- #1

EE18

- 112

- 13

$$\mu_j = RT \left[\phi_j(T) + \ln P + \ln x_j \right].$$

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?