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Heat capacity and its relation with internal energy

  1. Jun 25, 2012 #1
    hi everyone,
    in thermodynamics, when we calculate the heat capacity in constant volume, we assume Cv=dQ/dT..

    well, but at isothermal condition suddenly they came up with Cv=dU/dT...
    so i am getting stuck with this concept how they replace dQ with dU?

    i know U= internal energy is only a function of T, but what is the explanation behind it? plz help with the derivation and clarification...

    advanced regards.
     
  2. jcsd
  3. Jun 25, 2012 #2

    Rap

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    You have to be careful with that "dQ" - there is no such thing as "Q" so there is no such thing as "dQ". Usually it is written "[itex]\delta Q[/itex]", which is an abbreviation for "T dS" to emphasize this fact. You only use the "d" in front of state variables, and there is no "Q" state variable. A state variable X obeys [itex]\oint dX=0[/itex]. In other words the integral around a closed path of a state variable is zero, and the integral between two points is not dependent on the path. [itex]\oint \delta Q[/itex] is really [itex]\oint T dS[/itex] and that's not necessarily zero.

    The correct way to define the C's are [itex]TdS=C_V dT[/itex] at constant volume and [itex]TdS=C_P dT[/itex] at constant pressure.

    You can write the fundamental equation (at constant number of particles) as [tex]dU=TdS-PdV[/tex] and since, by the chain rule, [tex]dU=\left(\frac{\partial U}{\partial T}\right)_V dT + \left(\frac{\partial U}{\partial V}\right)_T dV[/tex] you can see that at constant volume, [tex]TdS= \left(\frac{\partial U}{\partial T}\right)_V dT[/tex] so that [tex]C_V=\left(\frac{\partial U}{\partial T}\right)_V[/tex]

    You can also write the fundamental equation (at constant number of particles) as [tex]dH=TdS+VdP[/tex] where H is the enthalpy (H=U+PV) and since, by the chain rule, [tex]dH=\left(\frac{\partial H}{\partial T}\right)_P dT + \left(\frac{\partial H}{\partial P}\right)_T dP[/tex] you can see that at constant pressure, [tex]TdS= \left(\frac{\partial H}{\partial T}\right)_P dT[/tex] so that [tex]C_P=\left(\frac{\partial H}{\partial T}\right)_P[/tex]
     
  4. Jun 25, 2012 #3
    I think you are mixing up several things you have been told.
    Don't worry this is not uncommon there is a lot to get hold of in thermodynamics.

    So firstly your statement about U is only true for ideal gasses. It is one of the possible definitions of an ideal gas and was proved experimentally by Joule. However it is not relevant here.


    [tex]{\left( {\frac{{\partial U}}{{\partial V}}} \right)_T} = 0[/tex]

    As regards your query about Cv,

    It is not necessary to invoke entropy and considering your definition of Cv you may not have met entropy anyway.

    Since your version of the first law probably reads something like

    dU = δQ - pdV and your definition of Cv is


    [tex]{C_v} = {\left( {\frac{{\delta Q}}{{\partial T}}} \right)_V}[/tex]

    (RAP is correct it is not a good idea to use dQ)

    So δQ = CvdT

    Substituting into the first law

    dU = CvdT - pdV

    But at constant volume dV = 0

    dU = CvdT

    rearranging


    [tex]{C_v} = {\left( {\frac{{dU}}{{dT}}} \right)_V}[/tex]
     
  5. Jun 25, 2012 #4
    oh now i got it... i was mixing up the things really.. thanks for the great help and explanation rap and studiot...

    regards
     
  6. Jun 25, 2012 #5
    oh now i got it... i was mixing up the things really.. thanks for the great help and explanation rap and studiot...

    regards
     
  7. Jun 25, 2012 #6
    Glad you got it sorted

    :wink:
     
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