One mole of an ideal monoatomtic gas in an intial state a with pressure Pi and volume Vi is to be taken to a final state d with Pf=B2Pi and Vf=Vi/B through the path a→b→c→d as shown in figure below for a particular value of B(>1) . Here a→b and c→d are adiabatic paths while b→c is an isothermal process with temperature T0. States b and c correspond to (P1,V1) and (P2,V2) respectively.

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One mole of an ideal monoatomtic gas in an intial state a with pressure $P_{i}$ and volume $V_{i}$ is to be taken to a final state d with $P_{f} = B^{2} P_{i} a n d V_{f} = V_{i} / B$ through the path $a \to b \to c \to d$ as shown in figure below for a particular value of $B (> 1)$ . Here $a \to b a n d c \to d$ are adiabatic paths while $b \to c$ is an isothermal process with temperature $T_{0}$ . States b and c correspond to $(P_{1}, V_{1}) a n d (P_{2}, V_{2})$ respectively.

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The ratio of $\frac{V_{2}}{V_{1}} i s \sqrt{B}$

The ratio of $\frac{V_{2}}{V_{1}} i s \sqrt{B^{3}}$

Work done by the gas is $\frac{R T_{0}}{2} \ln B + \frac{3}{2} (R T_{0} - B P_{i} V_{i})$

Work done by the gas is $\frac{R T_{0}}{2} \ln B + \frac{3}{2} P_{i} V_{i} (1 - B)$

answer is A, C.

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Detailed Solution

$a b : p_{i} v_{i}^{γ} = p_{1} v_{1}^{γ} - - - (i) c d : p_{f} v_{f}^{γ} = p_{2} v_{2}^{γ} - - - (i i) b c : p_{1} v_{1} = p_{2} v_{2} - - - (i i i)$
Dividing (2) by (1), we get
$\frac{v_{2}}{v_{1}} = B^{\frac{2 - γ}{γ - 1}} = \sqrt{B} [γ = \frac{5}{3}]$
Work done by the gas
$W = W_{a b} + W_{b c} + W_{c d} = \frac{R T_{0}}{2} \ln B + \frac{3}{2} p_{i} v_{i} (1 - B)$