An ideal gas at 27^oC is compressed adiabatically to 8/27 of its original volume. If $γ$ = 513, then the rise in temperature is

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405 K

225 K

450 K

375 K

answer is B.

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Detailed Solution

For an adiabatic process ${TV}^{γ - 1} = constant$ . Therefore, $\frac{T_{1}}{T_{2}} = {[\frac{V_{2}}{V_{1}}]}^{γ - 1}$

$\begin{array}{l} \Rightarrow T_{2} = T_{1} {[\frac{V_{1}}{V_{2}}]}^{γ - 1} = 300 {[\frac{27}{8}]}^{5 / 3 - 1} = 300 {[\frac{27}{8}]}^{2 / 3} = 675 K \\ \Rightarrow ΔT = 675 - 300 = 375 K \end{array}$

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