# Relationship between pressure & chem potential in ideal gas

• thecommexokid
In summary, the homework statement is that the ideal gas law states that the pressure p of a gas is related to the molar chemical potential k_{\rm B}T through the equation pV = Nk_{\rm B}T.
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.

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...

Goddar said:
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).

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

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.

## What is the relationship between pressure and chemical potential in an ideal gas?

In an ideal gas, the chemical potential is directly proportional to the pressure. This means that as the pressure increases, the chemical potential also increases and vice versa. This relationship is known as the ideal gas law and is expressed as μ = RT ln(P), where μ is the chemical potential, R is the gas constant, T is the temperature, and P is the pressure.

## How does increasing pressure affect the chemical potential of an ideal gas?

As the pressure of an ideal gas increases, the molecules become more tightly packed and are forced closer together. This results in a higher chance of collisions between molecules, leading to an increase in the chemical potential. In other words, the higher the pressure, the more energetic the gas molecules and the greater their tendency to escape.

## What happens to the chemical potential of an ideal gas when the pressure is decreased?

When the pressure of an ideal gas is decreased, the molecules have more space to move around. This leads to a decrease in the number of collisions between molecules, resulting in a lower chemical potential. In other words, the lower the pressure, the less energetic the gas molecules and the less likely they are to escape.

## How does the temperature affect the relationship between pressure and chemical potential in an ideal gas?

The temperature of an ideal gas plays a significant role in the relationship between pressure and chemical potential. As the temperature increases, the average kinetic energy of the gas molecules also increases, leading to a higher chemical potential. This means that at higher temperatures, the same pressure can result in a higher chemical potential compared to lower temperatures.

## Is the relationship between pressure and chemical potential in an ideal gas always linear?

In an ideal gas, the relationship between pressure and chemical potential is linear only when the temperature is constant. If the temperature increases, the relationship becomes logarithmic, and as the temperature approaches absolute zero, the relationship becomes exponential. This is because at lower temperatures, the molecules have less kinetic energy and are less likely to escape, resulting in a non-linear relationship between pressure and chemical potential.

Replies
1
Views
866
Replies
2
Views
1K
Replies
7
Views
2K
• Thermodynamics
Replies
6
Views
1K
• Classical Physics
Replies
6
Views
965
• Thermodynamics
Replies
19
Views
1K
Replies
1
Views
1K
• Thermodynamics
Replies
3
Views
789