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Question
An ideal gas is subjected to cyclic process involving four thermodynamic states, the amounts of heat (Q) and work (W) involved in each of these states are
Q1=6000J,
Q2=−5500J;
Q3=−3000J;
Q4=3500J
W1=2500J;
W2=−1000J;
W3=−1200J;
W4=xJ.
The ratio of the net work done by the gas to the total heat absorbed by the gas is . The values of × and η respectively are
An ideal gas is subjected to cyclic process involving four thermodynamic states, the amounts of heat (Q) and work (W) involved in each of these states are
Q1=6000J,
Q2=−5500J;
Q3=−3000J;
Q4=3500J
W1=2500J;
W2=−1000J;
W3=−1200J;
W4=xJ.
The ratio of the net work done by the gas to the total heat absorbed by the gas is . The values of × and η respectively are
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Solution
The correct option is D 700; 10.5%
From first law of thermodynamics
Q=ΔU+W
or Δ=Q−W
∴ΔU1−W1=6000−2500=3500J
ΔU2=Q1−W2=−5500+1000=−4500J
ΔU3=Q3−W3=−3000+1200=−1800J
ΔU4=Q4−W4=3500−X=0
For cyclic process ΔU=0
∴3500−4500−1800+3500−X=0
or X=700J
Efficiency, η=outputinput×100
=W1+W2+W3+W4Q1+Q4×100
=2500−1000−12000+7006000+3500×100
=10009500×100
η=10.5
From first law of thermodynamics
Q=ΔU+W
or Δ=Q−W
∴ΔU1−W1=6000−2500=3500J
ΔU2=Q1−W2=−5500+1000=−4500J
ΔU3=Q3−W3=−3000+1200=−1800J
ΔU4=Q4−W4=3500−X=0
For cyclic process ΔU=0
∴3500−4500−1800+3500−X=0
or X=700J
Efficiency, η=outputinput×100
=W1+W2+W3+W4Q1+Q4×100
=2500−1000−12000+7006000+3500×100
=10009500×100
η=10.5
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