# Relationship between pressure & chem potential in ideal gas

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1. Feb 2, 2015

### thecommexokid

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

(Excerpted from a longer, multipart problem but essentially)

Show that for an ideal gas,
$$\frac{\partial p}{\partial T}\bigg)_\mu = \frac{S}{V}.$$

2. Relevant equations

• The ideal gas law, of course
$$pV = Nk_{\rm B}T$$
• Pressure, temperature, and chemical potential are related in an ideal gas via
$$\mu-\mu_0=k_{\rm B}T\ln\frac{p}{p_0}$$
• The Gibbs free energy
$$G = E - TS + pV = \mu N$$
• Hopefully not the Sackur–Tetrode equation.

3. The attempt at a solution

Thoughts regarding left-hand side:

So I rearranged the $\mu$ equation for $p$:
$$p(T,\mu) = p_0e^{(\mu-\mu_0)/k_{\rm B}T}.$$
Computing the partial derivative,
$$\frac{\partial p}{\partial T}\bigg)_\mu = -\frac{\mu-\mu_0}{k_{\rm B}T^2}p.$$
By the ideal gas law $p/k_{\rm B}T = N/V$, so
$$\frac{\partial p}{\partial T}\bigg)_\mu = \frac{(\mu_0-\mu)N}{VT}.$$

Thoughts regarding the right-hand side:

From the definition of the Gibbs free energy,
$$\frac{S}{V} = \frac{E + pV - \mu N}{VT}.$$

These two would be equal if
$$E + pV = \mu_0N$$
in an ideal gas. But I don't really understand what the definition of the "reference" pressure and chem. potential $p_0$ and $\mu_0$ are. So maybe this is true, but if it is I don't know why!

…Or perhaps there's an altogether different starting point or process that gives the result. I am by no means married to my approach here.

2. Feb 2, 2015

### Goddar

Hi. Chemical potential is the partial molar Gibbs free energy, so for a pure substance (ideal gas) it is equal to the molar Gibbs free energy and you have:
dμ = dG = –SdT + VdP

Now examine this for fixed μ...

3. Feb 2, 2015

### thecommexokid

I agree. I knew I was going to get myself in trouble for excerpting this out of a larger question.

The greater context of the problem is:

(a) Derive the formula $N{\rm d}\mu=-S{\rm d}T+V{\rm d}p$.

(b) Derive the general formulas $\dfrac{\partial p}{\partial T}\bigg)_{\mu}=\dfrac{S}{V}$ and $\dfrac{\partial p}{\partial\mu}\bigg)_{T}=\dfrac{N}{V}$.

(c) Find the pressure p of an ideal gas as a function of µ and T. Use this to verify the two equations from part (b) for the specific case of an ideal gas.​

So your response is very relevant to parts (a) and (b), but I am now working on part (c).

4. Feb 2, 2015

### Goddar

Ok. To get to (a) and (b), remember that: G = μN = U–TS + PV and also: dU = TdS–PdV + μdN
Play with the differential dG...
(Isn't that the relation you were trying to use in your first post? How did you obtain it then?)
Hint: use the relation in (a) at fixed temperature and the equation of state for an ideal gas...

5. Feb 3, 2015

### thecommexokid

Okay, for posterity's sake, here's what I ultimately came up with. I suspect I go around in circles a little and could streamline some parts, but it'll do.$$\require{cancel}$$(a) The Gibbs free energy is
$$G = E - TS + pV = \mu N,$$
Consider its total differential:
$${\rm d}G = {\rm d}E - T{\rm d} S - S{\rm d} T + p{\rm d} V + V{\rm d} p = \mu{\rm d} N + N{\rm d}\mu.$$
Solve for $N{\rm d}\mu$:
$$N{\rm d}\mu = {\rm d} E - T{\rm d}S - S{\rm d}T + p{\rm d} V + V{\rm d}p - \mu{\rm d}N.$$
Substituting for ${\rm d}E$,
$${\rm d} E = T{\rm d}S - p{\rm d} V + \mu{\rm d} N,$$
we find
$$N{\rm d}\mu = (\cancel{T{\rm d} S} - \cancel{p{\rm d} V} + \cancel{\mu{\rm d} N}) - \cancel{T{\rm d} S} - S{\rm d} T + \cancel{p{\rm d} V} + V{\rm d} p - \cancel{\mu{\rm d}N} = -S{\rm d} T + V{\rm d} P.$$

