Relationship between pressure & chem potential in ideal gas

[thecommexokid]

Homework Statement

(Excerpted from a longer, multipart problem but essentially)

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

Homework 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.

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.

 

[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...

 

[thecommexokid]

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...

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).

 

[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...

(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.

(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...

 

[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,
$$
\frac{\partial p}{\partial\mu}\bigg)_{T} = V\frac{1}{k_{\rm B}T}p = N.
$$

 

[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.