Question

Ilustrated is a completely isolated system comprised of two compartments A & B) containing two monovalent molecular species (1 & 2). The compartments are separated by an ultrathin, freely moving and thermally conducting membrane that is permeable to only species 1. NA and N denote the number of molecules of species 1 occupying, respecively, compartments A and B at time t and NA and NB denote the number of molecules of species 2 occupying, respectively, compartments Freely moving membrane permeable to only molecular species 1 Compartment A Compartment B A (1 and Flow of molecular species 1 Molecules

Answer 1

$S=S_A+S_B$

$S_A\equiv S(U_A,V_A,N_A)$ & $S_B\equiv S(U_B,V_B,N_B)$

$dS=dS_A+dS_B\equiv dS_{mix}$

The entropy of mixing per molecule at fixed V, T is .

As only molecule 1 can pass through the membrane, ds_mix has only contribution from N₁.

So,

$\frac{\partial S_A}{\partial N_A}=-k \, lnx_1^A\,\frac{\partial N_1^{A}}{\partial N_A}$     -----(1)

$dS_A=\frac{\partial S_A}{\partial U_A}dU_A+\frac{\partial S_A}{\partial V_A}dV_A+\frac{\partial S_A}{\partial N_A}dN_A$ -----(2)

Given that,

$\frac{\partial S_A}{\partial U_A}=\frac{1}{T_A}$ ; $\frac{\partial S_A}{\partial V_A}=\frac{P_A}{T_A}$ ; $C_1=\frac{N_1^A}{ V_A} \Rightarrow \frac{\partial V_A}{\partial N_1^A}=C_1$ ------------(3)

Using Eq. (3) and (1) in Eq. (2) one can have :

$\frac{\partial S_A}{\partial N_1^A}=-k\,lnx_1^A+\frac{1}{T_A}\frac{\partial U_A}{\partial N_1^A}+\frac{P_A}{T_AC_1}$