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Two cylinders $A$ and $B$ fitted with pistons, contain equal number of moles of an ideal monoatomic gas at $400 \mathrm{~K}$. The piston of $A$ is free to move while that of $B$ is held fixed. Same amount of heat energy is given to the gas in each cylinder. If the rise in temperature of the gas in $A$ is $42 \mathrm{~K}$, the rise in temperature of the gas in $B$ is $(\gamma=5 / 3)$
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The correct answer is:
$25.2 \mathrm{~K}$
Given, $T=400 \mathrm{~K}, V=$ constant,
$\Delta T=42 \mathrm{~K}$
Heat supplied
$\Delta Q=\Delta U+\Delta W$
or
$\Delta Q=n \times \frac{3 R}{2} \times 42+0$
In second process, when piston is free to move
$\begin{aligned} \Delta Q & =n C_p \Delta T^{\prime} \\ n \times \frac{3 R}{2} \times 42 & =n \times \frac{5 R}{2} \Delta T^{\prime} \\ \Delta T^{\prime} & =\frac{126}{5}=25.2 \mathrm{~K}\end{aligned}$
$\Delta T=42 \mathrm{~K}$
Heat supplied
$\Delta Q=\Delta U+\Delta W$
or
$\Delta Q=n \times \frac{3 R}{2} \times 42+0$
In second process, when piston is free to move
$\begin{aligned} \Delta Q & =n C_p \Delta T^{\prime} \\ n \times \frac{3 R}{2} \times 42 & =n \times \frac{5 R}{2} \Delta T^{\prime} \\ \Delta T^{\prime} & =\frac{126}{5}=25.2 \mathrm{~K}\end{aligned}$
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