Revise Thermodynamics for IB Physics (B.4)

The first law is simply conservation of energy applied to a gas. Heat supplied to a gas cannot vanish: it either raises the internal energy of the gas or is spent by the gas doing work on its surroundings, or both. Writing $Q$ for the heat added to the gas, $\Delta U$ for the change in internal energy and $W$ for the work done BY the gas, the budget balances exactly.

$Q = \Delta U + W$

This is the form given in the IB data booklet. Rearranged it reads $\Delta U = Q - W$, which some prefer, but both say the same thing. The internal energy $U$ is the total random kinetic energy of the molecules, and for an ideal gas it depends only on the absolute temperature — so a rise in $U$ always means a rise in temperature. Everything in this topic follows from applying this one equation with a firm, consistent sign convention.

When a gas expands it pushes its boundary outward and does work on the surroundings. If the pressure stays constant while the volume changes by $\Delta V$, the work done by the gas is the product of pressure and volume change.

$W = p\Delta V$

For this to give joules, put the pressure in pascals and the volume change in cubic metres. More generally the pressure need not be constant, and then $W = p\Delta V$ no longer applies directly. The powerful idea is that on a pressure–volume (p–V) diagram the work done by the gas is always the area under the curve for the process. A p–V diagram plots the state of the gas as a point, a process as a path between states, and the area beneath that path as the work exchanged. For a complete cycle the gas returns to its starting point and the net work is the area enclosed by the loop.

Four idealised processes appear again and again, and the first law $Q = \Delta U + W$ takes a simple form in each. Recognising the process is the first move in almost every thermodynamics question, because it immediately tells you which term is zero.

The first law tells you energy is conserved, but not which way a process will actually go. Drop an ice cube in warm water and heat always flows from the water to the ice, never the reverse — yet both directions conserve energy. The second law supplies the missing arrow. It is stated using entropy $S$, a measure of how spread out, or disordered, the energy of a system is. The more ways the energy can be arranged among the molecules, the higher the entropy.

$\Delta S = \dfrac{Q}{T}$

For heat $Q$ transferred at a constant absolute temperature $T$ (in kelvin), the entropy change is $\Delta S = Q/T$, measured in J K⁻¹. Heat entering a body raises its entropy; heat leaving lowers it. The second law then states that for any real process the total entropy of an isolated system — the object together with everything it exchanges energy with — tends to increase, and can at best stay constant in an ideal reversible process.

$\Delta S_{universe} = \Delta S_{system} + \Delta S_{surroundings} \ge 0$

This is why heat flows from hot to cold: the entropy lost by the hot body ($Q/T_{hot}$) is smaller than the entropy gained by the cold body ($Q/T_{cold}$) because dividing by the smaller temperature gives the larger change, so the total rises. A local decrease in entropy — water freezing, a crystal forming — is always allowed provided the surroundings gain even more. The second law forbids only a decrease of the isolated whole, and in doing so it sets the direction of the arrow of time.

A heat engine is any device that takes in heat $Q_{in}$ from a hot reservoir, converts part of it into useful work $W$, and rejects the remainder $Q_{out}$ to a cold reservoir. Over a complete cycle the gas returns to its starting state, so $\Delta U = 0$ and the first law gives $W = Q_{in} - Q_{out}$. The thermal efficiency is the fraction of the incoming heat that is turned into useful work.

$\eta = \dfrac{W}{Q_{in}} = 1 - \dfrac{Q_{out}}{Q_{in}}$

Because some heat must always be rejected ($Q_{out} > 0$) for the engine to keep running without decreasing the entropy of the universe, the efficiency is always less than 1. The second law forbids a perfect engine that turns all its incoming heat into work. The interesting question is: for two given reservoir temperatures, what is the highest efficiency physically allowed?

Sadi Carnot answered that question with an idealised, perfectly reversible cycle. The Carnot cycle consists of four reversible steps between a hot reservoir at $T_{hot}$ and a cold reservoir at $T_{cold}$: an isothermal expansion (absorbing heat at $T_{hot}$), an adiabatic expansion (cooling to $T_{cold}$), an isothermal compression (rejecting heat at $T_{cold}$) and an adiabatic compression (warming back to $T_{hot}$). Because every step is reversible, the total entropy change over the cycle is zero, and this gives the highest efficiency any engine can achieve between those two temperatures.

$\eta_{max} = 1 - \dfrac{T_{cold}}{T_{hot}}$

Both temperatures must be in kelvin. No real engine, with its friction and finite-rate heat flow, can reach the Carnot limit — it is an unattainable ceiling. But the formula is powerful: it shows efficiency rises as the reservoirs are pushed further apart in temperature, and it lets you judge instantly whether a claimed engine performance is even possible. Any engine claiming to beat $1 - T_{cold}/T_{hot}$ is claiming to break the second law.

In Paper 2 the marks are analytic: each is tied to a specific line of working — a method mark (M) or an answer mark (A) — and error-carried-forward (ECF) means a wrong number early on does not have to cost you the marks that follow, provided your later steps follow correctly from it. But that protection only exists if the method is written down. Study how each mark below is earned by a specific line.

Thermodynamics ties the microscopic picture of the previous topics — molecules with kinetic energy and the ideal gas law — to the large-scale behaviour of engines, refrigerators and the cosmos itself. The first law is conservation of energy you have met before, now sharpened for gases; the second law is the deeper statement, the one physical principle that distinguishes past from future. Master the habit — name the process, write $Q = \Delta U + W$ with firm signs, work in kelvin, and show every line — and both the calculations and the conceptual questions in this topic become variations on a method you already own.