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  • 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?

    1 AnswerPhysics8 years ago