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A text problem

  1. Jun 3, 2003 #1
    I tried to solve it by knowing that a, b, n, and R are constants, so only V, T are variables.

    So I did this:

    P = nR(dT/d(V-nb)) - ((an^2)*(-2V^-3))

    but I still didn't get the correct answer. I believe I did something wrong, could someone help me out?
  2. jcsd
  3. Jun 3, 2003 #2
    The question says "If gas in a cylinder is maintained at a constant temperature T". So I don't think T is a variable. Does the answer contain somthing like dT/dV? I don't think so because T isn't a variable.
    Last edited: Jun 3, 2003
  4. Jun 3, 2003 #3
    not only does the answer including T, it has a, n, in it too.
  5. Jun 3, 2003 #4
    Is the answer
    -nRTV/(V-nb)2 + (2an2)/V3 ?
  6. Jun 3, 2003 #5
    If T isn't a constant but a variable, I would expect (dT/dV) as part of the answer. (chain rule)

    By the way, remember you need to use quotient rule when differentiate (nRT/(V - nb)) with respect to V as V is in the denominator
    Last edited: Jun 3, 2003
  7. Jun 3, 2003 #6

    dT/dV will not be in the answer, as T is assumed to be constant, so therefore does not depend upon V.
  8. Jun 3, 2003 #7
    Yep how did you get that?
  9. Jun 3, 2003 #8
    T is constant in this question

    =d/dV [nRT/(V - nb) - an2/V2]
    =d/dV [(nRT/(V - nb)] - d/dV (an2/V2)
    now take all the constants out to the left hand side of d/dV
    =nRT*d/dV [1/(V-nb)] - an2* d/dV (1/V2) .......................(1)

    The blue part:
    [1/(V-nb)] = (V-nb)-1
    d/dV [1/(V-nb)] = -1*(V-nb)-2 = - 1/(V-nb)2
    (the power rule)

    d/dV [1/(V-nb)]
    = [(V-nb)d/dV (1) - 1*d/dV (V-nb)]/(V-nb)2
    (the quotient rule)
    = (0-1)/(V-nb)2
    = - 1/(V-nb)2

    the green part
    d/dV (1/V2)
    = -2V-3

    I think you can do it because you got it right in your first post

    Substitute the blue part and green part back to (1), then you'll get the answer.
  10. Jun 4, 2003 #9
  11. Jun 4, 2003 #10
    Is that yoda guy smart or what? WOW
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