Daily Physicist

Today is the birthday of

• Émile Clapeyron (1799-1864).
• Polykarp Kusch (1911-1993).

Today’s Problem

There is a sudden adiabatic decrease in the pressure of an ideal gas such that \(P_i/P_f=\alpha\). How does the temperature of the gas change?

Answer

From a previous problem we have

\[\frac{V_f}{V_i}=\frac{\gamma -1}{\gamma}+\frac{P_i}{\gamma P_f}\]

For an ideal gas

\[\frac{V_f}{V_i}=\frac{nRT_f/P_j}{nRT_i/P_i}=\frac{T_f}{T_i}\frac{P_i}{P_f}=\frac{T_f}{T_i}\alpha\]

Substituting this into the above equation we get

\[\frac{T_f}{T_i}=\frac{\gamma -1}{\gamma\alpha}+\frac{1}{\gamma}\]

If \(\alpha =4\) the ratio of temperatures for a monatomic gas (\(\gamma =5/3\)) is \(T_f/T_i=7/10\), and for a diatomic gas is \(T_f/T_i=11/14\).