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Ideal Gas Problem

  1. Sep 28, 2013 #1
    1. The problem statement, all variables and given/known data
    Air that initially occupies 0.140m3 (V1) at a gauge pressure of 103.0 kPa (p1) is expanded isothermally to a pressure of 101.3 kPa (p2) and then cooled at constant pressure until it reaches its initial volume. Compute the work done by the air.


    2. Relevant equations
    Wisothermal = nRTln(V2/V1)
    Wisobaric = p2ΔV = p2(V1 - V2)
    Wisothermal + Wisobaric = Wtotal


    3. The attempt at a solution
    We can find V2 easily enough by multiplying p1 and V1 and dividing by p2 because both p1V1 and p2V2 equal nRT. I can now fill in most of the formula:

    nRTln(V2/V1) + p2(V1 - V2) = Wtotal

    Here, I thought I could then just back substitute nRT for either p1V1 or p2V2 and I'd be golden but the answer I'm getting is different from the back of the book (5.6 kJ). I'm getting .001 kJ with plugging these in:

    Wtotal = (103kPa)(0.140m3)ln(0.140m3/.141m3) + (101.3 kPa)(.140m3 - .141m3)

    Unfortunately, if I use the actual pressure, I still don't get the desired answer:

    Wtotal = (204kPa)(0.140m3)ln(0.140m3/.141m3) + (202.3 kPa)(.140m3 - .141m3)

    What am I fundamentally missing here? Thanks in advance.

    *** ANSWER ***

    Wtotal = (204kPa)(0.140m3)ln(0.140m3/.141m3) + (101.3 kPa)(.140m3 - .141m3) = 5.6 kPa
     
    Last edited: Sep 28, 2013
  2. jcsd
  3. Sep 28, 2013 #2

    Doc Al

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    Staff: Mentor

    For one thing, how does gauge pressure relate to absolute pressure?
     
  4. Sep 28, 2013 #3
    Gauge pressure is the difference between the atmospheric pressure and the actual pressure in the object.
     
  5. Sep 28, 2013 #4

    Doc Al

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    Staff: Mentor

    Right. But in your calculation, you used the gauge pressure. (At least that's what you wrote.)
     
  6. Sep 28, 2013 #5
    You're absolutely right, Doc Al. I've updated my original post to reflect that. Thanks! :)
     
  7. Sep 28, 2013 #6

    Doc Al

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    Staff: Mentor

    They may be pulling a sneaky one on you. Only the first pressure given is specified as gauge pressure. Treat the second pressure as actual pressure. (That would make this a much more realistic problem!)
     
  8. Sep 28, 2013 #7
    That was it! Man, it feels like this is a reading comprehension problem more than an actual physics problem. I was really worried that I wasn't understanding something very fundamental about pressures, temperatures and volumes.

    Thanks Doc Al!!!
     
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