An ideal gas undergoes a four-step cycle as s | vedclass

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(C) According to the first law of thermodynamics,$\Delta Q = \Delta U + W$.Process $1$: Isobaric expansion ($P = 	ext{constant}$,$V$ increases). Sinc

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steps $1$ and $4$

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(C) According to the first law of thermodynamics,$\Delta Q = \Delta U + W$.Process $1$: Isobaric expansion ($P = 	ext{constant}$,$V$ increases). Since $V$ increases,$W > 0$. Since $T \propto V$ at constant $P$,$T$ increases,so $\Delta U > 0$. Thus,$\Delta Q = \Delta U + W > 0$ (Heat is absorbed).Process $2$: Isochoric compression ($V = 	ext{constant}$,$P$ decreases). Since $V$ is constant,$W = 0$. Since $P \propto T$ at constant $V$,$P$ decreases implies $T$ decreases,so $\Delta U Process $3$: Isobaric compression ($P = 	ext{constant}$,$V$ decreases). Since $V$ decreases,$W Process $4$: Isochoric expansion ($V = 	ext{constant}$,$P$ increases). Since $V$ is constant,$W = 0$. Since $P \propto T$ at constant $V$,$P$ increases implies $T$ increases,so $\Delta U > 0$. Thus,$\Delta Q = \Delta U > 0$ (Heat is absorbed).Therefore,heat is absorbed in steps $1$ and $4$.

An ideal gas undergoes a four-step cycle as shown in the $P-V$ diagram below. During this cycle,in which step is heat absorbed by the gas?

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