The Adiabatic Expansion of an Ideal Gas — Question 3

Question 3

An adiabatic compression ratio sets a target temperature

Use $pV=nRT$, $R=8.314$ J/(mol K), and $\Delta U=Q-W$ with $W$ positive for work done by the gas. A .200 mol ideal gas with $\gamma=1.40$ and constant $C_V=5R/2$ starts at 300 K. It is compressed reversibly and adiabatically to one-fifth of its initial volume.

1. Find final temperature and work done on the gas.

2. Find the compression ratio $V_i/V_f$ needed to reach 600 K from the same initial state. Explain the distinction between work on and work by the gas.

Question 3 - Solution

An adiabatic compression ratio sets a target temperature

(a) Method and calculation

$$T_f=300(5)^{\gamma-1}=300(5)^{.400}\approx571.096\,\mathrm K.$$$$W_{\mathrm{on}}=\Delta U=nC_V(T_f-T_i)\approx(.200)(\tfrac52)(8.314)(271.096)=1126.947\,\mathrm J.$$

(b) Interpretation and further calculation

$$\frac{T_f}{T_i}=\left(\frac{V_i}{V_f}\right)^{.400}\ \Rightarrow\frac{V_i}{V_f}=\left(\frac{600}{300}\right)^{1/.400}=2^{5/2}\approx5.65685.$$$$W_{\mathrm{by}}=-W_{\mathrm{on}}<0.$$The temperature increase comes from mechanical work input; zero heat transfer does not imply zero energy transfer.

Physical check

The stronger compression needed for 600 K is consistent with the fivefold compression producing a slightly smaller temperature.