( C_V \Delta T \)
• Q3: Heat rejected during isothermal compression is?
  • A) Zero
  • B) Equal to work done on gas
  • C) Greater than work done
  • D) Less than work done

Answer Key

• Q1: B - Temperature remains constant.
• Q2: A - Work done is negative of expansion formula.
• Q3: B - Heat rejected equals work done on gas.

Memory Aids

• Isothermal compression: \( Q = W = RT \ln \frac{V_3}{V_4} \)
• Work done on gas is negative of expansion work.

Adiabatic Compression

Process Details

• No heat exchange (Q=0).
• Work done on gas ( W₄ = R(T₁ - T₂)/(γ - 1) ).
• Temperature rises from ( T₂ ) to ( T₁ ).

Solved Examples

• Ex 1: Calculate work done if ( R=8.31, T₁=400 K, T₂=300 K, γ=1.4 ).
  • ( W₄ = 8.31(400-300)/(1.4-1) = 2077.5 J )

Practice Set

• Q1: Heat exchange in adiabatic compression is?
  • A) Positive
  • B) Negative
  • C) Zero
  • D) Variable
• Q2: Work done in adiabatic compression is?
  • A) Zero
  • B) ( R(T₁ - T₂)/(γ - 1) )
  • C) ( RT ln V₂/V₁ )
  • D) ( C_P ΔT )
• Q3: Temperature change in adiabatic compression is?
  • A) Increases
  • B) Constant
  • C) Decreases
  • D) No change

Answer Key

• Q1: C - No heat exchange in adiabatic process.
• Q2: B - Work done formula for adiabatic compression.
• Q3: A - Temperature increases during compression.

Memory Aids

• Adiabatic: ( Q=0 ), ( W= R(T₁ - T₂)/(γ - 1) )
• Temperature rises during compression.

Process Details

• No heat exchange (Q=0).
• Work done on gas ( W₄ = R(T₁ - T₂)/(γ - 1) ).
• Temperature rises from ( T₂ ) to ( T₁ ).

Solved Examples

• Ex 1: Calculate work done if ( R=8.31, T₁=400 K, T₂=300 K, γ=1.4 ).
  • ( W₄ = 8.31(400-300)/(1.4-1) = 2077.5 J )

Practice Set

• Q1: Heat exchange in adiabatic compression is?
  • A) Positive
  • B) Negative
  • C) Zero
  • D) Variable
• Q2: Work done in adiabatic compression is?
  • A) Zero
  • B) ( R(T₁ - T₂)/(γ - 1) )
  • C) ( RT ln V₂/V₁ )
  • D) ( C_P ΔT )
• Q3: Temperature change in adiabatic compression is?
  • A) Increases
  • B) Constant
  • C) Decreases
  • D) No change

Answer Key

• Q1: C - No heat exchange in adiabatic process.
• Q2: B - Work done formula for adiabatic compression.
• Q3: A - Temperature increases during compression.

Memory Aids

• Adiabatic: ( Q=0 ), ( W= R(T₁ - T₂)/(γ - 1) )
• Temperature rises during compression.

Process Details

• No heat exchange (Q=0).
• Work done on gas W₄ = R(T₁ - T₂)/(γ - 1).
• Temperature rises from T₂ to T₁.

Solved Examples

• Ex 1: Calculate work done if R=8.31, T₁=400 K, T₂=300 K, γ=1.4.
  • W₄ = (8.31(400-300))/(1.4-1) = 2077.5 J

Practice Set

• Q1: Heat exchange in adiabatic compression is?
  • A) Positive
  • B) Negative
  • C) Zero
  • D) Variable
• Q2: Work done in adiabatic compression is?
  • A) Zero
  • B) R(T₁ - T₂)/(γ - 1)
  • C) RT ln V₂/V₁
  • D) C_P ΔT
• Q3: Temperature change in adiabatic compression is?
  • A) Increases
  • B) Constant
  • C) Decreases
  • D) No change

Answer Key

• Q1: C - No heat exchange in adiabatic process.
• Q2: B - Work done formula for adiabatic compression.
• Q3: A - Temperature increases during compression.

