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Homework Help: Thermo: isothermal, reversible expansion of ideal gas

  1. Nov 7, 2016 #1
    1. The problem statement, all variables and given/known data
    Two moles of a monatomic ideal gas are at a temperature of 300K. The gas expands reversibly and isothermally to twice its original volume. Calculate the work done by the gas, the heat supplied and the change in internal energy.
    T = 300K; ΔT = 0
    n = 2; R = 8.314 J K-1 mol-1
    V2 = 2V1
    Reversible process ⇒ work is path independent.

    2. Relevant equations
    PV = nRT
    W = -PdV
    U = W + Q

    3. The attempt at a solution
    PV = nRT ⇒ P = nRT / V
    dW = -PdV ⇒ W = - ∫ PdV = - nRT ∫ [dV / V] = nRT [ln |V2 - V1|] = nRT ln V1

    This isn't very helpful as I have no actual figure for what V1 is.

    I've tried to come up with something based on how the work is path independent (so W = P2V2 - P1V1 = V1[P2 - P1]) but I've just confused myself, as ΔT = 0 means that P1V1 = P2V2 = nRT, so that would mean that W = 0 which isn't right. I must be going wrong with my assumptions somewhere - any pointers would be much appreciated!

    Edit: I'm confident that once I get past the first stumbling block & calculate the work done I can fire ahead & calculate Q and ΔU. Thanks in advance!
  2. jcsd
  3. Nov 7, 2016 #2
    You integrated incorrectly. Try again.
  4. Nov 8, 2016 #3
    Aha! I see what you mean & I get the right answer when evaluating the integral like this:
    nRT [ln |V2 - V1|] = nRT [ln |V2 / V1| ] = nRT [ln | 2V1 / V1| ] = nRT ln[2]

    Thanks for the nudge in the right direction. May I ask: is it ever correct to write
    nRT [ln |V2 - V1|]
    or should it be
    nRT [ln |V2| - ln |V1|]
    when substituting end points in an integral of this form? Are these two expressions equivalent? (I'm now thinking they're not..?)
  5. Nov 8, 2016 #4
    They're not equivalent. You need to review some of the math related to this. When you evaluate results of a definite integral at the limits of integration, you write ##[f(x)]_{x=a}^{x=b}=f(b)-f(a)##, not ##f(b-a)##. Also, the argument of a combined natural log term should have no units.
  6. Nov 8, 2016 #5
    Thanks, that makes it very clear. It's more that I have some uncertainty around logarithms, which I need to revise / practice again; i.e. ln(a) - ln(b) ≠ ln(a-b), which is obvious to me now but was the mistake I made yesterday without questioning.

    Appreciate the help, thank you!
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