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Entropy Question

  1. Sep 3, 2010 #1
    This question appears in "Schaums Outline of College Physics". Please click the link:

    http://www.4shared.com/photo/jbJ3lZwT/ScreenHunter_01_Sep_04_1256.html"

    My question is how can they use this equation (delta S=(delta Q)/T) when the temperature is not constant? They treat it as being constant by taking the average temp.
     
    Last edited by a moderator: Apr 25, 2017
  2. jcsd
  3. Sep 3, 2010 #2

    Andrew Mason

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    It is just an approximation to use the mean temperature over such a small temperature change. The point is to show that total entropy increases.

    To do it accurately involves a bit of calculus:

    [tex]\Delta S_h = \int_{340}^{338} dS = \int_{340}^{338} dQ/T = \int_{340}^{338} n_1C_v dt/T = n_1C_v\ln\left(\frac{338}{340}\right) \approx n_1C_v\left(\frac{2}{339}\right) [/tex]

    AM
     
    Last edited by a moderator: Apr 25, 2017
  4. Sep 4, 2010 #3
    Thanks for the reply. I suspected that was the case.
     
  5. Sep 4, 2010 #4
    Elegant. I assume n1 is the mass.
     
    Last edited: Sep 4, 2010
  6. Sep 5, 2010 #5

    Andrew Mason

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    I intended n1 to be the number of moles of the gas in the first compartment and C_v to have units of J/mole K. I probably should have used m and Cv in units of J/kg K, since that is what is used in this example.

    AM
     
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