First slide

Thermodynamic processes

Question

Two identical containers A and B with frictionless pistons contain the same ideal gas at the same temperature and the same volume. 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 isothermal to the same final volume 2v. The change in the pressure in A and B are found to be $Δ P$ and $1.5 Δ P$ respectively. Then

Moderate

A

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

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

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

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

Solution

In isothermal process $P \propto \frac{1}{V}$
In chamber A : $Δ P = P_{i} - P_{f} = \frac{μ_{A} R T}{V} - \frac{μ_{A} R T}{2 V} = \frac{μ_{A} R T}{2 V} - - (1)$
In chamber B : $1.5 Δ P = P_{i} - P_{f} = \frac{μ_{B} R T}{V} - \frac{μ_{B} R T}{2 V} = \frac{μ_{B} R T}{2 V} - - - (2)$
divide 1 by 2 we get
From equal A and B $\frac{μ_{A}}{μ_{B}} = \frac{1}{1.5} = \frac{2}{3}$
$\frac{m_{A / M}}{m_{B / M}} = \frac{2}{3} \to 3 m_{A} = 2 m_{B}$

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