Memory Aids

• Adiabatic: Q=0, W= R(T₁ - T₂)/(γ - 1)
• Temperature rises during compression.

Process Details

• No heat exchange (Q=0).
• Work done on gas ( W₄ = R(T₁ - T₂)/(γ - 1) ).
• Temperature rises from ( T₂ ) to ( T₁ ).

Solved Examples

• Ex 1: Calculate work done if ( R=8.31, T₁=400 K, T₂=300 K, γ=1.4 ).
  • ( W₄ = 8.31(400-300)/(1.4-1) = 2077.5 J )

Practice Set

• Q1: Heat exchange in adiabatic compression is?
  • A) Positive
  • B) Negative
  • C) Zero
  • D) Variable
• Q2: Work done in adiabatic compression is?
  • A) Zero
  • B) ( R(T₁ - T₂)/(γ - 1) )
  • C) ( RT ln V₂/V₁ )
  • D) ( C_P ΔT )
• Q3: Temperature change in adiabatic compression is?
  • A) Increases
  • B) Constant
  • C) Decreases
  • D) No change

Answer Key

• Q1: C - No heat exchange in adiabatic process.
• Q2: B - Work done formula for adiabatic compression.
• Q3: A - Temperature increases during compression.

Memory Aids

• Adiabatic: ( Q=0 ), ( W= R(T₁ - T₂)/(γ - 1) )
• Temperature rises during compression.

Process Details

• No heat exchange (Q=0).
• Work done on gas ( W₄ = R(T₁ - T₂)/(γ - 1) ).
• Temperature rises from ( T₂ ) to ( T₁ ).

Solved Examples

• Ex 1: Calculate work done if ( R=8.31, T₁=400 K, T₂=300 K, γ=1.4 ).
  • ( W₄ = 8.31(400-300)/(1.4-1) = 2077.5 J )

Practice Set

• Q1: Heat exchange in adiabatic compression is?
  • A) Positive
  • B) Negative
  • C) Zero
  • D) Variable
• Q2: Work done in adiabatic compression is?
  • A) Zero
  • B) ( R(T₁ - T₂)/(γ - 1) )
  • C) ( RT ln V₂/V₁ )
  • D) ( C_P ΔT )
• Q3: Temperature change in adiabatic compression is?
  • A) Increases
  • B) Constant
  • C) Decreases
  • D) No change

Answer Key

• Q1: C - No heat exchange in adiabatic process.
• Q2: B - Work done formula for adiabatic compression.
• Q3: A - Temperature increases during compression.

Memory Aids

• Adiabatic: ( Q=0 ), ( W= R(T₁ - T₂)/(γ - 1) )
• Temperature rises during compression.

Process Details

• No heat exchange (Q=0).
• Work done on gas W₄ = R(T₁ - T₂)/(γ - 1).
• Temperature rises from T₂ to T₁.

Solved Examples

• Ex 1: Calculate work done if R=8.31, T₁=400 K, T₂=300 K, γ=1.4.
  • W₄ = 8.31(400-300)/(1.4-1) = 2077.5 J

Practice Set

• Q1: Heat exchange in adiabatic compression is?
  • A) Positive
  • B) Negative
  • C) Zero
  • D) Variable
• Q2: Work done in adiabatic compression is?
  • A) Zero
  • B) R(T₁ - T₂)/(γ - 1)
  • C) RT ln V₂/V₁
  • D) C_P ΔT
• Q3: Temperature change in adiabatic compression is?
  • A) Increases
  • B) Constant
  • C) Decreases
  • D) No change

Answer Key

• Q1: C - No heat exchange in adiabatic process.
• Q2: B - Work done formula for adiabatic compression.
• Q3: A - Temperature increases during compression.

Memory Aids

• Adiabatic: Q=0, W= R(T₁ - T₂)/(γ - 1)
• Temperature rises during compression.

