Basics of thermodynamics

Thermodynamics studies how energy passes from one body to another as heat and as work. It explains heat engines (cars, power-station turbines), refrigerators, air conditioners and the energy balance of any process involving a gas. Unlike kinetic theory, thermodynamics says nothing about molecules; it deals only with macroscopic quantities (p, V, T, Q, A, U). Exam papers include demanding multi-step problems with graphs from this section.

§1. Internal energy

The internal energy U of a body is the total kinetic energy of the random motion of its particles plus the potential energy of their interaction. In an ideal gas the molecules do not interact, so the internal energy depends only on temperature:

For a diatomic gas the coefficient is 5/2. Internal energy is a function of state: its change depends only on the initial and final states, not on the path the gas took. Round a closed cycle ΔU = 0.
Internal energy can be changed in two ways:

• by doing work: compressing a gas, friction (rubbing your hands, a pump warming up while inflating a tyre);
• by heat transfer: conduction, convection, radiation.

§2. Work done by a gas

A gas that expands and pushes a piston does work. At constant pressure:

In general the work equals the area under the p–V graph.

• During expansion (ΔV > 0) the work done by the gas is positive; during compression it is negative (external forces then do work on the gas: A_ext = −A).
• In an isochoric process ΔV = 0, so A = 0.
• Work depends on the process: between the same initial and final states, different paths give different work.
• Round a closed cycle the net work is the area enclosed by the cycle; if the cycle runs clockwise on a p–V diagram, the work is positive.

Worked example

Helium under a 20 kg piston in a cylinder of base area 1 dm² is heated at constant pressure and the piston rises 30 cm; p₀ = 100 kPa. How much work does the gas do?

1. Gas pressure: p = p₀ + mg/S = 10⁵ + 200/0.01 = 1.2·10⁵ Pa.
2. Change in volume: ΔV = S·h = 0.01·0.3 = 3·10⁻³ m³.
3. A = pΔV = 1.2·10⁵·3·10⁻³ = 360 J.

§3. The first law of thermodynamics

Conservation of energy for thermal processes:

Signs: if the gas takes in heat, Q > 0; if it gives out heat, Q < 0; if it expands, A > 0.

• Isothermal (T = const): ΔU = 0, Q = A; all the heat becomes work.
• Isochoric (V = const): A = 0, Q = ΔU.
• Isobaric (monatomic gas): Q = ΔU + A = (3/2)νRΔT + νRΔT = (5/2)νRΔT = (5/2)pΔV.
• Adiabatic (Q = 0, no heat exchange or a very fast process): A = −ΔU; an expanding gas cools, a compressed gas heats up. In a diesel engine air is compressed so quickly that it becomes hot enough to ignite the fuel; gas rushing out of a cylinder cools down.

The first law rules out a perpetual motion machine (of the first kind): no machine can do work without using energy.

Worked example

2 mol of a monatomic gas is heated by 100 K at constant pressure. Find ΔU, A and Q.

1. ΔU = (3/2)νRΔT = 1.5·2·8.31·100 ≈ 2490 J.
2. A = νRΔT = 2·8.31·100 ≈ 1660 J.
3. Q = ΔU + A ≈ 4150 J (check: (5/2)νRΔT = 2.5·1662 ≈ 4150 J).

§4. Cycles and graph problems

In a graph problem, work out for each stage separately: which quantity is constant, A (the area), ΔU ((3/2)Δ(pV) for a monatomic gas) and Q = ΔU + A. Stages where the gas takes in heat have Q > 0; stages where it gives out heat have Q < 0.

Worked example

A monatomic gas goes round the cycle 1(P₀; 5V₀) → 2(P₀; V₀) → 3(4P₀; V₀) → 1 (3→1 is a straight line). Find the work per cycle and the heat in stages 1→2 and 3→1.

1. Work per cycle is the area of the triangle: A = ½·4V₀·3P₀ = 6P₀V₀.
2. 1→2 is an isobaric compression: Q = (5/2)P₀·(V₀ − 5V₀) = −10P₀V₀, so the gas gives out 10P₀V₀.
3. 3→1: ΔU = (3/2)(5P₀V₀ − 4P₀V₀) = 1.5P₀V₀; the work is the trapezium area ½(4P₀ + P₀)·4V₀ = 10P₀V₀; Q = 11.5P₀V₀.

Worked example

A monatomic gas receives 6 kJ of heat in a process where V = k√T. Find the work done and ΔU.

1. T ∝ V² and pV ∝ T → p ∝ V: the graph is a straight line through the origin, so A = ½Δ(pV) = ½νRΔT.
2. Q = ΔU + A = (3/2 + 1/2)νRΔT = 2νRΔT = 6 kJ → νRΔT = 3 kJ.
3. A = 1.5 kJ, ΔU = 4.5 kJ.

§5. Heat engines and the second law

A heat engine turns internal energy into mechanical work. Every engine has three parts: a hot reservoir (temperature T₁), a working substance (gas or steam) and a cold reservoir (T₂, usually the atmosphere). The working substance takes Q₁ from the hot reservoir, gives Q₂ to the cold one and does work A = Q₁ − Q₂:

The Carnot formula needs temperatures in kelvins; real engines are much less efficient (roughly 25–30 % for a petrol engine and about 35–40 % for a diesel).
The second law of thermodynamics: heat flows by itself only from a hotter body to a colder one; no cyclic engine can turn all the heat it receives into work (η < 1). A refrigerator or air conditioner moves heat from cold to hot “by force”, and an external source (the compressor) must do work for that.

Worked example

An engine takes 10 kJ of heat from the hot reservoir and is 25 % efficient. How much work does it do, and how much heat goes to the cold reservoir? What is the Carnot efficiency for 500 K and 300 K?

1. A = ηQ₁ = 0.25·10 = 2.5 kJ.
2. Q₂ = Q₁ − A = 10 − 2.5 = 7.5 kJ.
3. η_max = (500 − 300)/500 = 0.4 = 40 %; the real efficiency is lower, so the result makes sense.

Check yourself

1. A gas receives 500 J of heat and does 200 J of work. ΔU?

2. 2 mol of a monatomic gas is heated by 100 K. ΔU?

3. In isochoric heating a gas receives 400 J. Work done by the gas?

4. Hot reservoir 500 K, cold reservoir 300 K. Maximum efficiency?

5. How does the temperature of a gas change in adiabatic expansion?

6. At 10⁵ Pa a gas expands from 2 l to 5 l. Work?

7. In an isothermal process a gas receives 800 J of heat. How much work does it do?

8. An engine is 30 % efficient and does 600 J of work per cycle. Heat taken from the hot reservoir?

Common mistakes

• Forgetting to add the weight of the piston and atmospheric pressure to the gas pressure.
• Taking the work per cycle as the area under the whole graph: only the enclosed area counts.
• Writing Q = ΔU for an isobaric process, i.e. forgetting A.
• Using °C in the Carnot efficiency.
• Mixing up the signs of Q and A: a gas giving out heat has Q < 0, a gas being compressed has A < 0.