Three processes compose a thermodynamic cycle | vedclass

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(D) For a complete thermodynamic cycle,the change in internal energy is zero,so $\Delta U_{cycle} = \Delta U_{1\rightarrow 2} + \Delta U_{2\rightarrow

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$30$

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(D) For a complete thermodynamic cycle,the change in internal energy is zero,so $\Delta U_{cycle} = \Delta U_{1\rightarrow 2} + \Delta U_{2\rightarrow 3} + \Delta U_{3\rightarrow 1} = 0$. Given: $\Delta U_{2\rightarrow 3} = -20\,J$ (decrease in internal energy). For process $3\rightarrow 1$ (adiabatic),$Q_{3\rightarrow 1} = 0$. From the first law of thermodynamics,$\Delta U_{3\rightarrow 1} = Q_{3\rightarrow 1} - W_{3\rightarrow 1}$. Work done on the system is $20\,J$,so $W_{3\rightarrow 1} = -20\,J$. Thus,$\Delta U_{3\rightarrow 1} = 0 - (-20\,J) = 20\,J$. Now,$\Delta U_{1\rightarrow 2} + (-20\,J) + 20\,J = 0$,which gives $\Delta U_{1\rightarrow 2} = 0$. Total work done in the cycle $W_{cycle} = W_{1\rightarrow 2} + W_{2\rightarrow 3} + W_{3\rightarrow 1} = 10\,J$. Since process $2\rightarrow 3$ is isochoric,$W_{2\rightarrow 3} = 0$. So,$W_{1\rightarrow 2} + 0 + (-20\,J) = 10\,J$,which gives $W_{1\rightarrow 2} = 30\,J$. For process $1\rightarrow 2$ (isothermal),$\Delta U_{1\rightarrow 2} = 0$. Applying the first law: $Q_{1\rightarrow 2} = \Delta U_{1\rightarrow 2} + W_{1\rightarrow 2} = 0 + 30\,J = 30\,J$.

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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