Kinetic Theory

NEET Ug . NEET Physics . Physics . Kinetic Theory

Chapter 13: Kinetic Theory

Ch 13

NEET Physics

NEET Ug

Introduction

Overview

Gases consist of rapidly moving atoms or molecules.
Kinetic theory explains gas behavior by neglecting inter-atomic forces.
Developed in 19ᵗʰ century by Maxwell, Boltzmann, and others.
Explains molecular basis of pressure and temperature.
Consistent with gas laws and Avogadro’s hypothesis.
Explains specific heat capacities of gases.

Solved Examples

Ex 1: Calculate pressure from molecular collisions.
- Use formula: Pressure = Force/Area = (Change in momentum × collisions per second)/Area.
- Apply kinetic theory assumptions for ideal gas.

Practice Set

Q1: Kinetic theory assumes gas molecules are:
- A) Stationary
- B) Rapidly moving
- C) Strongly interacting
- D) At rest
Q2: Kinetic theory was developed by:
- A) Newton and Galileo
- B) Maxwell and Boltzmann
- C) Dalton and Avogadro
- D) Einstein and Planck
Q3: Kinetic theory explains:
- A) Molecular basis of pressure and temperature
- B) Only chemical reactions
- C) Solid state properties
- D) Nuclear forces

Answer Key

Q1: B - Gas molecules are rapidly moving particles.
Q2: B - Maxwell and Boltzmann developed kinetic theory.
Q3: A - It explains molecular interpretation of pressure and temperature.

Memory Aids

Gas molecules = Rapid motion + negligible forces.
Maxwell & Boltzmann = Kinetic theory pioneers.
Pressure & temperature = Molecular collisions.

Molecular Nature of Matter

Key Concepts

Feynman’s atomic hypothesis: Atoms are tiny particles in constant motion, attracting at a distance, repelling when close.
Dalton’s Atomic Theory: Atoms combine in fixed mass ratios obeying simple laws.
First law: Fixed mass proportion in compounds.
Second law: Masses of elements in multiple compounds are in small integer ratios.
Gay Lussac’s law: Volumes of gases combine in small integer ratios.
Avogadro’s hypothesis: Equal volumes of gases at same T and P contain equal molecules.
Mean free path: Average distance a molecule travels without collision.

Solved Examples

Ex 1: Calculate mean free path.
- Use formula: \( \lambda = \frac{1}{\sqrt{2} \pi d^2 \frac{N}{V}} \)
- Where \(d\) = molecular diameter, \(N/V\) = number density.

Practice Set

Q1: According to Feynman, atoms:
- A) Are static
- B) Move perpetually
- C) Do not attract
- D) Are large particles
Q2: Dalton’s first law states:
- A) Variable mass ratios in compounds
- B) Fixed mass proportion in compounds
- C) Gases combine by volume
- D) Molecules are indivisible
Q3: Mean free path is:
- A) Time between collisions
- B) Average distance between collisions
- C) Molecular speed
- D) Volume of gas

Answer Key

Q1: B - Atoms move in perpetual motion.
Q2: B - Fixed mass proportion in compounds.
Q3: B - Mean free path is average distance traveled without collision.

Memory Aids

Feynman: Atoms always moving, attract & repel.
Dalton: Fixed mass ratios in compounds.
Mean free path = distance between collisions.

Behaviour of Gases

Gas Behavior at Low Temperature and High Pressure

For gases at low temperature and high pressure, (PV = KT)
Constant K = Nk_B, where k_B is Boltzmann constant (1.38 × 10^-23 J/K)
N is the number of molecules in the gas
At same P, V, T, number of molecules N is same for all gases
Number of molecules in 22.4 litres of any gas = Avogadro number N_A = 6.02 × 10²³

Perfect Gas Equation

Equation: (PV = μRT) where μ = number of moles, R = N_Ak_B = 8.314 J/mol·K
Relation: μ = M/M₀ = N/N_A, where M = mass, M₀ = molar mass
Alternate form: (PV = μRT = k_BNT) or (P = k_BnT), where n = number density
Real gases approach ideal gas behavior at low pressure and high temperature
Pressure relation: (P = M/M₀ RT/V = ρRT/M₀), where ρ = mass density
Ideal gas perfectly satisfies (PV = μRT) at all pressures and temperatures (theoretical model)

Solved Examples

Ex 1: Calculate pressure P if N = 2 × 10²³, T = 300 K, and number density n = 5 × 10²⁵ m^-3 using (P = k_BnT).
- P = 1.38 × 10^-23 × 5 × 10²⁵ × 300 = 2070 Pa
Ex 2: Find volume V of 1 mole of gas at P = 1 atm, T = 273 K using (PV = μRT).
- V = μRT/P = (1 × 8.314 × 273)/(1.013 × 10⁵) = 0.0224 m³ or 22.4 litres

Practice Set

Q1: What is the value of Boltzmann constant k_B?
- A. 1.38 × 10^-23 J/K
- B. 6.02 × 10²³ mol^-1
- C. 8.314 J/mol·K
- D. 22.4 litres
Q2: For an ideal gas, which equation is correct?
- A. PV = nRT
- B. PV = μRT
- C. P = k_BnT
- D. All of the above
Q3: Number of molecules in 22.4 litres of any gas at STP is?
- A. 6.02 × 10²³
- B. 1.38 × 10^-23
- C. 8.314
- D. 22.4

Answer Key

Q1: A - Boltzmann constant k_B = 1.38 × 10^-23 J/K is universal for gases.
Q2: D - All equations represent ideal gas relations in different forms.
Q3: A - Avogadro number 6.02 × 10²³ molecules in 22.4 litres at STP.

Memory Aids

PV = μRT is the perfect gas equation.
Boltzmann constant k_B = 1.38 × 10^-23 J/K.
Avogadro number N_A = 6.02 × 10²³.
Real gases behave ideally at low pressure, high temperature.

Gas Behavior at Low Temperature and High Pressure

For gases at low temperature and high pressure, (PV = KT)
Constant K = Nk_B, where k_B is Boltzmann constant (1.38 × 10^-23 J/K)
N is the number of molecules in the gas
At same P, V, T, number of molecules N is same for all gases
Number of molecules in 22.4 litres of any gas = Avogadro number N_A = 6.02 × 10²³

Perfect Gas Equation

Equation: (PV = μRT) where μ = number of moles, R = N_Ak_B = 8.314 J/mol·K
Relation: μ = M/M₀ = N/N_A, where M = mass, M₀ = molar mass
Alternate form: (PV = μRT = k_BNT) or (P = k_BnT), where n = number density
Real gases approach ideal gas behavior at low pressure and high temperature
Pressure relation: (P = M/M₀ RT/V = ρRT/M₀), where ρ = mass density
Ideal gas perfectly satisfies (PV = μRT) at all pressures and temperatures (theoretical model)

Solved Examples

Ex 1: Calculate pressure P if N = 2 × 10²³, T = 300 K, and number density n = 5 × 10²⁵ m^-3 using (P = k_BnT).
- P = 1.38 × 10^-23 × 5 × 10²⁵ × 300 = 2070 Pa
Ex 2: Find volume V of 1 mole of gas at P = 1 atm, T = 273 K using (PV = μRT).
- V = μRT/P = (1 × 8.314 × 273)/(1.013 × 10⁵) = 0.0224 m³ or 22.4 litres

Practice Set

Q1: What is the value of Boltzmann constant k_B?
- A. 1.38 × 10^-23 J/K
- B. 6.02 × 10²³ mol^-1
- C. 8.314 J/mol·K
- D. 22.4 litres
Q2: For an ideal gas, which equation is correct?
- A. PV = nRT
- B. PV = μRT
- C. P = k_BnT
- D. All of the above
Q3: Number of molecules in 22.4 litres of any gas at STP is?
- A. 6.02 × 10²³
- B. 1.38 × 10^-23
- C. 8.314
- D. 22.4

Answer Key

Q1: A - Boltzmann constant k_B = 1.38 × 10^-23 J/K is universal for gases.
Q2: D - All equations represent ideal gas relations in different forms.
Q3: A - Avogadro number 6.02 × 10²³ molecules in 22.4 litres at STP.

Memory Aids

PV = μRT is the perfect gas equation.
Boltzmann constant k_B = 1.38 × 10^-23 J/K.
Avogadro number N_A = 6.02 × 10²³.
Real gases behave ideally at low pressure, high temperature.

Gas Behavior at Low Temperature and High Pressure

For gases at low temperature and high pressure, PV = KT
Constant K = Nk_B, where k_B is Boltzmann constant (1.38 × 10^-23 J/K)
N is the number of molecules in the gas
At same P, V, T, number of molecules N is same for all gases
Number of molecules in 22.4 litres of any gas = Avogadro number N_A = 6.02 × 10²³

Perfect Gas Equation

Equation: PV = μRT where μ = number of moles, R = N_Ak_B = 8.314 J/mol·K
Relation: μ = M/M₀ = N/N_A, where M = mass, M₀ = molar mass
Alternate form: PV = μRT = k_BNT or P = k_BnT, where n = number density
Real gases approach ideal gas behavior at low pressure and high temperature
Pressure relation: P = M/M₀ RT/V = ρRT/M₀, where ρ = mass density
Ideal gas perfectly satisfies PV = μRT at all pressures and temperatures (theoretical model)

Solved Examples

Ex 1: Calculate pressure P if N = 2 × 10²³, T = 300 K, and number density n = 5 × 10²⁵ m^-3 using P = k_BnT.
- P = 1.38 × 10^-23 × 5 × 10²⁵ × 300 = 2070 Pa
Ex 2: Find volume V of 1 mole of gas at P = 1 atm, T = 273 K using PV = μRT.
- V = μRT/P = (1 × 8.314 × 273)/(1.013 × 10⁵) = 0.0224 m³ or 22.4 litres

Practice Set

Q1: What is the value of Boltzmann constant k_B?
- A. 1.38 × 10^-23 J/K
- B. 6.02 × 10²³ mol^-1
- C. 8.314 J/mol·K
- D. 22.4 litres
Q2: For an ideal gas, which equation is correct?
- A. PV = nRT
- B. PV = μRT
- C. P = k_BnT
- D. All of the above
Q3: Number of molecules in 22.4 litres of any gas at STP is?
- A. 6.02 × 10²³
- B. 1.38 × 10^-23
- C. 8.314
- D. 22.4

Answer Key

Q1: A - Boltzmann constant k_B = 1.38 × 10^-23 J/K is universal for gases.
Q2: D - All equations represent ideal gas relations in different forms.
Q3: A - Avogadro number 6.02 × 10²³ molecules in 22.4 litres at STP.

Memory Aids

PV = μRT is the perfect gas equation.
Boltzmann constant k_B = 1.38 × 10^-23 J/K.
Avogadro number N_A = 6.02 × 10²³.
Real gases behave ideally at low pressure, high temperature.

