Does quantum gases obey ideal gas equation $ PV= nRT$?

Question

At extremely low temperature, does an ideal gas of bosons or fermions obey the ideal gas equation, $PV= nRT$?

Answer 1

Generally, bosons and fermions are described by Bose-Einstein and Fermi-Dirac statistics respectively. I'm not going to do the whole derivation but to find quantum corrections to the ideal gas law, you can calculate the grand canonical ensemble $\Xi$ which is related to the pressure of the gas: $$ \frac{PV}{k_bT} = \log \Xi = \pm \frac{V}{\lambda_T^3}\sum_{l=1}^{+\infty}(\pm1)^l \frac{z^l}{l^{5/2}}.$$ The upper sign, or the ($+$) stands for the bosons and the lower sign ($-$) for fermions.

You can check for yourself if this equation returns you the ideal gas law when you only take the lowest order (so l=1). In the equation $z= e^{\beta \mu}$ and $\lambda_T$ is the thermal wavelength.

EDIT: To check the lowest order, you will also need the equation $$ n \lambda_T^3 =\pm \frac{V}{\lambda_T^3}\sum_{l=1}^{+\infty}(\pm1)^l \frac{z^l}{l^{3/2}} $$ which can be found using the identity $$ N = \frac{1}{\beta}\frac{\partial \log \Xi}{\partial \mu}. $$ It is also interesting to calculate the next order term. If you work that out you will find indeed that there is a higher pressure for fermions at low temperature due to the Pauli exclusion principle.

Answer 2

To zeroth order: yes!

Let us take a non-relativistic ideal gas, in which each particle has an energy of

$$ \epsilon=\frac{p^2}{2m}. $$

The ideal gas law follows from

$$ N(T,V,\mu)=\frac{1}{\beta}\frac{\partial\ln Y}{\partial\mu}, $$

so we need an expression for $\ln Y$. For bosons, we can write:

$\begin{align} Y(T,V,\mu)&=\sum_r\exp\left(-\beta\left(E_r\left(V,N_r\right)-\mu N_r\right)\right)\\ &=\prod_i\sum_{n_{p_i}}\exp(-\beta(\epsilon_{p_i}-\mu)n_{p_i})\\ &=\prod_p\frac{1}{1-\exp(-\beta(\epsilon_p-\mu))}\\ \Rightarrow\ln Y&=-\sum_p\ln(1-\exp(-\beta(\epsilon_p-\mu))) \end{align}$

A similar calculation yields $$\ln Y=2\sum_p\ln(1+\exp(-\beta(\epsilon_p-\mu)))$$ for fermions.

To find the ideal gas law and its quantum mechanical corrections we need to do a series expansion of those expressions:

$$ \ln Y = (2s+1)\sum_p\left(\exp(-\beta(\epsilon_p-\mu))\pm\frac{1}{2}\exp(-2\beta(\epsilon_p-\mu))+...\right) $$

with $s=0$ for bosons and $s=1/2$ for fermions.

The possible momentum states are very dense, therefore we can treat the sum as an integral, which yields:

$\begin{align} \ln Y &= (2s+1)\frac{V}{(2\pi\hbar)^3}\int\!\mathrm{d}^3p\,\exp(\beta\mu)\left(\exp\left(-\frac{p^2}{2mk_BT}\right)\pm\frac{1}{2}\exp(\beta\mu)\exp\left(-\frac{p^2}{mk_BT}\right)+...\right)\\ &=(2s+1)\frac{V}{(2\pi\hbar)^3}\exp(\beta\mu)\frac{1}{2}\sqrt{\pi}\sqrt{mk_BT}\left(2\sqrt{2}+\exp{\beta\mu}+...\right)\\ &:=(2s+1)\frac{V}{\lambda^3}\left(\exp(\beta\mu)\pm2^{-5/2}\exp(2\beta\mu)+...\right) \end{align}$

in the last step I just summed together some constants, to get the thermal wavelength $\lambda$, as mentioned in another answer.

Now we use $N(T,V,\mu)=\frac{1}{\beta}\frac{\partial\ln Y}{\partial\mu}$. In the zeroth-order approximation we find:

$$ N(T,V,\mu)=\frac{1}{\beta}\frac{\partial\ln Y}{\partial\mu}=\ln Y = \frac{PV}{k_BT}, $$

the ideal gas law. For higher orders we find certain corrections, such that

$$ \ln Y = N\mp\frac{2s+1}{2^{5/2}}\frac{V}{\lambda^3}\exp(2\beta\mu)+... $$