# Diatomic Gas, Grand Canonical Ensemble

## Homework Statement

Part (a): Using grand canonical distribution, show ideal gas law ##P = nkT## holds, where ##n = \frac{\overline{N}}{V}##.

Part (b): Find chemical potential of diatomic classical ideal gas in terms of ##P## and ##T##. The rotational levels are excited, but not the vibrational. Mass of each molecule is ##m##, separation is ##r##.

## The Attempt at a Solution

Part(a)

$$Z = \sum_{N=0}^{\infty} e^{\beta \mu N} Z_N = \sum_{N=0}^{\infty}\frac{1}{N!} \left( \frac{Ve^{\beta \mu}}{\lambda_{th}^3}\right)$$

Where ##Z_N = \frac{1}{N!} (\frac{V}{\lambda_{th}^3})^N## and ##\beta = \frac{1}{kT}##.

$$Z = exp \left( \frac{Ve^{\beta \mu}}{\lambda_{th}^3}\right)$$

Starting with Gibb's Entropy, where probability ##P_i = \frac{e^{-\beta(E_i - \mu N_i)}}{Z}##

$$S = -k \sum_i P_i ln(P_i)$$
$$S = -k \sum_i P_i \left[ -\beta (E_i - \mu N_i) - ln Z\right]$$
$$S = \frac{U}{T} - \frac{1}{T}\mu N + k ln Z$$

Grand Potential is defined as ##\Phi_G = -kT ln (Z)##.

$$\Phi_G = U - TS - \mu N$$
$$d\Phi_G = -SdT - pdV - Nd\mu$$

To prove the ideal gas law, we must find a relation between P, V and T. We do this using grand potential.

Starting:
$$P = -\left( \frac{\partial \Phi_G}{\partial V} \right)_{T,\mu}$$
$$N = -\left( \frac{\partial \Phi_G}{\partial \mu} \right)_{V,T}$$

Then by differentiating, I showed that ##PV = NkT##.

Part(b)

Attempt #1: Using grand partition function

We must first find the rotational partition function. Energies are given by ##E_a = \frac{\hbar ^2}{2I}j(j+1)##.

$$Z_{rot} = \sum (2j+1) e^{-\beta(E_j -\mu N)}$$

Converting sum into integral:

$$\int_0^{\infty} (2j+1) e^{-j(j+1)\frac{\theta_{rot}}{T}} dj \int_0^{\infty} e^{-\beta \mu N} dN$$
$$Z_{rot} = \frac{2I}{\hbar \mu} (kT)^2$$

Together:
$$Z = Z_{rot}Z_{trans} = \frac{2I}{\hbar \mu} (\frac{1}{\beta})^2 exp \left( \frac{Ve^{\beta \mu}}{\lambda_{th}^3}\right)$$

$$ln Z = ln(\frac{2I}{\hbar \mu}) - 2ln (\frac{1}{\beta}) + Z_1 e^{\beta \mu}$$

Gibbs Potential ##\Phi_G = -kt ln (Z) = -\frac{1}{\beta} ln (Z)##

$$\Phi_G = \frac{1}{\beta} ln \frac{2I}{\hbar \mu} - \frac{1}{\beta} ln(\frac{1}{\beta}) + \frac{1}{\beta} Z_1 e^{\beta \mu}$$

Using ##N = -(\frac{\partial \Phi_G}{\partial \mu})_{V,T}##:

$$N = -\frac{1}{\beta \mu} + Z_1 e^{\beta \mu}$$

Attempt #2: Using single-particle partition function

$$Z = Z_{rot}Z_{trans}$$

Translational one-particle Partition Function:
$$Z_{trans} = \int e^{-\beta E} g_{(\vec{k})} d^3\vec{k} = \frac{V}{\lambda_{th}^3}$$

Rotational one-particle Partition Function:
$$Z_{rot} = \sum (2j+1)e^{-j(j+1)\frac{\theta_{rot}}{T}}$$

where ##\theta_{rot} = \frac{\hbar^2}{2Ik}##

Converting sum into integral:

$$Z_{rot} = \frac{T}{\theta_{rot}}$$

Together, the total partition function is:

$$Z = Z_{rot}Z_{trans} = \frac{V}{\lambda_{th}^3}\frac{T}{\theta_{rot}}$$

To find ##\mu = -\left(\frac{\partial F}{\partial N}\right)_{T,V}##, how do I find ##F## from partition function ##Z##?

Last edited:

bumpp

part (b) anyone?

Part (b), Attempt #1: If I assume that the rotational partition function is simply ##Z_{rot} = \sum (2j+1)e^{-j(j+1)\frac{\theta_{rot}}{T}}##, the chemical expression comes out much nicer.

bump part (b) anyone?

bump part (b) anyone?

bumpp

bumpp part (b) anyone?

bumppp part (b) anyone?

bump part (b)??

bumpp part (b)?

bump part (b)?

I think I have figured it out. If total energy is ##U_{tot} = U_{trans} + U_{rot}##, the chemical potential can be accounted for either in ##U_{trans}## or ##U_{rot}## but not both.

Bearing this in mind, the partition function is ##Z = Z_{trans}Z_{rot}## where the ##e^{\beta \mu N}## only appears once in the ##Z_{trans}## and not in the ##Z_{rot}##. Hence the final equation should be less a ##-\frac{1}{\beta \mu}## term:

$$N = Z_1 e^{\beta \mu}$$

So this gives chemical potential as:

$$\mu = kT ln\left(\frac{N}{Z_1}\right)$$

Which is the same expression when derived using ##\mu = \left(\frac{\partial F}{\partial N}\right)_{T,V}##