1 mole of ${NH}_{3}$ gas at $27^{\circ} C$ is expanded under reversible adiabatic conditions to make volume 8 times $(γ = 1.33)$ . Final temperature and work done respectively are:

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$150 K, 900 cal$

$150 K, 400 cal$

$250 K, 1000 cal$

$200 K, 800 cal$

answer is A.

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Detailed Solution

For adiabatic process,

$\begin{aligned} T_{1} V_{1}^{γ - 1} = T_{2} V_{2}^{γ - 1} \Rightarrow T_{2} = \frac{300}{8^{0.33}} = 150 K \\ w = \frac{n R Δ T}{γ - 1} = \frac{- 1 \times 2 \times 150}{0.33} = - 900 cal \end{aligned}$

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