Q138 / 160 · TS EAMCET 2023 (13 May Shift 1) · Physics
Two identical containers A and B with frictionless pistons contain the same ideal gas at the same temperature and same volume $\mathrm{V}$. The mass of the gas in $\mathrm{A}$ is $\mathrm{m}_{\mathrm{A}}$ and that in $B$ is $m_B$. The gas in each cylinder is now allowed to expand isothermally to the same final volume $2 \mathrm{~V}$. The changes in the pressures of the gases in A and B are found to be $2 \Delta \mathrm{P}$ and $3 \Delta \mathrm{P}$ respectively. Then the relation between $\mathrm{m}_{\mathrm{A}}$ and $\mathrm{m}_{\mathrm{B}}$ is
- A$3 \mathrm{~m}_{\mathrm{A}}=4 \mathrm{~m}_{\mathrm{B}}$
- Correct answerB$3 \mathrm{~m}_{\mathrm{A}}=2 \mathrm{~m}_{\mathrm{B}}$
- C$2 \mathrm{~m}_{\mathrm{A}}=3 \mathrm{~m}_{\mathrm{B}}$
- D$4 \mathrm{~m}_{\mathrm{A}}=3 \mathrm{~m}_{\mathrm{B}}$
Solution
Here process is isothermal
So, $p \propto \frac{1}{V}$
for containers, A
$$
\begin{aligned}
& 2 \Delta \mathrm{P}=\left(\mathrm{P}_{\mathrm{A}}\right)_{\mathrm{j}}-\left(\mathrm{P}_{\mathrm{A}}\right)_{\mathrm{f}} \\
& =\frac{\mathrm{n}_{\mathrm{A}} \mathrm{RT}}{\mathrm{V}}-\frac{\mathrm{n}_{\mathrm{A}} \mathrm{RT}}{2 \mathrm{~V}}
\end{aligned}
$$
$2 \Delta \mathrm{P}=\frac{\mathrm{n}_{\mathrm{A}} \mathrm{RT}}{2 \mathrm{~V}}$ ...(1)
For containers, B
$$
\begin{aligned}
& 3 \Delta \mathrm{P}=\left(\mathrm{P}_{\mathrm{B}}\right)_{\mathrm{j}}-\left(\mathrm{P}_{\mathrm{B}}\right)_{\mathrm{f}} \\
& 3 \Delta \mathrm{P}=\frac{\mathrm{n}_{\mathrm{B}} \mathrm{RT}}{\mathrm{V}}-\frac{\mathrm{n}_{\mathrm{B}} \mathrm{RT}}{2 \mathrm{~V}}
\end{aligned}
$$
$3 \Delta \mathrm{P}=\frac{2 \mathrm{n}_{\mathrm{B}} \mathrm{RT}}{2 \mathrm{~V}}$ ...(3)
Divide equation (1) by (2), we have
$$
\begin{aligned}
& \frac{2 \Delta \mathrm{P}}{3 \Delta \mathrm{P}}=\frac{\mathrm{n}_{\mathrm{A}} \mathrm{RT}}{2 \mathrm{~V}} \times \frac{2 \mathrm{~V}}{\mathrm{n}_{\mathrm{B}} \mathrm{RT}} \\
& \frac{\mathrm{n}_{\mathrm{A}}}{\mathrm{n}_{\mathrm{B}}}=\frac{2}{3}=\frac{\frac{\mathrm{m}_{\mathrm{A}}}{\mathrm{m}_{\mathrm{B}}}}{\frac{\mathrm{m}_{\mathrm{B}}}{\mathrm{m}}}=\frac{2}{3} \\
& \frac{\mathrm{m}_{\mathrm{A}}}{\mathrm{m}_{\mathrm{B}}}=\frac{2}{3} \\
& \Rightarrow 3 \mathrm{~m}_{\mathrm{A}}=2 \mathrm{~m}_{\mathrm{B}} \\
&
\end{aligned}
$$
Hence, the correct answer is option: 'B'.