Question

Consider a box with two gases separated by an impermeable membrane. The membrane can move back and forth, but the gases cannot penetrate the membrane. The left side is filled with gas A and the right side is filled with gas B. We will assume that equipartition applies to both gases, but gas A has an excluded volume due to large molecules so its entropy has a different formula.
SA=NAkln(VA+ bNA)+f(UA,NA)
SB=NBkln(VB)+f(UB,NB)

Required:
If NA= 1 moles, NB = 2 moles, the total volume of the box is 1 m3, and b= 4 × 10-4 m3/mole, then find the equilibrium value of VA by maximizing the total entropy.

Answer 1

Answer:

The answer is "$0.3336\ m^3$"

Explanation:

Using the Promideal gas law:

$P_A=P_B\\\\P_A(V_A-\eta_A b)= \eta_A RT......(1)\\\\P_B V_B=\eta_B \bar{R}T........(2)\\\\From (1) \zeta (2)$  

$\frac{\eta_A}{V_A-\eta_A b}=\frac{\eta B}{V B}\\\\ \frac{V A- \eta_A b}{V B}=\frac{\eta A}{\eta B }\\\\ \frac{V A-b}{V B}=\frac{1}{2}\\\\V A+V B=1\\\\V B =1- V A\\\\\frac{V A-b}{1-V A}=\frac{1}{2}\\\\2V A-2b=1-V A\\\\3 V A=1+2b\\\\V A=\frac{1+2b}{3}$

      $=\frac{1+2(4\times 10^{-4})}{3}\\\\=0.3336\ m^3$