$P - T$ diagram of one mole of an ideal monoatomic gas is shown. Processes AB and CD are adiabatic. Work done in the complete cycle is

2.5 R T

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- 2 R T

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1.5 R T

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- 3.5 R T

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Solution

The correct option is B $2.5 R T$
let $A = 1, B = 2, C = 3, D = 4$
now $P_{1} = P_{4} = P$ & $P_{2} = P_{3} = 2 P$

so by ideal gas equation we have
$V_{1}$ = $\frac{3 R T}{P}$ $V_{2}$ = $\frac{R T}{2 P}$

now,
1) for process 1-2
$W_{1 - 2}$ = $\frac{P_{1} V_{1} - P_{2} V_{2}}{γ - 1}$
so $W_{1 - 2} = \frac{2 P \frac{R T}{2 P} - P \frac{3 R T}{P}}{0.4} = - 5 R T$

2)for process 3-4
$W_{3 - 4} = \frac{2 P \frac{2 R T}{P} - P \frac{5 R T}{P}}{0.4} = - 2.5 R T$

3) for process 4-1
$W_{4 - 1} = P_{1} (V_{4} - V_{1})$
$W_{4 - 1} = P [\frac{5 R T}{P} - \frac{3 R T}{P}] = 2 R T$

4) for process 2-3
$W_{2 - 3} = 2 P [\frac{2 R T}{P} - \frac{R T}{2 P}] = 3 R T$

now adding[1,2,3,4] we get the net work done
$W_{t o t a l}$ = $W_{1 - 2} + W_{2 - 3} + W_{3 - 4} + W_{4 - 1}$
= $- 5 R T + 3 R T + - 2.5 R T + 2 R T = - 2.5 R T$

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