Process Details

• No heat exchange (Q=0).
• Work done on gas ( W₄ = R(T₁ - T₂)/(γ - 1) ).
• Temperature rises from ( T₂ ) to ( T₁ ).

Solved Examples

• Ex 1: Calculate work done if ( R=8.31, T₁=400 K, T₂=300 K, γ=1.4 ).
  • ( W₄ = 8.31(400-300)/(1.4-1) = 2077.5 J )

Practice Set

• Q1: Heat exchange in adiabatic compression is?
  • A) Positive
  • B) Negative
  • C) Zero
  • D) Variable
• Q2: Work done in adiabatic compression is?
  • A) Zero
  • B) ( R(T₁ - T₂)/(γ - 1) )
  • C) ( RT ln V₂/V₁ )
  • D) ( C_P ΔT )
• Q3: Temperature change in adiabatic compression is?
  • A) Increases
  • B) Constant
  • C) Decreases
  • D) No change

Answer Key

• Q1: C - No heat exchange in adiabatic process.
• Q2: B - Work done formula for adiabatic compression.
• Q3: A - Temperature increases during compression.

Memory Aids

• Adiabatic: ( Q=0 ), ( W= R(T₁ - T₂)/(γ - 1) )
• Temperature rises during compression.

Process Details

• No heat exchange (Q=0).
• Work done on gas ( W₄ = R(T₁ - T₂)/(γ - 1) ).
• Temperature rises from ( T₂ ) to ( T₁ ).

Solved Examples

• Ex 1: Calculate work done if ( R=8.31, T₁=400 K, T₂=300 K, γ=1.4 ).
  • ( W₄ = 8.31(400-300)/(1.4-1) = 2077.5 J )

Practice Set

• Q1: Heat exchange in adiabatic compression is?
  • A) Positive
  • B) Negative
  • C) Zero
  • D) Variable
• Q2: Work done in adiabatic compression is?
  • A) Zero
  • B) ( R(T₁ - T₂)/(γ - 1) )
  • C) ( RT ln V₂/V₁ )
  • D) ( C_P ΔT )
• Q3: Temperature change in adiabatic compression is?
  • A) Increases
  • B) Constant
  • C) Decreases
  • D) No change

Answer Key

• Q1: C - No heat exchange in adiabatic process.
• Q2: B - Work done formula for adiabatic compression.
• Q3: A - Temperature increases during compression.

Memory Aids

• Adiabatic: ( Q=0 ), ( W= R(T₁ - T₂)/(γ - 1) )
• Temperature rises during compression.

Process Details

• No heat exchange (Q=0).
• Work done on gas W₄ = R(T₁ - T₂)/(γ - 1).
• Temperature rises from T₂ to T₁.

Solved Examples

• Ex 1: Calculate work done if R=8.31, T₁=400 K, T₂=300 K, γ=1.4.
  • W₄ = (8.31(400-300))/(1.4-1) = 2077.5 J

Practice Set

• Q1: Heat exchange in adiabatic compression is?
  • A) Positive
  • B) Negative
  • C) Zero
  • D) Variable
• Q2: Work done in adiabatic compression is?
  • A) Zero
  • B) R(T₁ - T₂)/(γ - 1)
  • C) RT ln V₂/V₁
  • D) C_P ΔT
• Q3: Temperature change in adiabatic compression is?
  • A) Increases
  • B) Constant
  • C) Decreases
  • D) No change

Answer Key

• Q1: C - No heat exchange in adiabatic process.
• Q2: B - Work done formula for adiabatic compression.
• Q3: A - Temperature increases during compression.

Memory Aids

• Adiabatic: Q=0, W= R(T₁ - T₂)/(γ - 1)
• Temperature rises during compression.

Process Details

• No heat exchange (Q=0).
• Work done on gas ( W₄ = R(T₁ - T₂)/(γ - 1) ).
• Temperature rises from ( T₂ ) to ( T₁ ).

