One mole of an ideal monoatomic gas is taken through a cyclic process ABCA as shown in the T-V diagram. In the process BC, TV=constant. The efficiency of the cycle in percentage is

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$44.2$

$11.2$

$33.2$

$22.2$

answer is B.

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Detailed Solution

$q_{A B} = n C_{v} Δ T = 1 \times \frac{3 R}{2} \times 900 = 1350 R$

In the process BC,
TV=constant
$P V^{2} = constant$
Molar heat capacity of the gas in the process BC,
$C = C_{v} + \frac{R}{(1 - x)} C = \frac{3 R}{2} + \frac{R}{(1 - 2)} = \frac{3 R}{2} - R = \frac{R}{2} C = \frac{R}{2} q_{B C} = n C Δ T = 1 \times \frac{R}{2} \times (600 - 1200) = - 300 R q_{C A} = n C_{P} Δ T = 1 \times \frac{5 R}{2} \times (300 - 600) = - 750 R - W_{c y c l e} = Q_{c y c l e} = q_{A B} + q_{B C} + q_{C A} - W_{c y c l e} = 1350 R - 300 R - 750 R = 300 R η = \frac{- W_{c y c l e}}{q_{supplied}} \times 100 = \frac{300 R}{1350 R} \times 100 η = \frac{200}{9} = 22.2 %$