$p _1=\frac{ n _1 RT }{ N _{ A }} \quad p _2=\frac{ n _2 RT }{ N _{ A }}$

$F =\left( n _1- n _2\right) k _{ B } T S=\Delta nk _{ B } T S(A)$

$V=\frac{\Delta n k_B T S}{\beta}$

Force balance $\Rightarrow$ Pressure $\times$ Area $=$ Total number of molecules $\times \beta v$

$\Delta k _{ B } TS =\ell n _1 S \beta v$

$\Rightarrow  n _1 \beta v \ell=\Delta k _{ B } T$

Total number of molecules $/ sec =\frac{\left( n _1 vdt \right) S }{ dt }$

$= n _1 vS =\frac{\Delta k _{ B } TvS }{\beta v \ell}$

$=\left(\frac{\Delta n }{\ell}\right)\left(\frac{ k _{ B } T }{\beta}\right) S$   ($C$)

As $\Delta n$ will decrease with time therefore rate of molecules getting transfer decreases with time.