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

The gas is cooled at a constant pressure, making this an isobaric process.

The work done in an isobaric process is $W = nR(T_3 - T_2)$.

From the initial state, the gas constant factor can be evaluated as $nR = \dfrac{P_1 V_1}{T_1} = \dfrac{100}{300} = \dfrac{1}{3} \text{ J/K}$.

The final temperature is $T_3 = 300 \text{ K}$ and the initial temperature for this phase is $T_2 = 300\sqrt{2} \text{ K}$.

Substituting these values gives $W = \dfrac{1}{3} (300 - 300\sqrt{2}) = 100(1 - \sqrt{2}) \text{ J}$.

Using $\sqrt{2} \approx 1.414$, the work done is $100(1 - 1.414) = -41.4 \text{ J}$.