Isobaric Process

Question

An ideal gas expands at a constant pressure of 2 × 10^5 Pa, increasing its volume from 0.01 m^3 to 0.03 m^3, so how much work does the gas do?

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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 $\int P\,dV$. 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 $2	imes10^5$ Pa. Substituting, $\Delta W = 2	imes10^5 	imes 0.02 = 4000$ 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: $	ext{Pa} 	imes 	ext{m}^3 = (	ext{N/m}^2)(	ext{m}^3) = 	ext{N·m} = 	ext{J}$, and the magnitude 4000 J is reasonable for a 20-litre expansion at roughly two atmospheres of pressure.

Isobaric Process

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