Question

Two spheres - 1,2 with flexible walls are connected by a tube of negligible volume. Their initial and final conditions are given by $P_1,V_1,n_1$ and $P_2,V_2,n_2$ respectively. The radius of these spheres depends on the temperature as $4\pi r^{2}dr=kdT$. Initially, sphere 1 was at $100K$ and sphere 2 was at $200K$. As the temperature of sphere 1 is increased to $300K$, find the final mole ratio $\left(\frac{n_1^{\prime }}{n_2^{\prime }}\right)$.

Answer 1

Since we know, $n_{total}=\text{constant}$
$n_1+n_2=n_1^{\prime }+n_2^{\prime }$

$4\pi r^{2}dr=kdT$
Integrating above equation, $\int 4\pi r^{2}dr=\frac{4}{3}\pi r^3{|}_{r_1}^{r_2}=kT{|}_{T_1}^{T_2}$
Volume of the sphere, $V=\frac{4}{3}\pi r^3$
so $dV=4\pi r^{2}dr=kdT$.
Hence we get, $\frac{dV}{dT}=\text{constant}$

Using ideal gas equation, $PV=nRT$
$\frac{n_1}{n_2}=\frac{P_1{V}_1}{T_1}\times \frac{T_2}{P_2{V}_2}=\frac{P_1}{P_2}\because \frac{V}{T}=\text{constant}$

When the stop cork is open, pressure equilises i.e. $P_1^{\prime }=P_2^{\prime }$
Hence $\frac{n_1^{\prime }}{n_2^{\prime }}$ becomes 1.