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 $2$V. The changes in the pressure in A and B are found to be $\Delta$P and $1.5\Delta$P respectively. Then.

• A. $4m_A=9m_B$
• B. $2m_A=3m_B$
• C. $3m_A=2m_B$
• D. $9m_A=4m_B$

Answer 1

The correct option is C $3m_A=2m_B$

Since the process is isothermal, hence T will be constant.

From the gas laws we know that,

PV = nRT

P = nRT/V

In container A,

$dP=P_1-P_2$

$dP=n_{a}RT/V-n_{a}RT/2V$

$dP=n_{a}RT/2V$

$dP=m_{A}RT/2MV$ ..(1)

In container B,

$1.5dP=P_1-P_2$

$1.5dP=n_{b}RT/V-n_{b}RT/2V$

$1.5dP=n_{b}RT/2V$

$1.5dP=m_{B}RT/2MV$ .....(2)

dividing both the eqn we get,

$1.5dP/dP=\frac{m_{B}RT/2MV}{m_{A}RT/2MV}$

$m_B/m_A=1.5=3/2$

$3m_A=2m_B$