An ideal gas ( γ = 1.5 ) starting at 1 litre, 300 K , and 100 kPa...

The gas undergoes an isothermal expansion at $T = 300 \text{ K}$ to reach its original volume $V_4 = 1 \text{ litre}$.

The initial volume for this process is found from the previous isobaric cooling where $V_3 = V_2 \dfrac{T_3}{T_2} = 0.5 \times \dfrac{300}{300\sqrt{2}} = \dfrac{1}{2\sqrt{2}} \text{ litre}$.

The work done in an isothermal process is $W = nRT \ln\\!\ \left(\dfrac{V_4}{V_3}\right)$.

Since $nRT = P_1 V_1 = 100 \text{ J},$the work equation becomes W = 100 \ln\\!\left(\$\dfrac{1}{1 / (2\sqrt{2})}$\right) = 100 \ln(2\$\sqrt{2}).<br><br>This simplifies to$W = 100 \ln(2^{1.5}) = 150 \ln 2$.<br><br>Using$\ln 2 \approx0.693$, the work done is$150 \$\times$0.693 \approx103.9 \text{ J}.