The Adiabatic Expansion of an Ideal Gas — Question 6

Question 6

Free expansion is adiabatic without obeying the reversible power law

Use $pV=nRT$, $R=8.314$ J/(mol K), and $\Delta U=Q-W$ with $W$ positive for work done by the gas. One mole of monatomic ideal gas at 300 K occupies one-third of a rigid insulated enclosure; the rest is vacuum. A partition is removed, and the gas eventually equilibrates throughout the enclosure. Neglect wall heat capacity and other energy stores.

1. Find heat, work, final temperature and pressure ratio for this free expansion.

2. Compare with a reversible adiabatic expansion from the same initial state to the same final volume. Explain why applying $pV^{5/3}=\text{constant}$ to the actual free expansion fails.

Question 6 - Solution

Free expansion is adiabatic without obeying the reversible power law

(a) Method and calculation

$$Q=0,\quad W=\int p_{\mathrm{ext}}\,dV=0\ \Rightarrow\Delta U=0.$$$$U=\tfrac32RT\ \Rightarrow\ T_f=300\,\mathrm K,\quad p_f/p_i=V_i/V_f=1/3.$$

(b) Interpretation and further calculation

$$T_{f,\mathrm{rev}}=300(1/3)^{2/3}\approx144.225\,\mathrm K,\quad p_{f,\mathrm{rev}}/p_i=3^{-5/3}\approx.160250.$$$$W_{\mathrm{rev}}=\tfrac32R(300-144.225)\approx1942.671\,\mathrm J.$$The actual gas does not pass through a sequence of equilibrium states pushing against a matching pressure. Its zero external work differs from reversible expansion work.

Physical check

Adiabatic specifies heat transfer; reversibility is an additional condition.