Match the following:Column IColumn IIA. Ratio | vedclass

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(A) For an isobaric process,the heat supplied is $\Delta Q = C_p \Delta T$ and the change in internal energy is $\Delta U = C_V \Delta T$.Thus,$\frac{

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$4 \quad 3 \quad 2 \quad 1$

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(A) For an isobaric process,the heat supplied is $\Delta Q = C_p \Delta T$ and the change in internal energy is $\Delta U = C_V \Delta T$.Thus,$\frac{\Delta Q}{\Delta U} = \frac{C_p \Delta T}{C_V \Delta T} = \frac{C_p}{C_V} = \gamma$. So,$A \rightarrow 4$.For an isobaric process,$\Delta Q = \Delta U + \Delta W$,so $\Delta W = \Delta Q - \Delta U = (C_p - C_V) \Delta T$.Thus,$\frac{\Delta Q}{\Delta W} = \frac{C_p \Delta T}{(C_p - C_V) \Delta T} = \frac{C_p}{C_p - C_V} = \frac{C_p/C_V}{(C_p/C_V) - 1} = \frac{\gamma}{\gamma - 1}$. So,$B \rightarrow 3$.For a refrigerator,the coefficient of performance $\beta = \frac{Q_2}{W} = \frac{Q_2}{Q_1 - Q_2} = \frac{T_2}{T_1 - T_2}$. So,$C \rightarrow 2$.For a heat pump,the coefficient of performance $\alpha = \frac{Q_1}{W} = \frac{Q_1}{Q_1 - Q_2} = \frac{T_1}{T_1 - T_2}$. So,$D \rightarrow 1$.Therefore,the correct match is $A \rightarrow 4, B \rightarrow 3, C \rightarrow 2, D \rightarrow 1$.

Column $I$	Column $II$
$A$. Ratio of $\frac{\Delta Q}{\Delta U}$ in an isobaric process	$1$. $\frac{T_1}{T_1-T_2}$
$B$. Ratio of $\frac{\Delta Q}{\Delta W}$ in an isobaric process	$2$. $\frac{T_2}{T_1-T_2}$
$C$. Coefficient of performance of a refrigerator	$3$. $\frac{\gamma}{\gamma-1}$
$D$. Coefficient of performance of a heat pump	$4$. $\gamma$

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