Isobaric Process

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Inside a frictionless piston the gas expands from a volume of 2.0 litres to 5.0 litres while the surrounding atmosphere keeps the pressure steady at $1.0	imes10^{5}	ext{ Pa}$. How much work does the gas perform?

RankGuru · Accepted Answer

For a process carried out at constant pressure, the work done by the gas simplifies to $W = P\,\Delta V$, because pressure can be pulled outside the integral $\int P\,dV$. The volume change must be expressed in SI units: $\Delta V = 5.0 - 2.0 = 3.0	ext{ L} = 3.0	imes10^{-3}	ext{ m}^3$. Multiplying gives $W = (1.0	imes10^{5})(3.0	imes10^{-3}) = 300	ext{ J}$. The value 150 J arises from halving the volume change without justification. The value 500 J wrongly uses the final volume of 5 litres instead of the difference. The value 30 J results from forgetting to convert litres to cubic metres, dropping a factor of ten. This constant-pressure expansion is also called isobaric, and on a pressure-volume diagram its work equals the rectangular area beneath the horizontal line joining the initial and final states. As a plausibility check, the gas expands so it does positive work on the surroundings, and the magnitude of a few hundred joules is reasonable for an everyday atmospheric expansion of three litres.

Isobaric Process

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