Thermodynamics
🎯What you need to be able to do
- Apply the first law of thermodynamics, \( Q = \Delta U + W \), with correct signs.
- Calculate work done by or on a gas, including as the area under a \(pV\) curve.
- Identify and analyse isovolumetric, isobaric, isothermal and adiabatic processes.
- Interpret \(pV\) diagrams and closed cycles.
- State the second law in terms of entropy, and calculate entropy change.
- Calculate the efficiency of a heat engine and the Carnot limit.
📚The physics
The first law is conservation of energy for gases:
Thermal energy supplied to a gas either raises its internal energy or is used by the gas doing work on its surroundings, or both. The signs are where marks are lost, so fix them once: \(Q\) is positive when energy goes in to the gas; \(W\) is positive when the gas expands and does work on the surroundings; \(\Delta U\) is positive when the temperature rises. For an ideal gas \(\Delta U\) depends only on temperature, which is a shortcut worth remembering.
Work done by a gas at constant pressure is \( W = p\Delta V \). When pressure varies, the work is the area under the curve on a \(pV\) diagram — the same “area under a graph” idea you met with force–distance. Expansion moves right on the diagram and the gas does positive work; compression moves left and work is done on the gas.
The four processes are best learned by asking which term of the first law vanishes.
- Isovolumetric (constant \(V\)): no volume change, so \( W = 0 \) and \( Q = \Delta U \). All the energy goes into internal energy. Vertical line on a \(pV\) diagram.
- Isobaric (constant \(p\)): \( W = p\Delta V \) and all three terms are generally non-zero. Horizontal line.
- Isothermal (constant \(T\)): for an ideal gas \( \Delta U = 0 \), so \( Q = W \) — every joule supplied leaves as work. Curve following \( pV = \) constant. Requires slow change and good thermal contact.
- Adiabatic (no thermal transfer): \( Q = 0 \), so \( \Delta U = -W \). An expanding gas does work at the expense of its own internal energy, so it cools. Steeper than an isotherm on a \(pV\) diagram. Requires rapid change or good insulation.
That adiabatic cooling is not abstract: it is why a rapidly released aerosol feels cold, and why rising air cools and forms cloud.
Closed cycles. Round a complete cycle the gas returns to its starting state, so \( \Delta U = 0 \) over the cycle and \( Q = W \). The net work is the area enclosed by the loop. Clockwise loops do net work on the surroundings — that is an engine. Anticlockwise loops require work to be put in — that is a refrigerator or heat pump.
The second law and entropy. Entropy is a measure of the number of ways the microscopic state can be arranged — loosely, of disorder. The second law says that in any real process the total entropy of the system and its surroundings never decreases. Entropy change is \( \Delta S = \Delta Q / T \). A local decrease is entirely allowed — a freezer makes ice — provided a larger increase happens elsewhere, in that case in the kitchen.
Heat engines. An engine takes \(Q_H\) from a hot reservoir, does useful work \(W\), and dumps \(Q_C\) into a cold reservoir.
The second law forbids \( \eta = 1 \): some energy must be rejected to the cold reservoir. The best possible efficiency, achieved only by an idealised reversible Carnot cycle, uses temperatures in kelvin. No engine between the same two temperatures can beat it, whatever it is made of.
✏️Worked example
(a) Actual efficiency. \( \eta = 520/1600 = 0.33 \), or 33%.
(b) Maximum efficiency allowed by the second law. \( \eta_{\text{Carnot}} = 1 - 300/810 = 1 - 0.370 = 0.63 \), or 63%.
(c) Rate of energy rejection to the cooling tower. \( 1600 - 520 = 1080 \) MW.
(d) Rate of entropy change of the cold reservoir. \( \Delta S/\Delta t = 1080 \times 10^{6}/300 = 3.6 \times 10^{6} \) W K\(^{-1}\).
📝Practise
Work through these, then reveal the answer. Each question targets a different objective from the list above.
1. 500 J of thermal energy is supplied to a gas, which does 200 J of work on its surroundings. Find the change in internal energy.
2. A gas expands from 0.020 m\(^{3}\) to 0.050 m\(^{3}\) at a constant pressure of \( 2.0 \times 10^{5} \) Pa. Find the work done by the gas.
3. An ideal gas absorbs 800 J isothermally. How much work does it do, and what is the change in internal energy?
4. A gas expands adiabatically, doing 450 J of work. Find the change in internal energy and state what happens to the temperature.
5. A heat engine operates between reservoirs at 600 K and 300 K. Find the maximum possible efficiency.
6. 2000 J of energy is transferred into a reservoir held at 400 K. Find the entropy change of that reservoir, and explain how a freezer can reduce entropy locally without breaking the second law.
🔗Go deeper — other people’s work
These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.
- The Physics Hypertextbook — thermodynamics and heat engines
- HyperPhysics — first and second laws, Carnot cycle