# How to write equation of state in terms of partition function?

While studying quantum gases (fermions, bosons), equation of state written were $$PV = k_B T Z_{gr}$$, where $$Z_{gr}$$ is the partition function of grand canonical ensemble. $$P$$ and $$V$$ are pressure and volume, respectively. How can we derive this equation of state? Is it same equation that we use for ideal gases ($$PV=NRT$$) or is it different?

• Did you find an expression for $Z_{gr}$? Jan 12, 2019 at 10:19

When dealing with quantum gases you have to apply a correction to your result from the ideal gas!

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

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

When you got the expression of the grand canonical partition function, which depends on whether you are studying bosons or fermions, you can use the link with the grand potential $$\Phi = -k_BT \ln (Z_{gr})$$.

From this you can use the relation $$P = -\frac{\partial \Phi}{\partial V},$$ which will give you the pressure in function of the partition function. From here it should be little calculation to get the form $$PV = k_BTZ_{gr}.$$

• Are you sure it's $F$? I think it's the grand canonical potential. Jan 12, 2019 at 11:52
• Yes,I also found a source where that is the grand canonical potential. But it doesn't really matter because it differs by a factor which does not depend on $V$. So when taking the partial derivative to $V$ it vanishes. I could be wrong but in my class it was 'derived' like that. Jan 12, 2019 at 11:56
• It's true that it generates the same formula, but I think that saying "it doesn't matter" is a bit bold. It can lead to confusion, and other magnitudes might differ. I'm pretty sure that $\Phi=-k_BT\ln(\mathcal{Z}_{gc})$. See Callen, for example. Jan 12, 2019 at 12:22
• I've edited it. I wonder, does the same relation $P = - \frac{\partial \Phi}{\partial V}$ hold? I'm starting to doubt now. Jan 12, 2019 at 13:25
• $\Phi=U-TS-\mu N = F - \mu N$, so yes, it does. Jan 12, 2019 at 13:33

All your questions are related to the choice of the independent variables. The shape of the functions might vary if you work with the canonical ensemble $$(T,V,N)$$ or the grand-canonical ensemble $$(T,V,\mu)$$. However, in the thermodynamic limit, all ensembles are equivalent and you should be able to go from one to the other through a change of variables. However, I get to a different expression for the equation of state, are you sure about its validity?

How can we derive this equation of state?

Say you have a very general system in which the canonical partition function in Boltzman's approximation (a.k.a. classical limit) reads:

$$Z(T,N,V)=\frac{(V \varphi(T))^N}{N!}$$

Where $$\varphi$$ depends on the system in question, for a three dimensional ideal gas, $$\varphi(T)=(\frac{2\pi m k_B T}{h^2})$$.

The grand-canonical partition function will then be:

$$Z_{gr}(T,\mu,V)=\sum_N e^{\frac{\mu N}{k_B T}}Z(T,N,V)=e^{e^{\frac{\mu }{k_B T}}V \varphi(T)}$$.

The associated thermodynamic potential is:

$$J(T,\mu,V)=k_BT\log(Z_{gr})=k_B T e^{\frac{\mu }{k_B T}}V \varphi(T)$$.

Therefore, pressure is:

$$P(T,\mu,V)=(\frac{\partial J}{\partial V})_{\mu,T}= k_B T e^{\frac{\mu }{k_B T}} \varphi(T)$$

Now the only thing is to get $$\varphi(T)$$. From the canonical partition function of one particle: $$\varphi(T)=\frac{Z_1(T,V)}{V}$$, from the grand-canonical partition function: $$\varphi(T)=\frac{\log(Z_{gr}(T,V,\mu))}{V }e^{-\frac{\mu }{k_B T}}$$ or from the grand-canonical potential: $$\varphi(T)=\frac{J(T,V,\mu)}{k_B T V }e^{-\frac{\mu }{k_B T}}$$. The one that approaches the most to your expression is:

$$P(T,\mu,V) V=k_B T \log (Z_{gr})$$.

One important thing to notice is that all possible ways of getting $$\varphi$$ lead to the same result (through a different functional expression).

Is it same equation that we use for ideal gases $$(PV=NRT)$$ or is it different?

Is the same! In order to check it, you have to go from $$P(T,\mu,V)$$ to $$P(T,N,V)$$.

$$N(T,\mu,V)=(\frac{\partial J(T,\mu,V)}{\partial \mu})_{_{V,T}}=\frac{J(T,\mu,V)}{k_B T}$$.

Since $$P(T,\mu,V)=\frac{J(T,\mu,V)}{V}$$ in one of its forms, combining the last two equations we get:

$$P(T,N,V) =\frac{N k_B T}{V}$$