Expansion of a gas in vacuum is called free expansion. Calculate the work done and...

Work done in vacuum is calculated by:
$w=-P_{\text{ext}}(V_{\text{final}}-V_{\text{initial}})$
For the given problem: $P_{\text{ext}}=0$, because it is a free expansion (in the vacuum).
Hence,
$$\begin{gathered} w=-(0\times (5-1)) \\ w=0 \end{gathered}$$
According to joule experimental observation, For an isothermal expansion of ideal gas, no heat will be absorbed or evolved, hence q=0.
By the 1st law of thermodynamics,
$$\begin{gathered} q+w=\mathrm{△}U \\ \mathrm{△}U=0 \end{gathered}$$
Since, w=0 and q=0, then the change in internal energy $\mathrm{△}U$ will also be 0.