= Solution
For a monatomic <ideal gas>, $\gamma=C_P/C_V=5/3$. If the expansion is reversible adiabatic, $TV^{\gamma-1}$ is constant, so
$$
\boxed{T_f=300\times8^{-2/3}=75\,\mathrm K\quad\text{(isentropic expansion).}}
$$
Adiabatic isolation alone does not imply this <temperature> law. For example, <free expansion> into vacuum has $Q=W=0$, so an <ideal gas> remains at $300\,\mathrm K$. Thus $75\,\mathrm K$ is the usual intended answer with the additional isentropic assumption; the literal adjective “adiabatic” does not uniquely select it.<Thermodynamics and Statistical Mechanics by Richard Fitzpatrick>[“Isothermal and Adiabatic Expansion”, Eqs. (6.53) and (6.61); “Free Expansion of Gas”, Eqs. (6.144)–(6.148)]{ref}
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