An ideal monoatomic gas is carried around the cycle ABCDA as shown in the figure. The efficiency of the gas cycle is

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$\frac{4}{21}$

$\frac{2}{21}$

$\frac{2}{31}$

$\frac{4}{31}$

answer is A.

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Detailed Solution

$A \to B : V = (constant)$

$\Rightarrow Q_{A \to B} = {nC}_{v} \cdot ΔT = n (\frac{3}{2} R) (ΔT) = \frac{3}{2} nR (ΔT) = \frac{3}{2} (ΔP) V$

$\Rightarrow Q_{A \to B} = 3 P_{0} V_{0}$

$B \to C : P = (constant)$

$\Rightarrow Q_{B \to C} = n C_{P} (∆ T) = n (\frac{5}{2} R) (∆ T) = \frac{5}{2} n R (∆ T) = \frac{5}{2} \times P (∆ V)$

$\Rightarrow Q_{B \to C} = \frac{15}{2} P_{0} V_{0}$

$W = a r e a (□ A B C D) = 2 P_{0} V_{0}$

$\Rightarrow η = \frac{W}{Q_{i n}} = \frac{W}{Q_{A \to B} + Q_{B \to C}} = \frac{4}{21}$

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