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(B) For process $A \rightarrow B$: This is an isobaric compression ($P$ is constant,$V$ decreases).Since $V$ decreases,work is done on the gas $(t)$.

Vedclass · Accepted Answer

$(A) ightarrow p, r, 	ext{ and } t, (B) ightarrow p 	ext{ and } r, (C) ightarrow q, 	ext{ and } s, (D) ightarrow r 	ext{ and } t$

Vedclass · Answer

(B) For process $A \rightarrow B$: This is an isobaric compression ($P$ is constant,$V$ decreases).Since $V$ decreases,work is done on the gas $(t)$. Since $T$ decreases,internal energy decreases $(p)$. Since $Q = \Delta U + W$,both $\Delta U$ and $W$ are negative,so heat is lost $(r)$. Thus,$(A) \rightarrow p, r, t$.For process $B \rightarrow C$: This is an isochoric process ($V$ is constant,$P$ decreases).Since $V$ is constant,$W = 0$. Since $P$ decreases,$T$ decreases,so internal energy decreases $(p)$. Since $\Delta U For process $C \rightarrow D$: This is an isobaric expansion ($P$ is constant,$V$ increases).Since $V$ increases,work is done by the gas. Since $T$ increases,internal energy increases $(q)$. Since $Q = \Delta U + W$,both are positive,so heat is gained $(s)$. Thus,$(C) \rightarrow q, s$.For process $D \rightarrow A$: This is an isothermal compression ($T$ is constant,$P$ increases,$V$ decreases).Since $V$ decreases,work is done on the gas $(t)$. Since $T$ is constant,$\Delta U = 0$. Since $W Comparing with the options,option $(B)$ is correct.

Column $I$	Column $II$
$(A)$ Process $A \rightarrow B$	$(p)$ Internal energy decreases.
$(B)$ Process $B \rightarrow C$	$(q)$ Internal energy increases.
$(C)$ Process $C \rightarrow D$	$(r)$ Heat is lost.
$(D)$ Process $D \rightarrow A$	$(s)$ Heat is gained.
	$(t)$ Work is done on the gas.

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