Thermodynamics and Kinetic Theory of Gases Notes – Complete Chapter

System, Boundaries and Laws

Types of Thermodynamic Systems

Open system: both mass and energy can be exchanged with surroundings
Closed system: only energy (no mass) can be exchanged
Isolated system: neither mass nor energy exchanged

Types of Boundaries

Adiabatic boundary: no heat can be exchanged across it
Diathermal (conducting) boundary: heat can be exchanged

Zeroth Law of Thermodynamics

If T_A = T_B and T_B = T_C, then T_A = T_C.

When bodies A, B, C are in thermal contact → T_A = T_B = T_C (thermal equilibrium). This is the basis of thermometry.

Thermal equilibrium: heat transfer = 0, temperature = constant

Types of Thermodynamic Processes

Reversible: path followed can be exactly retraced; quasi-static process
Irreversible: cannot be retraced exactly (most real processes)
Cyclic process: initial and final state same; ΔU = 0; Q = W

Work done by gas: dW = PdV → W = ∫PdV = Area under P-V graph

Expansion (V_f > V_i): W = +ve | Compression (V_f < V_i): W = −ve

CW loop on P-V diagram: W = +ve (net work done by gas)

ACW loop on P-V diagram: W = −ve (net work done on gas)

First Law of Thermodynamics

dQ = dU + dW

Heat supplied = Change in internal energy + Work done by gas

dQ +ve → heat given TO system | dU +ve → internal energy increases | dW +ve → work done BY gas

Sign Conventions

dQ = +ve → heat given to system; dQ = −ve → heat taken from system
dU = +ve → temperature increases; dU = −ve → temperature decreases
dW = +ve → gas expands; dW = −ve → gas compressed

Internal Energy

Internal energy U depends on state, not on path
For ideal gas: U = KE only (PE = 0, no intermolecular forces)
For real gas: U = KE + PE
Internal energy is temperature-dependent: U ∝ T
Gas has internal energy; heat and work are NOT properties of gas but of process

Thermodynamic Processes

Process	Condition	Work W	ΔU	Q	Graph (PV)
Isochoric (Gay-Lussac's Law)	V = const, dV = 0	0	nC_vΔT	nC_vΔT	Vertical line
Isobaric (Charles' Law)	P = const	PΔV = nRΔT	nC_vΔT	nC_pΔT	Horizontal line
Isothermal (Boyle's Law)	T = const, ΔU = 0	nRT ln(V_f/V_i) = 2.303nRT log(P_i/P_f)	0	= W	Rectangular hyperbola
Adiabatic	Q = 0, dQ = 0	nC_v(T₁−T₂) = (P₁V₁−P₂V₂)/(γ−1)	nC_vΔT	0	Steeper than isothermal

Adiabatic Process Relations

PV^γ = constant (P-V)
TV^γ−1 = constant (T-V)
T^γP^1−γ = constant (T-P)
Expansion: V↑, W = +ve, U = −ve, T↓
Compression: V↓, W = −ve, U = +ve, T↑
Slope of adiabat on P-V graph = γ × slope of isotherm at same point

⚡ Work Done Comparison (Expansion)

W_isobaric > W_isothermal > W_adiabatic > W_isochoric (= 0)
For compression: W_isochoric(=0) > W_adiabatic > W_isothermal > W_isobaric
Adiabatic curve is steeper than isothermal at same point

Polytropic Process

PVⁿ = constant (general process)
n = 0 → isobaric; n = 1 → isothermal; n = γ → adiabatic; n = ∞ → isochoric
W = nR(T₁−T₂)/(n−1) | C_m = C_v − R/(n−1)

Q: In a P-V diagram, a gas undergoes: A→B (P = const, V₁→V₂), B→C (V = const), C→D (P = const), D→A (V = const). Find total work done.

