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Question

A figure show a closed cycle for a gas. The change in internal energy along the path ca is 160 J. The energy transferred to the gas as heat is 200 J along path ab and 40 J along path bc. What is the ratio of work done by the gas along the path abc and ab?


A
1 : 1
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B
2 : 1
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C
1 : 2
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D
3 : 2
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Solution

The correct option is A 1 : 1
Given that
ΔUca=UaUc=160 J
ΔUac=UcUa=160 J
Qbc=40 J
Qab=200 J

Since internal energy only depends on initial and final states,
ΔUac=ΔUab+ΔUbc
160=ΔUab+ΔUbc(1)

Using first law of thermodynamics,
ΔUab=QabWab;ΔUbc=QbcWbc
From (1),
160=QbcWbc+QabWab
160=400+200Wab
Wab=80 J
[Since there is no change in volume in path bc and work done is zero]

So, Wab=Wabc=80 J
WabWbc=80 J80 J=1
Hence, option (a) is correct.

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