Problem 11.7*
It follows from equation (11.34) of Problem 11.6 that
$\Omega = -kT \log \Xi = U - TS - \mu N$.
Show that $[-kT \log \Xi (T, \mu, V)]$ is an extensive variable, so that if the
intensive variables $T$ and $\mu$ are kept constant, $(-kT \log \Xi)$ is proportional
to the volume $V$. Hence show that
$(\partial \log \Xi/\partial V)_{T, \mu} = (\log \Xi)/V$.
(11.35)
Show that equation (11.29) for the pressure $p$ can be rewritten in the form
$pV = kT \log \Xi = -\Omega$.
(11.32)
