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Question

# Three processes form a thermodynamic cycle as shown on P−V diagram for an ideal gas. Process 1→2 takes place at constant temperature (300K). Process 2→3 takes place at constant volume. During this process 40 J of heat leaves the system. Process 3→1 is adiabatic and temperature T3 is 275K. Work done by the gas during the process 3→1 is

A
40 J
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B
20 J
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C
+40 J
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D
+20 J
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Solution

## The correct option is B −40 JIn the process 1→2,ΔQ1→2=ΔW1→2In the process 2→3,ΔW2→3=0In the process 3→1,ΔQ3→1=0The complete process being cyclic, ΔU=0Hence ΔQ=ΔW⇒ΔQ1→2+ΔQ2→3+ΔQ3→1=ΔW1→2+ΔW2→3+ΔW3→1⇒ΔQ2→3=ΔW3→1=−40J

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