 Solution

For a reversible adiabatic process, an ideal gas obeys $T{V}^{\gamma -1}=\text{constant}$. A monatomic gas has heat capacity ratio $\gamma =5/3$. Taking the stated expansion factor to mean $V_f=9V_i$,This assumes the usual quasistatic isentropic process. Heat insulation alone does not determine the final temperature: an ideal-gas free expansion is also adiabatic but leaves the temperature unchanged. The numerical answer uses nine times the original volume, the conventional intended meaning of the source's “nine times larger”.