Gas Behavior at Low Temperature and High Pressure

For gases at low temperature and high pressure, (PV = KT)
Constant K = Nk_B, where k_B is Boltzmann constant (1.38 × 10^-23 J/K)
N is the number of molecules in the gas
At same P, V, T, number of molecules N is same for all gases
Number of molecules in 22.4 litres of any gas = Avogadro number N_A = 6.02 × 10²³

Perfect Gas Equation

Equation: (PV = μRT) where μ = number of moles, R = N_Ak_B = 8.314 J/mol·K
Relation: μ = M/M₀ = N/N_A, where M = mass, M₀ = molar mass
Alternate form: (PV = μRT = k_BNT) or (P = k_BnT), where n = number density
Real gases approach ideal gas behavior at low pressure and high temperature
Pressure relation: (P = M/M₀ RT/V = ρRT/M₀), where ρ = mass density
Ideal gas perfectly satisfies (PV = μRT) at all pressures and temperatures (theoretical model)

Solved Examples

Ex 1: Calculate pressure P if N = 2 × 10²³, T = 300 K, and number density n = 5 × 10²⁵ m^-3 using (P = k_BnT).
- P = 1.38 × 10^-23 × 5 × 10²⁵ × 300 = 2070 Pa
Ex 2: Find volume V of 1 mole of gas at P = 1 atm, T = 273 K using (PV = μRT).
- V = μRT/P = (1 × 8.314 × 273)/(1.013 × 10⁵) = 0.0224 m³ or 22.4 litres

Practice Set

Q1: What is the value of Boltzmann constant k_B?
- A. 1.38 × 10^-23 J/K
- B. 6.02 × 10²³ mol^-1
- C. 8.314 J/mol·K
- D. 22.4 litres
Q2: For an ideal gas, which equation is correct?
- A. PV = nRT
- B. PV = μRT
- C. P = k_BnT
- D. All of the above
Q3: Number of molecules in 22.4 litres of any gas at STP is?
- A. 6.02 × 10²³
- B. 1.38 × 10^-23
- C. 8.314
- D. 22.4

Answer Key

Q1: A - Boltzmann constant k_B = 1.38 × 10^-23 J/K is universal for gases.
Q2: D - All equations represent ideal gas relations in different forms.
Q3: A - Avogadro number 6.02 × 10²³ molecules in 22.4 litres at STP.

Memory Aids

PV = μRT is the perfect gas equation.
Boltzmann constant k_B = 1.38 × 10^-23 J/K.
Avogadro number N_A = 6.02 × 10²³.
Real gases behave ideally at low pressure, high temperature.

Gas Behavior at Low Temperature and High Pressure

For gases at low temperature and high pressure, \( PV = K T \)
Constant \( K = N k_B \), where \( k_B \) is Boltzmann constant (\(1.38 \times 10^{-23} \) J/K)
\( N \) is the number of molecules in the gas
At same \( P, V, T \), number of molecules \( N \) is same for all gases
Number of molecules in 22.4 litres of any gas = Avogadro number \( N_A = 6.02 \times 10^{23} \)

Perfect Gas Equation

Equation: \( PV = \mu RT \) where \( \mu \) = number of moles, \( R = N_A k_B = 8.314 \) J/mol·K
Relation: \( \mu = \frac{M}{M_0} = \frac{N}{N_A} \), where \( M \) = mass, \( M_0 \) = molar mass
Alternate form: \( PV = \mu RT = k_B N T \) or \( P = k_B n T \), where \( n \) = number density
Real gases approach ideal gas behavior at low pressure and high temperature
Pressure relation: \( P = \frac{M}{M_0} \frac{RT}{V} = \frac{\rho RT}{M_0} \), where \( \rho \) = mass density
Ideal gas perfectly satisfies \( PV = \mu RT \) at all pressures and temperatures (theoretical model)

Solved Examples

Ex 1: Calculate pressure \( P \) if \( N = 2 \times 10^{23} \), \( T = 300 K \), and number density \( n = 5 \times 10^{25} \) m^-3 using \( P = k_B n T \).
- \( P = 1.38 \times 10^{-23} \times 5 \times 10^{25} \times 300 = 2070 \) Pa
Ex 2: Find volume \( V \) of 1 mole of gas at \( P = 1 \) atm, \( T = 273 K \) using \( PV = \mu RT \).
- \( V = \frac{\mu RT}{P} = \frac{1 \times 8.314 \times 273}{1.013 \times 10^{5}} = 0.0224 \) m³ or 22.4 litres

Practice Set

Q1: What is the value of Boltzmann constant \( k_B \)?
- A. \(1.38 \times 10^{-23} \) J/K
- B. \(6.02 \times 10^{23} \) mol^-1
- C. \(8.314 \) J/mol·K
- D. \(22.4 \) litres
Q2: For an ideal gas, which equation is correct?
- A. \( PV = nRT \)
- B. \( PV = \mu RT \)
- C. \( P = k_B n T \)
- D. All of the above
Q3: Number of molecules in 22.4 litres of any gas at STP is?
- A. \(6.02 \times 10^{23} \)
- B. \(1.38 \times 10^{-23} \)
- C. \(8.314 \)
- D. \(22.4 \)

Answer Key

Q1: A - Boltzmann constant \( k_B = 1.38 \times 10^{-23} \) J/K is universal for gases.
Q2: D - All equations represent ideal gas relations in different forms.
Q3: A - Avogadro number \( 6.02 \times 10^{23} \) molecules in 22.4 litres at STP.

Memory Aids

\( PV = \mu RT \) is the perfect gas equation.
Boltzmann constant \( k_B = 1.38 \times 10^{-23} \) J/K.
Avogadro number \( N_A = 6.02 \times 10^{23} \).
Real gases behave ideally at low pressure, high temperature.

Gas Behavior at Low Temperature and High Pressure

For gases at low temperature and high pressure, (PV = KT)
Constant K = Nk_B, where k_B is Boltzmann constant (1.38 × 10^-23 J/K)
N is the number of molecules in the gas
At same P, V, T, number of molecules N is same for all gases
Number of molecules in 22.4 litres of any gas = Avogadro number N_A = 6.02 × 10²³

Perfect Gas Equation

Equation: (PV = μRT) where μ = number of moles, R = N_Ak_B = 8.314 J/mol·K
Relation: μ = M/M₀ = N/N_A, where M = mass, M₀ = molar mass
Alternate form: (PV = μRT = k_BNT) or (P = k_BnT), where n = number density
Real gases approach ideal gas behavior at low pressure and high temperature
Pressure relation: (P = M/M₀ RT/V = ρRT/M₀), where ρ = mass density
Ideal gas perfectly satisfies (PV = μRT) at all pressures and temperatures (theoretical model)

Solved Examples

Ex 1: Calculate pressure P if N = 2 × 10²³, T = 300 K, and number density n = 5 × 10²⁵ m^-3 using (P = k_BnT).
- P = 1.38 × 10^-23 × 5 × 10²⁵ × 300 = 2070 Pa
Ex 2: Find volume V of 1 mole of gas at P = 1 atm, T = 273 K using (PV = μRT).
- V = μRT/P = (1 × 8.314 × 273)/(1.013 × 10⁵) = 0.0224 m³ or 22.4 litres

Practice Set

Q1: What is the value of Boltzmann constant k_B?
- A. 1.38 × 10^-23 J/K
- B. 6.02 × 10²³ mol^-1
- C. 8.314 J/mol·K
- D. 22.4 litres
Q2: For an ideal gas, which equation is correct?
- A. PV = nRT
- B. PV = μRT
- C. P = k_BnT
- D. All of the above
Q3: Number of molecules in 22.4 litres of any gas at STP is?
- A. 6.02 × 10²³
- B. 1.38 × 10^-23
- C. 8.314
- D. 22.4

Answer Key

Q1: A - Boltzmann constant k_B = 1.38 × 10^-23 J/K is universal for gases.
Q2: D - All equations represent ideal gas relations in different forms.
Q3: A - Avogadro number 6.02 × 10²³ molecules in 22.4 litres at STP.

Memory Aids

PV = μRT is the perfect gas equation.
Boltzmann constant k_B = 1.38 × 10^-23 J/K.
Avogadro number N_A = 6.02 × 10²³.
Real gases behave ideally at low pressure, high temperature.

Gas Behavior at Low Temperature and High Pressure

For gases at low temperature and high pressure, PV = KT
Constant K = Nk_B, where k_B is Boltzmann constant (1.38 × 10^-23 J/K)
N is the number of molecules in the gas
At same P, V, T, number of molecules N is same for all gases
Number of molecules in 22.4 litres of any gas = Avogadro number N_A = 6.02 × 10²³

Perfect Gas Equation

Equation: PV = μRT where μ = number of moles, R = N_Ak_B = 8.314 J/mol·K
Relation: μ = M/M₀ = N/N_A, where M = mass, M₀ = molar mass
Alternate form: PV = μRT = k_BNT or P = k_BnT, where n = number density
Real gases approach ideal gas behavior at low pressure and high temperature
Pressure relation: P = M/M₀ RT/V = ρRT/M₀, where ρ = mass density
Ideal gas perfectly satisfies PV = μRT at all pressures and temperatures (theoretical model)

Solved Examples

Ex 1: Calculate pressure P if N = 2 × 10²³, T = 300 K, and number density n = 5 × 10²⁵ m^-3 using P = k_BnT.
- P = 1.38 × 10^-23 × 5 × 10²⁵ × 300 = 2070 Pa
Ex 2: Find volume V of 1 mole of gas at P = 1 atm, T = 273 K using PV = μRT.
- V = μRT/P = (1 × 8.314 × 273)/(1.013 × 10⁵) = 0.0224 m³ or 22.4 litres

Practice Set

Q1: What is the value of Boltzmann constant k_B?
- A. 1.38 × 10^-23 J/K
- B. 6.02 × 10²³ mol^-1
- C. 8.314 J/mol·K
- D. 22.4 litres
Q2: For an ideal gas, which equation is correct?
- A. PV = nRT
- B. PV = μRT
- C. P = k_BnT
- D. All of the above
Q3: Number of molecules in 22.4 litres of any gas at STP is?
- A. 6.02 × 10²³
- B. 1.38 × 10^-23
- C. 8.314
- D. 22.4

Answer Key

Q1: A - Boltzmann constant k_B = 1.38 × 10^-23 J/K is universal for gases.
Q2: D - All equations represent ideal gas relations in different forms.
Q3: A - Avogadro number 6.02 × 10²³ molecules in 22.4 litres at STP.

Memory Aids

PV = μRT is the perfect gas equation.
Boltzmann constant k_B = 1.38 × 10^-23 J/K.
Avogadro number N_A = 6.02 × 10²³.
Real gases behave ideally at low pressure, high temperature.

Gas Behavior at Low Temperature and High Pressure

For gases at low temperature and high pressure, (PV = KT)
Constant K = Nk_B, where k_B is Boltzmann constant (1.38 × 10^-23 J/K)
N is the number of molecules in the gas
At same P, V, T, number of molecules N is same for all gases
Number of molecules in 22.4 litres of any gas = Avogadro number N_A = 6.02 × 10²³

Perfect Gas Equation

Equation: (PV = μRT) where μ = number of moles, R = N_Ak_B = 8.314 J/mol·K
Relation: μ = M/M₀ = N/N_A, where M = mass, M₀ = molar mass
Alternate form: (PV = μRT = k_BNT) or (P = k_BnT), where n = number density
Real gases approach ideal gas behavior at low pressure and high temperature
Pressure relation: (P = M/M₀ RT/V = ρRT/M₀), where ρ = mass density
Ideal gas perfectly satisfies (PV = μRT) at all pressures and temperatures (theoretical model)

Solved Examples

Ex 1: Calculate pressure P if N = 2 × 10²³, T = 300 K, and number density n = 5 × 10²⁵ m^-3 using (P = k_BnT).
- P = 1.38 × 10^-23 × 5 × 10²⁵ × 300 = 2070 Pa
Ex 2: Find volume V of 1 mole of gas at P = 1 atm, T = 273 K using (PV = μRT).
- V = μRT/P = (1 × 8.314 × 273)/(1.013 × 10⁵) = 0.0224 m³ or 22.4 litres

Practice Set

Q1: What is the value of Boltzmann constant k_B?
- A. 1.38 × 10^-23 J/K
- B. 6.02 × 10²³ mol^-1
- C. 8.314 J/mol·K
- D. 22.4 litres
Q2: For an ideal gas, which equation is correct?
- A. PV = nRT
- B. PV = μRT
- C. P = k_BnT
- D. All of the above
Q3: Number of molecules in 22.4 litres of any gas at STP is?
- A. 6.02 × 10²³
- B. 1.38 × 10^-23
- C. 8.314
- D. 22.4

Answer Key

Q1: A - Boltzmann constant k_B = 1.38 × 10^-23 J/K is universal for gases.
Q2: D - All equations represent ideal gas relations in different forms.
Q3: A - Avogadro number 6.02 × 10²³ molecules in 22.4 litres at STP.

