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 300 K. (R = 2 cal/mol-degree)

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360 cal

720 cal

800 cal

1000 cal

answer is B.

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Detailed Solution

Work done,

$\begin{array}{l} w = - P_{ext} (V_{2} - V_{1}) = - 2 [\frac{2 \times R \times T}{2} - \frac{2 \times R \times 300}{1}] \\ Δ U = n C_{v} (T_{2} - T_{1}) = 2 \times \frac{3}{2} \times R [T - 300] \\ Δ U = w \\ \Rightarrow 3 T - 900 = - 2 T + 1200 \\ \Rightarrow T = 420 K \\ w = \frac{2 \times 2 \times 120}{\frac{2}{3}} (∵ w = \frac{+ n R Δ T}{γ - 1}) \\ = 720 cal \end{array}$

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