One mole of a monoatomic ideal gas goes through a thermodynamic cycle, as shown in the volume versus temperature (V-T) diagram. The correct statement(s) is/are:
[R is the gas constant]

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The ratio of heat transfer during processes $1 \to 2$ and $2 \to 3$ is $| \frac{Q_{1 \to 2}}{Q_{2 \to 3}} | = \frac{5}{3}$

The ratio of heat transfer during processes $1 \to 2$ and $3 \to 4$ is $| \frac{Q_{1 \to 2}}{Q_{3 \to 4}} | = \frac{1}{2}$

Work done in this thermodynamic cycle $(1 \to 2 \to 3 \to 4 \to 1)$ is $| W | = \frac{1}{2} R T_{0}$

The above thermodynamic cycle exhibits only isochoric and adiabatic processes.

answer is A, C.

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Detailed Solution

$A) W = W_{1 \to 2} + W_{2 \to 3} + W_{3 \to 4} + W_{4 \to 1} = R (T_{2} - T_{1}) + 0 + R (T_{4} - T_{3}) + 0 = R (2 T_{0} - T_{0}) + R (\frac{T_{0}}{2} - T_{0}) = R T_{0} - \frac{R T_{0}}{2} W = \frac{R T_{0}}{2}$

$C) Q_{1 \to 2} = n C_{p} (T_{2} - T_{1}) = 1 \times \frac{5 R}{2} (2 T_{0} - T_{0}) = \frac{5}{2} R T_{0} Q_{2 \to 3} = n C_{v} Δ T = 1 \times \frac{3 R}{2} (T_{3} - T_{2}) = \frac{3 R}{2} (T_{0} - 2 T_{0}) = \frac{- 3}{2} R T_{0} \frac{| Q_{1 \to 2} |}{| Q_{2 \to 3} |} = \frac{5}{3} Q_{3 \to 4} = n C_{p} (T_{4} - T_{3}) = 1 \times \frac{5 R}{2} (\frac{T_{0}}{2} - T_{0}) = \frac{- 5 R}{2} \frac{T_{0}}{2} = \frac{- 5}{4} R T_{0} \frac{| Q_{1 \to 2} |}{| Q_{3 \to 4} |} = \frac{5 / 2}{5 / 4} = 2$