Memory Aids

PV = μRT is the perfect gas equation.
Boltzmann constant k_B = 1.38 × 10^-23 J/K.
Avogadro number N_A = 6.02 × 10²³.
Real gases behave ideally at low pressure, high temperature.

Gas Behavior at Low Temperature and High Pressure

For gases at low temperature and high pressure, \( PV = K T \)
Constant \( K = N k_B \), where \( k_B \) is Boltzmann constant (\(1.38 \times 10^{-23} \) J/K)
\( N \) is the number of molecules in the gas
At same \( P, V, T \), number of molecules \( N \) is same for all gases
Number of molecules in 22.4 litres of any gas = Avogadro number \( N_A = 6.02 \times 10^{23} \)

Perfect Gas Equation

Equation: \( PV = \mu RT \) where \( \mu \) = number of moles, \( R = N_A k_B = 8.314 \) J/mol·K
Relation: \( \mu = \frac{M}{M_0} = \frac{N}{N_A} \), where \( M \) = mass, \( M_0 \) = molar mass
Alternate form: \( PV = \mu RT = k_B N T \) or \( P = k_B n T \), where \( n \) = number density
Real gases approach ideal gas behavior at low pressure and high temperature
Pressure relation: \( P = \frac{M}{M_0} \frac{RT}{V} = \frac{\rho RT}{M_0} \), where \( \rho \) = mass density
Ideal gas perfectly satisfies \( PV = \mu RT \) at all pressures and temperatures (theoretical model)

Solved Examples

Ex 1: Calculate pressure \( P \) if \( N = 2 \times 10^{23} \), \( T = 300 K \), and number density \( n = 5 \times 10^{25} \) m^-3 using \( P = k_B n T \).
- \( P = 1.38 \times 10^{-23} \times 5 \times 10^{25} \times 300 = 2070 \) Pa
Ex 2: Find volume \( V \) of 1 mole of gas at \( P = 1 \) atm, \( T = 273 K \) using \( PV = \mu RT \).
- \( V = \frac{\mu RT}{P} = \frac{1 \times 8.314 \times 273}{1.013 \times 10^{5}} = 0.0224 \) m³ or 22.4 litres

Practice Set

Q1: What is the value of Boltzmann constant \( k_B \)?
- A. \(1.38 \times 10^{-23} \) J/K
- B. \(6.02 \times 10^{23} \) mol^-1
- C. \(8.314 \) J/mol·K
- D. \(22.4 \) litres
Q2: For an ideal gas, which equation is correct?
- A. \( PV = nRT \)
- B. \( PV = \mu RT \)
- C. \( P = k_B n T \)
- D. All of the above
Q3: Number of molecules in 22.4 litres of any gas at STP is?
- A. \(6.02 \times 10^{23} \)
- B. \(1.38 \times 10^{-23} \)
- C. \(8.314 \)
- D. \(22.4 \)

Answer Key

Q1: A - Boltzmann constant \( k_B = 1.38 \times 10^{-23} \) J/K is universal for gases.
Q2: D - All equations represent ideal gas relations in different forms.
Q3: A - Avogadro number \( 6.02 \times 10^{23} \) molecules in 22.4 litres at STP.

Memory Aids

\( PV = \mu RT \) is the perfect gas equation.
Boltzmann constant \( k_B = 1.38 \times 10^{-23} \) J/K.
Avogadro number \( N_A = 6.02 \times 10^{23} \).
Real gases behave ideally at low pressure, high temperature.

Boyle’s Law

Pressure-Volume Relationship

At constant temperature (μ, T), PV = constant
Pressure varies inversely with volume: P ∝ 1/V
Experimental P–V curves for steam match Boyle’s law at high temperature and low pressure

Solved Examples

Ex 1: If volume halves at constant temperature, what happens to pressure?
- Pressure doubles (P ∝ 1/V)
Ex 2: Calculate new pressure if initial P₁ = 2 atm, V₁ = 4 L, and volume changes to V₂ = 2 L.
- P₂ = P₁ V₁/V₂ = 2 × 4/2 = 4 atm

Practice Set

Q1: Boyle’s law states that at constant temperature, pressure is?
- A. Directly proportional to volume
- B. Inversely proportional to volume
- C. Independent of volume
- D. Proportional to temperature
Q2: If volume is doubled at constant temperature, pressure will?
- A. Double
- B. Halve
- C. Remain same
- D. Become zero
Q3: For a gas at constant temperature, if pressure is 3 atm and volume is 2 L, what is volume when pressure is 6 atm?
- A. 1 L
- B. 4 L
- C. 3 L
- D. 6 L

Answer Key

Q1: B - Pressure varies inversely with volume at constant temperature.
Q2: B - Doubling volume halves the pressure.
Q3: A - Using P₁V₁ = P₂V₂, V₂ = 1 L.

Memory Aids

Boyle’s law: PV = constant
Pressure and volume are inversely related at constant temperature

Pressure-Volume Relationship

At constant temperature \( (\mu, T) \), \( PV = \text{constant} \)
Pressure varies inversely with volume: \( P \propto \frac{1}{V} \)
Experimental P–V curves for steam match Boyle’s law at high temperature and low pressure

Solved Examples

Ex 1: If volume halves at constant temperature, what happens to pressure?
- Pressure doubles (\( P \propto \frac{1}{V} \))
Ex 2: Calculate new pressure if initial \( P_1 = 2 \) atm, \( V_1 = 4 \) L, and volume changes to \( V_2 = 2 \) L.
- \( P_2 = \frac{P_1 V_1}{V_2} = \frac{2 \times 4}{2} = 4 \) atm

Practice Set

Q1: Boyle’s law states that at constant temperature, pressure is?
- A. Directly proportional to volume
- B. Inversely proportional to volume
- C. Independent of volume
- D. Proportional to temperature
Q2: If volume is doubled at constant temperature, pressure will?
- A. Double
- B. Halve
- C. Remain same
- D. Become zero
Q3: For a gas at constant temperature, if pressure is 3 atm and volume is 2 L, what is volume when pressure is 6 atm?
- A. 1 L
- B. 4 L
- C. 3 L
- D. 6 L

Answer Key

Q1: B - Pressure varies inversely with volume at constant temperature.
Q2: B - Doubling volume halves the pressure.
Q3: A - Using \( P_1 V_1 = P_2 V_2 \), \( V_2 = 1 \) L.

Memory Aids

Boyle’s law: \( PV = \text{constant} \)
Pressure and volume are inversely related at constant temperature

Charles’ Law

Volume-Temperature Relationship

At constant pressure, volume V is proportional to absolute temperature T: V ∝ T
Experimental T–V curves for CO₂ match Charles’ law at high temperature and low pressure

Solved Examples

Ex 1: If volume is 2 L at 300 K, find volume at 600 K at constant pressure.
- V₂ = V₁ × T₂/T₁ = 2 × 600/300 = 4 L
Ex 2: Calculate temperature if volume changes from 1 L to 3 L at constant pressure and initial temperature 273 K.
- T₂ = T₁ × V₂/V₁ = 273 × 3/1 = 819 K

Practice Set

Q1: Charles’ law states volume is proportional to?
- A. Pressure
- B. Temperature (absolute)
- C. Number of moles
- D. Mass
Q2: If temperature doubles, volume will?
- A. Halve
- B. Double
- C. Remain same
- D. Become zero
Q3: Volume of gas is 3 L at 300 K. Find volume at 450 K at constant pressure.
- A. 4.5 L
- B. 2 L
- C. 3 L
- D. 6 L

Answer Key

Q1: B - Volume is proportional to absolute temperature at constant pressure.
Q2: B - Doubling temperature doubles volume.
Q3: A - V₂ = 3 × 450/300 = 4.5 L.

Memory Aids

Charles’ law: V ∝ T at constant pressure
Volume increases with temperature

Volume-Temperature Relationship

At constant pressure, volume \( V \) is proportional to absolute temperature \( T \): \( V \propto T \)
Experimental T–V curves for CO₂ match Charles’ law at high temperature and low pressure

Solved Examples

Ex 1: If volume is 2 L at 300 K, find volume at 600 K at constant pressure.
- \( V_2 = V_1 \times \frac{T_2}{T_1} = 2 \times \frac{600}{300} = 4 \) L
Ex 2: Calculate temperature if volume changes from 1 L to 3 L at constant pressure and initial temperature 273 K.
- \( T_2 = T_1 \times \frac{V_2}{V_1} = 273 \times \frac{3}{1} = 819 \) K

Practice Set

Q1: Charles’ law states volume is proportional to?
- A. Pressure
- B. Temperature (absolute)
- C. Number of moles
- D. Mass
Q2: If temperature doubles, volume will?
- A. Halve
- B. Double
- C. Remain same
- D. Become zero
Q3: Volume of gas is 3 L at 300 K. Find volume at 450 K at constant pressure.
- A. 4.5 L
- B. 2 L
- C. 3 L
- D. 6 L

Answer Key

Q1: B - Volume is proportional to absolute temperature at constant pressure.
Q2: B - Doubling temperature doubles volume.
Q3: A - \( V_2 = 3 \times \frac{450}{300} = 4.5 \) L.

Memory Aids

Charles’ law: \( V \propto T \) at constant pressure
Volume increases with temperature

Dalton’s Law of Partial Pressures

Partial Pressure Concept

Total pressure of gas mixture = sum of partial pressures of each gas
At constant T and P, P_total = P₁ + P₂ + P₃ + ...
Partial pressure of i^th gas: P_i = μ_iRT / V
Total pressure: P_total = (RT / V)(μ₁ + μ₂ + ...) = (RT / V) μ_total
Partial pressure relation: P_i = P_total × μ_i/μ_total, where mole fraction μ_i/μ_total

Solved Examples

Ex 1: Calculate total pressure if μ₁ = 1 mol, μ₂ = 2 mol, R = 8.314 J/mol·K, T = 300 K, V = 0.01 m³.
- P_total = RT/V (μ₁ + μ₂) = 8.314 × 300/0.01 × 3 = 7.48 × 10⁵ Pa
Ex 2: Find partial pressure of gas 1 if total pressure is 7.48 × 10⁵ Pa and mole fraction μ₁/μ_total = 1/3.
- P₁ = P_total × 1/3 = 2.49 × 10⁵ Pa

Practice Set

Q1: Dalton’s law states total pressure is?
- A. Product of partial pressures
- B. Sum of partial pressures
- C. Difference of partial pressures
- D. Ratio of partial pressures
Q2: Partial pressure of a gas is proportional to?
- A. Volume
- B. Mole fraction
- C. Temperature
- D. Mass
Q3: If total pressure is 1 atm and mole fraction of gas A is 0.4, partial pressure of A is?
- A. 0.4 atm
- B. 0.6 atm
- C. 1 atm
- D. 0.8 atm

Answer Key

Q1: B - Total pressure equals sum of partial pressures of gases.
Q2: B - Partial pressure proportional to mole fraction.
Q3: A - Partial pressure = total pressure × mole fraction = 0.4 atm.

