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Q.

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

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a

1315 J

b

275 J

c

765 J

d

675 J

answer is C.

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Detailed Solution

Q=Q1+Q2+Q3+Q4 =5960−5585−2980+3645=1040JW=W1+W2+W3+W4 =2200−825−1100+W4=275+W4 For a cyclic process, Uf=Uf         ΔU=Uf−Ui=0From the first law of thermodynamics,            Q=ΔU+W1040=0=275+W4  or  W4=765J
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