1. Tardigrade
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  4. Two identical containers A and B having the same volume of an ideal gas at the same temperature have the mass of the gas as m1 and m2 respectively and 2m1=3m2 . The gas in each cylinder expands isothermally to double of its volume. If the change in pressure in A is 300 Pa , then the change in pressure in B is

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Q. Two identical containers $A$ and $B$ having the same volume of an ideal gas at the same temperature have the mass of the gas as $m_{1}$ and $m_{2}$ respectively and $2m_{1}=3m_{2}$ . The gas in each cylinder expands isothermally to double of its volume. If the change in pressure in $A$ is $300 \, Pa$ , then the change in pressure in $B$ is

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Solution:

$2m_{1}=3m_{2}$
$P_{1}V=\frac{m_{1}}{M}RT \, \, \, and \, \, \, P_{2}V=\frac{m_{2}}{M}RT=\frac{2}{3}\frac{m_{1}}{M}RT$
$\frac{P_{1}}{P_{2}}=\frac{3}{2}$ ………(i)
The process is isothermal $\left(P_{1} V_{1} = P_{2} V_{2}\right)$
Hence, $\Delta P_{1}=P_{1}-\frac{P_{1}}{2}\Rightarrow P_{1}=2\Delta P_{1}$
$\Delta P_{2}=P_{2}-\frac{P_{2}}{2}=\frac{P_{2}}{2}=\frac{1}{2}\frac{2}{3}P_{1}=\frac{P_{1}}{3}$
$\Delta P_{2}=\frac{2}{3}\Delta P_{1}=\frac{2}{3}.300$
$\Delta P_{2}=200 \, Pa$