One mole of a monatomic ideal gas goes through a thermodynamics 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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Work done in this thermodynamic cycle $(1 \to 2 \to 3 \to 4 \to 1)$ is $|W| = \frac{1}{2} R T_{0}$

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

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 above thermodynamic cycle exhibits only isochoric and adiabatic processes.

answer is A, B.

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Detailed Solution

$(i) W = Area$ $= (\frac{n R T_{0}}{2 V_{0}}) V_{0} = \frac{1}{2} R T_{0}$

$(ii) \frac{|Q_{12}|}{|Q_{13}|} = \frac{n C_{p} Δ T}{n C_{V} Δ T} = \frac{5}{3}$

$(iii) |\frac{Q_{12}}{Q_{34}}| = \frac{n C_{p} Δ T_{1}}{n C_{p} Δ T_{2}} = 2$

(iv) The above thermodynamic cycle exhibits only isochoric and isobaric processes.

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