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Homework Help: Work Done by Air pressure difference

  1. May 4, 2010 #1
    Work Done by Air "pressure difference"

    1. The problem statement, all variables and given/known data
    Air that initially occupies .14 m^3 at a gauge pressure of 103.0kPa is expanded isothermally to a pressure to 101.3 kPa and then cooled at constant pressure until it reaches its initial volume. Compute the work done by the air. (Gauge pressure is the difference between the actual pressure and atmospheric pressure.)


    2. Relevant equations
    W = V2 [tex]\int[/tex] V1 (pdv)

    pV=nRT


    3. The attempt at a solution
    The professor did some simplification on the board and it came down to nRT(ln)(V_2 / V_1)

    then I solved for V2 and V1 V=(nRT)/p "the nRT cancels" and then I have this equation

    piV1(ln)(pi / pf) = .239 kJ =Work

    The answer in the back of the book is 5.6 kJ, so if someone could help me out where I'm going wrong this would be highly appreciated.
     
  2. jcsd
  3. May 4, 2010 #2
    Re: Work Done by Air "pressure difference"

    help? please?
     
  4. May 6, 2010 #3

    Redbelly98

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    Staff Emeritus
    Science Advisor
    Homework Helper

    Re: Work Done by Air "pressure difference"

    239 J is the work done during isothermal expansion. I would think we need to account for the constant pressure compression as well.

    (Even so, 5.6 kJ doesn't look right to me.)
     
  5. May 7, 2010 #4
    Re: Work Done by Air "pressure difference"

    For a polytropic process of an ideal gas with isothermal expansion (n=1) you can also use the expression for work=P1*V1*ln(V2/V1) which gives a similar result of 284.2 J However that is the total area under the curve from pt 1 to pt 2. There is also the area (work) under the curve when the air is cooled at constant pressure to the initial volume that must be subtracted from the above work to get the area of the complete process. If that is the case it get a very small total work of about 1.4 J. In any case 5.6 KJ seems way to large.
     
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