One mole of an ideal monoatomic gas is taken along the path ABCA as… | JEE Main 2018 (16 Apr, Shift 1)

Q66JEE Main 2018 · 16 APR · SHIFT 1Difficult

P-V Diagrams & Cyclic Processes

One mole of an ideal monoatomic gas is taken along the path ABCA as shown in the PV diagram. The maximum temperature attained by the gas along the path BC is given by:

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Answer: (b)

Line BC through $(V_0, 3P_0)$ and $(2V_0, P_0)$ : $P = 5P_0 - \frac{2P_0}{V_0}V$ .
$T = \frac{PV}{R} = \frac{1}{R}\left(5P_0V - \frac{2P_0}{V_0}V^2\right)$
$\frac{dT}{dV} = 0 \Rightarrow V = \frac{5}{4}V_0$ , $P = \frac{5}{2}P_0$ .
$T_{max} = \frac{1}{R} \times \frac{5}{2}P_0 \times \frac{5}{4}V_0 = \frac{25}{8}\frac{P_0V_0}{R}$
So the answer is (b).

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