A Closer Look at Heat and Work — Question 3

Question 3

External pressure determines work in an irreversible expansion

A gas initially occupies 2.00 L at 400 kPa. It expands against a constant external pressure of 100 kPa until its volume is 6.00 L. The intermediate gas pressure is not uniform. For boundary work use $W=\int p_{\mathrm{ext}}\,dV$.

1. Find work done by the gas. Assess the calculation $p_i(V_f-V_i)$.

2. Compare with a separate quasistatic route having $p(V)=800/V$ kPa when $V$ is numerically in litres, over the same volume interval. Must its final pressure equal the irreversible run's final pressure?

Question 3 - Solution

External pressure determines work in an irreversible expansion

(a) Method and calculation

$$W_{\mathrm{irr}}=100(6-2)=400\,\mathrm J.$$$$p_i\Delta V=400(4)=1600\,\mathrm J\ne W_{\mathrm{irr}}.$$The initial internal pressure is not the resisting pressure throughout the displacement.

(b) Interpretation and further calculation

$$W_{\mathrm{qs}}=\int_2^6\frac{800}{V}dV\,\mathrm J=800\ln3\,\mathrm J\approx878.890\,\mathrm J.$$$$p_{f,\mathrm{qs}}=800/6=133.333\,\mathrm{kPa}.$$The endpoints share volumes but need not share final thermodynamic states; the irreversible run can finish at 100 kPa if mechanical equilibrium is reached.

Physical check

An equilibrium $pV$ curve is not automatically available during a strongly irreversible process.