One mole of a monoatomic ideal gas goes through a thermodynamic cycle, as shown in the volume versus…

1. The ratio of heat transfer during processes $1\to 2$ and $2\to 3$ is $\left|\frac{{\mathrm{Q}}_{1\to 2}}{{\mathrm{Q}}_{2\to 3}}\right|=\frac{5}{3}$
2. The ratio of heat transfer during processes $1\to 2$ and $3\to 4$ is $\left|\frac{{\mathrm{Q}}_{1\to 2}}{{\mathrm{Q}}_{3\to 4}}\right|=\frac{1}{2}$
3. The above thermodynamic cycle exhibits only isochoric and adiabatic processes.
4. Work done in this thermodynamic cycle $\left(1\to 2\to 3\to 4\to 1\right)$ is $\left|\mathrm{W}\right|=\frac{1}{2}\mathrm{R}{\mathrm{T}}_0$

Solution

$\left(\mathrm{A}\right)\Rightarrow \left|\frac{\Delta Q_{1\to 2}}{\Delta Q_{2\to 3}}\right|=\left|\frac{N{C}_P\Delta T_{1\to 2}}{N{C}_V\Delta T_{2\to 3}}\right|=\frac{C_P}{C_V}=\frac{5}{3}\Rightarrow$ correct
$\left(\mathrm{B}\right)\Rightarrow \left|\frac{\Delta Q_{1\to 2}}{\Delta Q_{3\to 4}}\right|=\frac{T_0}{T_0/2}=2\Rightarrow$ Incorrect
$\left(\mathrm{C}\right)\Rightarrow$ No adiabatic process is involved $\Rightarrow$ incorrect
$\left(\mathrm{D}\right)\Rightarrow$ Work done $=$ area inside $\mathrm{P}-\mathrm{V}$ diagram $=P_0{V}_0=nR\frac{T_0}{2}$ [at point $4$ in diagram] $\Rightarrow$ correct.
$1\to 2\Rightarrow$ isobaric at pressure $2P_0$
$2\to 3\Rightarrow$ isobaric at volume $2V_0$
$3\to 4\Rightarrow$ isobaric at pressure $P_0$
$4\to 1\Rightarrow$ isobaric at volume $V_0$
Given, number of moles, $n=1$
and $r=\frac{C_P}{C_V}=\frac{5}{3}$ (for mono atomic gas)
Now, $\Delta Q_{1\to 2}{=}^n{C}_P(T_2-T_1)$
$=n\cdot C_p\cdot T_0$
$\Delta Q_{2\to 3}{=}^n{C}_V\left(T_3-T_2\right)=n\cdot C_V\left(-T_2\right)$
$\Delta Q_{3\to 4}=n\cdot C_P\left(T_4-T_3\right)=n\cdot C_P\left(\frac{-T_0}{2}\right)$

Asked in: JEE Advanced 2019 (Paper 1)

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