Memory Aids

Dalton’s law: P_total = ∑P_i
Partial pressure P_i = P_total × mole fraction

Partial Pressure Concept

Total pressure of gas mixture = sum of partial pressures of each gas
At constant \( T \) and \( P \), \( P_{total} = P_1 + P_2 + P_3 + \ldots \)
Partial pressure of \( i^{th} \) gas: \( P_i = \mu_i RT / V \)
Total pressure: \( P_{total} = (RT / V)(\mu_1 + \mu_2 + \ldots) = (RT / V) \mu_{total} \)
Partial pressure relation: \( P_i = P_{total} \times \frac{\mu_i}{\mu_{total}} \), where mole fraction \( \frac{\mu_i}{\mu_{total}} \)

Solved Examples

Ex 1: Calculate total pressure if \( \mu_1 = 1 \) mol, \( \mu_2 = 2 \) mol, \( R = 8.314 \) J/mol·K, \( T = 300 K \), \( V = 0.01 \) m³.
- \( P_{total} = \frac{RT}{V} (\mu_1 + \mu_2) = \frac{8.314 \times 300}{0.01} \times 3 = 7.48 \times 10^{5} \) Pa
Ex 2: Find partial pressure of gas 1 if total pressure is \( 7.48 \times 10^{5} \) Pa and mole fraction \( \frac{\mu_1}{\mu_{total}} = \frac{1}{3} \).
- \( P_1 = P_{total} \times \frac{1}{3} = 2.49 \times 10^{5} \) Pa

Practice Set

Q1: Dalton’s law states total pressure is?
- A. Product of partial pressures
- B. Sum of partial pressures
- C. Difference of partial pressures
- D. Ratio of partial pressures
Q2: Partial pressure of a gas is proportional to?
- A. Volume
- B. Mole fraction
- C. Temperature
- D. Mass
Q3: If total pressure is 1 atm and mole fraction of gas A is 0.4, partial pressure of A is?
- A. 0.4 atm
- B. 0.6 atm
- C. 1 atm
- D. 0.8 atm

Answer Key

Q1: B - Total pressure equals sum of partial pressures of gases.
Q2: B - Partial pressure proportional to mole fraction.
Q3: A - Partial pressure = total pressure × mole fraction = 0.4 atm.

Memory Aids

Dalton’s law: \( P_{total} = \sum P_i \)
Partial pressure \( P_i = P_{total} \times \) mole fraction

Law of Equipartition of Energy

Concept

In thermal equilibrium, total energy is equally divided among all energy modes.
Each mode has average energy 1/2 k_BT, where k_B is Boltzmann constant and T is temperature in Kelvin.
Each translational and rotational degree of freedom contributes 1/2 k_BT energy.
Each vibrational frequency contributes 2 × 1/2 k_BT = k_BT due to kinetic and potential energy.

Monoatomic Molecules

Translational kinetic energy per molecule: E = 1/2 m v̄_x² + 1/2 m v̄_y² + 1/2 m v̄_z²
Or simplified: E = 3/2 k_BT

Diatomic Molecules (No Vibration)

Kinetic energy includes translational and rotational parts:
E = E_t + E_r = 1/2 m (v_x² + v_y² + v_z²) + 1/2 I₁ω₁² + 1/2 I₂ω₂² = 5/2 k_BT
Assumes no vibrational energy.

Diatomic Molecules (With Vibration)

Vibrational energy term: E_v = 1/2 m (dy/dt)² + 1/2 ky²
Total energy: E = E_t + E_r + E_v

Solved Examples

Ex 1: Calculate average KE of a monoatomic gas molecule at temperature T = 300 K.
- Use E = 3/2 k_BT
- Substitute k_B = 1.38 × 10^-23 J/K
- E = 3/2 × 1.38 × 10^-23 × 300 = 6.21 × 10^-21 J
Ex 2: Find total energy of a diatomic molecule (no vibration) at 400 K.
- Use E = 5/2 k_BT
- E = 5/2 × 1.38 × 10^-23 × 400 = 1.38 × 10^-20 J

Practice Set

Q1: What is the average energy per degree of freedom at temperature T?
- A) k_BT
- B) 1/2 k_BT
- C) 2 k_BT
- D) 3/2 k_BT
Q2: The kinetic energy of a monoatomic gas molecule at temperature 300 K is approximately:
- A) 1.38 × 10^-21 J
- B) 6.21 × 10^-21 J
- C) 4.14 × 10^-21 J
- D) 9.0 × 10^-21 J
Q3: For a diatomic molecule with vibration considered, total energy is:
- A) 5/2 k_BT
- B) 3 k_BT
- C) 7/2 k_BT
- D) 4 k_BT

Answer Key

Q1: B) 1/2 k_BT - Each degree of freedom contributes half k_BT energy.
Q2: B) 6.21 × 10^-21 J - Using 3/2 k_BT at 300 K.
Q3: D) 4 k_BT - Vibration adds k_BT to 5/2 k_BT making total 4 k_BT.

Memory Aids

Energy per mode = 1/2 k_BT
Monoatomic KE = 3/2 k_BT
Diatomic (no vibration) KE = 5/2 k_BT
Vibration adds k_BT energy

Concept

In thermal equilibrium, total energy is equally divided among all energy modes.
Each mode has average energy 1/2 k_BT, where k_B is Boltzmann constant and T is temperature in Kelvin.
Each translational and rotational degree of freedom contributes 1/2 k_BT energy.
Each vibrational frequency contributes 2 × 1/2 k_BT = k_BT due to kinetic and potential energy.

Monoatomic Molecules

Translational kinetic energy per molecule: E = 1/2 m v̄_x² + 1/2 m v̄_y² + 1/2 m v̄_z²
Or simplified: E = 3/2 k_BT

Diatomic Molecules (No Vibration)

Kinetic energy includes translational and rotational parts:
E = E_t + E_r = 1/2 m (v_x² + v_y² + v_z²) + 1/2 I₁ω₁² + 1/2 I₂ω₂² = 5/2 k_BT
Assumes no vibrational energy.

Diatomic Molecules (With Vibration)

Vibrational energy term: E_v = 1/2 m (dy/dt)² + 1/2 ky²
Total energy: E = E_t + E_r + E_v

Solved Examples

Ex 1: Calculate average KE of a monoatomic gas molecule at temperature T = 300 K.
- Use E = 3/2 k_BT
- Substitute k_B = 1.38 × 10^-23 J/K
- E = 3/2 × 1.38 × 10^-23 × 300 = 6.21 × 10^-21 J
Ex 2: Find total energy of a diatomic molecule (no vibration) at 400 K.
- Use E = 5/2 k_BT
- E = 5/2 × 1.38 × 10^-23 × 400 = 1.38 × 10^-20 J

Practice Set

Q1: What is the average energy per degree of freedom at temperature T?
- A) k_BT
- B) 1/2 k_BT
- C) 2 k_BT
- D) 3/2 k_BT
Q2: The kinetic energy of a monoatomic gas molecule at temperature 300 K is approximately:
- A) 1.38 × 10^-21 J
- B) 6.21 × 10^-21 J
- C) 4.14 × 10^-21 J
- D) 9.0 × 10^-21 J
Q3: For a diatomic molecule with vibration considered, total energy is:
- A) 5/2 k_BT
- B) 3 k_BT
- C) 7/2 k_BT
- D) 4 k_BT

Answer Key

Q1: B) 1/2 k_BT - Each degree of freedom contributes half k_BT energy.
Q2: B) 6.21 × 10^-21 J - Using 3/2 k_BT at 300 K.
Q3: D) 4 k_BT - Vibration adds k_BT to 5/2 k_BT making total 4 k_BT.

Memory Aids

Energy per mode = 1/2 k_BT
Monoatomic KE = 3/2 k_BT
Diatomic (no vibration) KE = 5/2 k_BT
Vibration adds k_BT energy

Concept

In thermal equilibrium, total energy is equally divided among all energy modes.
Each mode has average energy 1/2 k_BT, where k_B is Boltzmann constant and T is temperature in Kelvin.
Each translational and rotational degree of freedom contributes 1/2 k_BT energy.
Each vibrational frequency contributes 2 × 1/2 k_BT = k_BT due to kinetic and potential energy.

Monoatomic Molecules

Translational kinetic energy per molecule: E = 1/2 m v_x² + 1/2 m v_y² + 1/2 m v_z²
Or simplified: E = 3/2 k_BT

Diatomic Molecules (No Vibration)

Kinetic energy includes translational and rotational parts:
E = E_t + E_r = 1/2 m (v_x² + v_y² + v_z²) + 1/2 I₁ω₁² + 1/2 I₂ω₂² = 5/2 k_BT
Assumes no vibrational energy.

Diatomic Molecules (With Vibration)

Vibrational energy term: E_v = 1/2 m (dy/dt)² + 1/2 ky²
Total energy: E = E_t + E_r + E_v

Solved Examples

Ex 1: Calculate average KE of a monoatomic gas molecule at temperature T = 300 K.
- Use E = 3/2 k_BT
- Substitute k_B = 1.38 × 10^-23 J/K
- E = 3/2 × 1.38 × 10^-23 × 300 = 6.21 × 10^-21 J
Ex 2: Find total energy of a diatomic molecule (no vibration) at 400 K.
- Use E = 5/2 k_BT
- E = 5/2 × 1.38 × 10^-23 × 400 = 1.38 × 10^-20 J

Practice Set

Q1: What is the average energy per degree of freedom at temperature T?
- A) k_BT
- B) 1/2 k_BT
- C) 2 k_BT
- D) 3/2 k_BT
Q2: The kinetic energy of a monoatomic gas molecule at temperature 300 K is approximately:
- A) 1.38 × 10^-21 J
- B) 6.21 × 10^-21 J
- C) 4.14 × 10^-21 J
- D) 9.0 × 10^-21 J
Q3: For a diatomic molecule with vibration considered, total energy is:
- A) 5/2 k_BT
- B) 3 k_BT
- C) 7/2 k_BT
- D) 4 k_BT

Answer Key

Q1: B) 1/2 k_BT - Each degree of freedom contributes half k_BT energy.
Q2: B) 6.21 × 10^-21 J - Using 3/2 k_BT at 300 K.
Q3: D) 4 k_BT - Vibration adds k_BT to 5/2 k_BT making total 4 k_BT.

Memory Aids

Energy per mode = 1/2 k_BT
Monoatomic KE = 3/2 k_BT
Diatomic (no vibration) KE = 5/2 k_BT
Vibration adds k_BT energy

Concept

In thermal equilibrium, total energy is equally divided among all energy modes.
Each mode has average energy 1/2 k_BT, where k_B is Boltzmann constant and T is temperature in Kelvin.
Each translational and rotational degree of freedom contributes 1/2 k_BT energy.
Each vibrational frequency contributes 2 × 1/2 k_BT = k_BT due to kinetic and potential energy.

Monoatomic Molecules

Translational kinetic energy per molecule: E = 1/2 m v̄_x² + 1/2 m v̄_y² + 1/2 m v̄_z²
Or simplified: E = 3/2 k_BT

Diatomic Molecules (No Vibration)

Kinetic energy includes translational and rotational parts:
E = E_t + E_r = 1/2 m (v_x² + v_y² + v_z²) + 1/2 I₁ω₁² + 1/2 I₂ω₂² = 5/2 k_BT
Assumes no vibrational energy.

Diatomic Molecules (With Vibration)

Vibrational energy term: E_v = 1/2 m (dy/dt)² + 1/2 ky²
Total energy: E = E_t + E_r + E_v

Solved Examples

Ex 1: Calculate average KE of a monoatomic gas molecule at temperature T = 300 K.
- Use E = 3/2 k_BT
- Substitute k_B = 1.38 × 10^-23 J/K
- E = 3/2 × 1.38 × 10^-23 × 300 = 6.21 × 10^-21 J
Ex 2: Find total energy of a diatomic molecule (no vibration) at 400 K.
- Use E = 5/2 k_BT
- E = 5/2 × 1.38 × 10^-23 × 400 = 1.38 × 10^-20 J

Practice Set

Q1: What is the average energy per degree of freedom at temperature T?
- A) k_BT
- B) 1/2 k_BT
- C) 2 k_BT
- D) 3/2 k_BT
Q2: The kinetic energy of a monoatomic gas molecule at temperature 300 K is approximately:
- A) 1.38 × 10^-21 J
- B) 6.21 × 10^-21 J
- C) 4.14 × 10^-21 J
- D) 9.0 × 10^-21 J
Q3: For a diatomic molecule with vibration considered, total energy is:
- A) 5/2 k_BT
- B) 3 k_BT
- C) 7/2 k_BT
- D) 4 k_BT

Answer Key

Q1: B) 1/2 k_BT - Each degree of freedom contributes half k_BT energy.
Q2: B) 6.21 × 10^-21 J - Using 3/2 k_BT at 300 K.
Q3: D) 4 k_BT - Vibration adds k_BT to 5/2 k_BT making total 4 k_BT.

