One mole of an ideal monoatomic gas initially at 800 K and 64 atm is expanded to a final state at 200 K and 1 atm. To achieve the above change a reversible path is constructed which involves an adiabatic expansion in the beginning followed by an isothermal expansion to the final state. The magnitude of net work done by the gas (in cal) is
$(ln2=0.7), ({R=2 cal mol}^{-1} k^{-1})$

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answer is 2080.

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Detailed Solution

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For AB path:
$P_{2} = P_{1} {(\frac{T_{1}}{T_{2}})}^{\frac{1}{\frac{1}{7} - 1}} = 64 {(\frac{800}{200})}^{(\frac{1}{\frac{3}{5} - 1})} = 2 a t m {Total W}_{done} {=W}_{AB} {+W}_{BC} = n C_{v} Δ T + (- n R T \ln \frac{P_{B}}{P_{C}}) = 1 \times \frac{3}{2} R \times (200 - 800) + [- 1 \times R \times 200 \ln \frac{2}{1}] = (- 900 R) + (- 140 R)$
=2080.00 cal