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Question

One mole of a diatomic ideal gas (γ=1.4) is taken through a cyclic process starting from point A. The process AB is an adiabatic compression, BC is an isobaric expansion, CD is an adibatic expansion and DA is an isochoric process. The volume ratios are VAVB=16,VCVB=2 and temperature TA=27C. Find the efficiency of the cycle.

A
32.84 %
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B
61.40 %
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C
28.56 %
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D
35.61 %
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Solution

The correct option is B 61.40 %

From the figure, for adiabatic compression process AB,
TAVγ1A=TBVγ1B
From this, we can write that
TB=(VAVB)γ1TA TB=(16)25×300909 K
For an isobaric process BC,
VBTB=VCTC
TC=TB(VCVB)=909×21818 K
For adiabatic expansion process CD
VAVB=16 and VCVB=2
VAVC=VDVC=8
Also, TCVγ1C=TDVγ1DTD=TC(VCVD)γ1=1818(18)25791 K

Heat absorbed from BC
Q1=nCP(TCTB)=nγRγ1(TCTB)=7R525(1818909)=7R2×9093182R J
For process DA, heat released
Q2=nCV(TDTA)=nRγ1(TDTA)=5R2(791300)=5R2×4911227.5R J

Work done during process AB
WAB=nRγ1(TBTA)=5R2(909300)=5R2×609=1522.5R J
Similarly,
WBC=nR(TCTB)=1×R(1818909)=909R J
WCD=nRγ1(TCTD)=5R2×1027=2567.5R J
Therefore,
Wnet=WAB+WBC+WCD=1522.5R+909R+2567.5R
=1954R J
Thus, the efficiency of cycle (η)=WnetQ1×100=1954 R3182 R×10061.4 %
Hence, option (b) is the correct answer.

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