Solved Examples

• Ex 1: Calculate work done if ( R=8.31, T₁=400 K, T₂=300 K, γ=1.4 ).
  • ( W₄ = 8.31(400-300)/(1.4-1) = 2077.5 J )

Practice Set

• Q1: Heat exchange in adiabatic compression is?
  • A) Positive
  • B) Negative
  • C) Zero
  • D) Variable
• Q2: Work done in adiabatic compression is?
  • A) Zero
  • B) ( R(T₁ - T₂)/(γ - 1) )
  • C) ( RT ln V₂/V₁ )
  • D) ( C_P ΔT )
• Q3: Temperature change in adiabatic compression is?
  • A) Increases
  • B) Constant
  • C) Decreases
  • D) No change

Answer Key

• Q1: C - No heat exchange in adiabatic process.
• Q2: B - Work done formula for adiabatic compression.
• Q3: A - Temperature increases during compression.

Memory Aids

• Adiabatic: ( Q=0 ), ( W= R(T₁ - T₂)/(γ - 1) )
• Temperature rises during compression.

Process Details

• No heat exchange (Q=0).
• Work done on gas ( W₄ = R(T₁ - T₂)/(γ - 1) ).
• Temperature rises from ( T₂ ) to ( T₁ ).

Solved Examples

• Ex 1: Calculate work done if ( R=8.31, T₁=400 K, T₂=300 K, γ=1.4 ).
  • ( W₄ = 8.31(400-300)/(1.4-1) = 2077.5 J )

Practice Set

• Q1: Heat exchange in adiabatic compression is?
  • A) Positive
  • B) Negative
  • C) Zero
  • D) Variable
• Q2: Work done in adiabatic compression is?
  • A) Zero
  • B) ( R(T₁ - T₂)/(γ - 1) )
  • C) ( RT ln V₂/V₁ )
  • D) ( C_P ΔT )
• Q3: Temperature change in adiabatic compression is?
  • A) Increases
  • B) Constant
  • C) Decreases
  • D) No change

Answer Key

• Q1: C - No heat exchange in adiabatic process.
• Q2: B - Work done formula for adiabatic compression.
• Q3: A - Temperature increases during compression.

Memory Aids

• Adiabatic: ( Q=0 ), ( W= R(T₁ - T₂)/(γ - 1) )
• Temperature rises during compression.

Process Details

• No heat exchange (Q=0).
• Work done on gas \( W_4 = \frac{R(T_1 - T_2)}{\gamma - 1} \).
• Temperature rises from \( T_2 \) to \( T_1 \).

Solved Examples

• Ex 1: Calculate work done if \( R=8.31, T_1=400 K, T_2=300 K, \gamma=1.4 \).
  • \( W_4 = \frac{8.31(400-300)}{1.4-1} = 2077.5 \mathrm{J} \)

Practice Set

• Q1: Heat exchange in adiabatic compression is?
  • A) Positive
  • B) Negative
  • C) Zero
  • D) Variable
• Q2: Work done in adiabatic compression is?
  • A) Zero
  • B) \( \frac{R(T_1 - T_2)}{\gamma - 1} \)
  • C) \( RT \ln \frac{V_2}{V_1} \)
  • D) \( C_P \Delta T \)
• Q3: Temperature change in adiabatic compression is?
  • A) Increases
  • B) Constant
  • C) Decreases
  • D) No change

Answer Key

• Q1: C - No heat exchange in adiabatic process.
• Q2: B - Work done formula for adiabatic compression.
• Q3: A - Temperature increases during compression.

Memory Aids

• Adiabatic: \( Q=0 \), \( W= \frac{R(T_1 - T_2)}{\gamma - 1} \)
• Temperature rises during compression.

Net Work Done

Calculation

• Total work ( W = W₁ + W₂ + W₃ + W₄ ).
• Since ( W₂ ) and ( W₄ ) cancel, ( W = W₁ - W₃ ).
• Net work done equals area enclosed by ABCD in P-V diagram.

