Analysis
The gas expands against a constant external atmospheric pressure of $10^5$ Pa. The process ends when the piston stops, implying mechanical equilibrium is restored ($P_{gas} = P_{atm}$). Thus, the expansion can be treated as an Isobaric Process ($P = \text{constant}$) for the purpose of calculating energy changes associated with the volume change.
Calculation
For an isobaric process, the heat supplied $Q$ is given by the change in enthalpy:
$$ Q = n C_P \Delta T $$Using the ideal gas law $PV = nRT$, at constant pressure $P \Delta V = nR \Delta T$.
We can rewrite the heat equation as:
$$ Q = \frac{C_P}{R} (nR \Delta T) = \frac{C_P}{R} (P \Delta V) $$For a mono-atomic gas, $C_P = \frac{5R}{2}$, so $\frac{C_P}{R} = 2.5$.
Given: $P = 10^5 \text{ Pa}$ and $\Delta V = 0.2 \text{ L} = 0.2 \times 10^{-3} \text{ m}^3$.
$$ Q = 2.5 \times (10^5 \times 0.2 \times 10^{-3}) $$ $$ Q = 2.5 \times 20 = 50 \text{ J} $$Correct Option: (c) 50 J