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Heat integral and molar heat capacity?

  1. Aug 24, 2011 #1
    dQ = nCvdT if volume is constant.
    However, n = pV/RT.
    What I don't understand is, why are we thinking n as constant when doing the integral?
    I had two problems that involved this on a test I had today. At first I kept it constant and then changed n. But then I thought, wait... isn't there a T in n? then that T should be in the integral.
    I understand the point, heat capacity per mole. But mathematically, the T that is in the equation for n should matter, right?
    dS = dQ / T, if we substitute dQ in that equation we should get 1 / T2 in the integral also.
    I know I'm wrong however, so if someone could tell me what's wrong?
     
  2. jcsd
  3. Aug 24, 2011 #2

    Mapes

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    The equation dQ = nCvdT also assumes a closed system (i.e., constant n). Otherwise you could effect a temperature change by simply removing gas molecules at constant volume without heating or cooling the system, and this would violate the equation.
     
  4. Aug 24, 2011 #3
    That's true... higher temperature with lower amount of moles doesn't sound right. When you say that it assumes a closed system, is that a result of the second law?
    Or do we always speak of closed systems when talking about heat capacity?
     
  5. Aug 24, 2011 #4

    Mapes

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    It's not a result of the second law. It's a typical assumption of the definition of heat capacity; that is, we mean

    [tex]C_{V,N}=T\left(\frac{\partial S}{\partial T}\right)_{V,N}[/tex]

    but we generally just write [itex]C_V[/itex].

    (In some esoteric circumstances, we want to work with systems at constant chemical potential rather than constant matter, but that's an advanced topic.)
     
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