ThermodynamicsHard
Question
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
Options
A.8: 1
B.1: 4
C.1: 2
D.4: 1
Solution
$T_{1}V_{1}^{\gamma - 1} = T_{2}V_{2}^{\gamma - 1} \Rightarrow \frac{T_{2}}{T_{1}} = \left( \frac{V_{1}}{V_{2}} \right)^{\gamma - 1} = \left( \frac{1}{8} \right)^{5/3 - 1} = \frac{1}{4}$
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