The Adiabatic Expansion of an Ideal Gas — Question 1

Question 1

The adiabatic power law follows from the first law and the equation of state

Use $pV=nRT$, $R=8.314$ J/(mol K), and $\Delta U=Q-W$ with $W$ positive for work done by the gas. A fixed amount of ideal gas expands reversibly with no heat transfer. The molar $C_V$ is constant and $C_p=C_V+R$. Let $\gamma=C_p/C_V$.

1. Derive $TV^{\gamma-1}=\text{constant}$ and $pV^\gamma=\text{constant}$ without assuming either relation.

2. Find the local slope $dp/dV$ and compare its magnitude with that of an isotherm through the same state. State which assumptions are needed to replace work by $p\,dV$.

Question 1 - Solution

The adiabatic power law follows from the first law and the equation of state

(a) Method and calculation

$$\delta Q=0:\quad nC_VdT=-p\,dV=-\frac{nRT}{V}dV.$$$$\frac{dT}T=-\frac R{C_V}\frac{dV}V=-(\gamma-1)\frac{dV}V\ \Rightarrow\ TV^{\gamma-1}=\text{constant}.$$$$T=pV/(nR)\ \Rightarrow\ pV^\gamma=\text{constant}.$$

(b) Interpretation and further calculation

$$\left(\frac{dp}{dV}\right)_{\mathrm{ad}}=-\gamma\frac pV,\quad\left(\frac{dp}{dV}\right)_T=-\frac pV.$$$$\frac{|dp/dV|_{\mathrm{ad}}}{|dp/dV|_T}=\gamma>1.$$The gas must pass through equilibrium states with negligible mechanical imbalance and no dissipative work. Adiabatic alone does not guarantee the reversible work formula.

Physical check

A reversible adiabatic expansion cools the gas because its positive work reduces internal energy.