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Question

An ideal gas is taken through a cyclic thermo dynamical process through four steps. The amounts of heat involved in steps are
Q1=5960J,Q2=−5585J,Q3=−2980J,Q4=3645J; respectively, the corresponding works involved are W1=2200,W2=−825J,W3=−1100J and W4 respectively. Find the value of W4 and efficiency of the cycle

A
1315 J, 10%
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B
765 J, 11%
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C
both of them
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D
none of these
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Solution

The correct option is B 765 J, 11%
Given;-
Q1=5960J,W1=2200JQ2=5585J,W2=825JQ3=2980J,W3=1100JQ4=3645J,W4=findη=find
In cycic process ΔU=0
A/q 1st law of thermodynamics
ΔQ=ΔU+W
+1040=275+W4
W4=1040275
=765
η%=outputinput×100

=10405960+3645

=2081321×100
Hence, we get
η=10.82%



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