First slide

Different type of processes

Question

Two identical containers A and B with frictionless pistons contain the same ideal gas at the same temperature and the same volume V. The mass of the gas in A is $m_{A}$ and that in B is $m_{B}$ . The gas in each cylinder is now allowed to expand isothermally to the same final volume 2V.The changes in the pressure in A and.B are found to be P and 1.5 P respectively. Then

Moderate

A

$4 m_{A} = 9 m_{B}$
B

$2 m_{A} = 3 m_{B}$
C

$3 m_{A} = 2 m_{B}$
D

$9 m_{A} = 3 m_{B}$

Solution

Process is isothermal. Therefore, T = constant, $(P \propto \frac{1}{V})$ volume is increasing, therefore pressure will decreases.
In chamber A :
$∆ P = P_{i} - P_{f} = \frac{μ_{A} RT}{V} - \frac{μ_{A} RT}{2 V} = \frac{μ_{A} RT}{2 V} - - (i)$
In chamber B:
$1.5 ∆ P = P_{i} - P_{f} = \frac{μ_{B} RT}{V} - \frac{μ_{B} RT}{2 V} = \frac{μ_{B} RT}{2 V} (ii)$
from equations (i) and (ii) $\frac{μ_{A}}{μ_{B}} = \frac{1}{1.5} = \frac{2}{3}$
$\Rightarrow \frac{\frac{m_{A}}{M}}{\frac{m_{B}}{M}} = \frac{2}{3} \Rightarrow 3 m_{A} = 2 m_{B}$

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