In the reversible adiabatic expansion of an ideal monoatomic gas, the final volume is 8 times the initial volume. The ratio of final temperature to initial temperature is

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$8 : 1$

$4 : 1$

$1 : 2$

$1 : 4$

answer is B.

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Detailed Solution

$\begin{array}{l} T_{1} V_{1}^{γ - 1} = T_{2} V_{2}^{γ - 1} \\ \frac{T_{2}}{T_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ - 1} \Rightarrow {(\frac{1}{8})}^{\frac{5}{3} - 1} \\ \Rightarrow {(\frac{1}{8})}^{3 / 2} \Rightarrow \frac{1}{4} \end{array}$

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