(b) Our result from part (a) above reads
$$N{\rm d}\mu = -S{\rm d}T + V{\rm d}p.$$
If we suppose $\mu$ to be held constant, this reads
$$0 = -S{\rm d} T + V{\rm d} p \qquad\Rightarrow\qquad S = V\frac{\partial p}{\partial T}\bigg)_{\mu}.$$
Or if we suppose instead that $T$ is held constant,
$$N{\rm d}\mu = 0 + V{\rm d} p \qquad\Rightarrow\qquad N = V\frac{\partial p}{\partial\mu}\bigg)_{T}.$$

(c) The energy of a monatomic ideal gas is
$$E = \frac32Nk_{\rm B}T.$$
The entropy of a monatomic ideal gas is
$$S = Nk_{\rm B}\bigg(\ln\frac{V}{Nh^3}(2\pi mk_{\rm B}T)^{\frac32}+\frac52\bigg).$$
The volume of an ideal gas is
$$V = \frac{Nk_{\rm B}T}{p}.$$
Combine all these expressions into the relation $E - TS + pV = \mu N$:
$$\frac32Nk_{\rm B}T - TNk_{\rm B}\bigg(\ln\frac{k_{\rm B}T}{h^3p}(2\pi mk_{\rm B}T)^{\frac32}+\frac52\bigg) + Nk_{\rm B} T = \mu N.$$
Divide through by $Nk_{\rm B}T$:
$$\cancel{\frac32} - \ln\frac{k_{\rm B}T}{h^3p}(2\pi mk_{\rm B} T)^{\frac32} - \cancel{\frac52} + \cancel{1} = \frac{\mu}{k_{\rm B}T}.$$
Solving for $p$,
$$p(\mu, T) = \frac{(2\pi m)^{\frac32}}{h^3}(k_{\rm B}T)^{\frac52}e^{\mu/k_{\rm B}T}.$$

The derivatives under consideration are
$$\frac{\partial p}{\partial T}\bigg)_{\mu} = \frac{(2\pi m)^{\frac32}}{h^3}\bigg(\frac52k_{\rm B}^{\frac52}T^{\frac32}\bigg)e^{\mu/k_{\rm B}T} + \frac{(2\pi m)^{\frac32}}{h^3}(k_{\rm B}T)^{\frac52}\bigg({-\frac{\mu}{k_{\rm B}T^2}}e^{\mu/k_{\rm B}T}\bigg)=\bigg(\frac52\frac1T - \frac{\mu}{k_{\rm B}T^2}\bigg)p$$
and
$$\frac{\partial p}{\partial\mu}\bigg)_{T} = \frac{(2\pi m)^{\frac32}}{h^3}(k_{\rm B}T)^{\frac52}\bigg(\frac{1}{k_{\rm B}T}e^{-\mu/k_{\rm B}T}\bigg) = \frac{1}{k_{\rm B}T}p.$$
Therefore
$$V \frac{\partial p}{\partial T}\bigg)_{\mu}= V\bigg(\frac52\frac1T - \frac{\mu}{k_{\rm B}T^2}\bigg)p = \frac52Nk_{\rm B}- \frac{\mu N}{T};$$
but we want to get the entropy back in there somehow, so
$$V \frac{\partial p}{\partial T}\bigg)_{\mu}= \frac52Nk_{\rm B}- \frac{E-TS+pV}{T} = \cancel{\frac52Nk_{\rm B}} - \cancel{\frac32Nk_{\rm B}} + S - \cancel{Nk_{\rm B}} = S.$$
And, more straightforwardly,
$$\frac{\partial p}{\partial\mu}\bigg)_{T} = V\frac{1}{k_{\rm B}T}p = N.$$

6. Feb 3, 2015

### Goddar

That's one way to go... i'm guessing there's a quicker way directly from the partition function to the derivatives of U or F (Helmholtz free energy), but it really comes down to the same calculations minus the detour so your answer works regardless.