The Adiabatic Expansion of an Ideal Gas — Question 2

Question 2

A large reversible expansion can be checked by two work formulas

Use $pV=nRT$, $R=8.314$ J/(mol K), and $\Delta U=Q-W$ with $W$ positive for work done by the gas. A .500 mol monatomic ideal gas initially at 600 K expands reversibly and adiabatically from $V_i$ to $8V_i$. Take $C_V=3R/2$ and $\gamma=5/3$.

1. Find final temperature, pressure ratio and work done by the gas.

2. Evaluate work again from $W=(p_iV_i-p_fV_f)/(\gamma-1)$ and show why pressure falls by a greater factor than the inverse volume ratio.

Question 2 - Solution

A large reversible expansion can be checked by two work formulas

(a) Method and calculation

$$T_f=T_i(V_i/V_f)^{\gamma-1}=600(1/8)^{2/3}=150\,\mathrm K.$$$$p_f/p_i=(1/8)^{5/3}=1/32,\quad W=nC_V(T_i-T_f)=2805.975\,\mathrm J.$$

(b) Interpretation and further calculation

$$W=\int_{V_i}^{V_f}p_iV_i^\gamma V^{-\gamma}dV=\frac{p_iV_i-p_fV_f}{\gamma-1}.$$$$W=\frac{nR(600-150)}{2/3}=2805.975\,\mathrm J,\quad\frac{p_f}{p_i}=\frac{T_f/T_i}{V_f/V_i}=\frac{1/4}{8}=\frac1{32}.$$Expansion reduces both number density and thermal energy per molecule.

Physical check

The final energy remains positive; the extracted work is less than the initial thermal energy.