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Quick Entropy Problem

  1. Jan 7, 2008 #1
    [SOLVED] Quick Entropy Problem

    1. The problem statement, all variables and given/known data

    1.124J of heat is added to a gas at 300K causing the temperature to increase to 600K. What is the change in entropy of the gas?

    2. Relevant equations
    (Delta)S=Q/T


    3. The attempt at a solution

    I know the simple equation i need. all i need to know is whether to use the Tfinal or the Tinitial in the equation.
    Thanks
     
  2. jcsd
  3. Jan 7, 2008 #2

    G01

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    You are going to need to use the more general version of the equation you posted:

    [tex]\Delta S = \int \frac{dQ}{T}[/tex]

    Start the problem like so: Get dQ in terms of dT and then integrate with respect to T. What will your integral's bounds be in this case?
     
  4. Jan 7, 2008 #3
    Im in a non-calculus based physics class. Any ideas what the teacher would want in this case if integrating is not an option?
     
  5. Jan 7, 2008 #4

    G01

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    O ok, well, I would assume you would use the initial temperature of the system, but you may want to check with your teacher, since I am not completely sure.
     
  6. Jan 7, 2008 #5
    thanks
     
  7. Jan 8, 2008 #6

    Andrew Mason

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    If you use either the initial or final temperature, you will be wrong. If you are in a non-calculus class, your teacher should either be giving you the general solution to the integral (which I expect is the case) or not be giving you these kinds of questions. The correct solution to the integal is:

    [tex]\Delta S = \int_{T_i}^{T_f}\frac{dQ}{T} = \int_{T_i}^{T_f}\frac{mC_vdT}{T} = mC_v(\ln{T_f} - \ln{T_i}) = mC_v\ln{\frac{{T_f}}{{T_i}}} = \frac{\Delta Q}{\Delta T}\ln{\frac{{T_f}}{{T_i}}} = \frac{124}{300}\ln{2}[/tex]

    AM
     
    Last edited: Jan 8, 2008
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