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 and$g=9.8m/s^2$. The mass of the piston is

Answer 1

Let $l_1$ and $l_2$ be the lengths of the upper part and the lower part of the cylinder respectively. Clearly, $l_1=50cm$ and $l_2=40cm$. 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 = 300 K. 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, $p_{2}A=p_{1}A+mg\dots \dots \left(i\right)$

Using pV = nRT for the upper and the lower parts

$p_1{l}_{1}A=nRT\dots \dots \left(ii\right)$

and $p_2{l}_{2}A=nRT\dots \dots \left(iii\right)$

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

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

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

$=\frac{\left(0.1mol\right)(8.3J/mol-K)\left(300K\right)}{9.8m/s^2}[\frac{1}{0.4m}-\frac{1}{0.5m}]$

$=12.7kg$