A monoatomic ideal gas is taken through a cycle $A B C A$ as shown. The efficiency of the cycle is

7

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10

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8.7

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15

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Solution

The correct option is C $8.7$ %
Work done in the cycle = Area under $P V$ graph = $\frac{1}{2} \times P \times 2 V = P V$

As process $A B$ is isochoric $Q_{A B} = n C_{v} Δ T = \frac{C_{v}}{R} V (2 P - P) = \frac{3 P V}{2}$

As process $B C$ is isobaric $Q_{B C} = n C_{P} Δ T = \frac{C_{p}}{R} 2 P (3 V - V) = 10 P V$
For process $C A, Δ U$ decreases as temperature decreases and $Δ W$ decreases as Volume decreases. Hence $Q_{C A}$ decreases(heat is rejected in this process)
Total heat supplied is $Q = Q_{A B} + Q_{B C} = 23 P V / 2$

Efficiency = $\frac{W o r k d o n e}{Q}$ $η = \frac{W}{Q} = \frac{2 P V}{23 P V} \times 100 = 8.7 %$

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