HomeLearning HubIB DP PhysicsB.4 Thermodynamics
B.4

Thermodynamics

Theme B · The particulate nature of matter · HL only

Another topic promoted out of the retired Option B. Every HL candidate now studies it.

🎯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:

\[ Q = \Delta U + W \]

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.

\( \eta = \dfrac{W}{Q_H} = 1 - \dfrac{Q_C}{Q_H} \)
\( \eta_{\text{Carnot}} = 1 - \dfrac{T_C}{T_H} \)

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 power station operates between a boiler at 810 K and a cooling tower at 300 K. It receives 1600 MW of thermal power and generates 520 MW of electrical power.

(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\(^{-1}\).

What the numbers mean. The station achieves about half the theoretical maximum, which is typical of real plant — friction, turbulence and finite-rate heat transfer all cost efficiency. Note that more than twice as much energy leaves as waste heat as leaves as electricity, and that this is not bad engineering but a requirement of the second law. Building a 100% efficient version is not a hard problem; it is a forbidden one.

📝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.
\( \Delta U = Q - W = 500 - 200 = 300 \) J. Positive, so the temperature rises.
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.
\( W = p\Delta V = 2.0 \times 10^{5} \times 0.030 = 6.0 \times 10^{3} \) J. It is positive because the gas expanded, doing work on the surroundings.
3. An ideal gas absorbs 800 J isothermally. How much work does it do, and what is the change in internal energy?
For an ideal gas \(\Delta U\) depends only on temperature, and the temperature is constant, so \( \Delta U = 0 \). The first law then gives \( Q = W \), so the gas does 800 J of work — every joule supplied leaves again as work.
4. A gas expands adiabatically, doing 450 J of work. Find the change in internal energy and state what happens to the temperature.
Adiabatic means \( Q = 0 \), so \( \Delta U = -W = -450 \) J. The gas does work at the expense of its own internal energy, so the temperature falls. This is why a rapidly released aerosol feels cold.
5. A heat engine operates between reservoirs at 600 K and 300 K. Find the maximum possible efficiency.
\( \eta_{\text{Carnot}} = 1 - T_C/T_H = 1 - 300/600 = 0.50 \), or 50%. No engine working between these two temperatures can beat this, whatever it is made of.
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.
\( \Delta S = \Delta Q/T = 2000/400 = 5.0 \) J K\(^{-1}\). A freezer decreases the entropy of the water it turns to ice, but it dumps a larger amount of entropy into the kitchen via its warm coils. The second law constrains the total for system plus surroundings, not any one part of it.

🔗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