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Calculation of work (P.dV) in an isothermal system

  1. Oct 25, 2012 #1
    Here's the textbook way of calculating the work done by an ideal gas in an isothermal case.

    ∴ ∫P.dV=nRT∫dV/V
    → W2-W1=nRT*ln(V2/V1)

    My question.
    Consider a cylindrical (or of any other shape) container of surface area A and a frictionless movable piston attached.
    Let the coordinate x run along the length of this container, with origin at the bottom of the container.
    Let the piston be at some x at some point of time.

    At this x, P.V= constant = nRT
    This holds for all x.
    Now since V(x)=A.x
    P(x)=constant/A.x=K/x .... for some contant k (=nRT/A)

    Now P and V are both functions of x.
    While calculating the infinitesimal work P.dV, how can we treat P(x) constant over the range dx?

    Instead, since P(x)V(x)=constant
    differentiating both sides, d[P(x)V(x)]=0
    → V(x).dP + P(x).dV=0
    ∴ P(x).dV= -V(x).dP

    Shouldn't this relation hold?

    I ask this because I was solving a similar problem, although which wasn't isothermal, involved P and V that depended on x and the approach used there was the second one.
  2. jcsd
  3. Oct 25, 2012 #2
    The relationship holds and if you integrate -VdP instead of pdV you get the same result, don't you?
    I am not sure what is the question...
  4. Oct 28, 2012 #3
    Usually, in calculus, functions of x are taken as being constant along any differential displacement dx.
  5. Oct 28, 2012 #4
    Any rigorous mathematical explanation you could point me to? I know that if dx is infinitesimally small, f(x) will not change much in dx, but it will change neverthless. So i am looking for a rigorous mathematical explanation of why we can treat it as constant.
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