Answer
$$
33.3 \mathrm{\ kJ} .
$$
Work Step by Step
We use $$Q=-\lambda_{F} m_{i c e}=W+\Delta E_{\text {int. }}$$ In this case $$\Delta E_{\text {int }}=0 .$$ since $\Delta T=0$
for the ideal gas, then the work done on the gas is
$$
W^{\prime}=-W=\lambda_{F} m_{i}=(333 \mathrm{J} / \mathrm{g})(100 \mathrm{g})=33.3 \mathrm{\ kJ} .
$$