Solved Examples

• Ex 1: If ( W₁ = 500 J ) and ( W₃ = 300 J ), find net work.
  • ( W = 500 - 300 = 200 J )

Practice Set

• Q1: Net work done in Carnot cycle is?
  • A) ( W₁ + W₃ )
  • B) ( W₁ - W₃ )
  • C) ( W₂ + W₄ )
  • D) Zero
• Q2: Which works cancel out in net work calculation?
  • A) ( W₁ ) and ( W₃ )
  • B) ( W₂ ) and ( W₄ )
  • C) ( W₁ ) and ( W₄ )
  • D) ( W₃ ) and ( W₂ )
• Q3: Area representing net work done is?
  • A) ABGEA
  • B) BCHGB
  • C) ABCD
  • D) CHFDC

Answer Key

• Q1: B - Net work is ( W₁ - W₃ ).
• Q2: B - ( W₂ ) and ( W₄ ) cancel each other.
• Q3: C - Area ABCD represents net work done.

Memory Aids

• Net work = area enclosed by cycle.
• Adiabatic works cancel out.

Calculation

• Total work ( W = W₁ + W₂ + W₃ + W₄ ).
• Since ( W₂ ) and ( W₄ ) cancel, ( W = W₁ - W₃ ).
• Net work done equals area enclosed by ABCD in P-V diagram.

Solved Examples

• Ex 1: If ( W₁ = 500 J ) and ( W₃ = 300 J ), find net work.
  • ( W = 500 - 300 = 200 J )

Practice Set

• Q1: Net work done in Carnot cycle is?
  • A) ( W₁ + W₃ )
  • B) ( W₁ - W₃ )
  • C) ( W₂ + W₄ )
  • D) Zero
• Q2: Which works cancel out in net work calculation?
  • A) ( W₁ ) and ( W₃ )
  • B) ( W₂ ) and ( W₄ )
  • C) ( W₁ ) and ( W₄ )
  • D) ( W₃ ) and ( W₂ )
• Q3: Area representing net work done is?
  • A) ABGEA
  • B) BCHGB
  • C) ABCD
  • D) CHFDC

Answer Key

• Q1: B - Net work is ( W₁ - W₃ ).
• Q2: B - ( W₂ ) and ( W₄ ) cancel each other.
• Q3: C - Area ABCD represents net work done.

Memory Aids

• Net work = area enclosed by cycle.
• Adiabatic works cancel out.

Calculation

• Total work W = W₁ + W₂ + W₃ + W₄.
• Since W₂ and W₄ cancel, W = W₁ - W₃.
• Net work done equals area enclosed by ABCD in P-V diagram.

Solved Examples

• Ex 1: If W₁ = 500 J and W₃ = 300 J, find net work.
  • W = 500 - 300 = 200 J

Practice Set

• Q1: Net work done in Carnot cycle is?
  • A) W₁ + W₃
  • B) W₁ - W₃
  • C) W₂ + W₄
  • D) Zero
• Q2: Which works cancel out in net work calculation?
  • A) W₁ and W₃
  • B) W₂ and W₄
  • C) W₁ and W₄
  • D) W₃ and W₂
• Q3: Area representing net work done is?
  • A) ABGEA
  • B) BCHGB
  • C) ABCD
  • D) CHFDC

Answer Key

• Q1: B - Net work is W₁ - W₃.
• Q2: B - W₂ and W₄ cancel each other.
• Q3: C - Area ABCD represents net work done.

Memory Aids

• Net work = area enclosed by cycle.
• Adiabatic works cancel out.

Calculation

• Total work ( W = W₁ + W₂ + W₃ + W₄ ).
• Since ( W₂ ) and ( W₄ ) cancel, ( W = W₁ - W₃ ).
• Net work done equals area enclosed by ABCD in P-V diagram.

