A monoatomic gas consisting of n moles is taken from state A to state C as shown in the figure. Process $A \to B$ is isochoric which of following is/are CORRECT?

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Temperature of gas at state B is minimum for $P_{1} = \frac{3 P_{0}}{2}$ and $V_{1} = \frac{3 V_{0}}{2} .$

In process $B \to C,$ temperature of gas first decreases and then increases if $P_{1} = \frac{3 P_{0}}{2}, V_{1} = \frac{3 V_{0}}{2} .$

For $P_{1} = \frac{7 P_{0}}{4}, V_{1} = \frac{3 V_{0}}{2},$ heat absorbed/released from $A \to C$ will be more compare to heat released/absorbed from $A \to C$ when $P_{1} = \frac{3 P_{0}}{2}, V_{1} = \frac{7 V_{0}}{4} .$

For $P_{1} = \frac{3 P_{0}}{2}, V_{1} = \frac{3 V_{0}}{2},$ minimum internal energy of gas will be $3 P_{0} V_{0} .$

answer is B, C.

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Detailed Solution

Temperature of gas at C is minimum as temperature of gas from $B \to C$ continuously decreases.
Question Image
Heat dQ, in process,
$d Q_{1} = \frac{V_{0}}{2} (\frac{P_{0} + \frac{7 P_{0}}{4}}{2}) + n \frac{3}{2} R [\frac{2 P_{0} V_{0}}{n R} - \frac{3 P_{0} V_{0}}{n R}] = \frac{- 13 P_{0} V_{0}}{16}$
In process–2
$d Q_{2} = \frac{V_{0}}{4} [\frac{P_{0} + \frac{3 P_{0}}{2}}{2}] + n \times \frac{3}{2} R [\frac{2 P_{0} V_{0}}{n R} - \frac{21 P_{0} V_{0}}{8 n R}]$
= $\frac{5 P_{0} V_{0}}{16} - \frac{15 P_{0} V_{0}}{16} = \frac{- 5 P_{0} V_{0}}{8}$