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Homework Help: Work Done (Thermodynamics)

  1. May 4, 2010 #1
    1. The problem statement, all variables and given/known data
    A gaseous system has an internal energy U which can be expressed

    U=2.5pV +constant
    The system is initially in A. (attatchement)
    The system goes from A to B ,B to C, C to A.
    How can we find the work done on the Gas and the heat input to the gas in each for the three processes.
    The total work done for the cycle is zero?

    2. Relevant equations
    W(Intergral from A to B)[pdV] is a possible candidate equation. Can also the general equationU=2.5pV+constant be related with the equation U=W+Q ?!


    3. The attempt at a solution
    I suspect that the total work done i.e Wabca must be zero. Also to find the work from B to C ,maybe the area under the line B C can be used.My current physics experience does not allow me to think anything else that can help.
     

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    Last edited: May 4, 2010
  2. jcsd
  3. May 4, 2010 #2
    Please show relevant equations and an attempt at the solution.
     
  4. May 4, 2010 #3
    You know that the work is the area under the curve. So figure out all the areas under the curves you're taking. As a hint, work isn't zero, but what is?
     
  5. May 4, 2010 #4
    If work is area under the corresponding curves the half problem is soilved
    In this case im not stupid,(well in every case im not stupid :) ) the TOTAL WORK DONE should be equal with the surface area of the triangle in attachement
    Also work from A to B must be conventionally negative.

    Thanks

    What about the heat input and the equation u=2.5pV+ ct ?
    Any hints here? Obviously we MUST use this equation in this problem
     
  6. May 4, 2010 #5
    Right, so the work done is the area of the triangle, and make sure you get the sign right.

    My old hint still applies. The total work done isn't zero, but is something else zero?
     
  7. May 4, 2010 #6
    Just fot the record and not to confuse other readers the total work done is negative ,since it moves in the counterclockwise(anticlockwise) direction.

    The first law of thermodynamics dictates that the net heat input is equal to the net work output over any cycle.(wikipedia)

    Therefore total work done=total heat input

    Also W (C to A) is zero .
     
  8. May 5, 2010 #7
    Well, the key that you're looking for is that you know that the energy change for the complete cycle in this case is zero because it returns to the same state. Net heat input is equal to net work output is another way of saying [itex]\Delta U = 0[\itex].
     
  9. May 5, 2010 #8

    physicsworks

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    Gold Member

    The most interesting thing in this problem is to calculate the input heat in the BC process.
    In fact, there is a point (that is, some p and V) on the BC, before which the heat is extracted and after which the heat is absorbed.
    To do this, use the first law of thermodynamics in its original form, substitute the equation of the line BC (that is some linear function [tex]p=p(V)[/tex]), and rewrite [tex]dU = C_v dT[/tex] using ideal gas law
    [tex]d(pV) \equiv pdV + V dp = \nu R dT[/tex]
    You'll get something like
    [tex]\delta Q = (c_1V + c_2)dV[/tex]
    where
    [tex]c_1[/tex],[tex]c_2[/tex]
    are constants.
     
    Last edited: May 5, 2010
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