Question

An ideal monatomic gas at 300 k expands adiabatically and reversibly to twice its volume. what is its final temperature?

Answer 1

In an adiabatic process, the following relationship holds:
$TV^{\gamma -1} = cost.$
where T is the gas temperature, V is the volume and $\gamma$ is the adiabatic index, which is equal to $\gamma = \frac{5}{3}$ for a monoatomic gas.

We can re-write the equation as
$T_1 V_1^{\gamma -1} = T_2 V_2^{\gamma -1}$
where the labels 1,2 refer to the initial and final conditions of the gas.
Let's rewrite it for $T_2$, the final temperature:
$T_2 = T_1 ( \frac{V_1}{V_2} )^{\gamma-1}$

We can now substitute the initial temperature, T1=300 K, and $V_2 = 2V_1$, because the final volume is twice the initial one. So we find the value of the final temperature:
$T_2 = 300 K( \frac{1}{2})^{ \frac{2}{3} } =189 K$