Question

Two identical glass spheres filled with air are connected by a horizontal glass tube. The glass tube contains a pellet of mercury at its mid-point. Air in one sphere is at $0^o$C and the other is at ${20}^o$C. If both the vessels are heated through ${10}^o$C, then neglecting the expansions of the bulbs and the tube.

• A. The mercury pellet gets displaced towards the sphere at lower temperature
• B. The mercury pellet gets displaced towards the sphere at higher temperature
• C. The mercury pellet does not get displaced at all
• D. The temperature rise causes the pellet to expand without any displacement

Answer 1

Answer: (C)

The mercury pellet does not get displaced at all

Let $n_1$ and $n_2$ be the number of moles in the bulbs at $0^o$C and ${20}^o$C, respectively. Then,
$pV=273n_{1}R=293n_{2}R$
$\therefore \frac{n_1}{n_2}=\frac{293}{273}$ ........(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
$V_{1}p=283n_{1}R$ and $V_{2}p=303n_{2}R$
$\therefore \frac{V_1}{V_2}=\frac{283n_1}{303n_2}$ .........(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.