Question

Two identical containers P and Q with frictionless piston contain the same ideal gas at the same temperature and the same volume. The mass of gas in P is $m_1$ and that in Q is $m_2$. The gas in each cylinder is now allowed to expand isothermally to the same final volume $1.5$V. The changes pressure in P and Q are found to be $\Delta$p and $2\Delta$p respectively then.

• A. $m_2=2m_1$
• B. $m_1=2m_2$
• C. $m_1=\frac{m_2}{4}$
• D. $m_1{m}_2=1$

Answer 1

The correct option is A $m_2=2m_1$

From ideal gas equation, we have
$pV=mRT$
So $\frac{p}{m}=\frac{RT}{V}$
As both gases expand isothermally to the same volume
So, $RT/V=$ constant
$\therefore \frac{p}{m}=$ constant
$\therefore \frac{p_1}{p_2}=\frac{m_1}{m_2}\Rightarrow \frac{\Delta p}{2\Delta p}=\frac{m_1}{m_2}$
$\therefore m_2=2m_1$.