A cyclic process is shown in the p-v diagram. | vedclass

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(A, B, C, D) In a $p-v$ diagram:$1$. The process $1 \rightarrow 2$ is a horizontal line, meaning pressure $p$ is constant. Thus, it is an isobaric pro

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The heat absorbed by the system in the complete cyclic process is non-zero

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(A, B, C, D) In a $p-v$ diagram:$1$. The process $1 \rightarrow 2$ is a horizontal line, meaning pressure $p$ is constant. Thus, it is an isobaric process.$2$. The process $3 \rightarrow 1$ is a vertical line, meaning volume $v$ is constant. Thus, it is an isochoric process.$3$. The process $2 \rightarrow 3$ is a curve, which represents an adiabatic process in this cycle.$4$. For any complete cyclic process, the change in internal energy $\Delta U = 0$. According to the first law of thermodynamics, $Q = \Delta U + W$. Since $\Delta U = 0$, $Q = W$. The work done $W$ in a cyclic process is equal to the area enclosed by the cycle, which is non-zero. Therefore, both work done and heat absorbed are non-zero.Thus, statements $A$, $B$, $C$, and $D$ are all correct.

Column-$I$	Column-$II$
$(a)$ Heat engine	$(i)$ $\text{COP} = \frac{Q_2}{Q_1 - Q_2}$
$(b)$ Heat pump	$(ii)$ $\eta = \frac{Q_1 - Q_2}{Q_1}$
	$(iii)$ $\eta = \frac{T_1 - T_2}{T_1}$

$A$ cyclic process is shown in the $p-v$ diagram. Which of the following statement$(s)$ is/are true?

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Column-$I$ Column-$II$

$(a)$ Heat engine $(i)$ $\text{COP} = \frac{Q_2}{Q_1 - Q_2}$

$(b)$ Heat pump $(ii)$ $\eta = \frac{Q_1 - Q_2}{Q_1}$

$(iii)$ $\eta = \frac{T_1 - T_2}{T_1}$

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