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Typo in PR book?

  1. May 12, 2007 #1
    Typo in PR book??

    Personally I think PR has tons of mistakes. I think I've found one in the book and it confuses me. Can someone see if it is a mistake? Thanks

    PR says: "The expanding gas did negative work against its surroundings, pushing the piston upward."

    --I think it should be: The expanding gas did positive work against its surroundings, pushing the piston upward. I think the book meant that the surroundings did negative work on the system.

    This is a typo right?
  2. jcsd
  3. May 12, 2007 #2
    Also, can [tex] Q=nC_{p}\Delta T[/tex] be written as [tex]\Delta U=nC_{p}\Delta T[/tex]?? Why or why not?

    Likewise, can [tex] \Delta U=nC_{v}\Delta T[/tex] be written as [tex] Q=nC_{v}\Delta T[/tex]?? Please explain. I thought that Internal energy and heat are different things but they can be used interchangably??

    I have two books and they use them interchangably so I don't know which I need to use since they all equal the same amount of joules but they have different variables which are suppoes to be different right?

    HyperPhysics also says that [tex]\Delta U= Q=nC_{v}\Delta T[/tex]. I'm not sure about C_p though. Thanks for your help!
  4. May 12, 2007 #3

    Andrew Mason

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    The gas performs positive work on the surroundings. Or you could say that the work done by the surroundings on the gas is negative.

    But this is just a convention. 20-30 years ago the convention was that work done on the gas was positive and work done by the gas was negative. So if you have an old book that may be the problem.

    Last edited: May 12, 2007
  5. May 13, 2007 #4
    Yeah, but the book was using the newer convention earlier. lol, I think the author is getting mixed up. -_- Mabye he was too used to the older convention haha
  6. May 13, 2007 #5

    Andrew Mason

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    No. dQ and dU are the same only if W = 0 (no change in volume).

    No. Heat flow is equal to change in internal energy + work done by the gas (first law).

    The first law of thermodynamics always applies:

    [tex]\Delta Q = \Delta U + W[/tex]

    where W is the work done by the system

    If volume is constant, no work is done (PdV = 0). So, for constant volume
    [tex]\Delta U = Q = nC_{V}\Delta T[/tex]

    If pressure is constant (volume is not), [itex]d(PV) = PdV + VdP = PdV = d(nRT) = nR\Delta T[/itex]

    [tex]\Delta Q = nC_{P}\Delta T = nC_{V}\Delta T + W = nC_{V}\Delta T + P\Delta V = nC_{V}\Delta T + nR\Delta T = n(C_{V} + R)\Delta T[/tex]

    So Cp and Cv are always related this way: [itex]C_p - C_v = R[/itex]

  7. May 13, 2007 #6

    So basically, stating what you've said: [tex] \Delta U=nC_{v}\Delta T[/tex]
    is equal to [tex] Q=nC_{v}\Delta T[/tex] when there's no work done on or by the gas. (when the graph is a vertical line)

    However, [tex] Q=nC_{p}\Delta T[/tex] can never be equal to [tex]\Delta U=nC_{p}\Delta T[/tex] because work is always done on the system when the pressure remains constant. (horizontal line)

    Are these assumptions correctly stated? Thanks for the help! :smile:
    Last edited: May 13, 2007
  8. May 13, 2007 #7

    Andrew Mason

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    Correct. Except when dT = 0, of course.

    Last edited: May 13, 2007
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