Discussion
A certain mass of gas at 0°C is expanded to 81 times its original volume under adiabatic conditions. If ratio of specific heats of the gas, \(\gamma = 1.25\), the final temperature of the gas is
Answer: B
The final temperature is -182°C.
For a reversible adiabatic process \(TV^{\gamma - 1}\) is constant:
\[ \begin{aligned} T_{2} &= T_{1}\left(\frac{V_{1}}{V_{2}}\right)^{\gamma - 1} = 273\left(\frac{1}{81}\right)^{0.25} \\ &= \frac{273}{3} = 91\ \text{K} \end{aligned} \]
since \(81^{0.25} = 3\). In Celsius, 91 - 273 = -182°C.
-91°C comes from writing 91 K as -91°C without converting, and 0°C would mean no cooling, which an adiabatic expansion cannot give. -235°C does not follow from the adiabatic relation.
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