The ideal gas equation in higher dimensions Basically I wanted to know whether or not the ideal gas equation, $PV=NkT$ would hold in higher dimensions? If so, how would you go about proving this? 
I can't see any reason as to why it shouldn't hold, all that changes would be that the volume element would be raised to the $d$th power, $V^d$, where $d$ being the number of dimensions, right?
 A: I stumbled upon this and, because I had to deal with this recently, can share my solution:
Let's start with a microcanonical ensemble in $D$ dimensions.
$$
S = k \log \Omega
$$
with $S$ being the entropy, $\Omega$ the statistical weight, $k$ the Boltzmann constant.
$$
\Omega = \lim_{\Delta E \rightarrow 0} \frac{1}{\Delta E}
\int_{E \leq H \leq E+\Delta E} d\Gamma
$$
with Hamiltonian
$$
H = \sum_{n=1}^{N} \frac{\vec p_n^2}{2m}
$$
and
$$
d\Gamma = \frac{1}{N!} \prod_{n=1}^N \frac{d^Dp_n\,d^D\!q_n}{(2\pi\hbar)^D}
$$
results in
$$
\begin{align}
\Omega &= \frac{1}{N!\,(2\pi\hbar)^{DN}}
  \lim_{\Delta E \rightarrow 0} \frac{1}{\Delta E}
  \int_{E \leq H \leq E+\Delta E}
  \prod_{n=1}^N d^Dp_n\,d^D\!q_n \\
&= \frac{1}{N!\,(2\pi\hbar)^{DN}}
  \lim_{\Delta E \rightarrow 0} \frac{1}{\Delta E}
  \int_{E \leq H \leq E+\Delta E}
  \prod_{n=1}^N (d^Dp_n) \cdot V^N \\
&= \frac{1}{N!\,(2\pi\hbar)^{DN}}
  S_{DN}(\sqrt{2mE}) \cdot V^N \propto V^N\\
\end{align}
$$
I used the independence of $p$ and $q$ to split the integral and used that the resulting integral is the surface of a $DN$-dimensional hypersphere with radius $\sqrt{2mE}$: $S_{DN}(\sqrt{2mE})$. For this calculation it is sufficient to realize that $\Omega \propto V^N$ (see following).
$$
\begin{align}
&&dE &= TdS-PdV = 0 \\
\Leftrightarrow&&\frac{P}{T} &= \frac{dS}{dV}
  = \frac{\partial S}{\partial \Omega} \frac{d\Omega}{dV} \\
&&&=\frac{k}{\Omega} \frac{d\Omega}{dV} \\
&&&=\frac{kN}{V} \\
\Leftrightarrow&&PV&=NkT
\end{align}
$$
So the dimension does not matter.
