# Diatomic Gas, Grand Canonical Ensemble

1. Mar 22, 2014

### unscientific

1. The problem statement, all variables and given/known data

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 $r$.

2. Relevant equations

3. 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: Mar 22, 2014
2. Mar 23, 2014

### unscientific

bumpp

3. Mar 24, 2014

### unscientific

part (b) anyone?

4. Mar 25, 2014

### unscientific

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.

5. Apr 2, 2014

### unscientific

bump part (b) anyone?

6. Apr 6, 2014

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bump part (b) anyone?

7. Apr 8, 2014

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bumpp

8. Apr 12, 2014

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bumpp part (b) anyone?

9. Apr 14, 2014

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bumppp part (b) anyone?

10. Apr 18, 2014

### unscientific

bump part (b)??

11. Apr 21, 2014

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bumpp part (b)?

12. Apr 24, 2014

### unscientific

bump part (b)?

13. Apr 24, 2014

### unscientific

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 $\mu = \left(\frac{\partial F}{\partial N}\right)_{T,V}$