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Ideal gases, Thermodynamics

  1. Mar 19, 2007 #1
    I'm trying to show a formula for an ideal gas, but I don't get the right results.

    1. The problem statement, all variables and given/known data
    For an ideal gas PV = nRT where n is the number of momles. Show that the heat transferred in an infinitesimal quasistatic process of an ideal gas can be written as

    [tex]dQ = \frac{C_V}{nR}VdP + \frac{C_P}{nR}PdV[/tex]


    2. Relevant equations
    [tex]
    dU = dQ + dW
    [/tex]

    [tex]
    C_P = C_V + R
    [/tex]

    [tex]
    dU = nC_VdT
    [/tex]

    [tex]
    dW = -PdV
    [/tex]


    3. The attempt at a solution

    I differented the formula for the ideal gas PV = nRT so it becomes

    PdV + VdP = nRdT

    [tex]
    dT = \frac{PdV + VdP}{nR}
    [/tex]

    [tex]
    dU = C_V\frac{PdV + VdP}{R}
    [/tex]

    [tex]
    dQ = C_V\frac{PdV + VdP}{R} + PdV = \left(\frac{C_V}{R} + 1\right)PdV + \frac{C_V}{R}VdP = \frac{C_P}{R}PdV + \frac{C_V}{R}VdP
    [/tex]

    What have I done wrong? There is no dependens on n in my final equation.
    I know there should be bars on dW and dQ but i didn't got it to work in latex :/
     
    Last edited: Mar 19, 2007
  2. jcsd
  3. Mar 19, 2007 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    Your answer is correct. The solution posed by the question is wrong. There is no "n" in the denominator. dQ must have the same dimensions as VdP or PdV, which has dimensions of energy. C_v/R is dimensionless.

    AM
     
  4. Mar 18, 2011 #3
    I have just come across the same problem in an exercise book (no solution unfortunately). Its very unlikely two sources are incorrect?
     
  5. Mar 18, 2011 #4

    Andrew Mason

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    Science Advisor
    Homework Helper

    How do you know they are two different sources?

    The dimensions of the suggested answer are dimensions of energy per mole. If the dQ was the specific heat flow per mole the suggested answer would be correct, which is maybe what the OP was saying.


    Consider an expansion at constant pressure. By definition:

    (1) [tex]dQ = nC_pdT[/tex]

    where [itex]C_p[/itex] is the molar heat capacity at constant pressure.

    The solution of the OP gives:

    (2) [tex]dQ = \frac{C_P}{R}PdV[/tex]

    since VdP = 0 (constant pressure).

    You can see that (2) is equivalent to (1) if:

    [tex]nRdT = PdV[/tex]

    This, of course, follows from the ideal gas law for a constant pressure process.

    AM
     
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