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Homework Help: Work, reversible expansion

  1. Dec 6, 2011 #1
    1. The problem statement, all variables and given/known data

    Okay i feel like i'm going crazy. Here is a sample problem my professor gave me. But I cannot figure out for my life how he got his answer and it hsould be very simple.


    A 10.0 mol sample of a perfect gas having Cv = 2.0R undergoes a reversible expanion form 5.00 L to 20.0 L at 5.00 atm of pressure. Calculate q, w, delta U, and delta H for this process..

    Solving for work, he has...


    w = -PdV = -5(20-5) = -22.5 L atm = -22.8 J

    HOW ON EARTH does he get those numbers?







    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 6, 2011 #2
    dV is the change in volume, which is (20.0-5.00 L)=15.0 L
    -P is the given pressure x(-1), -P = -5.00 atm
    w = -PdV= -5.00*15.0 L atm = -75.0 L atm
    Since L atm is a unit of work, it can be converted to J:
    1 L atm = 101.325 J, so -75.0(101.325) J = -7.60x10^3 J = w

    I'm not sure what else your professor did to get to his answer, but this is what I got based on what was given.
     
  4. Dec 6, 2011 #3
    dV is the change in volume, which is (20.0-5.00 L)=15.0 L
    -P is the given pressure x(-1), -P = -5.00 atm
    w = -PdV= -5.00*15.0 L atm = -75.0 L atm
    Since L atm is a unit of work, it can be converted to J:
    1 L atm = 101.325 J, so -75.0(101.325) J = -7.60x10^3 J = w

    I'm not sure what else your professor did to get to his answer, but this is what I got based on what was given.
     
  5. Dec 6, 2011 #4
    thank you! I just needed confirmation that I wasn't missing something. That was exactly what I thought the answer should be
     
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