The First Law of Thermodynamics

There are only two ways to change the internal energy of a gas: heat it, or do work on it. The first law of thermodynamics is the statement that energy is conserved when you do either or both: $\Delta U = q + W$. This note shows where the work done $p\Delta V$ comes from, sets out the sign convention Cambridge uses (which catches out many students), and works through the classic exam questions: single processes, $p$–$V$ graphs and the cycle table that appears in Paper 4 almost every year.

Two ways to transfer energy

Think of a gas in a cylinder with a movable piston.

• Heating. Place the cylinder on a hot plate. Energy flows into the gas because of the temperature difference. Faster-moving molecules in the hot wall pass kinetic energy to the gas molecules that collide with it. This is a transfer of energy by heating.
• Doing work. Push the piston in. The molecules bouncing off the inward-moving piston rebound faster than they arrived, so the gas gains kinetic energy. This is a transfer of energy by doing work: a force moves its point of application.

Both raise the internal energy. The result can be identical: from the final state alone you cannot tell which method was used, because internal energy depends only on the state.

Work done when a gas changes volume

Consider gas at pressure $p$ in a cylinder with a piston of area $A$. The gas pushes on the piston with force $F = pA$. If the gas expands slowly at constant pressure so the piston moves out a distance $\Delta x$, the work done by the gas is

$$W = F\Delta x = pA\Delta x = p\Delta V$$

since $A\Delta x$ is the increase in volume $\Delta V$.

Work done by the gas and work done on the gas

The same quantity of work can be described from either side:

• When a gas expands, it pushes the surroundings back: work is done by the gas. Equivalently, the work done on the gas is negative.
• When a gas is compressed, the surroundings push the piston in: work is done on the gas. Equivalently, the work done by the gas is negative.

Work done on the gas $= -($work done by the gas$)$. Always say which one you mean.

Work as an area on a $p$–$V$ graph

If the pressure changes during the expansion, split it into small steps, each with nearly constant pressure. The work in each step is $p\Delta V$, the area of a thin strip under the curve. So:

The shaded area is the work done by an ideal gas as it expands isothermally from $1$ to $5$ units of volume. A constant-pressure change appears as a horizontal line, and its area is the rectangle $p\Delta V$. A constant-volume change is a vertical line, with zero area: no work is done when the volume does not change.

The first law of thermodynamics

Energy is conserved. The increase in a system's internal energy must equal the energy put in by heating plus the energy put in by doing work.

The sign convention

Every quantity is measured as energy going into the gas:

Quantity	Positive when	Negative when
$\Delta U$	internal energy increases (temperature of an ideal gas rises)	internal energy decreases
$q$	energy is transferred to the gas by heating	energy is transferred from the gas (it cools its surroundings)
$W$	work is done on the gas (it is compressed)	work is done by the gas (it expands)

So for a gas expanding against a constant pressure, $W = -p\Delta V$, with $\Delta V$ positive. For a compression, $\Delta V$ is negative and $W = -p\Delta V$ is positive. Many students find it easier to decide the sign from the physics (expanding means $W$ negative) and then put in the magnitude $p|\Delta V|$.

Applying the first law to an ideal gas

For an ideal gas, internal energy depends only on temperature (for a monatomic gas, $U = \tfrac{3}{2}nRT$). This makes four special processes easy to analyse.

Process	What is fixed	Consequence	First law becomes
Constant volume	$V$	$W = 0$	$\Delta U = q$: all heating goes to internal energy
Constant pressure	$p$	$W = -p\Delta V$	$\Delta U = q - p\Delta V$: some heating is used to do work
Isothermal	$T$	$\Delta U = 0$	$q = -W$: heating supplied equals work done by the gas
Adiabatic (no heat transfer)	thermal isolation, or a very fast change	$q = 0$	$\Delta U = W$

Physical meanings to recognise in questions:

• Rapid compression (a bicycle pump, a diesel engine): there is no time for energy to escape by heating, so $q \approx 0$. Work done on the gas increases its internal energy, so its temperature rises. That is why a pump gets hot.
• Rapid expansion (gas escaping from an aerosol can, air rushing out of a tyre valve): $q \approx 0$ and $W$ is negative, so $\Delta U$ is negative and the gas cools.
• Slow expansion in good thermal contact with surroundings: the temperature stays constant (isothermal). The gas does work, so it must absorb an equal amount of energy by heating.
• Heating at constant pressure needs more energy than at constant volume for the same temperature rise, because some of the energy supplied is used for the work done by the expanding gas.

