The Adiabatic Expansion of an Ideal Gas — Question 7

Question 7

Expansion against a fixed load requires external-pressure work

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 starts at $(p_i,V_i,500\,\mathrm K)$. A latched insulating piston is released against a constant external pressure $p_e=p_i/2$. It settles at $p_f=p_e$ with zero final piston kinetic energy. Any transient mechanical energy is dissipated inside the gas; the piston and walls retain no energy. The gas is insulated.

1. Find $T_f$, $V_f/V_i$ and work done by the gas using $W=p_e(V_f-V_i)$.

2. Compare with reversible adiabatic expansion to the same final pressure and test whether the actual endpoints obey $pV^{5/3}=\text{constant}$.

Question 7 - Solution

Expansion against a fixed load requires external-pressure work

(a) Method and calculation

$$nC_V(T_f-T_i)=-p_e(V_f-V_i),\quad p_eV_f=nRT_f,\quad p_eV_i=\tfrac12nRT_i.$$$$(C_V+R)T_f=(C_V+\tfrac12R)T_i\ \Rightarrow\ T_f=\tfrac45T_i=400\,\mathrm K.$$$$V_f/V_i=(T_f/T_i)/(p_f/p_i)=1.60,\quad W=\tfrac32R(100)=1247.10\,\mathrm J.$$

(b) Interpretation and further calculation

$$T_{f,\mathrm{rev}}=500(p_f/p_i)^{(\gamma-1)/\gamma}=500(1/2)^{2/5}\approx378.929\,\mathrm K.$$$$W_{\mathrm{rev}}\approx1509.875\,\mathrm J,\quad\frac{p_fV_f^{5/3}}{p_iV_i^{5/3}}=\tfrac12(1.60)^{5/3}\approx1.09438\ne1.$$Finite mechanical imbalance makes the actual path irreversible, despite zero heat transfer.

Physical check

The reversible and fixed-load paths have the same final pressure but different final temperatures and volumes.