Question

Figure shows a vertical cylindrical vessel separated in two parts by a frictionless piston free to move along the length of the vessel. The length of the cylinder is 90 cm and the piston divides the cylinder in the ratio of 5:4. Each of the two parts of the vessel contains 0.1 mole of an ideal gas. The temperature of the gas is 300 K in each part. Calculate the mass of the piston

• A. 10.3 kg
• B. 5.7 kg
• C. 12.7 kg
• D. 15.5 kg

Answer 1

The correct option is C

12.7 kg

Let $l_1$ and $l_2$ be the lengths of the upper part and the lower part of the cylinder respectively. Clearly, $l_1=50cmand{l}_2=40cm$. Let the pressures in the upper and lower parts be $P_1$ and $P_2$ respectively. Let the area of cross section of the cylinder be A. The temperature in both parts is T=300K.

Consider the equilibrium of the piston. The forces acting on the piston are

(a) its weight mg

(b) $P_{1}A$ downward, by the upper part of the gas and

(c) $P_{2}A$ upward, by the lower part of the gas.

Thus,

$P_{2}A=P_{1}A+mg.$ .....(i)

Using PV = nRT for the upper and the lower parts

$P_1{l}_{1}A=nRT$ .....(ii)

and $P_2{l}_{2}A=nRT$ ....(iii)

Putting $P_{1}A$ and $P_{2}A$ from (ii) and (iii) into (i),

$\frac{nRT}{l_2}=\frac{nRT}{l_1}+mg$.

Thus,

$m=\frac{nRT}{g}[\frac{1}{l_2}-\frac{1}{l_1}]$

$=\left[\frac{\left(0.1mol\right)\times \left(8.3J{K}^{-1}mo{l}^{-1}\right)\times \left(300K\right)}{9.8m{s}^{-2}}\right][\frac{1}{0.4m}-\frac{1}{0.5m}]=12.7kg.$

Do you see the similarities with the previous problem, where we had the mercury column?