W_AB = P₁(V₂−V₁) = +ve; W_BC = 0; W_CD = P₂(V₁−V₂) = −ve; W_DA = 0; W_total = (P₁−P₂)(V₂−V₁) = Area of rectangle on P-V diagram

Ans: W = (P₁−P₂)(V₂−V₁) = Area enclosed in P-V diagram ✓

Specific Heat Capacities of Gases

C_v = (dQ/dT) at constant volume = (f/2)R (f = degrees of freedom)

C_p = (dQ/dT) at constant pressure = C_v + R

C_p − C_v = R (Mayer's relation)

γ = C_p/C_v = 1 + 2/f = 1 + R/C_v

C_v = R/(γ−1) | C_p = γR/(γ−1)

Gas Type	Examples	f (DOF)	C_v	C_p	γ
Monoatomic	He, Ne, Ar	3 (3T)	3R/2	5R/2	5/3 ≈ 1.67
Diatomic (room temp)	H₂, O₂, N₂	5 (3T+2R)	5R/2	7R/2	7/5 = 1.4
Triatomic linear	CO₂	5 (3T+2R)	5R/2	7R/2	7/5 = 1.4
Triatomic non-linear	H₂O, SO₂, NH₃	6 (3T+3R)	3R	4R	4/3 ≈ 1.33
Diatomic (high temp, vibrational)	H₂ at very high T	7 (3T+2R+2V)	7R/2	9R/2	9/7 ≈ 1.29

Gas Mixture Specific Heats

(C_v)_mix = (n₁C_v1 + n₂C_v2) / (n₁ + n₂)
(C_p)_mix = (n₁C_p1 + n₂C_p2) / (n₁ + n₂)
γ_mix = (C_p)_mix / (C_v)_mix

Degrees of Freedom and Internal Energy

Degrees of Freedom (f)

Number of independent ways in which a molecule can possess energy
T = translational, R = rotational, V = vibrational
Monoatomic (He, Ne, Ar): f = 3T + 0R = 3
Diatomic (N₂, O₂, H₂): f = 3T + 2R = 5 (at room temp)
Triatomic linear (CO₂): f = 3T + 2R = 5
Triatomic non-linear (H₂O, SO₂, CH₄, NH₃): f = 3T + 3R = 6
At high temperature, vibrational modes activated: each vibrational mode adds 2 DOF
Diatomic at high T: f = 3T + 2R + 2V = 7
Triatomic linear at high T: f = 3T + 2R + 2V = 7
Triatomic non-linear at high T: f = 3T + 3R + 2V = 8

Law of Equipartition of Energy: each DOF contributes ½k_BT to average energy

U per molecule = (f/2)k_BT | U per mole = (f/2)RT

U for n moles = n·(f/2)RT = nC_vT

k_B = Boltzmann constant = R/N_A = 1.38×10⁻²³ J/K

Q: At 300 K, find average translational KE per molecule of any gas.

KE_trans = (3/2)k_BT = (3/2)×1.38×10⁻²³×300 = 6.21×10⁻²¹ J (same for all gases at same T)

Ans: 6.21×10⁻²¹ J ✓

Heat Engine, Carnot Engine and Refrigerator

Heat Engine

Converts heat into work: Q₁ (from hot source) → W (work) + Q₂ (to cold sink)
Q₁ = W + Q₂ → W = Q₁ − Q₂
Efficiency η = W/Q₁ = 1 − Q₂/Q₁

Carnot Engine — Most Efficient Heat Engine

Four-step cycle between two temperature reservoirs T₁ (source, hot) and T₂ (sink, cold):

A→B: Isothermal expansion at T₁ (absorbs Q₁)
B→C: Adiabatic expansion (T₁→T₂)
C→D: Isothermal compression at T₂ (releases Q₂)
D→A: Adiabatic compression (T₂→T₁)

Carnot Efficiency: η = 1 − T₂/T₁ = 1 − Q₂/Q₁

η is maximum possible for engines operating between T₁ and T₂

η can never be 100% (T₂ cannot be 0 K; Second Law)

T₂ = sink temperature (lower) | T₁ = source temperature (higher) | Both in Kelvin

Device	What it does	Efficiency / COP
Heat Engine	Converts heat (Q₁) to work (W); rejects Q₂ to cold sink	η = 1 − T₂/T₁ = W/Q₁
Refrigerator	Takes heat Q₂ from cold body; rejects Q₁ to hot; needs work W	β = Q₂/W = T₂/(T₁−T₂)
Heat Pump	Delivers heat Q₁ to hot body using work W	β' = Q₁/W = T₁/(T₁−T₂) = β+1

⚡ Key Relations

β (refrigerator) + 1 = β' (heat pump)
η + β/(β+1) = 1 (efficiency + 1/(β+1) = 1)
Higher T₁ or lower T₂ → better Carnot efficiency
Carnot's theorem: no engine can be more efficient than Carnot engine between same temperatures

Q: A Carnot engine operates between 500 K and 300 K. Find efficiency and if Q₁ = 1000 J, find W and Q₂.

