The Adiabatic Expansion of an Ideal Gas — Question 8

Question 8

Variable heat capacity replaces the constant-gamma power law by an integral

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 has $C_V(T)=R[\tfrac32+T/(600\,\mathrm K)]$. It expands reversibly and adiabatically from 600 K to 300 K. The supplied caloric law is valid throughout that interval.

1. Derive an implicit relation between temperature and volume and calculate $V_f/V_i$.

2. Calculate work done and compare the volume ratio with one obtained by incorrectly freezing $\gamma$ at its initial value.

Question 8 - Solution

Variable heat capacity replaces the constant-gamma power law by an integral

(a) Method and calculation

$$C_V(T)\frac{dT}T=-R\frac{dV}V\ \Rightarrow\ \frac32\ln\frac{T_f}{T_i}+\frac{T_f-T_i}{600\,\mathrm K}=-\ln\frac{V_f}{V_i}.$$$$\ln(V_f/V_i)=\tfrac32\ln2+\tfrac12\ \Rightarrow\ V_f/V_i=2^{3/2}e^{1/2}\approx4.66329.$$

(b) Interpretation and further calculation

$$W=U_i-U_f=R\left[\frac32(600-300)+\frac{600^2-300^2}{1200}\right]\mathrm K=5611.95\,\mathrm J.$$$$\gamma_i=1+1/2.50=1.40,\quad (V_f/V_i)_{\gamma_i}=2^{1/.400}\approx5.65685.$$The initial exponent overpredicts the expansion ratio because $C_V$ decreases during cooling and $\gamma$ is not constant.

Physical check

Integrating $C_V/T$ determines the reversible adiabat; integrating $C_V$ determines its work through the energy change.