One mole of ideal monoatomic gas is taken along a cyclic process as shown in the figure. Process $1 \to 2$ shown is 1/4 th part of a circle as shown by dotted line process. $2 \to 3$ is isochoric while $3 \to 1$ is isobaric. If efficiency of the cycle is $n %$ where $n$ is an integer, find n.

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Solution

Area enclosed by the $P - V$ graph gives the net work done.
$W_{1 \to 2} = 1.22 P_{0} V_{0} W_{2 \to 3} = 0 W_{3 \to 1} = - P_{0} V_{0}$

$∴ W_{n e t} = 0.22 P_{0} V_{0}$

Now,
$T_{1} = \frac{P_{0} V_{0}}{R}, T_{2} = \frac{4 P_{0} V_{0}}{R}, T_{3} = \frac{2 P_{0} V_{0}}{R}$

Thus,
$Δ U_{1 \to 2} = 1 \times \frac{3 R}{2} [T_{2} - T_{1}] = \frac{9}{2} P_{0} V_{0}$
$Δ U_{2 \to 3} = 1 \times \frac{3 R}{2} [T_{3} - T_{2}] = - 3 P_{0} V_{0}$
$Δ U_{3 \to 1} = 1 \times \frac{3 R}{2} [T_{1} - T_{3}] = \frac{- 3}{2} P_{0} V_{0}$

$Δ Q_{1 \to 2} = (4.5) (P_{0} V_{0}) + (1.22) (P_{0} V_{0}) = (5.72) (P_{0} V_{0})$
$Δ Q_{2 \to 3} = - 3 P_{0} V_{0} + 0 = - 3 (P_{0} V_{0})$
$Δ Q_{3 \to 1} = - 1.5 P_{0} V_{0} - (P_{0} V_{0}) = - 2.5 (P_{0} V_{0})$

Thus efficiency $η = \frac{W_{n e t}}{Q_{+ v e}}$
$η = \frac{0.22 (P_{0} V_{0})}{5.72 (P_{0} V_{0})} = 0.04$
Thus efficiency is $4 %$ .
Hence n = 4.

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