Solved Examples

• Ex 1: If ( W₁ = 500 J ) and ( W₃ = 300 J ), find net work.
  • ( W = 500 - 300 = 200 J )

Practice Set

• Q1: Net work done in Carnot cycle is?
  • A) ( W₁ + W₃ )
  • B) ( W₁ - W₃ )
  • C) ( W₂ + W₄ )
  • D) Zero
• Q2: Which works cancel out in net work calculation?
  • A) ( W₁ ) and ( W₃ )
  • B) ( W₂ ) and ( W₄ )
  • C) ( W₁ ) and ( W₄ )
  • D) ( W₃ ) and ( W₂ )
• Q3: Area representing net work done is?
  • A) ABGEA
  • B) BCHGB
  • C) ABCD
  • D) CHFDC

Answer Key

• Q1: B - Net work is ( W₁ - W₃ ).
• Q2: B - ( W₂ ) and ( W₄ ) cancel each other.
• Q3: C - Area ABCD represents net work done.

Memory Aids

• Net work = area enclosed by cycle.
• Adiabatic works cancel out.

Calculation

• Total work ( W = W₁ + W₂ + W₃ + W₄ ).
• Since ( W₂ ) and ( W₄ ) cancel, ( W = W₁ - W₃ ).
• Net work done equals area enclosed by ABCD in P-V diagram.

Solved Examples

• Ex 1: If ( W₁ = 500 J ) and ( W₃ = 300 J ), find net work.
  • ( W = 500 - 300 = 200 J )

Practice Set

• Q1: Net work done in Carnot cycle is?
  • A) ( W₁ + W₃ )
  • B) ( W₁ - W₃ )
  • C) ( W₂ + W₄ )
  • D) Zero
• Q2: Which works cancel out in net work calculation?
  • A) ( W₁ ) and ( W₃ )
  • B) ( W₂ ) and ( W₄ )
  • C) ( W₁ ) and ( W₄ )
  • D) ( W₃ ) and ( W₂ )
• Q3: Area representing net work done is?
  • A) ABGEA
  • B) BCHGB
  • C) ABCD
  • D) CHFDC

Answer Key

• Q1: B - Net work is ( W₁ - W₃ ).
• Q2: B - ( W₂ ) and ( W₄ ) cancel each other.
• Q3: C - Area ABCD represents net work done.

Memory Aids

• Net work = area enclosed by cycle.
• Adiabatic works cancel out.

Calculation

• Total work \( W = W_1 + W_2 + W_3 + W_4 \).
• Since \( W_2 \) and \( W_4 \) cancel, \( W = W_1 - W_3 \).
• Net work done equals area enclosed by ABCD in P-V diagram.

Solved Examples

• Ex 1: If \( W_1 = 500 J \) and \( W_3 = 300 J \), find net work.
  • \( W = 500 - 300 = 200 J \)

Practice Set

• Q1: Net work done in Carnot cycle is?
  • A) \( W_1 + W_3 \)
  • B) \( W_1 - W_3 \)
  • C) \( W_2 + W_4 \)
  • D) Zero
• Q2: Which works cancel out in net work calculation?
  • A) \( W_1 \) and \( W_3 \)
  • B) \( W_2 \) and \( W_4 \)
  • C) \( W_1 \) and \( W_4 \)
  • D) \( W_3 \) and \( W_2 \)
• Q3: Area representing net work done is?
  • A) ABGEA
  • B) BCHGB
  • C) ABCD
  • D) CHFDC

Answer Key

• Q1: B - Net work is \( W_1 - W_3 \).
• Q2: B - \( W_2 \) and \( W_4 \) cancel each other.
• Q3: C - Area ABCD represents net work done.

Memory Aids

• Net work = area enclosed by cycle.
• Adiabatic works cancel out.

Efficiency of Carnot Engine

Formula and Explanation

• Efficiency ( η = W/Q₁ = 1 - Q₂/Q₁ ).
• Using volumes and temperatures, ( η = 1 - T₂/T₁ ).
• All reversible engines between same temperatures have same efficiency.
• No engine can be more efficient than Carnot engine.

