# Thermodynamics pV work?

So in my thermo class, we recently found that work is equal to the integral of pdV (pressure multiplied by the differential of volume). This makes sense to me. However, if we want to measure a change in the values of pV, which is essentially what we're doing by taking the integral of pdV, it would also make sense to me that work would equal the integral of Vdp (volume multiplied by the differential of pressure). Why is this not the case?

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• Anonymous
7 years ago

because pdV work is done in the context of the closed cylinder system.

vdp work appears in open system, steady flow analysis -- like in work done on a turbine by a steady flow through it.

in particular, that's why you consider the Maxwell relation

du = T dS - P dv

when working with piston-cylinder arrangements, but use the Maxwell relation

dh = S dT - v dP

when doing steady flow analysis -- like through a throttle or a turbine or a steady flow axial compressor

what relates them is the definition h = u + Pv

enthalpy = (internal energy) + pressure * (specific volume)

if you do the math, you should find that you can use one Maxwell relation plus the definition of enthalpy to get the other Maxwell relation.