Thermodynamics and kinetic theory for JEE Main

Thermodynamics rewards a small number of ideas used very carefully. Nearly every question is the first law, Q = ΔU + W, applied to one of four processes, plus the fact that the internal energy of an ideal gas depends only on its temperature. Kinetic theory then explains where the specific heats come from. If you keep your sign convention fixed and read P–V graphs fluently, this is one of the most reliable chapters in the physics paper.

The syllabus covers thermal equilibrium and the zeroth law, heat, work and internal energy, the first law, isothermal and adiabatic processes, the second law with reversible and irreversible processes, and kinetic theory: the ideal gas equation, pressure, rms speed, degrees of freedom, equipartition, specific heats and mean free path.

Basic ideas

• Zeroth law: if A and B are each in thermal equilibrium with C, they are in equilibrium with each other. It is what makes temperature a meaningful quantity.
• Internal energy U is a state function: it depends only on the state, not on how the gas got there. For an ideal gas, ΔU = nC_vΔT in any process.
• Heat Q and work W are path functions. Their values depend on the process.

The first law

Q = ΔU + W, where Q is the heat given to the gas and W is the work done by the gas.

Work done by a gas is W = ∫P dV, the area under the curve on a P–V diagram.

The four standard processes

For an adiabatic process, you can also write TVᵞ⁻¹ = constant. On a P–V diagram, the adiabatic curve through a point is γ times steeper than the isothermal one. An adiabatic expansion therefore cools the gas, and an adiabatic compression heats it.

Derivation: isothermal work. P = nRT/V with T fixed, so W = ∫nRT dV/V from V₁ to V₂ = nRT ln(V₂/V₁).

Cyclic processes

In a complete cycle the gas returns to its starting state, so ΔU = 0 and the net heat absorbed equals the net work done. On a P–V diagram (V along the horizontal axis):

• The net work equals the area enclosed by the cycle.
• A clockwise cycle does positive net work (an engine); an anticlockwise one does negative work (a refrigerator).

The second law and efficiency

The second law says heat cannot flow on its own from a colder body to a hotter one, and no engine working in a cycle can turn all the heat it absorbs into work. Real processes are irreversible; a reversible process is an idealised, infinitely slow one.

For an engine that absorbs Q₁ from the hot reservoir and rejects Q₂:

η = W/Q₁ = 1 − Q₂/Q₁

A Carnot engine, working reversibly between T₁ and T₂ (in kelvin), has the highest possible efficiency: η = 1 − T₂/T₁. A refrigerator's coefficient of performance is Q₂/W, which for a Carnot refrigerator is T₂/(T₁ − T₂).

Kinetic theory of gases

Treat a gas as many tiny molecules in random motion, colliding elastically with the walls. This gives:

• Pressure: P = ⅓ρv_rms², or PV = ⅓Nmv_rms².
• Temperature: the average translational KE per molecule is (3/2)kT. Temperature measures the average kinetic energy of the molecules.
• Speeds: v_rms = √(3RT/M), v_avg = √(8RT/πM), v_mp = √(2RT/M), where M is the molar mass in kg/mol. So v_rms > v_avg > v_mp, roughly in the ratio 1.73 : 1.60 : 1.41.

Equipartition: each degree of freedom has an average energy of ½kT per molecule. With f degrees of freedom:

• U = (f/2)nRT, so C_v = (f/2)R.
• C_p − C_v = R (Mayer's relation).
• γ = C_p/C_v = 1 + 2/f.

For a mixture, C_v = (n₁C_v1 + n₂C_v2)/(n₁ + n₂).

Mean free path: λ = 1/(√2 π d² n), where d is the molecular diameter and n is the number of molecules per unit volume.

Worked problems

Problem 1: heating at constant pressure. 2 mol of a monatomic ideal gas is heated at constant pressure through 10 K. Find Q, ΔU and W (R = 8.3 J/mol K).

Q = nC_pΔT = 2 × 2.5 × 8.3 × 10 = 415 J.
ΔU = nC_vΔT = 2 × 1.5 × 8.3 × 10 = 249 J.
W = Q − ΔU = 166 J, which matches nRΔT = 2 × 8.3 × 10.

Problem 2 (numerical answer): adiabatic compression. A diatomic gas (γ = 1.4) at 300 K is compressed adiabatically to 1/32 of its volume. Find the final temperature in kelvin.

TVᵞ⁻¹ is constant, so T₂ = T₁ × 32 raised to the power 0.4.
Since 32 = 2⁵, that factor is 2 raised to (5 × 0.4), which is 2² = 4. T₂ = 300 × 4 = 1,200 K.

Problem 3: a cycle. A gas goes round a rectangle on the P–V diagram: (1 L, 100 kPa) → (1 L, 300 kPa) → (3 L, 300 kPa) → (3 L, 100 kPa) → back to the start. Find the net work and net heat absorbed.

Area = (3 − 1) × 10⁻³ m³ × (300 − 100) × 10³ Pa = 400 J. The cycle runs clockwise, so the gas does +400 J of net work and absorbs 400 J of net heat.

Problem 4: rms speed. At what temperature is the rms speed of oxygen molecules twice its value at 27 °C?

v_rms ∝ √T. Doubling v needs 4 × 300 = 1,200 K, which is 927 °C.

Practice set

1. Find the efficiency of a Carnot engine working between 600 K and 300 K.
2. What is γ for nitrogen at room temperature?
3. Find the ratio of the rms speeds of hydrogen and oxygen molecules at the same temperature.
4. 1 mol of an ideal gas expands isothermally at 300 K to twice its volume. Find the work done (R = 8.3 J/mol K, ln 2 = 0.693).
5. 1 mol of helium is mixed with 1 mol of oxygen. Find γ for the mixture.
6. 1,000 J of heat is given to a diatomic ideal gas at constant pressure. How much goes into internal energy, and how much into work?
7. A Carnot engine works between 500 K and 300 K and absorbs 1,000 J per cycle. Find the work done and the heat rejected per cycle.
8. On a P–V diagram, find the ratio of the slope of an adiabatic curve to that of an isothermal curve through the same point, for a monatomic gas.

Answers

1. 1 − 300/600 = 50%.
2. N₂ is diatomic (f = 5), so 7/5.
3. v ∝ 1/√M: √(32/2) = 4 : 1.
4. W = 8.3 × 300 × 0.693 ≈ 1,726 J.
5. C_v = (1.5R + 2.5R)/2 = 2R, C_p = 3R, so γ = 1.5.
6. Fraction to ΔU = C_v/C_p = 5/7, so about 714 J to internal energy and about 286 J as work.
7. η = 1 − 300/500 = 0.4, so W = 400 J and 600 J is rejected.
8. The ratio is γ = 5/3.

What to do next

• Fix one sign convention (W done by the gas) and never change it.
• Draw the four processes on one P–V diagram and label W, ΔU and Q for each.
• Solve 25 problems, including at least five cyclic-process graphs.
• Revise thermal properties of matter, and compare with the chemistry view in chemical thermodynamics, which uses a different sign convention for work.