# How does the volume change affect an adiabatic process?

A gas consisting of rigid diatomic molecules was initially under standard conditions. Then the gas was compressed adiabatically, 5 times the original volume. Find the mean KE of rotating molecules in the final state.

We know from the law of equipartition of energy that for a gas at equilibrium at a certain temperature, the average KE per particle associated with each degree of freedom is $\frac{1}{2}kT$. Since for a diatomic gas, degree of freedom due to rotation only is 2, I can say that the mean KE for rotation of one molecule is $\frac{1}{2}(2)kT = kT$.

I have also thought that, since the process is adiabatic, temperature remains constant. So the rms velocities of the molecules should not be affected. Also, $v=\omega r$, so $v$ and $r$ being constant for the diatomic molecule, $\omega$ should not change either. So KE doesn’t change.

The problem is that I can’t seem to involve the volume change in any way.

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adiabatic isn't constant temperature unless the adiabatic index is 1, and I don't think it ever is. – Alan Rominger May 21 '13 at 17:04
ok. noted. but can you tell me whether the volume change will have any effect here? – Pallavi Roy May 21 '13 at 17:06
This might sound dismissive, but honestly, the first Google hit seems to solve your problem. Wikipedia gives $P V^{\gamma} = \operatorname{constant}$, and this is enough information to find the new $T$ in your case. It also gives a formula for $\gamma$, although it outright gives you the value for a diatomic gas, so there's hardly any work left for you to do. en.wikipedia.org/wiki/Adiabatic_process – Alan Rominger May 21 '13 at 17:47