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At 27^oC two moles of an ideal monoatomic gas occupy a volume V. The gas expands adiabatically to a volume 2 V. Calculate the final temperature of the gas.
[Given 2^2/3 = 1.59]

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190 K

180 K

200 K

170 K

answer is B.

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Detailed Solution

$\begin{array}{l} T_{1} {V_{1}}^{γ - 1} = T_{2} V_{2}^{γ - 1} \\ 300 \times V^{2 / 3} = T (2 V)^{2 / 3} \\ T = \frac{300}{2^{2 / 3}} = \frac{300}{1 .59} \approx 190 K \end{array}$

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