η = 1 − 300/500 = 1 − 0.6 = 0.4 = 40%; W = ηQ₁ = 0.4×1000 = 400 J; Q₂ = Q₁ − W = 600 J

Ans: η = 40%, W = 400 J, Q₂ = 600 J ✓

Kinetic Theory of Gases (KTG)

Assumptions of Kinetic Theory

A gas consists of a large number of identical molecules (order of Avogadro's number)
Molecules move randomly in all directions
Intermolecular interactions are negligible (ideal gas: no PE)
Molecules obey Newton's Laws of Motion
All collisions are perfectly elastic (momentum and KE conserved)
Real gases approach ideal behaviour at low pressure and high temperature

Pressure of ideal gas: P = (1/3)nmv²_rms = (1/3)ρv²_rms

n = number density = N/V; m = mass of each molecule

v²_rms = 3P/ρ = 3RT/M (M = molar mass)

KE of 1 mole = (3/2)RT = (3/2)PV

Energy density (KE per unit volume) = (3/2)P

Molecular Speeds

RMS speed: v_rms = √(3RT/M) = √(3P/ρ) = √(3k_BT/m)

Average speed: v_avg = √(8RT/πM) = 0.921 v_rms

Most probable speed: v_mp = √(2RT/M) = 0.816 v_rms

Order: v_rms > v_avg > v_mp (ratio ≈ 1 : 0.921 : 0.816)

All speeds ∝ √T (double T → speed increases by √2 times)

Maxwell's Speed Distribution

N_v = number of molecules per unit velocity range at speed v
Area under Maxwell distribution curve = total number of molecules (N = constant)
At higher temperature: peak (v_mp) shifts to higher speed; curve becomes broader and flatter
v_mp corresponds to the peak of the distribution
Speed distribution is temperature dependent

Mean Free Path

Definition

Mean free path (λ) = average distance a molecule travels between two successive collisions
d = diameter of molecule = 2r
n = number density = N/V (proportional to pressure)

λ = 1/(√2 × π × d² × n)

λ = RT/(√2 × π × d² × N_A × P)

λ ∝ T/P (increases with temperature, decreases with pressure)

Ideal Gas Law and Gas Density

Ideal Gas Equation: PV = nRT = NkT

n = number of moles; N = number of molecules; k_B = R/N_A

Gas density: ρ = PM/(RT) = Pm/(k_BT)

M = molar mass | m = mass of one molecule | At fixed T: ρ ∝ P (denser gas at higher pressure)

Dalton's Law of Partial Pressures

For a mixture of non-reacting gases: P_total = P₁ + P₂ + P₃ + ...
Each gas exerts pressure independently of others
f_mix = (n₁f₁ + n₂f₂)/(n₁ + n₂) (weighted average degrees of freedom)

Q: Find RMS speed of oxygen molecules at 27°C. (M = 32×10⁻³ kg/mol, R = 8.314 J/mol·K)

v_rms = √(3RT/M) = √(3×8.314×300/32×10⁻³) = √(233793.75) ≈ 483.5 m/s

Ans: v_rms ≈ 484 m/s ✓

Q: An adiabatic cylinder with piston. Gas compressed to 1/8 of volume (γ = 5/3). Find ratio T_f/T_i.

TV^γ−1 = const; T_iV_i^2/3 = T_fV_f^2/3; T_f/T_i = (V_i/V_f)^2/3 = 8^2/3 = 4

Ans: T_f/T_i = 4 (temperature quadruples) ✓

Q: Mixture of 2 moles monoatomic and 3 moles diatomic gas. Find C_v of mixture.

(C_v)_mix = (n₁C_v1 + n₂C_v2)/(n₁+n₂) = (2×3R/2 + 3×5R/2)/5 = (3R + 7.5R)/5 = 10.5R/5 = 2.1R

Ans: (C_v)_mix = 2.1R ✓

Q: A Carnot refrigerator works between 0°C and 27°C. COP?

T₂ = 273 K, T₁ = 300 K; β = T₂/(T₁−T₂) = 273/(300−273) = 273/27 ≈ 10.1

Ans: COP β ≈ 10.1 ✓

Thermodynamics & Kinetic Theory