# How to see this form for the chemical potential of an ideal gas?

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• EE18
In summary, Callen states that the chemical potential with respect to the ##j##th component of an ideal gas can be written as $$\mu_j = RT \left[\phi_j(T) + \ln P + \ln x_j \right].$$To calculate ##\mu_j##, we need to use the free energy and differentiate the second term on the right side.
EE18
In Chapter 13.2 of his text, Callen states that the chemical potential with respect to the ##j##th component of an ideal gas can be written as
$$\mu_j = RT \left[\phi_j(T) + \ln P + \ln x_j \right].$$
He states this outright and doesn't prove it, and I am trying to do so now. Based on what has been developed in the text thus far, I am trying to do it by using the free energy.

We have
$$\mu_j \equiv \left( \frac{\partial F}{\partial N_j} \right)_{T,V,N_i; \, i \neq j} \stackrel{(1)}{=} \left( \frac{\partial F_j}{\partial N_j} \right)_{T,V,N_i; \, i \neq j} = \left( \frac{\partial U_j}{\partial N_j} \right)_{T,V,N_i; \, i \neq j} + T\left( \frac{\partial S_j}{\partial N_j} \right)_{T,V,N_i; \, i \neq j}.$$
where (1) follows from the additivity of the free energy/Helmholtz potential for an ideal gas. Thus it remains to evaluate the unknown derivatives.

I try to do this below but get lost. Callen shows that
$$U_j = N_ju_{j0} + N_j\int_{T_0}^T c_{vj} \, dT'$$
and
$$S_j = N_js_{j0} + N_j\int_{T_0}^T \frac{c_{vj}}{T'} \, dT' + N_j R \ln \left(\frac{V}{V_0}\right)$$.
Thus, differentiating even just the first one I get:
$$\left( \frac{\partial U_j}{\partial N_j} \right)_{T,V,N_i; \, i \neq j} = u_{j0} +\int_{T_0}^T c_{vj} \, dT' +N_j\int_{T_0}^T \left( \frac{\partial c_{vj}}{\partial N_j} \right)_{T',V,N_i; \, i \neq j}\,dT'$$
$$= u_{j0} +\int_{T_0}^T c_{vj} \, dT' +N_j\int_{T_0}^T \left( \frac{\partial \left(\frac{T'}{N}\left( \frac{\partial S_j}{\partial T'} \right)_{V,N_j} \right)}{\partial N_j} \right)_{T',V,N_i; \, i \neq j} \,dT'$$
$$= u_{j0} +\int_{T_0}^T c_{vj} \, dT' - \frac{N_j}{N^2}\int_{T_0}^T T' \left(\frac{\left( \frac{\partial S_j}{\partial T} \right)_{V,N_j}}{\partial N_j} \right)_{T,V,N_i; \, i \neq j}$$
but I can't go any further, and surely the derivative of the entropy will get even uglier. Can anyone provide some help as to how to proceed?

EE18 said:
In Chapter 13.2 of his text, Callen states that the chemical potential with respect to the ##j##th component of an ideal gas can be written as
$$\mu_j = RT \left[\phi_j(T) + \ln P + \ln x_j \right].$$
Not sure why it's not allowing me to edit the OP, but here ##\phi_j## is some function of temperature ##T##, ##P## is the pressure of the overall system, and ##x_j \equiv N_j/N## is the mole fraction of component ##j##.

EE18 said:
We have
$$\mu_j \equiv \left( \frac{\partial F}{\partial N_j} \right)_{T,V,N_i; \, i \neq j} \stackrel{(1)}{=} \left( \frac{\partial F_j}{\partial N_j} \right)_{T,V,N_i; \, i \neq j} = \left( \frac{\partial U_j}{\partial N_j} \right)_{T,V,N_i; \, i \neq j} + T\left( \frac{\partial S_j}{\partial N_j} \right)_{T,V,N_i; \, i \neq j}.$$
The ##+## for the last term on the right should be ##-##.

EE18 said:
I try to do this below but get lost. Callen shows that
$$U_j = N_ju_{j0} + N_j\int_{T_0}^T c_{vj} \, dT'$$
Note that since ##c_{vj}## is a function of ##T## alone, the integral is just a function of ##T## alone. So, we may write ##U_j = N_j f(T)## for some function ##f(T)## (which includes the ##u_{j0}##).

##U_j## has the form we expect for an ideal gas.

Then, ##\large \left( \frac{\partial U_j}{\partial N_j} \right)_{T,V,N_i; \, i \neq j}## ## = f(T)##.

