The first law and ideal gas processes

By the end of this topic you will be able to

State the first law for a closed system and the sign convention used.
Calculate work, heat and internal-energy change for isobaric, isochoric, isothermal and adiabatic processes of an ideal gas.
Interpret work as the area under a process curve on a P–V diagram.

The first law

Energy is conserved. For a closed system, the heat added $Q$ goes partly into raising the internal energy $U$ and partly into work $W$ done by the system on its surroundings:

Q = \Delta U + W

Sign convention used here: $Q$ is positive when heat is added to the gas, and $W$ is positive when the gas does work on its surroundings (expands).

The ideal gas

For an ideal gas, $PV = nRT$ , with $R = 8.314$ J/(mol·K), and the internal energy depends only on temperature:

\Delta U = n c_v \Delta T, \qquad c_v = \frac{R}{\gamma - 1}

where $\gamma = c_p/c_v$ is about 1.4 for air and other diatomic gases, and 1.67 for monatomic gases such as argon.

Work on a P–V diagram

When a gas expands slowly, the work it does is

W = \int P\,\mathrm{d}V

which is the area under the process curve on a P–V diagram. Expansion (volume rising) gives positive work; compression, negative.

Four standard processes

Process	Condition	Work	$\Delta U$	$Q$
Isobaric	$P$ constant	$P(V_2 - V_1)$	$n c_v \Delta T$	$n c_p \Delta T$
Isochoric	$V$ constant	0	$n c_v \Delta T$	$= \Delta U$
Isothermal	$T$ constant	$P_1V_1\ln(V_2/V_1)$	0	$= W$
Adiabatic	$Q = 0$	$(P_1V_1 - P_2V_2)/(\gamma - 1)$	$= -W$	0

In an adiabatic process $PV^\gamma$ is constant, so the curve falls more steeply than the isothermal one.

Interactive modelPick a process and move the volume (or pressure) ratio. The shaded area is the work.

This interactive model needs JavaScript. The theory, worked example and problems on this page work without it.

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Worked example

One mole of air ( $\gamma = 1.4$ ) is at 100 kPa and 25 L (so $T_1 = PV/nR = 301$ K). The volume doubles. Compare an isothermal and an adiabatic expansion.

Isothermal. $P_2 = 50$ kPa. $W = P_1V_1\ln2 = 100\times25\times0.693 = 1733$ J. $\Delta U = 0$ , so $Q = 1733$ J of heat must flow in.

Adiabatic. $P_2 = 100\times2^{-1.4} = 37.9$ kPa, and $T_2 = 301\times2^{-0.4} = 228$ K. $W = (100\times25 - 37.9\times50)/0.4 = 1513$ J. $Q = 0$ , so $\Delta U = -1513$ J: the gas cools as it does work.

The adiabatic expansion does less work because it is not supplied with heat; the gas pays for the work from its own internal energy.

Practice problems

A gas expands at a constant 200 kPa from 10 L to 30 L. Find the work.

Show answer

$W = P\Delta V = 200\times20 = 4000$ J = 4 kJ.
Why is no work done in an isochoric process?

Show answer

The volume does not change, so there is no area under the curve ( $\mathrm{d}V = 0$ ).
In an isothermal process, 500 J of heat is added to an ideal gas. How much work does it do?

Show answer

$\Delta U = 0$ so $W = Q = 500$ J.
In the model, for the same volume ratio, compare the work done in isothermal and adiabatic expansion. Which is bigger and why?

Show answer

Isothermal. It has heat input, so the pressure stays higher along the way (the curve lies above the adiabatic one), giving more area under it.

Key formulae

First law (closed system): $Q = \Delta U + W$
Ideal gas law: $PV = nRT$
Isothermal work: $W = P_1 V_1 \ln\frac{V_2}{V_1}$
Adiabatic relation: $P V^{\gamma} = \text{constant}, \quad \gamma = \frac{c_p}{c_v}$
Adiabatic work: $W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1}$

Lecturer notes: suggested plan, activity and assessment

Suggested plan (50 min). 10 min: bicycle pump demonstration (the barrel warms: compression work, nearly adiabatic); 10 min: first law and sign convention; 15 min: four processes with the model; 15 min: worked example.

Activity. Students find a combination of two processes that returns the gas to its start (a cycle) and sketch it: this leads into engine cycles.

Discussion. Why is the sign convention for work different in some textbooks (work done on the system)?

Assessment idea. A three-process cycle with a table of Q, W and ΔU, checking that the cycle sums satisfy ΣQ = ΣW.