Two cylinders $A$ and $B$ fitted with pistons contain equal number of moles of an ideal monoatomic gas at…
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
$21 \mathrm{~K}$
$35 \mathrm{~K}$
$42 \mathrm{~K}$
$70 \mathrm{~K}$
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
$\begin{aligned}
\mu & =1 \\
\gamma & =\frac{5}{3} \\
W & =\frac{1 \times R}{\frac{5}{3}-1}(442-400) \\
& =\frac{3}{2} R \times 42
\end{aligned}$
$or \quad$ $W=63 R$
But $\Delta U=0$, for cylinder $A$
$\begin{aligned}
\therefore \quad Q & =0+63 R \\
Q & =63 R
\end{aligned}$
For cylinder $B$ volume is constant,
$\begin{aligned}
\therefore W =0 \\
\text { and } Q=\mu C_V \Delta T
\end{aligned}$
For monoatomic gas
$\begin{aligned}
C_V & =\frac{3}{2} R \\
Q & =1 \times \frac{3}{2} R \Delta T
\end{aligned}$
As heat given to both cylinder is same
$\begin{aligned}
\therefore \quad 63 R & =\frac{3}{2} R \Delta T \\
\Delta T & =42 \mathrm{~K}
\end{aligned}$