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Need Timely Help with Unique Work by a Variable Force Problem

  1. Sep 6, 2006 #1
    Hi, I'm taking a university extension program, so I can't talk with my professor. The Calc teacher at my high school doesn't understand the problem either.

    The Problem:

    Suppose that the gas in a circular cylinder of cross section area A is being compressed by a piston. If p is the pressure of the gas in psi, and V is the volume in cubic inches, the work done in compressing the gas from state (P1,V1) **(P sub 1, V sub 1)** to state (P2,V2) is given by the equation:

    Work= [definite integral from (P2,V2) to (P1,V1)] p*dV


    Find the work done in compressing the gas from V1=243 cubic inches to V2=32 cubic inches if P1=50 lb/cubic inch AND p and V obey the gas law pV^(1.4)=constant.


    I cannot make heads or tails of this problem. I have never seen a coordinate pair used as a limit of integration.
    I first thought that I should use the gas constant to make those limits into usable numbers, but P2 is not given.
    I then thought that maybe I should use the gas constant (using P1 and V1) to find P2 (by setting P2(V2)^1.4 equal to the number given by plugging in P1 and V1). This creates a problem, making the upper and lower limit the same (109,350 is the exact number). This would defeat the purpose of integrating the equation, because the result would obviously be 0.
    Any thoughts? Any help is greatly appreciated. My email is M_Daun@msn.com
     
  2. jcsd
  3. Sep 10, 2006 #2

    Astronuc

    User Avatar

    Staff: Mentor

    It is a matter of putting one variable in terms of the other.

    Starting with the definition of work, which is the integral of force applied over distance

    [tex]W\,=\,\int_{x_1}^{x_2}\,F\,dx[/tex], but F = PA, where P is the pressure and A is area. Remember that F, P and x are vectors (x is the displacement vector.

    Now the Work equation becomes

    [tex]W\,=\,\int_{x_1}^{x_2}\,PA\,dx[/tex]

    Now Volume V = Ax, and dV = A dx, and Vi corresponds to xi

    [tex]W\,=\,\int_{V_1}^{V_2}\,P\,dV[/tex]

    Now one is given that

    [tex]PV^\gamma\,=\,k[/tex] or

    [tex]P\,=\,kV^{-\gamma}[/tex], and substituting into the Work equation

    [tex]W\,=\,\int_{V_1}^{V_2}\,kV^{-\gamma}\,dV[/tex]

    k is known since one knows one of the state points P1, V1. Knowing either P2 or V2, one can solve for the other using the relationship between P and V.
     
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