Solved Examples

• Ex 1: Calculate efficiency if ( T₁ = 600 K, T₂ = 300 K ).
  • ( η = 1 - 300/600 = 0.5 = 50% )

Practice Set

• Q1: Efficiency formula of Carnot engine is?
  • A) ( 1 - T₁/T₂ )
  • B) ( 1 - T₂/T₁ )
  • C) ( T₂/T₁ )
  • D) ( T₁/T₂ )
• Q2: Which engine has maximum efficiency?
  • A) Diesel engine
  • B) Carnot engine
  • C) Otto engine
  • D) Real engine
• Q3: If ( T₁ = 500 K, T₂ = 400 K ), efficiency is?
  • A) 0.2
  • B) 0.1
  • C) 0.3
  • D) 0.4

Answer Key

• Q1: B - Efficiency is ( 1 - T₂/T₁ ).
• Q2: B - Carnot engine is ideal with maximum efficiency.
• Q3: B - ( η = 1 - 400/500 = 0.2 = 20% ).

Memory Aids

• Efficiency ( η = 1 - T₂/T₁ )
• Carnot engine is ideal and most efficient.

Formula and Explanation

• Efficiency ( η = W/Q₁ = 1 - Q₂/Q₁ ).
• Using volumes and temperatures, ( η = 1 - T₂/T₁ ).
• All reversible engines between same temperatures have same efficiency.
• No engine can be more efficient than Carnot engine.

Solved Examples

• Ex 1: Calculate efficiency if ( T₁ = 600 K, T₂ = 300 K ).
  • ( η = 1 - 300/600 = 0.5 = 50% )

Practice Set

• Q1: Efficiency formula of Carnot engine is?
  • A) ( 1 - T₁/T₂ )
  • B) ( 1 - T₂/T₁ )
  • C) ( T₂/T₁ )
  • D) ( T₁/T₂ )
• Q2: Which engine has maximum efficiency?
  • A) Diesel engine
  • B) Carnot engine
  • C) Otto engine
  • D) Real engine
• Q3: If ( T₁ = 500 K, T₂ = 400 K ), efficiency is?
  • A) 0.2
  • B) 0.1
  • C) 0.3
  • D) 0.4

Answer Key

• Q1: B - Efficiency is ( 1 - T₂/T₁ ).
• Q2: B - Carnot engine is ideal with maximum efficiency.
• Q3: B - ( η = 1 - 400/500 = 0.2 = 20% ).

Memory Aids

• Efficiency ( η = 1 - T₂/T₁ )
• Carnot engine is ideal and most efficient.

Formula and Explanation

• Efficiency η = W/Q₁ = 1 - Q₂/Q₁.
• Using volumes and temperatures, η = 1 - T₂/T₁.
• All reversible engines between same temperatures have same efficiency.
• No engine can be more efficient than Carnot engine.

Solved Examples

• Ex 1: Calculate efficiency if T₁ = 600 K, T₂ = 300 K.
  • η = 1 - 300/600 = 0.5 = 50%

Practice Set

• Q1: Efficiency formula of Carnot engine is?
  • A) 1 - T₁/T₂
  • B) 1 - T₂/T₁
  • C) T₂/T₁
  • D) T₁/T₂
• Q2: Which engine has maximum efficiency?
  • A) Diesel engine
  • B) Carnot engine
  • C) Otto engine
  • D) Real engine
• Q3: If T₁ = 500 K, T₂ = 400 K, efficiency is?
  • A) 0.2
  • B) 0.1
  • C) 0.3
  • D) 0.4

Answer Key

• Q1: B - Efficiency is 1 - T₂/T₁.
• Q2: B - Carnot engine is ideal with maximum efficiency.
• Q3: B - η = 1 - 400/500 = 0.2 = 20%.

Memory Aids

• Efficiency η = 1 - T₂/T₁
• Carnot engine is ideal and most efficient.

Formula and Explanation

• Efficiency ( η = W/Q₁ = 1 - Q₂/Q₁ ).
• Using volumes and temperatures, ( η = 1 - T₂/T₁ ).
• All reversible engines between same temperatures have same efficiency.
• No engine can be more efficient than Carnot engine.

