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Gas expands in a cylinder

  1. Jun 7, 2008 #1
    1. The problem statement, all variables and given/known data
    When gas expands in a cylinder with radius r, the pressure at any given time is a function of the volume: P = P (V). The force exerted by the gas on the piston is the product of the pressure and the area: F = π(r^2)P. Show that the work done by the gas when the volume expands from Volume V1 to Volume V2 is




    2. Relevant equations
    [​IMG]


    3. The attempt at a solution

    I was having a problem doing this question so I just sort of trudged ahead to see if I could get something and I ended up using substitution with u = n(r^2)x and du = nr^2 and then adjusted the limits of integration which were x+a and x. I am fairly certain this is wrong though and would appreciate help.
     
  2. jcsd
  3. Jun 7, 2008 #2

    Redbelly98

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    What is the usual definition of work?
     
  4. Jun 7, 2008 #3
    Integral of Force times distance
     
    Last edited: Jun 7, 2008
  5. Jun 7, 2008 #4

    Redbelly98

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    Yes.

    Can you make substitutions and rearrange terms to get that integral in terms of P and V?
     
  6. Jun 7, 2008 #5
    [​IMG]

    Ok so thats the integral. So I replace (pi)(r^2)x with V ?
     
    Last edited: Jun 7, 2008
  7. Jun 7, 2008 #6

    Redbelly98

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    Um, wait a moment.

    If a force F moves an object over a distance dx, then the work done in that small distance is F*dx.

    For extended distances, we would integrate F * dx

    Try working with that expression ... you were on the right track otherwise.
     
  8. Jun 7, 2008 #7
    Hmm, so I am confused. This is my best guess.

    f * dx = W

    πr^2P * dx = W

    PV = W

    P*dv = W'

    Then you integrate it using the change in volume as the limits of integration?
     
  9. Jun 8, 2008 #8

    Redbelly98

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    Try this:

    [tex]W = \int F dx[/tex]

    and do the same substitution F = P * pi * r^2
     
  10. Jun 8, 2008 #9
    Okay, when I do that I end up, obviously, W = integral P * pi * r^2 * dx. I know I have to get V's in here, but I am unsure if I substitute pi * r^2 * dx with a V or a dv since I am talking about the change in x.
     
  11. Jun 8, 2008 #10
    V = [itex]\pi r^2 x[/itex] where we take the origin to be wherever the piston starts. What's dV?
     
  12. Jun 8, 2008 #11
    So dv is pi*r^2*dx. Now, must I do anything to make the limits of intergration V1 to V2?
     
  13. Jun 8, 2008 #12

    Redbelly98

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    This is a change-of-variable as taught in integral calculus.

    With the substitution V = pi r^2 x, the limits simply change accordingly so that:
    x1 becomes V1, where V1 = pi r^2 x1
    and similarly for x2 and V2.

    In other words,

    [tex]
    \int^{x2}_{x1} ... \ dx
    [/tex]

    becomes


    [tex]
    \int^{\pi r^2 x2}_{\pi r^2 x1} ... \ dV
    [/tex]

    or just

    [tex]
    \int^{V2}_{V1} ... \ dV
    [/tex]
     
    Last edited: Jun 8, 2008
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