= Solution
For the intended reversible <adiabatic process>, an <ideal gas> obeys $TV^{\gamma-1}=\text{constant}$. A <monatomic gas> has <heat capacity ratio> $\gamma=5/3$, so
$$
T_f=T_i\left(\frac{V_i}{V_f}\right)^{\gamma-1}
=400\left(\frac18\right)^{2/3},\qquad
\boxed{T_f=100\ \mathrm K.}
$$
The required extra assumption is that “adiabatic” denotes the usual quasistatic <isentropic process>. Absence of heat transfer alone would not fix the final <temperature>: an <ideal gas> undergoing <free expansion> into vacuum is also adiabatic but leaves its <temperature> unchanged.
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