Two identical glass spheres filled with air are connected by a...

Let $n_1$ and $n_2$ be the number of moles in the bulbs at $0^{\circ} \mathrm{C}$ and $20^{\circ} \mathrm{C}$, respectively. Then, $p V=273 n_1 R=293 n_2 R$ $\therefore \frac{n_1}{n_2}=\frac{293}{273}...(i)$ When the vessels are heated, let the volume of low temperature be $V_1$ and that of the other be $V_2$. Since, pressure are same, hence $V_1 p=283 n_1 R$ and $V_2 p=303 n_2 R$ $\therefore \frac{V_1}{V_2}=\frac{283 n_1}{303 n_2}...(ii)$ From Eqs. (i) and (ii), we get $\frac{V_1}{V_2}=\frac{283}{303} \times \frac{293}{273}=1$ Thus, the mercury pellet remains at the same position.