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Thermodynamics and polytropic ideal gas (very simple theory question)

  1. Jul 7, 2013 #1


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    hey all!

    i have a question i was hoping some of you could unravel. specifically, in thermodynamics i understand in a quasi-static situation we can right work as:

    [tex]W=\int PdV[/tex] where [itex]W[/itex] is work, [itex]P[/itex] is pressure, and [itex]V[/itex] is volume.

    my book defines polytropic to be [tex]PV^n = constant[/tex]

    it then writes the following polytropic, quasi-static equality when [itex]n=1[/itex] , which is where i am lost:

    [tex]W=\int PdV = \int P \frac{{V_1}}{V} dV = P{V_1} \int \frac{{dV}}{V} = P{V_1} ln(\frac{{V_2}}{V_1}) [/tex]

    specifically, if [itex]PV_1[/itex] is constant, then if we pull it out of the integral how is it we integrate over [itex]\frac{1}{V}[/itex] ? why doesn't the equality implode here: [tex] \int P \frac{{V_1}}{V} dV = P{V_1} \int \frac{{dV}}{V}[/tex] if [itex]V_1[/itex] is a constant why isn't [itex]V[/itex]? any help is greatly appreciated!
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  3. Jul 7, 2013 #2


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    I think either you copied something wrong our your book is leaving out a small detail here. Namely, for the polytropic process with [itex]n=1[/itex], you have the relation simplified to
    [tex]pV = c.[/tex]
    So, if you have some base state that represents the starting conditions of your system, you can say
    [tex]p_1 V_1 = c.[/tex]
    There, [itex]p_1[/itex] and [itex]V_1[/itex] are both constants because they simply represent the pressure and volume of your system at its initial state. You can rearrange this for substitution into the work equation,
    [tex]p = \dfrac{c}{V} = \dfrac{p_1 V_1}{V}[/tex]
    They are initial conditions. Going back to your work equation,
    [tex]W = \int\limits_1^2 p\;dV = \int\limits_1^2 \dfrac{p_1 V_1}{V}\;dV = p_1 V_1 \int\limits_1^2 \dfrac{1}{V}\;dV = p_1 V_1 \left.\ln(V)\right|_1^2 = p_1 V_1 \left[\ln(V_2) - \ln(V_1)\right] = p_1 V_1 \ln\left(\dfrac{V_2}{V_1}\right)[/tex]
    which recovers the answer in your question with the exception that there is a subscript 1 in front of the pressure term. The key is that you know that [itex]pV[/itex] is equal to a constant, and you "know" the initial state of the system, so you can replace that constant with the values of [itex]p[/itex] and [itex]V[/itex] at that initial state, and both of those are constants and can be removed from the integral. In short, [itex]p_1[/itex] and [/itex]V_1[/itex] are constants while [itex]p[/itex] and [itex]V[/itex] are not because the former refer to specific values of the later at a specific time, and those don't change.
  4. Jul 7, 2013 #3

    Andrew Mason

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    The P is missing a subscript.

    If PV = constant then PV must equal P1V1, so P = P1V1/V.


    [tex]W=\int PdV = \int P_1 \frac{{V_1}}{V} dV = P_1V_1 \int \frac{{dV}}{V} = P_1V_1 ln(\frac{{V_2}}{V_1}) [/tex]

    PV = constant. So PV1 cannot be constant (ie. P cannot be constant) as V changes.

  5. Jul 7, 2013 #4


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    if this is true, how can we make the questionable integral equality? it seems if [itex]V_1 \neq V[/itex] then we cannot have the equality that was pointed out. can you explain?
  6. Jul 7, 2013 #5


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    never mind, it just "clicked". thanks guys! yay for PF saving the day again!!!
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