Cycles

In an engine, the gas goes through a sequence of changes and returns to its starting state. Because internal energy is a function of state, after a complete cycle

$$\Delta U_{\text{cycle}} = 0 \quad\Rightarrow\quad q_{\text{total}} = -W_{\text{total}}$$

The net work done by the gas in one cycle is the area enclosed by the loop on a $p$–$V$ graph. Going clockwise round the loop, the gas does more work expanding (at high pressure) than is done on it while it is compressed (at low pressure), so there is net work output: this is a heat engine.

The graph shows a rectangular cycle A$(1, 1)$ to B$(1, 3)$ to C$(3, 3)$ to D$(3, 1)$ and back to A, with pressure in units of $10^{5}\ \text{Pa}$ and volume in units of $10^{-3}\ \text{m}^3$. The gas expands along BC at high pressure and is compressed along DA at low pressure; the net work done by the gas is the area of the rectangle enclosed by the loop. The cycle is analysed in a worked example below.

Worked examples

Practice questions

Answers

1. The increase in internal energy of a system ($\Delta U$) is equal to the sum of the energy transferred to the system by heating ($q$) and the work done on the system ($W$).
2. $W = p|\Delta V| = 2.0 \times 10^{5} \times 4.0 \times 10^{-4} = 80\ \text{J}$, positive because the gas is compressed: $W = +80\ \text{J}$.
3. Work done by the gas $= 1.0 \times 10^{5} \times 0.010 = 1000\ \text{J}$, so $W = -1000\ \text{J}$. $\Delta U = 3500 - 1000 = +2500\ \text{J}$.
4. $W = 0$ (constant volume), so $q = \Delta U = \tfrac{3}{2}nR\Delta T = 1.5 \times 2.0 \times 8.31 \times 60 = 1.50 \times 10^{3}\ \text{J}$.
5. The air expands rapidly, so there is no time for heating: $q \approx 0$. The air does work pushing back the atmosphere, so $W$ is negative. Therefore $\Delta U$ is negative: the internal (kinetic) energy of the molecules falls, and the temperature of the air drops.
6. $\Delta U = 0$ (ideal gas at constant temperature). $q = \Delta U - W = -400\ \text{J}$: $400\ \text{J}$ is transferred from the gas to the surroundings by heating, equal to the work done on it.
7. The same temperature rise means the same $\Delta U$. At constant volume $W = 0$, so $q = \Delta U$. At constant pressure the gas expands and does work on the surroundings ($W$ negative), so $q = \Delta U - W$ is larger by the work done by the gas.
8. (a) $p\Delta V = 3.0 \times 10^{5} \times 3.0 \times 10^{-3} = 900\ \text{J}$. (b) $\Delta U = \tfrac{3}{2}p\Delta V = 1350\ \text{J}$. (c) $W = -900\ \text{J}$, so $q = \Delta U - W = 1350 + 900 = 2250\ \text{J}$.
9. P to Q: $q = 0$, $W = +600\ \text{J}$, $\Delta U = +600\ \text{J}$. Q to R: $q = +1000\ \text{J}$, $W = 0$, $\Delta U = +1000\ \text{J}$. R to P: $\Delta U = -(600 + 1000) = -1600\ \text{J}$ (cycle total zero); $W = -1300\ \text{J}$; $q = \Delta U - W = -1600 + 1300 = -300\ \text{J}$. Net work done by the gas $= 1300 - 600 = 700\ \text{J}$, which equals the net energy supplied by heating, $1000 - 300 = 700\ \text{J}$.
10. (a) $V = nRT/p = 0.20 \times 8.31 \times 300/1.0 \times 10^{5} = 4.99 \times 10^{-3}\ \text{m}^3$. (b) At constant volume $p \propto T$: $T = 600\ \text{K}$. $q = \Delta U = 1.5 \times 0.20 \times 8.31 \times 300 = 748\ \text{J}$. (c) At constant pressure, doubling $V$ doubles $T$ to $1200\ \text{K}$. Work done by the gas $= p\Delta V = 2.0 \times 10^{5} \times 4.99 \times 10^{-3} = 997\ \text{J}$ (equivalently $nR\Delta T = 0.20 \times 8.31 \times 600 = 997\ \text{J}$), so $W = -997\ \text{J}$. $\Delta U = 1.5 \times 0.20 \times 8.31 \times 600 = 1.50 \times 10^{3}\ \text{J}$. $q = \Delta U - W = 1496 + 997 = 2.49 \times 10^{3}\ \text{J}$.