Isobaric Process

Following NCERT Class 11, Chapter 12 (Thermodynamics), the work done by a gas in any process is the area under its path on a pressure-volume diagram, written generally as $\int PdV$. For an isobaric process the pressure stays constant, so this integral simplifies to $\Delta W=P\Delta V=P(V_f-V_i)$, the pressure multiplied by the change in volume. Here the change in volume is $\Delta V=0.03-0.01=0.02$ m${}^3$ at a fixed pressure of $2\times 10^5$ Pa. Substituting, $\Delta W=2\times 10^5\times 0.02=4000$ J. Because the gas expands against a constant external pressure, this positive work is done by the gas on the surroundings as it pushes the piston outward. The option 2000 J is wrong because it uses only half the volume change or the wrong endpoint. The option 8000 J is wrong because it doubles the correct volume change, perhaps by mistakenly using $0.04$ m${}^3$. The option 400 J is wrong by a factor of ten, a common decimal-place error in the volume term. A plausibility check on units confirms the result: $\text{Pa}\times {\text{m}}^3=({\text{N/m}}^2)({\text{m}}^3)=\text{Ncdotpm}=\text{J}$, and the magnitude 4000 J is reasonable for a 20-litre expansion at roughly two atmospheres of pressure.