One mole of an ideal diatomic gas undergoes a process AB as shown. Pressure at A and B are $5 P_{0}$ and $P_{0}$ while volumes are $V_{0}$ and $5 V_{0}$ respectively. On the basis of the given information answer the following question (s).
During the process from $A \to B$ , pressure and volume of the gas, where the direction of flow of heat is reversed, is

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$\frac{5}{2} P_{0}, \frac{7}{2} V_{0}$

$P_{0}, 5 V_{0}$

$4 P_{0}, 2 V_{0}$

$\frac{9}{2} P_{0}, \frac{3}{2} V_{2}$

answer is A.

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Detailed Solution

Temperature of gas at A and B is equal so $Δ U = 0.$
Equation of process

$P = - \frac{P_{0}}{V_{0}} V + 6 P_{0}$ ... (1)

Heat reverse its direction when slope of line should be equal to $\frac{γ P}{V} = \frac{P_{0}}{V_{0}}$ ... (2)
From equation (1) and (2)
$V = \frac{7}{2} V_{0}, P = \frac{5}{2} P_{0}$

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