(C) For cylinder $A$ (isobaric process): Heat given $Q = n C_P \Delta T_A$. Given $\Delta T_A = 42 ~K$ and for monoatomic gas $C_P = \frac{5}{2} R$. T

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(C) For cylinder $A$ (isobaric process): Heat given $Q = n C_P \Delta T_A$. Given $\Delta T_A = 42 ~K$ and for monoatomic gas $C_P = \frac{5}{2} R$. Thus, $Q = n \left(\frac{5}{2} R\right) (42) = 105 nR$.For cylinder $B$ (isochoric process): Heat given $Q = n C_V \Delta T_B$. For monoatomic gas $C_V = \frac{3}{2} R$.Since the heat given is the same, $n \left(\frac{3}{2} R\right) \Delta T_B = 105 nR$.$\frac{3}{2} \Delta T_B = 105 \implies \Delta T_B = 105 \times \frac{2}{3} = 70 ~K$.

Class 11 Chemistry Data Entry - Chemistry Level 0 of all question bank

Two cylinders $A$ and $B$ fitted with pistons contain equal number of moles of an ideal monoatomic gas at $400 ~K$. 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 $42 ~K$, the rise in temperature of the gas in $B$ is (in $~K$)

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A
$21$
B
$35$
C
$70$
D
$42$

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