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Quasi-static, adiabatic problem

  1. Nov 1, 2013 #1
    Two moles of a monatomic ideal gas are at a temperature of 0°C and a
    (dQ = 0) and quasi-static volume of 45 liters. The gas is expanded adiabatically
    until its temperature falls to - 50°C. What are its initial and final pressures and its final volume?

    I've been beating my head against a wall for a while now on this one.

    I know how to find the initial pressure. That's just an easy PV=nRT problem. I get an answer of 100,956 Pa (0.1MPa). I can't find the final pressures though, totally stuck.

    So what I've done thus far is to utilize the first law in that I know dQ = 0. So I know that dU = -dW. Solving the U = cNRT for initial and final temperatures and taking the difference, I get an energy which I know must be attributed to the work done. This is where I get stuck.

    Not sure what steps to take next. Need some assistance, thanks.

    I thought I could try
    [tex]
    W = - \int PdV
    [/tex]
    which I could do some substitution to get to (eventually),
    [tex]
    W = -nRTln\left(\frac{V_f}{V_i}\right)
    [/tex]
    But when I do that and I set W to the difference in energy I found earlier, I get an answer that doesn't match my book's.
     
  2. jcsd
  3. Nov 1, 2013 #2
    No, but you almost have it. Going back to your equation for the first law:

    dU=cNRdT=-PdV=-(NRT/V)dV

    where c = 3/2

    Cancel NR from both sides, divide both sides by T, and integrate.

    Chet
     
  4. Nov 1, 2013 #3

    vela

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    This doesn't work because that expression for the work done is for isothermal processes. That's not what you have here.
     
  5. Nov 1, 2013 #4
    So this is what I've done on your suggestion (asking for my work to be checked because my answer still isn't correct):

    [tex]
    cNRdT = -\frac{NRTdV}{V}
    [/tex]

    [tex]
    \frac{cdT}{T} = \frac{dV}{V}
    [/tex]

    [tex]
    c \int_{T_i}^{T_f} \frac{dT}{T} = - \int_{V_i}^{V_f} \frac{dV}{V}
    [/tex]

    [tex]
    c\,ln\left(\frac{T_f}{T_i}\right) = - ln\left(\frac{V_f}{V_i}\right)
    [/tex]

    [tex]
    e^{c} (T_f - T_i) = -(V_f - V_i)\,\,\,\,\,\,negatives\,cancel\,here
    [/tex]

    from which I get a result of 224.04. However, the answer is [itex]61E3[/itex]. Still unsure...
     
  6. Nov 1, 2013 #5

    haruspex

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    You exponentiated the LHS wrongly. eab is not eaeb.
     
  7. Nov 1, 2013 #6
    I'm insanely confused then. Is it,
    [tex]
    \left(\frac{T_f}{T_i}\right)^{c}
    [/tex]
     
  8. Nov 1, 2013 #7
    Now you're cookin'. Of course, you follow the same recipe on the RHS of the equation.

    Chet
     
  9. Nov 1, 2013 #8
    Now I'm becoming embarrassed...
    [tex]
    \left(\frac{T_f}{T_i}\right)^{c} = \left(\frac{V_f}{V_i}\right)^{-1}
    [/tex]
    [tex]
    \left(\frac{T_f}{T_i}\right)^{c} = \left(\frac{V_i}{V_f}\right)
    [/tex]
    [tex]
    V_i \left(\frac{T_i}{T_f}\right)^{c} = V_f
    [/tex]
    But I'm still not getting the correct answer.
     
  10. Nov 1, 2013 #9

    haruspex

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  11. Nov 1, 2013 #10
    The given answer (from Callen) is [itex]61 x 10^{3} m^{3}[/itex]. My answer is 0.06095. It almost seems like there's a conversion error but I don't see it. From the initial values: 45 liters = 0.045 cubic meters, 0 C = 273K, and -50 C = 223K.
     
  12. Nov 1, 2013 #11
    Although...

    I'm looking at my PDF (don't have the 'real' book) and it's entirely possible that the negative is smudged out. I've noticed a lot of cosmetic errors in this PDF. That seems reasonable if my equation is correct and I am off by that magnitude. Maybe someone could check my numbers to confirm?
     
  13. Nov 1, 2013 #12

    haruspex

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    No need to check the calculation in detail. It's quite clear that the volume would not expand by more than a factor of 10, probably much less. So 61 litres is reasonable.
     
  14. Nov 1, 2013 #13
    Yeah. That should be 10-3. You did a really nice job on this problem. Too bad the book had that typo.
     
  15. Nov 1, 2013 #14
    Ugh. I can't believe it was a typo. At least I got some lessons out of it. Thank you, both.
     
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