A container $\mathrm{AB}$ in the shape of a rectangular parallelopiped of length $5 \mathrm{~m}$ is divided internally by a movable partition $\mathrm{P}$ as shown in the figure. The left compartment is filled with a given mass of an ideal gas of molar mass 32 while the right compartment is filled with an equal mass of another ideal gas of molar mass 18 at same temperature. What will be the distance of P from the left wall A when equilibrium is established?

• A

  $2.5 \mathrm{~m}$
• Correct answer
  B

  $1.8 \mathrm{~m}$
• C

  $3.2 \mathrm{~m}$
• D

  $2.1 \mathrm{~m}$

Solution

Hint:
At equilibrium $\quad P_{1}=P_{2}, T_{1}=T_{2}$ (assuming wall to be conducting) $\frac{P_{1} V_{1}}{\mu_{1}}=\frac{P_{2} V_{2}}{\mu_{2}}=R T$ $\frac{(5-x) 32}{m}=\frac{x \times 18}{m}$ So, $x=3.2 \mathrm{~m}$ Hence the distance of the piston from $\mathrm{A}=5-3.2=1.8 \mathrm{~m}$

Hence, the correct answer is option: 'B'.