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Joule coefficient

  1. Feb 12, 2014 #1
    1. The problem statement, all variables and given/known data

    Hi, i have a little problem with a demostration, I hope you can help me.

    2. Relevant equations
    this said that we have a system, a gas is containing in a recipe, there's no heat exchange neither work with the enviroment, only an expansion v to 2v, we have to find that

    (dT/dv)u=-(T^2/Cv)(d/dT)(P/T)v

    3. The attempt at a solution

    I start with
    dU=(dU/dT)vdT+(dU/dv)t dU=0
    (dT/dv)u=-(dU/dv)(dT/dU)vdv

    (dT/du)v=1/Cv

    (dT/du)=-(1/cv)(T(dP/dT)-P)

    I don't know how continue!
     
  2. jcsd
  3. Feb 12, 2014 #2

    Simon Bridge

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    Welcome to PF;

    Have I understood you correctly:
    An ideal(?) gas in a container (recipe?!), no heat may enter or leave the container(?), and no work is done on or by the gas(?) ... yet there is an expansion? How can this be?

    It sounds like you are trying to describe an adiabatic expansion.
     
  4. Feb 13, 2014 #3

    yes! Sorry i have some problems lol! In fact it's a free expansion U=0 Q=0 and w=0
     
  5. Feb 13, 2014 #4

    Simon Bridge

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  6. Feb 13, 2014 #5
    A free expansion

    I try to find the Joule coefficient (dT/dv) constant U.
    I must find that it's equal to -(T^2/Cv)d/dT(P/T) constant V, but
    I found that it's equal to (-1/Cv)(T(dP/dT)v-P)
     
  7. Feb 13, 2014 #6

    Simon Bridge

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    Oh you mean - like in the title ?!

    All right: you got
    $$\left.\frac{dT}{dv}\right|_U = -\frac{1}{C_v}\left(T\left.\frac{dP}{dT}\right|_v -P \right)$$

    You need to get from there to:$$\left.\frac{dT}{dv}\right|_U = -\frac{T^2}{C_v}\left. \frac{d}{dT}\frac{P}{T}\right|_v$$

    ... it looks like you are almost there since you expression rearranges as:

    $$\left.\frac{dT}{dv}\right|_U = -\frac{T^2}{C_v}\left(\frac{1}{T}\left.\frac{dP}{dT}\right|_v -\frac{P}{T^2} \right)$$

    So what is $$\frac{d}{dT}\frac{P}{T}$$
     
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