Memory Aids

Energy per mode = 1/2 k_BT
Monoatomic KE = 3/2 k_BT
Diatomic (no vibration) KE = 5/2 k_BT
Vibration adds k_BT energy

Concept

In thermal equilibrium, total energy is equally divided among all energy modes.
Each mode has average energy \( \frac{1}{2} k_B T \), where \( k_B \) is Boltzmann constant and \( T \) is temperature in Kelvin.
Each translational and rotational degree of freedom contributes \( \frac{1}{2} k_B T \) energy.
Each vibrational frequency contributes \( 2 \times \frac{1}{2} k_B T = k_B T \) due to kinetic and potential energy.

Monoatomic Molecules

Translational kinetic energy per molecule: \( E = \frac{1}{2} m \overline{v_x^2} + \frac{1}{2} m \overline{v_y^2} + \frac{1}{2} m \overline{v_z^2} \)
Or simplified: \( E = \frac{3}{2} k_B T \)

Diatomic Molecules (No Vibration)

Kinetic energy includes translational and rotational parts:
\( E = E_t + E_r = \frac{1}{2} m (v_x^2 + v_y^2 + v_z^2) + \frac{1}{2} I_1 \omega_1^2 + \frac{1}{2} I_2 \omega_2^2 = \frac{5}{2} k_B T \)
Assumes no vibrational energy.

Diatomic Molecules (With Vibration)

Vibrational energy term: \( E_v = \frac{1}{2} m \left( \frac{dy}{dt} \right)^2 + \frac{1}{2} k y^2 \)
Total energy: \( E = E_t + E_r + E_v \)

Solved Examples

Ex 1: Calculate average KE of a monoatomic gas molecule at temperature \( T = 300 K \).
- Use \( E = \frac{3}{2} k_B T \)
- Substitute \( k_B = 1.38 \times 10^{-23} J/K \)
- \( E = \frac{3}{2} \times 1.38 \times 10^{-23} \times 300 = 6.21 \times 10^{-21} J \)
Ex 2: Find total energy of a diatomic molecule (no vibration) at \( 400 K \).
- Use \( E = \frac{5}{2} k_B T \)
- \( E = \frac{5}{2} \times 1.38 \times 10^{-23} \times 400 = 1.38 \times 10^{-20} J \)

Practice Set

Q1: What is the average energy per degree of freedom at temperature \( T \)?
- A) \( k_B T \)
- B) \( \frac{1}{2} k_B T \)
- C) \( 2 k_B T \)
- D) \( \frac{3}{2} k_B T \)
Q2: The kinetic energy of a monoatomic gas molecule at temperature \( 300 K \) is approximately:
- A) \( 1.38 \times 10^{-21} J \)
- B) \( 6.21 \times 10^{-21} J \)
- C) \( 4.14 \times 10^{-21} J \)
- D) \( 9.0 \times 10^{-21} J \)
Q3: For a diatomic molecule with vibration considered, total energy is:
- A) \( \frac{5}{2} k_B T \)
- B) \( 3 k_B T \)
- C) \( \frac{7}{2} k_B T \)
- D) \( 4 k_B T \)

Answer Key

Q1: B) \( \frac{1}{2} k_B T \) - Each degree of freedom contributes half \( k_B T \) energy.
Q2: B) \( 6.21 \times 10^{-21} J \) - Using \( \frac{3}{2} k_B T \) at 300 K.
Q3: D) \( 4 k_B T \) - Vibration adds \( k_B T \) to \( \frac{5}{2} k_B T \) making total \( 4 k_B T \).

Memory Aids

Energy per mode = \( \frac{1}{2} k_B T \)
Monoatomic KE = \( \frac{3}{2} k_B T \)
Diatomic (no vibration) KE = \( \frac{5}{2} k_B T \)
Vibration adds \( k_B T \) energy

Specific Heat Capacity

Key Concepts

Specific heat capacity is the amount of heat required to raise the temperature of a substance by 1 K.
For gases, specific heats at constant volume (C_v) and constant pressure (C_p) are important.
Relation for ideal gases: C_p - C_v = R (gas constant).

Solved Examples

Ex 1: Calculate C_p if C_v = 3/2R.
- Use C_p = C_v + R = 3/2R + R = 5/2R.

Practice Set

Q1: What is the relation between C_p and C_v for an ideal gas?
- A) C_p = C_v
- B) C_p - C_v = R
- C) C_p + C_v = R
- D) C_p < C_v
Q2: If C_v = 5/2R, find C_p.
- A) 7/2R
- B) 5/2R
- C) 3/2R
- D) 9/2R
Q3: For an ideal gas, if C_p = 20.8 J/mol·K and R = 8.3 J/mol·K, find C_v.
- A) 12.5 J/mol·K
- B) 28.9 J/mol·K
- C) 20.8 J/mol·K
- D) 8.3 J/mol·K

Answer Key

Q1: B - By definition for ideal gases, C_p - C_v = R.
Q2: A - C_p = C_v + R = 5/2R + R = 7/2R.
Q3: A - C_v = C_p - R = 20.8 - 8.3 = 12.5 J/mol·K.

Memory Aids

C_p - C_v = R for ideal gases.
Remember C_v = f/2R where f is degrees of freedom.
Ratio γ = C_p/C_v is key for gas behavior.

Key Concepts

Specific heat capacity is the amount of heat required to raise the temperature of a substance by 1 K.
For gases, specific heats at constant volume (C_v) and constant pressure (C_p) are important.
Relation for ideal gases: C_p - C_v = R (gas constant).

Solved Examples

Ex 1: Calculate C_p if C_v = \(\frac{3}{2}R\).
- Use C_p = C_v + R = \(\frac{3}{2}R + R = \frac{5}{2}R\).

Practice Set

Q1: What is the relation between C_p and C_v for an ideal gas?
- A) C_p = C_v
- B) C_p - C_v = R
- C) C_p + C_v = R
- D) C_p < C_v
Q2: If C_v = \(\frac{5}{2}R\), find C_p.
- A) \(\frac{7}{2}R\)
- B) \(\frac{5}{2}R\)
- C) \(\frac{3}{2}R\)
- D) \(\frac{9}{2}R\)
Q3: For an ideal gas, if C_p = 20.8 J/mol·K and R = 8.3 J/mol·K, find C_v.
- A) 12.5 J/mol·K
- B) 28.9 J/mol·K
- C) 20.8 J/mol·K
- D) 8.3 J/mol·K

Answer Key

Q1: B - By definition for ideal gases, C_p - C_v = R.
Q2: A - C_p = C_v + R = \(\frac{5}{2}R + R = \frac{7}{2}R\).
Q3: A - C_v = C_p - R = 20.8 - 8.3 = 12.5 J/mol·K.

Memory Aids

C_p - C_v = R for ideal gases.
Remember \(C_{v} = \frac{f}{2}R\) where f is degrees of freedom.
Ratio \(\gamma = \frac{C_{p}}{C_{v}}\) is key for gas behavior.

Monoatomic Gases

Key Concepts

Degrees of freedom, f = 3.
Total internal energy: U = 3/2 N_A k_B T = 3/2 RT.
Specific heat at constant volume: C_v = dU/dT = 3/2 R.
Specific heat at constant pressure: C_p = C_v + R = 5/2 R.
Ratio of specific heats: γ = C_p/C_v = 5/3.

Solved Examples

Ex 1: Calculate γ for monoatomic gas.
- γ = 5/3 = 1.67.

Practice Set

Q1: Degrees of freedom for monoatomic gases?
- A) 2
- B) 3
- C) 5
- D) 6
Q2: Value of C_v for monoatomic gases?
- A) 3/2 R
- B) 5/2 R
- C) R
- D) 3R
Q3: Calculate C_p if C_v = 3/2 R.
- A) 3/2 R
- B) 5/2 R
- C) 2R
- D) 3R

Answer Key

Q1: B - Monoatomic gases have 3 degrees of freedom (translational).
Q2: A - C_v = 3/2 R for monoatomic gases.
Q3: B - C_p = C_v + R = 5/2 R.

Memory Aids

Monoatomic gases: f = 3, C_v = 3/2 R, γ = 5/3.
Use γ = C_p/C_v to check values.

Key Concepts

Degrees of freedom, f = 3.
Total internal energy: \(U = \frac{3}{2} N_{A} k_{B} T = \frac{3}{2} RT\).
Specific heat at constant volume: \(C_{v} = \frac{dU}{dT} = \frac{3}{2} R\).
Specific heat at constant pressure: \(C_{p} = C_{v} + R = \frac{5}{2} R\).
Ratio of specific heats: \(\gamma = \frac{C_{p}}{C_{v}} = \frac{5}{3}\).

Solved Examples

Ex 1: Calculate \(\gamma\) for monoatomic gas.
- \(\gamma = \frac{5}{3} = 1.67\).

Practice Set

Q1: Degrees of freedom for monoatomic gases?
- A) 2
- B) 3
- C) 5
- D) 6
Q2: Value of \(C_{v}\) for monoatomic gases?
- A) \(\frac{3}{2} R\)
- B) \(\frac{5}{2} R\)
- C) R
- D) 3R
Q3: Calculate \(C_{p}\) if \(C_{v} = \frac{3}{2} R\).
- A) \(\frac{3}{2} R\)
- B) \(\frac{5}{2} R\)
- C) 2R
- D) 3R

Answer Key

Q1: B - Monoatomic gases have 3 degrees of freedom (translational).
Q2: A - \(C_{v} = \frac{3}{2} R\) for monoatomic gases.
Q3: B - \(C_{p} = C_{v} + R = \frac{5}{2} R\).

Memory Aids

Monoatomic gases: f = 3, \(C_{v} = \frac{3}{2} R\), \(\gamma = \frac{5}{3}\).
Use \(\gamma = \frac{C_{p}}{C_{v}}\) to check values.

Diatomic Gases

Key Concepts

Degrees of freedom, f = 5 (3 translational + 2 rotational).
Total internal energy: U = 5/2 RT.
Specific heat at constant volume: C_v = 5/2 R.
Specific heat at constant pressure: C_p = C_v + R = 7/2 R.
Ratio of specific heats: γ = 7/5 = 1.4.
If vibrational mode is active, U = 7/2 RT, C_v = 7/2 R, C_p = 9/2 R, γ = 9/7.

Solved Examples

Ex 1: Calculate γ for diatomic gas without vibration.
- γ = 7/5 = 1.4.
Ex 2: Calculate C_p if C_v = 7/2 R.
- C_p = C_v + R = 9/2 R.

Practice Set

Q1: Degrees of freedom for diatomic gases?
- A) 3
- B) 5
- C) 6
- D) 7
Q2: Value of γ for diatomic gases without vibration?
- A) 1.2
- B) 1.4
- C) 1.67
- D) 1.5
Q3: If vibrational mode is active, what is C_v?
- A) 5/2 R
- B) 7/2 R
- C) 9/2 R
- D) 4R

Answer Key

Q1: B - Diatomic gases have 5 degrees of freedom (3 translational + 2 rotational).
Q2: B - γ = 7/5 = 1.4 without vibration.
Q3: B - With vibration, C_v = 7/2 R.

Memory Aids

Diatomic gases: f = 5, γ = 7/5.
Vibrational mode adds 1 to f and increases C_v and C_p.

Key Concepts

Degrees of freedom, f = 5 (3 translational + 2 rotational).
Total internal energy: \(U = \frac{5}{2} RT\).
Specific heat at constant volume: \(C_{v} = \frac{5}{2} R\).
Specific heat at constant pressure: \(C_{p} = C_{v} + R = \frac{7}{2} R\).
Ratio of specific heats: \(\gamma = \frac{7}{5} = 1.4\).
If vibrational mode is active, \(U = \frac{7}{2} RT\), \(C_{v} = \frac{7}{2} R\), \(C_{p} = \frac{9}{2} R\), \(\gamma = \frac{9}{7}\).

