The Adiabatic Expansion of an Ideal Gas — Question 5

Question 5

An isothermal path and an adiabat from the same state do different 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 at 400 K expands reversibly to twice its initial volume. Compare an isothermal path with an adiabatic path having $\gamma=1.40$ and $C_V=5R/2$.

1. Calculate work and heat for each path, including final adiabatic temperature.

2. Use the pressure curves to explain the work ordering. Do these paths reach the same final thermodynamic state?

Question 5 - Solution

An isothermal path and an adiabat from the same state do different work

(a) Method and calculation

$$W_T=nRT_i\ln2\approx2305.130\,\mathrm J,\quad\Delta U_T=0\ \Rightarrow\ Q_T=W_T.$$$$T_{{\mathrm{ad}},f}=400(2)^{-.400}\approx303.143\,\mathrm K.$$$$Q_{\mathrm{ad}}=0,\quad W_{\mathrm{ad}}=\tfrac52R(400-303.143)\approx2013.166\,\mathrm J.$$

(b) Interpretation and further calculation

For $V>V_i$,$$p_T=p_i(V_i/V),\quad p_{\mathrm{ad}}=p_i(V_i/V)^\gamma<p_T\ \Rightarrow\ W_{\mathrm{ad}}<W_T.$$$$V_{f,T}=V_{f,\mathrm{ad}}=2V_i,\quad T_{f,T}\ne T_{f,\mathrm{ad}}.$$The final temperatures and pressures differ. Heat supplied along the isotherm maintains its higher pressure during expansion.

Physical check

The comparison holds fixed initial state and final volume, not fixed final temperature.