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Question

# A sample of an ideal gas is taken through a cycle as shown in the figure. It absorbs 50J of energy during the process AB, no heat during BC, rejects 70J during CA. 40J of work is done on the gas during BC. the internal energy of a gas at A is 1500J, the internal energy at C would be

A
1590J
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B
1620J
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C
1540J
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D
1570J
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Solution

## The correct option is D 1590JGiven,QAB = 50 JQBC = 0 JQCA =−70 JWBC =−40 JUA = 1500 JFor process A→B,ΔVAB=0 ⟹ ΔWAB=0WAC=WAB+WBC=0+(−40)=−40 JQAC=QAB+QBC=50+0=50 JUsing First law of Thermodynamics,ΔU=ΔQ−ΔWTherefore,UC−UA=UAC=QAC−WAC⇒ UC=UA+QAC−WAC=1500+50−(−40)=1590JTheanswerisoption(A).

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