Solved Examples

Ex 1: Calculate \(\gamma\) for diatomic gas without vibration.
- \(\gamma = \frac{7}{5} = 1.4\).
Ex 2: Calculate \(C_{p}\) if \(C_{v} = \frac{7}{2} R\).
- \(C_{p} = C_{v} + R = \frac{9}{2} R\).

Practice Set

Q1: Degrees of freedom for diatomic gases?
- A) 3
- B) 5
- C) 6
- D) 7
Q2: Value of \(\gamma\) for diatomic gases without vibration?
- A) 1.2
- B) 1.4
- C) 1.67
- D) 1.5
Q3: If vibrational mode is active, what is \(C_{v}\)?
- A) \(\frac{5}{2} R\)
- B) \(\frac{7}{2} R\)
- C) \(\frac{9}{2} R\)
- D) \(4R\)

Answer Key

Q1: B - Diatomic gases have 5 degrees of freedom (3 translational + 2 rotational).
Q2: B - \(\gamma = \frac{7}{5} = 1.4\) without vibration.
Q3: B - With vibration, \(C_{v} = \frac{7}{2} R\).

Memory Aids

Diatomic gases: f = 5, \(\gamma = \frac{7}{5}\).
Vibrational mode adds 1 to f and increases \(C_{v}\) and \(C_{p}\).

Polyatomic Gases

Key Concepts

Total internal energy: U = [(3/2)k_BT + (3/2)k_BT + f k_BT] N_A.
Specific heat at constant volume: C_v = (3 + f) R.
Specific heat at constant pressure: C_p = (4 + f) R.
Ratio of specific heats: γ = C_p/C_v = (4 + f)/(3 + f).

Solved Examples

Ex 1: Calculate γ if f = 3.
- γ = (4 + 3)/(3 + 3) = 7/6 = 1.17.

Practice Set

Q1: Expression for C_v in polyatomic gases?
- A) (3 + f) R
- B) (4 + f) R
- C) f R
- D) 3 R
Q2: Expression for γ in polyatomic gases?
- A) (3 + f)/(4 + f)
- B) (4 + f)/(3 + f)
- C) f/3
- D) 3/f
Q3: If f = 2, find γ.
- A) 1.2
- B) 1.25
- C) 1.3
- D) 1.4

Answer Key

Q1: A - C_v = (3 + f) R by definition.
Q2: B - γ = (4 + f)/(3 + f).
Q3: B - γ = (4 + 2)/(3 + 2) = 6/5 = 1.25.

Memory Aids

Polyatomic gases: C_v = (3 + f) R, γ = (4 + f)/(3 + f).
Remember to add 1 to numerator and denominator for C_p and ratio.

Key Concepts

Total internal energy: \(U = [(3/2)k_{B}T + (3/2)k_{B}T + f k_{B}T] N_{A}\).
Specific heat at constant volume: \(C_{v} = (3 + f) R\).
Specific heat at constant pressure: \(C_{p} = (4 + f) R\).
Ratio of specific heats: \(\gamma = \frac{C_{p}}{C_{v}} = \frac{4 + f}{3 + f}\).

Solved Examples

Ex 1: Calculate \(\gamma\) if \(f = 3\).
- \(\gamma = \frac{4 + 3}{3 + 3} = \frac{7}{6} = 1.17\).

Practice Set

Q1: Expression for \(C_{v}\) in polyatomic gases?
- A) \((3 + f) R\)
- B) \((4 + f) R\)
- C) \(f R\)
- D) \(3 R\)
Q2: Expression for \(\gamma\) in polyatomic gases?
- A) \(\frac{3 + f}{4 + f}\)
- B) \(\frac{4 + f}{3 + f}\)
- C) \(\frac{f}{3}\)
- D) \(\frac{3}{f}\)
Q3: If \(f = 2\), find \(\gamma\).
- A) 1.2
- B) 1.25
- C) 1.3
- D) 1.4

Answer Key

Q1: A - \(C_{v} = (3 + f) R\) by definition.
Q2: B - \(\gamma = \frac{4 + f}{3 + f}\).
Q3: B - \(\gamma = \frac{4 + 2}{3 + 2} = \frac{6}{5} = 1.25\).

Memory Aids

Polyatomic gases: \(C_{v} = (3 + f) R\), \(\gamma = \frac{4 + f}{3 + f}\).
Remember to add 1 to numerator and denominator for C_p and ratio.

Specific Heat Capacity of Solids

Key Concepts

Consider a solid of N atoms vibrating about mean positions.
Average energy per 1D oscillator: k_B T.
For 3D, average energy: 3 k_B T.
For 1 mole, total energy: U = 3 R T.
Specific heat capacity: C = ΔQ/ΔT = ΔU/ΔT = 3 R (no work done).

Solved Examples

Ex 1: Calculate specific heat capacity of a solid.
- C = 3 R = 3 × 8.314 = 24.94 ext{ J/mol·K}.

Practice Set

Q1: Average energy of 1D oscillator?
- A) 1/2 k_B T
- B) k_B T
- C) 2 k_B T
- D) 3 k_B T
Q2: Total energy for 3D oscillators per mole?
- A) R T
- B) 3 R T
- C) 3/2 R T
- D) 6 R T
Q3: Specific heat capacity of solids?
- A) R
- B) 2 R
- C) 3 R
- D) 4 R

Answer Key

Q1: B - Average energy per 1D oscillator is k_B T.
Q2: B - For 3D, total energy is 3 k_B T N_A = 3 R T.
Q3: C - Specific heat capacity of solids is 3 R.

Memory Aids

Specific heat of solids = 3R (Dulong-Petit law).
Energy per oscillator = k_B T in 1D.

Key Concepts

Consider a solid of N atoms vibrating about mean positions.
Average energy per 1D oscillator: \(k_{B} T\).
For 3D, average energy: \(3 k_{B} T\).
For 1 mole, total energy: \(U = 3 R T\).
Specific heat capacity: \(C = \frac{\Delta Q}{\Delta T} = \frac{\Delta U}{\Delta T} = 3 R\) (no work done).

Solved Examples

Ex 1: Calculate specific heat capacity of a solid.
- \(C = 3 R = 3 \times 8.314 = 24.94 \text{ J/mol·K}\).

Practice Set

Q1: Average energy of 1D oscillator?
- A) \(\frac{1}{2} k_{B} T\)
- B) \(k_{B} T\)
- C) \(2 k_{B} T\)
- D) \(3 k_{B} T\)
Q2: Total energy for 3D oscillators per mole?
- A) \(R T\)
- B) \(3 R T\)
- C) \(\frac{3}{2} R T\)
- D) \(6 R T\)
Q3: Specific heat capacity of solids?
- A) \(R\)
- B) \(2 R\)
- C) \(3 R\)
- D) \(4 R\)

Answer Key

Q1: B - Average energy per 1D oscillator is \(k_{B} T\).
Q2: B - For 3D, total energy is \(3 k_{B} T N_{A} = 3 R T\).
Q3: C - Specific heat capacity of solids is \(3 R\).

Memory Aids

Specific heat of solids = 3R (Dulong-Petit law).
Energy per oscillator = \(k_{B} T\) in 1D.

Mean Free Path

Concept and Formula

Mean free path (λ) is the average distance a gas molecule travels between collisions.
In time t, molecule travels distance vt and collides with molecules in a cylindrical volume V = πd²vt, where d is molecular diameter.
Number density n = N/V, where N is number of molecules in volume V.
Number of collisions N = nπd²vt.
Mean free path formula: [ λ = vt/N = 1/(nπd²) ]
Exact formula considering all molecular motion: [ λ = 1/(√2 nπd²) ]
Effective collision area A = πd².

Solved Examples

Ex 1: Calculate λ if n = 2 × 10²⁵ m^-3 and d = 3 × 10^-10 m.
- Use λ = 1/(√2 nπd²)
- λ = 1/(√2 × 2 × 10²⁵ × π × (3 × 10^-10)²) ≈ 4.2 × 10^-8 m

Practice Set

Q1: Mean free path depends inversely on
- A) Number density n
- B) Molecular diameter d
- C) Temperature
- D) Pressure
Q2: If molecular diameter doubles, mean free path
- A) Doubles
- B) Halves
- C) Quadruples
- D) Quartered
Q3: For n = 1 × 10²⁵ m^-3, d = 2 × 10^-10 m, calculate λ (use π = 3.14)
- A) 7.1 × 10^-8 m
- B) 1.1 × 10^-7 m
- C) 5.6 × 10^-8 m
- D) 3.2 × 10^-8 m

Answer Key

Q1: A - Mean free path λ = 1/(√2 nπd²) inversely depends on number density n.
Q2: D - λ ∝ 1/d², doubling d quarters λ.
Q3: A - λ = 1/(√2 × 1 × 10²⁵ × 3.14 × (2 × 10^-10)²) ≈ 7.1 × 10^-8 m.

Memory Aids

Mean free path λ inversely ∝ nd².
Exact formula includes √2 factor.
Think "collision cylinder" volume πd²vt.

Concept and Formula

Mean free path (λ) is the average distance a gas molecule travels between collisions.
In time t, molecule travels distance vt and collides with molecules in a cylindrical volume V = πd²vt, where d is molecular diameter.
Number density n = N/V, where N is number of molecules in volume V.
Number of collisions N = nπd²vt.
Mean free path formula: [ λ = vt/N = 1/(nπd²) ]
Exact formula considering all molecular motion: [ λ = 1/(√2 nπd²) ]
Effective collision area A = πd².

Solved Examples

Ex 1: Calculate λ if n = 2 × 10²⁵ m^-3 and d = 3 × 10^-10 m.
- Use λ = 1/(√2 nπd²)
- λ = 1/(√2 × 2 × 10²⁵ × π × (3 × 10^-10)²) ≈ 4.2 × 10^-8 m

Practice Set

Q1: Mean free path depends inversely on
- A) Number density n
- B) Molecular diameter d
- C) Temperature
- D) Pressure
Q2: If molecular diameter doubles, mean free path
- A) Doubles
- B) Halves
- C) Quadruples
- D) Quartered
Q3: For n = 1 × 10²⁵ m^-3, d = 2 × 10^-10 m, calculate λ (use π = 3.14)
- A) 7.1 × 10^-8 m
- B) 1.1 × 10^-7 m
- C) 5.6 × 10^-8 m
- D) 3.2 × 10^-8 m

Answer Key

Q1: A - Mean free path λ = 1/(√2 nπd²) inversely depends on number density n.
Q2: D - λ ∝ 1/d², doubling d quarters λ.
Q3: A - λ = 1/(√2 × 1 × 10²⁵ × 3.14 × (2 × 10^-10)²) ≈ 7.1 × 10^-8 m.

Memory Aids

Mean free path λ inversely ∝ nd².
Exact formula includes √2 factor.
Think "collision cylinder" volume πd²vt.

Concept and Formula

Mean free path (λ) is the average distance a gas molecule travels between collisions.
In time t, molecule travels distance vt and collides with molecules in a cylindrical volume V = πd²vt, where d is molecular diameter.
Number density n = N/V, where N is number of molecules in volume V.
Number of collisions N = nπd²vt.
Mean free path formula: [ λ = vt/N = 1/(nπd²) ]
Exact formula considering all molecular motion: [ λ = 1/(√2 nπd²) ]
Effective collision area A = πd².

Solved Examples

Ex 1: Calculate λ if n = 2 × 10²⁵ m^-3 and d = 3 × 10^-10 m.
- Use λ = 1/(√2 nπd²)
- λ = 1/(√2 × 2 × 10²⁵ × π × (3 × 10^-10)²) ≈ 4.2 × 10^-8 m

Practice Set

Q1: Mean free path depends inversely on
- A) Number density n
- B) Molecular diameter d
- C) Temperature
- D) Pressure
Q2: If molecular diameter doubles, mean free path
- A) Doubles
- B) Halves
- C) Quadruples
- D) Quartered
Q3: For n = 1 × 10²⁵ m^-3, d = 2 × 10^-10 m, calculate λ (use π = 3.14)
- A) 7.1 × 10^-8 m
- B) 1.1 × 10^-7 m
- C) 5.6 × 10^-8 m
- D) 3.2 × 10^-8 m

Answer Key

Q1: A - Mean free path λ = 1/(√2 nπd²) inversely depends on number density n.
Q2: D - λ ∝ 1/d², doubling d quarters λ.
Q3: A - λ = 1/(√2 × 1 × 10²⁵ × 3.14 × (2 × 10^-10)²) ≈ 7.1 × 10^-8 m.