EE18 said:
$$S_j = N_js_{j0} + N_j\int_{T_0}^T \frac{c_{vj}}{T'} \, dT' + N_j R \ln \left(\frac{V}{V_0}\right)$$
The integral in the second term on the right is again just some function of ##T##. So, the first two terms on the right side may be written as ##N_j g(T)## for some function ##g(T)##.

In the last term, the logarithm should be $$\ln \left(\frac{VN_0}{V_0N_j}\right)$$ See post #6 in this thread.

So, in calculating ##\mu_j##, we need $$\frac{\partial}{\partial N_j} \left[N_j\ln \left(\frac{VN_0}{V_0N_j}\right) \right] = \ln \left(\frac{VN_0}{V_0N_j}\right) - 1$$ Replace ##V## by ##\large \frac{NRT}{P}##, where ##N## is the total number of moles of all components and ##P## is the total pressure.
$$\frac{\partial}{\partial N_j} \left[N_j\ln \left(\frac{VN_0}{V_0N_j}\right) \right] = \ln \left(\frac{NRT}{P} \frac{N_0}{V_0N_j}\right) - 1 = -\ln P - \ln x_j +\ln \left(\frac{N_0RT}{V_0}\right) - 1$$ where ##x_j = \large \frac{N_j}{N}##.

Putting it all together should yield the result for ##\mu_j##.

TSny said:
Note that since ##c_{vj}## is a function of ##T## alone, the integral is just a function of ##T## alone.
How come we can say this? I see no a priori reason that it can't be a function of ##v## too, unless I am forgetting something crucial? Is the argument here that since ##c_{vj} = \left(\frac{\partial u}{\partial T} \right)_v## and since ##u## depends only on ##T## by definition of a general ideal gas, so too must ##c_{vj}##?

EE18 said:
How come we can say this? I see no a priori reason that it can't be a function of ##v## too, unless I am forgetting something crucial? Is the argument here that since ##c_{vj} = \left(\frac{\partial u}{\partial T} \right)_v## and since ##u## depends only on ##T## by definition of a general ideal gas, so too must ##c_{vj}##?
Yes. I think that's the right argument.

EE18
EE18 said:
Also, how did you arrive at this? When I do it out I get
$$= R \ln \left(\frac{V}{V_0}\frac{N_0}{N}\right) +N_jR\frac{V_0}{V}\frac{N_j}{N_0}(-N_j^2)$$
The mistake is in the second term on the right. In taking the partial of ##\ln\left( \frac{VN_0}{V_0N_j}\right)## with respect to ##N_j##, first write ##\ln\left( \frac{VN_0}{V_0N_j}\right)## as ## \ln\left( \frac{VN_0}{V_0}\right) -\ln N_j##.

TSny said:
The mistake is in the second term on the right. In taking the partial of ##\ln\left( \frac{VN_0}{V_0N_j}\right)## with respect to ##N_j##, first write ##\ln\left( \frac{VN_0}{V_0N_j}\right)## as ## \ln\left( \frac{VN_0}{V_0}\right) -\ln N_j##.
Thanks! I got sloppy with the chain rule.

If you have the time, I was reflecting on this question which I asked way back when (##dQ = dH## comes up again in Chapter 13) and have still not been able to convince myself why I can just treat the vessel "like a black box". If you get the chance, I would greatly appreciate your thoughts!

## What is the chemical potential of an ideal gas?

The chemical potential of an ideal gas is a measure of the change in the free energy of the system when an additional particle is introduced, keeping the temperature and volume constant. It is denoted by μ and can be expressed as a function of temperature, pressure, and volume.

## How is the chemical potential of an ideal gas derived?

The chemical potential of an ideal gas can be derived from the Gibbs free energy. For an ideal gas, the chemical potential μ is given by the equation: μ = μ⁰ + RT ln(P/P⁰), where μ⁰ is the standard chemical potential, R is the universal gas constant, T is the temperature, P is the pressure, and P⁰ is the standard pressure.

## What is the standard chemical potential (μ⁰) in the context of an ideal gas?

The standard chemical potential (μ⁰) is the chemical potential of the gas at a standard state, typically 1 bar of pressure and a specified temperature. It serves as a reference point for calculating the chemical potential under different conditions.

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

The chemical potential of an ideal gas decreases with increasing temperature. This is because the term RT ln(P/P⁰) becomes more negative as temperature (T) increases, assuming the pressure (P) remains constant. Hence, higher temperatures result in lower chemical potentials for the gas.

## Can the chemical potential of an ideal gas be negative?

Yes, the chemical potential of an ideal gas can be negative. This typically occurs when the gas is at a low pressure relative to the standard pressure. Since the term RT ln(P/P⁰) can be negative if P is less than P⁰, the overall chemical potential μ can also be negative.

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