A fixed amount of a monatomic ideal gas (C_V=tfrac32R) is taken through a rectangular cycle on a P-V diagram: (2P_0,V_0)->(2P_0,2V_0) (isobaric expansion) ->(P_0,2V_0) (isochoric cooling) ->(P_0,V_0) (isobaric compression) ->(2P_0,V_0) (isochoric heating). Find the efficiency of the cycle, defined as net work divided by total heat absorbed, as a percentage (two decimals).

JEE Advanced levelNumerical answerDifficulty 3 of 3td148

A fixed amount of a monatomic ideal gas ( $C_{V} = \frac{3}{2} R$ ) is taken through a rectangular cycle on a $P$ - $V$ diagram: $(2 P_{0}, V_{0}) \to (2 P_{0}, 2 V_{0})$ (isobaric expansion) $\to (P_{0}, 2 V_{0})$ (isochoric cooling) $\to (P_{0}, V_{0})$ (isobaric compression) $\to (2 P_{0}, V_{0})$ (isochoric heating). Find the efficiency of the cycle, defined as net work divided by total heat absorbed, as a percentage (two decimals).

Numerical answer type, answer in %, tolerance 0.05.

Net work

=

enclosed area

= (2 P_{0} - P_{0}) (2 V_{0} - V_{0}) = P_{0} V_{0}

. Heat is absorbed in the isobaric expansion,

q = \frac{5}{2} Δ (P V) = \frac{5}{2} (4 - 2) P_{0} V_{0} = 5 P_{0} V_{0}

, and in the isochoric heating,

q = \frac{3}{2} Δ (P V) = \frac{3}{2} (2 - 1) P_{0} V_{0} = 1.5 P_{0} V_{0}

. Total

= 6.5 P_{0} V_{0}

. Efficiency

= 1/6.5 = 15.38%