Memory Aids

Mean free path λ inversely ∝ nd².
Exact formula includes √2 factor.
Think "collision cylinder" volume πd²vt.

Concept and Formula

Mean free path (λ) is the average distance a gas molecule travels between collisions.
In time t, molecule travels distance vt and collides with molecules in a cylindrical volume V = πd²vt, where d is molecular diameter.
Number density n = N/V, where N is number of molecules in volume V.
Number of collisions N = nπd²vt.
Mean free path formula: [ λ = vt/N = 1/(nπd²) ]
Exact formula considering all molecular motion: [ λ = 1/(√2 nπd²) ]
Effective collision area A = πd².

Solved Examples

Ex 1: Calculate λ if n = 2 × 10²⁵ m^-3 and d = 3 × 10^-10 m.
- Use λ = 1/(√2 nπd²)
- λ = 1/(√2 × 2 × 10²⁵ × π × (3 × 10^-10)²) ≈ 4.2 × 10^-8 m

Practice Set

Q1: Mean free path depends inversely on
- A) Number density n
- B) Molecular diameter d
- C) Temperature
- D) Pressure
Q2: If molecular diameter doubles, mean free path
- A) Doubles
- B) Halves
- C) Quadruples
- D) Quartered
Q3: For n = 1 × 10²⁵ m^-3, d = 2 × 10^-10 m, calculate λ (use π = 3.14)
- A) 7.1 × 10^-8 m
- B) 1.1 × 10^-7 m
- C) 5.6 × 10^-8 m
- D) 3.2 × 10^-8 m

Answer Key

Q1: A - Mean free path λ = 1/(√2 nπd²) inversely depends on number density n.
Q2: D - λ ∝ 1/d², doubling d quarters λ.
Q3: A - λ = 1/(√2 × 1 × 10²⁵ × 3.14 × (2 × 10^-10)²) ≈ 7.1 × 10^-8 m.

Memory Aids

Mean free path λ inversely ∝ nd².
Exact formula includes √2 factor.
Think "collision cylinder" volume πd²vt.

Concept and Formula

Mean free path (λ) is the average distance a gas molecule travels between collisions.
In time t, molecule travels distance vt and collides with molecules in a cylindrical volume V = πd²vt, where d is molecular diameter.
Number density n = N/V, where N is number of molecules in volume V.
Number of collisions N = nπd²vt.
Mean free path formula: [ λ = vt/N = 1/(nπd²) ]
Exact formula considering all molecular motion: [ λ = 1/(√2 nπd²) ]
Effective collision area A = πd².

Solved Examples

Ex 1: Calculate λ if n = 2 × 10²⁵ m^-3 and d = 3 × 10^-10 m.
- Use λ = 1/(√2 nπd²)
- λ = 1/(√2 × 2 × 10²⁵ × π × (3 × 10^-10)²) ≈ 4.2 × 10^-8 m

Practice Set

Q1: Mean free path depends inversely on
- A) Number density n
- B) Molecular diameter d
- C) Temperature
- D) Pressure
Q2: If molecular diameter doubles, mean free path
- A) Doubles
- B) Halves
- C) Quadruples
- D) Quartered
Q3: For n = 1 × 10²⁵ m^-3, d = 2 × 10^-10 m, calculate λ (use π = 3.14)
- A) 7.1 × 10^-8 m
- B) 1.1 × 10^-7 m
- C) 5.6 × 10^-8 m
- D) 3.2 × 10^-8 m

Answer Key

Q1: A - Mean free path λ = 1/(√2 nπd²) inversely depends on number density n.
Q2: D - λ ∝ 1/d², doubling d quarters λ.
Q3: A - λ = 1/(√2 × 1 × 10²⁵ × 3.14 × (2 × 10^-10)²) ≈ 7.1 × 10^-8 m.

Memory Aids

Mean free path λ inversely ∝ nd².
Exact formula includes √2 factor.
Think "collision cylinder" volume πd²vt.

Concept and Formula

Mean free path (λ) is the average distance a gas molecule travels between collisions.
In time t, molecule travels distance vt and collides with molecules in a cylindrical volume V = πd²vt, where d is molecular diameter.
Number density n = N/V, where N is number of molecules in volume V.
Number of collisions N = nπd²vt.
Mean free path formula: [ λ = vt/N = 1/(nπd²) ]
Exact formula considering all molecular motion: [ λ = 1/(√2 nπd²) ]
Effective collision area A = πd².

Solved Examples

Ex 1: Calculate λ if n = 2 × 10²⁵ m^-3 and d = 3 × 10^-10 m.
- Use λ = 1/(√2 nπd²)
- λ = 1/(√2 × 2 × 10²⁵ × π × (3 × 10^-10)²) ≈ 4.2 × 10^-8 m

Practice Set

Q1: Mean free path depends inversely on
- A) Number density n
- B) Molecular diameter d
- C) Temperature
- D) Pressure
Q2: If molecular diameter doubles, mean free path
- A) Doubles
- B) Halves
- C) Quadruples
- D) Quartered
Q3: For n = 1 × 10²⁵ m^-3, d = 2 × 10^-10 m, calculate λ (use π = 3.14)
- A) 7.1 × 10^-8 m
- B) 1.1 × 10^-7 m
- C) 5.6 × 10^-8 m
- D) 3.2 × 10^-8 m

Answer Key

Q1: A - Mean free path λ = 1/(√2 nπd²) inversely depends on number density n.
Q2: D - λ ∝ 1/d², doubling d quarters λ.
Q3: A - λ = 1/(√2 × 1 × 10²⁵ × 3.14 × (2 × 10^-10)²) ≈ 7.1 × 10^-8 m.

Memory Aids

Mean free path λ inversely ∝ nd².
Exact formula includes √2 factor.
Think "collision cylinder" volume πd²vt.

Concept and Formula

Mean free path (λ) is the average distance a gas molecule travels between collisions.
In time t, molecule travels distance vt and collides with molecules in a cylindrical volume V = πd²vt, where d is molecular diameter.
Number density n = N/V, where N is number of molecules in volume V.
Number of collisions N = nπd²vt.
Mean free path formula: [ λ = vt/N = 1/(nπd²) ]
Exact formula considering all molecular motion: [ λ = 1/(√2 nπd²) ]
Effective collision area A = πd².

Solved Examples

Ex 1: Calculate λ if n = 2 × 10²⁵ m^-3 and d = 3 × 10^-10 m.
- Use λ = 1/(√2 nπd²)
- λ = 1/(√2 × 2 × 10²⁵ × π × (3 × 10^-10)²) ≈ 4.2 × 10^-8 m

Practice Set

Q1: Mean free path depends inversely on
- A) Number density n
- B) Molecular diameter d
- C) Temperature
- D) Pressure
Q2: If molecular diameter doubles, mean free path
- A) Doubles
- B) Halves
- C) Quadruples
- D) Quartered
Q3: For n = 1 × 10²⁵ m^-3, d = 2 × 10^-10 m, calculate λ (use π = 3.14)
- A) 7.1 × 10^-8 m
- B) 1.1 × 10^-7 m
- C) 5.6 × 10^-8 m
- D) 3.2 × 10^-8 m

Answer Key

Q1: A - Mean free path λ = 1/(√2 nπd²) inversely depends on number density n.
Q2: D - λ ∝ 1/d², doubling d quarters λ.
Q3: A - λ = 1/(√2 × 1 × 10²⁵ × 3.14 × (2 × 10^-10)²) ≈ 7.1 × 10^-8 m.

Memory Aids

Mean free path λ inversely ∝ nd².
Exact formula includes √2 factor.
Think "collision cylinder" volume πd²vt.

Concept and Formula

Mean free path (λ) is the average distance a gas molecule travels between collisions.
In time t, molecule travels distance vt and collides with molecules in a cylindrical volume V = πd²vt, where d is molecular diameter.
Number density n = N/V, where N is number of molecules in volume V.
Number of collisions N = nπd²vt.
Mean free path formula: [ λ = vt/N = 1/(nπd²) ]
Exact formula considering all molecular motion: [ λ = 1/(√2 nπd²) ]
Effective collision area A = πd².

Solved Examples

Ex 1: Calculate λ if n = 2 × 10²⁵ m^-3 and d = 3 × 10^-10 m.
- Use λ = 1/(√2 nπd²)
- λ = 1/(√2 × 2 × 10²⁵ × π × (3 × 10^-10)²) ≈ 4.2 × 10^-8 m

Practice Set

Q1: Mean free path depends inversely on
- A) Number density n
- B) Molecular diameter d
- C) Temperature
- D) Pressure
Q2: If molecular diameter doubles, mean free path
- A) Doubles
- B) Halves
- C) Quadruples
- D) Quartered
Q3: For n = 1 × 10²⁵ m^-3, d = 2 × 10^-10 m, calculate λ (use π = 3.14)
- A) 7.1 × 10^-8 m
- B) 1.1 × 10^-7 m
- C) 5.6 × 10^-8 m
- D) 3.2 × 10^-8 m

Answer Key

Q1: A - Mean free path λ = 1/(√2 nπd²) inversely depends on number density n.
Q2: D - λ ∝ 1/d², doubling d quarters λ.
Q3: A - λ = 1/(√2 × 1 × 10²⁵ × 3.14 × (2 × 10^-10)²) ≈ 7.1 × 10^-8 m.

Memory Aids

Mean free path λ inversely ∝ nd².
Exact formula includes √2 factor.
Think "collision cylinder" volume πd²vt.

Concept and Formula

Mean free path (\(\lambda\)) is the average distance a gas molecule travels between collisions.
In time \(t\), molecule travels distance \(vt\) and collides with molecules in a cylindrical volume \(V = \pi d^2 vt\), where \(d\) is molecular diameter.
Number density \(n = \frac{N}{V}\), where \(N\) is number of molecules in volume \(V\).
Number of collisions \(N = n \pi d^2 vt\).
Mean free path formula: \[ \lambda = \frac{vt}{N} = \frac{1}{n \pi d^2} \]
Exact formula considering all molecular motion: \[ \lambda = \frac{1}{\sqrt{2} n \pi d^2} \]
Effective collision area \(A = \pi d^2\).

Solved Examples

Ex 1: Calculate \(\lambda\) if \(n = 2 \times 10^{25} \text{ m}^{-3}\) and \(d = 3 \times 10^{-10} \text{ m}\).
- Use \(\lambda = \frac{1}{\sqrt{2} n \pi d^2}\)
- \(\lambda = \frac{1}{\sqrt{2} \times 2 \times 10^{25} \times \pi \times (3 \times 10^{-10})^2} \approx 4.2 \times 10^{-8} \text{ m}\)

Practice Set

Q1: Mean free path depends inversely on
- A) Number density \(n\)
- B) Molecular diameter \(d\)
- C) Temperature
- D) Pressure
Q2: If molecular diameter doubles, mean free path
- A) Doubles
- B) Halves
- C) Quadruples
- D) Quartered
Q3: For \(n = 1 \times 10^{25} \text{ m}^{-3}\), \(d = 2 \times 10^{-10} \text{ m}\), calculate \(\lambda\) (use \(\pi = 3.14\))
- A) \(7.1 \times 10^{-8} \text{ m}\)
- B) \(1.1 \times 10^{-7} \text{ m}\)
- C) \(5.6 \times 10^{-8} \text{ m}\)
- D) \(3.2 \times 10^{-8} \text{ m}\)

Answer Key

Q1: A - Mean free path \(\lambda = \frac{1}{\sqrt{2} n \pi d^2}\) inversely depends on number density \(n\).
Q2: D - \(\lambda \propto \frac{1}{d^2}\), doubling \(d\) quarters \(\lambda\).
Q3: A - \(\lambda = \frac{1}{\sqrt{2} \times 1 \times 10^{25} \times 3.14 \times (2 \times 10^{-10})^2} \approx 7.1 \times 10^{-8} \text{ m}\).

