Q96 / 120 · VITEEE 2009 · Physics
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 $\mathrm{A}$ is $42 \mathrm{~K}$, the rise in temperature of the gas in $B$ is
- A$21 \mathrm{~K}$
- B$42 \mathrm{~K}$
- Correct answerC$35 \mathrm{~K}$
- D$70 \mathrm{~K}$
Solution
From first law of thermodynamics
$$
Q=\Delta U+W
$$
For cylinder $A$ pressure remains constant.
$\therefore$ Work done by a system
$$
W=\frac{\mu R}{\gamma-1}\left(T_{1}-T_{2}\right)
$$
For monoatomic gases, $\mu=1 ; \gamma=\frac{5}{3}$
$$
\therefore W=\frac{1 \times R}{\frac{5}{3}-1}(442-400)=\frac{3}{2} R \times 42
$$
or $W=63 R$
But $\Delta U=0$, for cylinder $A$
$$
\therefore Q=0+63 R=63 R
$$
For cylinder $B$ volume is constant, $\therefore W=0$ and $Q=\mu C_{V} \Delta T$
For monoatomic gas
$$
C_{V}=\frac{3}{2} R \Rightarrow Q=1 \times \frac{3}{2} R \Delta T
$$
As heat given on both cylinder is same,
$$
\therefore 63 R=\frac{3}{2} R \Delta T \Rightarrow \Delta T=42 \mathrm{~K}
$$
Hence, the correct answer is option: 'C'.