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

[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]

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 $j$.

 

[TSny]

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 $-$.

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

$$S_j = N_js_{j0} + N_j\int_{T_0}^T \frac{c_{vj}}{T'} \, dT' + N_j R \ln \left(\frac{V}{V_0}\right)$$

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 $\mu_j$.

 

[EE18]

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 $c_{vj}$?

 

[TSny]

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.

 

[TSny]

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 $\ln\left( \frac{VN_0}{V_0}\right) -\ln N_j$.

 

[EE18]

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!