Solved Examples

• Ex 1: Calculate efficiency if ( T₁ = 600 K, T₂ = 300 K ).
  • ( η = 1 - 300/600 = 0.5 = 50% )

Practice Set

• Q1: Efficiency formula of Carnot engine is?
  • A) ( 1 - T₁/T₂ )
  • B) ( 1 - T₂/T₁ )
  • C) ( T₂/T₁ )
  • D) ( T₁/T₂ )
• Q2: Which engine has maximum efficiency?
  • A) Diesel engine
  • B) Carnot engine
  • C) Otto engine
  • D) Real engine
• Q3: If ( T₁ = 500 K, T₂ = 400 K ), efficiency is?
  • A) 0.2
  • B) 0.1
  • C) 0.3
  • D) 0.4

Answer Key

• Q1: B - Efficiency is ( 1 - T₂/T₁ ).
• Q2: B - Carnot engine is ideal with maximum efficiency.
• Q3: B - ( η = 1 - 400/500 = 0.2 = 20% ).

Memory Aids

• Efficiency ( η = 1 - T₂/T₁ )
• Carnot engine is ideal and most efficient.

Formula and Explanation

• Efficiency ( η = W/Q₁ = 1 - Q₂/Q₁ ).
• Using volumes and temperatures, ( η = 1 - T₂/T₁ ).
• All reversible engines between same temperatures have same efficiency.
• No engine can be more efficient than Carnot engine.

Solved Examples

• Ex 1: Calculate efficiency if ( T₁ = 600 K, T₂ = 300 K ).
  • ( η = 1 - 300/600 = 0.5 = 50% )

Practice Set

• Q1: Efficiency formula of Carnot engine is?
  • A) ( 1 - T₁/T₂ )
  • B) ( 1 - T₂/T₁ )
  • C) ( T₂/T₁ )
  • D) ( T₁/T₂ )
• Q2: Which engine has maximum efficiency?
  • A) Diesel engine
  • B) Carnot engine
  • C) Otto engine
  • D) Real engine
• Q3: If ( T₁ = 500 K, T₂ = 400 K ), efficiency is?
  • A) 0.2
  • B) 0.1
  • C) 0.3
  • D) 0.4

Answer Key

• Q1: B - Efficiency is ( 1 - T₂/T₁ ).
• Q2: B - Carnot engine is ideal with maximum efficiency.
• Q3: B - ( η = 1 - 400/500 = 0.2 = 20% ).

Memory Aids

• Efficiency ( η = 1 - T₂/T₁ )
• Carnot engine is ideal and most efficient.

Formula and Explanation

• Efficiency \( \eta = \frac{W}{Q_1} = 1 - \frac{Q_2}{Q_1} \).
• Using volumes and temperatures, \( \eta = 1 - \frac{T_2}{T_1} \).
• All reversible engines between same temperatures have same efficiency.
• No engine can be more efficient than Carnot engine.

Solved Examples

• Ex 1: Calculate efficiency if \( T_1 = 600 K, T_2 = 300 K \).
  • \( \eta = 1 - \frac{300}{600} = 0.5 = 50\% \)

Practice Set

• Q1: Efficiency formula of Carnot engine is?
  • A) \( 1 - \frac{T_1}{T_2} \)
  • B) \( 1 - \frac{T_2}{T_1} \)
  • C) \( \frac{T_2}{T_1} \)
  • D) \( \frac{T_1}{T_2} \)
• Q2: Which engine has maximum efficiency?
  • A) Diesel engine
  • B) Carnot engine
  • C) Otto engine
  • D) Real engine
• Q3: If \( T_1 = 500 K, T_2 = 400 K \), efficiency is?
  • A) 0.2
  • B) 0.1
  • C) 0.3
  • D) 0.4

Answer Key

• Q1: B - Efficiency is \( 1 - \frac{T_2}{T_1} \).
• Q2: B - Carnot engine is ideal with maximum efficiency.
• Q3: B - \( \eta = 1 - \frac{400}{500} = 0.2 = 20\% \).

Memory Aids

• Efficiency \( \eta = 1 - \frac{T_2}{T_1} \)
• Carnot engine is ideal and most efficient.