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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.
    So:
    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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