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Relation between P, V, T, Gamma in Adiabatic Proceses

The work done...

Question

The work done in adiabatic compression of 2 mole of an ideal monoatomic gas by constant external pressure of 2 atm starting from initial pressure of 1 atm and initial temperature of 300K is:
(Take $R = 2 c a l / K . m o l$ )

360 cal

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720 cal

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800 cal

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550 cal

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Solution

The correct option is B 720 cal
Given:
$n = 2 m o l P_{e x t} = 2 a t m = P_{2} P_{1} = 1 a t m T = 300 K, R = 2 c a l / K . m o l$

for monoatomic gas, $C_{v} = \frac{3}{2} R$
for adiabatic process $q = 0 ∴ f r o m I s t l a w d U = w$
so, $n C_{v} d T = - P_{e x t} d V$
$n C_{v} (T_{2} - T_{1}) = - P_{e x t} (V_{2} - V_{1})$
$n C_{v} (T_{2} - T_{1}) = - P_{e x t} (\frac{n R T_{2}}{P_{2}} - \frac{n R T_{1}}{P_{1}})$
$n \frac{3}{2} R (T_{2} - 300) = - 2 (\frac{n R T_{2}}{2} - \frac{n R \times 300}{1})$

$\frac{3}{2} (T_{2} - 300) = - T_{2} + 600$
$T_{2} = 420 K$

so, work done = $n C_{v} (T_{2} - T_{1})$
$= 2 \times \frac{3}{2} \times 2 \times (420 - 300)$
$w = 720 c a l$

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