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Heat transfer processes in ideal gas PV diagram

  1. Sep 6, 2011 #1
    1. The problem statement, all variables and given/known data
    Note: I know the wording of this post may be lengthy but the problem shouldn't take too long in theory and it's a basic problem from an introductory thermodynamics class so I desperately wish to understand this beginning part so I don't fall behind, so I'd REALLY appreciate any guidance. Thank you!
    https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0BwVa69357vahZGExMTU2NWItZTIyMi00NWU4LTg2N2UtODhhOTcwNDhmZjA4&hl=en_US [Broken]


    2. Relevant equations
    [itex]\Delta[/itex]E = Q + W
    PV = nRT


    3. The attempt at a solution
    I'll put up what my study group has worked through and if anyone can tell me if that's the right way to do it or if there is a simpler way, please do so!

    D to X : Because it's isothermal, TD = TX, and because in an ideal gas, energy only depends on temperature, [itex]\Delta[/itex]EDX = 0, thus QDX = -WDX
    C to X : Because adiabatic means no heat transfer, [itex]\Delta[/itex]ECX = W
    B to X: No change in volume, so no work, so [itex]\Delta[/itex]EBX = Q

    Now, we were wondering - why is it important to know that A - B is an adiabatic process? So we took it to be a cycle.. X to A to B back to X, and overall [itex]\Delta[/itex]E = 0. So..
    [itex]\Delta[/itex]E = -EXA + EAB + EBX = 0.
    EXA = EAB + EBX
    QXA + WXA = WAB + QBX
    QXA = WAB + QBX - WXA

    Since QXA equals all of that, QXA > QBX.

    Then... the more work you do, the more you heat you need to do the work. Comparing DX to BX, BX has no work and DX does have an "area under the curve" (i.e. work) so QDX > QBX

    Then... we took another loop, from X to D to B to A back to X, so that net energy is 0 as well, so:
    WAX + QAX + WXD - QXD + WDB + QDB + WAB = 0

    If we move QXD to the right side, it is just equal to everything above on left side. Because it is equal to all of that, it must mean that QXD > QAX (since QAX is on the left side).


    Thus... QDX > QXA > QXB > QXC

    Does that look correct? We couldn't find any similar examples online or in our textbooks and this is the first kind of this problem we've run into. We'd appreciate any guidance please :)
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Sep 8, 2011 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    Corrrect. W refers to the work done ON the gas. You are using the first law in the form:
    [itex]\Delta Q = \Delta U - W[/itex] where W is the work done ON the gas. So long as you remember that dW = -PdV, it is not a problem.
    Correct. W = work done ON the gas.
    Correct.
    Knowing that AB is an adiabatic process allows you to compare the change in internal energy in A-X to the change in internal energy in B-X. If the process from B to A involved heat flow into the gas you would not know whether the temperature at A was higher or lower than at B. So you would not be able to say whether the increase in internal energy plus the work done by the gas (-W done ON the gas) in the isobaric process from A to X (ie. QAX) was greater than the heat flow into the gas in the isochoric process from B to X. Since it is adiabatic the temperature at A is lower than at B so you know that the change in internal energy is greater from A to X than from B to X. Since there is also work done from A-X you know that the heat flow from A-X is greater than from B-X.
    Careful. What is the direction of the process? Is work being done on or by the gas? If work is being done on the gas in an isothermal process, what direction is the heat flow? Into or out of the gas? If it is into the gas it is positive but if it is out of the gas it is negative.

    Not correct. Be careful about the signs. You don't have to quantify the work. The signs will tell you the order after you figure out that Qax > Qbx

    AM
     
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