Q. Inside a cylinder closed at both ends is a movable piston. On one side of the piston is a mass $m$ of a gas, and on the other side a mass 2 $m$ of the same gas. What fraction of the volume of the cylinder will be occupied by the larger mass of the gas when the piston is in equilibrium? The temperature is the same throughout.
A) $\frac{2}{3}$
B) $\frac{1}{3}$
C) $\frac{1}{2}$
D) $\frac{1}{4}$
Explanation
(a) When the piston is in equilibrium, the pressure is same on both the sides of the piston. It is given that temperature and weight of gas on the two sides of piston not change. From ideal gas equation, $pV = n\ RT,\ $we have $V \propto$ mass of the gas.
So, $\frac{V_{1}}{V_{2}} = \frac{m_{1}}{m_{2}}\ or\ \frac{V_{1}}{V_{2}} + 1 = \frac{m_{1}}{m_{2}} + 1$
Or $\frac{V_{1} + V_{2}}{V_{2}}\ = \frac{m_{1} + m_{2}}{m_{2}}$
Or $\frac{V_{2}}{V_{1} + V_{2}}$ = $\frac{m_{2}}{m_{1} + m_{2}} = \ \frac{2m}{m + 2m} = \frac{2}{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.