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Homework Help: PV Diagram Help

  1. Mar 21, 2012 #1
    1. The problem statement, all variables and given/known data
    Untitled picture.png

    the diagram above of pressure P versus volume V shows the expansion of 2.0 moles of a monatomic ideal gas from state A to state B. As shown in the diagram PA = PB = 600 N/m2 , VA = 3.0 m3 and VB = 9.0 m3.

    i) Calculate the work done by the gas as it expands
    ii) Calculate the change in internal energy of the gas as it expands.
    iii) Calculate the heat added to or removed from the gas during this expansion.
    2. Relevant equations
    W = PΔV
    U = [itex]\frac{3}{2}[/itex]nRT
    PV = nRT

    3. The attempt at a solution

    i) W = PΔV
    W = 600(9.0 - 3.0)
    W = - 3600J
    I put negative to indicate work done by system

    ii) Here is where I have problems, I'm not sure how to calculate internal energy without given temperature. Here is my failed attempt.

    UA = [itex]\frac{3}{2}[/itex]nRTA
    Since PV = nRT
    UA = [itex]\frac{3}{2}[/itex]PAVA

    ΔU = UB - UA
    = [itex]\frac{3}{2}[/itex]PBVB - [itex]\frac{3}{2}[/itex]PAVA
    Since PA = PB = P
    = [itex]\frac{3}{2}[/itex]P(VB - VA)
    = [itex]\frac{3}{2}[/itex]PΔV
    = [itex]\frac{3}{2}[/itex]W
    = [itex]\frac{3}{2}[/itex](-3600)
    = -5400J

    I am not sure that this is correct because I don't know if I am allowed to rearrange the equations like that, although it is correct algebraically, I'm not sure if it is correct physically.
    I am unsure of whether the change in internal energy is supposed to be negative or positive, negative makes sense to me because work is being done by the gas and thus the internal energy should decrease.

    iii) ΔU = Q + W
    -5400 = Q -3600
    Q = -1800J

    Now I know there is something wrong since heat cannot be lost by the gas in an isobaric process.
  2. jcsd
  3. Mar 21, 2012 #2


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    Looks good. Strictly speaking, the answer to the question as asked is +3600 J, even though you are correct that W is -3600 J.

    Temperature could be calculated from n, P, and V. But your method is simpler and better.
    Is PΔV equal to W or -W?
    The problem with this argument is that you don't know Q (yet). It is possible that Q could offset the effect of the work done, and reverse the sign of ΔU.

    A better way to check the sign of ΔU is to think about whether the temperature increases or decreases during the process. You don't have to actually calculate T, just apply the ideal gas law to the case where P is constant and V increases.

    Yes it can. Or it could be gained. Q=0 for adiabatic processes, not isobaric.

    Hope this helps.
  4. Mar 21, 2012 #3
    Thanks it helps alot.

    PΔV is equal to W. I guess ii) is wrong since numerically the value for W is +3600J not -3600J
    Thus the ΔU should be +5400J not -5400J

    That would effect the answer in iii) so the Q is +9000J.

    Sorry but I'd just like to clarify the last line.
    Do you mean that, when heat is gained in an isobaric process then the gas is expanding and doing work, whereas when heat is lost in an isobaric process then the gas is shrinking and work is being done one it? And so technically heat can be lost or gained in an isobaric process?
  5. Mar 22, 2012 #4


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    You're correct that ΔU is +5400J.

    That being said, we should clarify what W is. It can be defined as either work done on the system, or work done by the system. Both definitions get used in practice, and it's good to be clear on which definition is used in your present course -- and to use that definition consistently.

    If W is work done on the system, then
    • W = -PΔV = -3600 J
    • ΔU = Q + W

    If W is work done by the system, then
    • W = +PΔV = +3600 J
    • ΔU = Q - W

    (Either way, you subtract 3600 J from Q to get ΔU.)

    For this type of problem, I like to think about what is likely happening in a real device. For this problem, I am imagining a gas in a sealed container with a moveable piston. It appears that the gas is being actively heated -- meaning it is heated by a flame or some other means. As it heats, it expands, as we know gases tend to do when heated. By expanding, some of the energy gained by heating is lost via the gas doing work on the surroundings. So ΔU is still positive -- as it must be if the temperature increased -- but less than the heat added to the gas.

    (Sorry if my rambling here isn't clear.)

    You're class may or may not yet have discussed processes where Q=0. For an ideal gas, it looks like processes DA and BC in this diagram:

  6. Mar 22, 2012 #5
    I'd just like to thank you, you've cleared up so much!
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