One mole of ideal monoatomic gas is taken along a cyclic process as shown in the figure. Process 1 → 2 shown is 1/4 th part of a circle as shown by dotted line process 2 → 3 is isochoric while 3 → 1 is isobaric. If efficiency of the cycle is n% where n is an integer. Find n.
Text Solution
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(4)
W net = (2P 0 V 0 ) – (P 0 V 0 ) – 
W net = 
P 0 V 0 = (0.22) (P 0 V 0 )
Now,
Δ Q 1 → 2 = (4.5) (P 0 V 0 ) + (1.22) (P 0 V 0 ) = (5.72) (P 0 V 0 )
Δ Q 3 → 2 = – 3P 0 V 0 + 0 = – 3.22) (P 0 V 0 )
Δ Q 3 → 1 = – 1.5(P 0 V 0 ) – (P 0 V 0 ) = – 2.5(P 0 V 0 )
Thus efficiency η = 
η = 
= 0.04
Thus, efficiency is 4%.
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