Memory Aids

Mean free path \(\lambda\) inversely \(\propto n d^2\).
Exact formula includes \(\sqrt{2}\) factor.
Think "collision cylinder" volume \(\pi d^2 vt\).

Avogadro’s Number

Definition and Value

Number of units (atoms, molecules, ions, electrons) in one mole of substance.
Value: 6.022140857 × 10²³ units/mole.

Solved Examples

Ex 1: Calculate number of molecules in 2 moles of oxygen.
- Number = moles × Avogadro’s number = 2 × 6.022 × 10²³ = 1.2044 × 10²⁴

Practice Set

Q1: Avogadro’s number represents
- A) Number of atoms in 1 gram of substance
- B) Number of units in 1 mole
- C) Number of molecules in 1 liter
- D) Number of ions in 1 kg
Q2: Number of molecules in 3 moles of water is
- A) 1.8 × 10²³
- B) 1.8 × 10²⁴
- C) 6.022 × 10²³
- D) 3 × 10²³
Q3: Avogadro’s number is approximately
- A) 6.02 × 10²²
- B) 6.02 × 10²³
- C) 6.02 × 10²⁴
- D) 6.02 × 10²⁵

Answer Key

Q1: B - Avogadro’s number is units per mole.
Q2: B - 3 × 6.022 × 10²³ = 1.8 × 10²⁴.
Q3: B - Standard value is 6.022 × 10²³.

Memory Aids

Avogadro’s number ≈ 6.02 × 10²³.
Represents units per mole.
Used to convert moles to number of particles.

Definition and Value

Number of units (atoms, molecules, ions, electrons) in one mole of substance.
Value: 6.022140857 × 10²³ units/mole.

Solved Examples

Ex 1: Calculate number of molecules in 2 moles of oxygen.
- Number = moles × Avogadro’s number = 2 × 6.022 × 10²³ = 1.2044 × 10²⁴

Practice Set

Q1: Avogadro’s number represents
- A) Number of atoms in 1 gram of substance
- B) Number of units in 1 mole
- C) Number of molecules in 1 liter
- D) Number of ions in 1 kg
Q2: Number of molecules in 3 moles of water is
- A) 1.8 × 10²³
- B) 1.8 × 10²⁴
- C) 6.022 × 10²³
- D) 3 × 10²³
Q3: Avogadro’s number is approximately
- A) 6.02 × 10²²
- B) 6.02 × 10²³
- C) 6.02 × 10²⁴
- D) 6.02 × 10²⁵

Answer Key

Q1: B - Avogadro’s number is units per mole.
Q2: B - 3 × 6.022 × 10²³ = 1.8 × 10²⁴.
Q3: B - Standard value is 6.022 × 10²³.

Memory Aids

Avogadro’s number ≈ 6.02 × 10²³.
Represents units per mole.
Used to convert moles to number of particles.

Definition and Value

Number of units (atoms, molecules, ions, electrons) in one mole of substance.
Value: 6.022140857 × 10²³ units/mole.

Solved Examples

Ex 1: Calculate number of molecules in 2 moles of oxygen.
- Number = moles × Avogadro’s number = 2 × 6.022 × 10²³ = 1.2044 × 10²⁴

Practice Set

Q1: Avogadro’s number represents
- A) Number of atoms in 1 gram of substance
- B) Number of units in 1 mole
- C) Number of molecules in 1 liter
- D) Number of ions in 1 kg
Q2: Number of molecules in 3 moles of water is
- A) 1.8 × 10²³
- B) 1.8 × 10²⁴
- C) 6.022 × 10²³
- D) 3 × 10²³
Q3: Avogadro’s number is approximately
- A) 6.02 × 10²²
- B) 6.02 × 10²³
- C) 6.02 × 10²⁴
- D) 6.02 × 10²⁵

Answer Key

Q1: B - Avogadro’s number is units per mole.
Q2: B - 3 × 6.022 × 10²³ = 1.8 × 10²⁴.
Q3: B - Standard value is 6.022 × 10²³.

Memory Aids

Avogadro’s number ≈ 6.02 × 10²³.
Represents units per mole.
Used to convert moles to number of particles.

Definition and Value

Number of units (atoms, molecules, ions, electrons) in one mole of substance.
Value: 6.022140857 × 10²³ units/mole.

Solved Examples

Ex 1: Calculate number of molecules in 2 moles of oxygen.
- Number = moles × Avogadro’s number = 2 × 6.022 × 10²³ = 1.2044 × 10²⁴

Practice Set

Q1: Avogadro’s number represents
- A) Number of atoms in 1 gram of substance
- B) Number of units in 1 mole
- C) Number of molecules in 1 liter
- D) Number of ions in 1 kg
Q2: Number of molecules in 3 moles of water is
- A) 1.8 × 10²³
- B) 1.8 × 10²⁴
- C) 6.022 × 10²³
- D) 3 × 10²³
Q3: Avogadro’s number is approximately
- A) 6.02 × 10²²
- B) 6.02 × 10²³
- C) 6.02 × 10²⁴
- D) 6.02 × 10²⁵

Answer Key

Q1: B - Avogadro’s number is units per mole.
Q2: B - 3 × 6.022 × 10²³ = 1.8 × 10²⁴.
Q3: B - Standard value is 6.022 × 10²³.

Memory Aids

Avogadro’s number ≈ 6.02 × 10²³.
Represents units per mole.
Used to convert moles to number of particles.

Definition and Value

Number of units (atoms, molecules, ions, electrons) in one mole of substance.
Value: 6.022140857 × 10²³ units/mole.

Solved Examples

Ex 1: Calculate number of molecules in 2 moles of oxygen.
- Number = moles × Avogadro’s number = 2 × 6.022 × 10²³ = 1.2044 × 10²⁴

Practice Set

Q1: Avogadro’s number represents
- A) Number of atoms in 1 gram of substance
- B) Number of units in 1 mole
- C) Number of molecules in 1 liter
- D) Number of ions in 1 kg
Q2: Number of molecules in 3 moles of water is
- A) 1.8 × 10²³
- B) 1.8 × 10²⁴
- C) 6.022 × 10²³
- D) 3 × 10²³
Q3: Avogadro’s number is approximately
- A) 6.02 × 10²²
- B) 6.02 × 10²³
- C) 6.02 × 10²⁴
- D) 6.02 × 10²⁵

Answer Key

Q1: B - Avogadro’s number is units per mole.
Q2: B - 3 × 6.022 × 10²³ = 1.8 × 10²⁴.
Q3: B - Standard value is 6.022 × 10²³.

Memory Aids

Avogadro’s number ≈ 6.02 × 10²³.
Represents units per mole.
Used to convert moles to number of particles.

Definition and Value

Number of units (atoms, molecules, ions, electrons) in one mole of substance.
Value: 6.022140857 × 10²³ units/mole.

Solved Examples

Ex 1: Calculate number of molecules in 2 moles of oxygen.
- Number = moles × Avogadro’s number = 2 × 6.022 × 10²³ = 1.2044 × 10²⁴

Practice Set

Q1: Avogadro’s number represents
- A) Number of atoms in 1 gram of substance
- B) Number of units in 1 mole
- C) Number of molecules in 1 liter
- D) Number of ions in 1 kg
Q2: Number of molecules in 3 moles of water is
- A) 1.8 × 10²³
- B) 1.8 × 10²⁴
- C) 6.022 × 10²³
- D) 3 × 10²³
Q3: Avogadro’s number is approximately
- A) 6.02 × 10²²
- B) 6.02 × 10²³
- C) 6.02 × 10²⁴
- D) 6.02 × 10²⁵

Answer Key

Q1: B - Avogadro’s number is units per mole.
Q2: B - 3 × 6.022 × 10²³ = 1.8 × 10²⁴.
Q3: B - Standard value is 6.022 × 10²³.

Memory Aids

Avogadro’s number ≈ 6.02 × 10²³.
Represents units per mole.
Used to convert moles to number of particles.

Definition and Value

Number of units (atoms, molecules, ions, electrons) in one mole of substance.
Value: 6.022140857 × 10²³ units/mole.

Solved Examples

Ex 1: Calculate number of molecules in 2 moles of oxygen.
- Number = moles × Avogadro’s number = 2 × 6.022 × 10²³ = 1.2044 × 10²⁴

Practice Set

Q1: Avogadro’s number represents
- A) Number of atoms in 1 gram of substance
- B) Number of units in 1 mole
- C) Number of molecules in 1 liter
- D) Number of ions in 1 kg
Q2: Number of molecules in 3 moles of water is
- A) 1.8 × 10²³
- B) 1.8 × 10²⁴
- C) 6.022 × 10²³
- D) 3 × 10²³
Q3: Avogadro’s number is approximately
- A) 6.02 × 10²²
- B) 6.02 × 10²³
- C) 6.02 × 10²⁴
- D) 6.02 × 10²⁵

Answer Key

Q1: B - Avogadro’s number is units per mole.
Q2: B - 3 × 6.022 × 10²³ = 1.8 × 10²⁴.
Q3: B - Standard value is 6.022 × 10²³.

Memory Aids

Avogadro’s number ≈ 6.02 × 10²³.
Represents units per mole.
Used to convert moles to number of particles.

Definition and Value

Number of units (atoms, molecules, ions, electrons) in one mole of substance.
Value: \(6.022140857 \times 10^{23}\) units/mole.

Solved Examples

Ex 1: Calculate number of molecules in 2 moles of oxygen.
- Number = moles \(\times\) Avogadro’s number = \(2 \times 6.022 \times 10^{23} = 1.2044 \times 10^{24}\)

Practice Set

Q1: Avogadro’s number represents
- A) Number of atoms in 1 gram of substance
- B) Number of units in 1 mole
- C) Number of molecules in 1 liter
- D) Number of ions in 1 kg
Q2: Number of molecules in 3 moles of water is
- A) \(1.8 \times 10^{23}\)
- B) \(1.8 \times 10^{24}\)
- C) \(6.022 \times 10^{23}\)
- D) \(3 \times 10^{23}\)
Q3: Avogadro’s number is approximately
- A) \(6.02 \times 10^{22}\)
- B) \(6.02 \times 10^{23}\)
- C) \(6.02 \times 10^{24}\)
- D) \(6.02 \times 10^{25}\)

Answer Key

Q1: B - Avogadro’s number is units per mole.
Q2: B - \(3 \times 6.022 \times 10^{23} = 1.8 \times 10^{24}\).
Q3: B - Standard value is \(6.022 \times 10^{23}\).

Memory Aids

Avogadro’s number \(\approx 6.02 \times 10^{23}\).
Represents units per mole.
Used to convert moles to number of particles.

Diagram 10 – NEET Physics Kinetic Theory

NEET Physics — ALL CHAPTERS

Basic Mathematics

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Units and Measurement

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Motion in a Straight Line

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Motion in a Plane

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Laws of Motion

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Work, Energy and Power

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System of Particles and Rotational Motion

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Gravitation

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Mechanical Properties of Solids

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Mechanical Properties of Fluids

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Thermal Properties of Matter

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Thermodynamics

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Kinetic Theory

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Oscillations

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Waves

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Electric Charges and Fields

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Electrostatic Potential and Capacitance

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Current Electricity

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Moving Charges and Magnetism

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Magnetism and Matter

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Electromagnetic Induction

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Alternating Current

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Electromagnetic Waves

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Ray Optics and Optical Instruments

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Wave Optics

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Dual Nature of Radiation and Matter

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Atoms

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Nuclei

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Semiconductor Electronics: Materials, Devices and Simple Circuits

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