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Entropy and Final Temperature

  1. Mar 16, 2012 #1
    1. The problem statement, all variables and given/known data

    http://img812.imageshack.us/img812/1874/entropy.jpg [Broken]

    3. The attempt at a solution

    First, not sure if this is an adequate explanation but I think for both gases in both partitions, regardless of whether it starts off colder or warmer, the total entropy will increase in order to reach thermal equilibrium.

    To find the final temprature, I have an equation in my notes for the final temprature when an object and its environment come in contact and reach a final temperature:

    [itex]T_f = \frac{C_oT_o + C_eT_e}{C_o+C_e}[/itex]

    Is this the correct equation for this situation? And what kind of heat capacities are the C's (Cv or Cp)?

    For the total entropy I would use the equation which gives the total change of entropy by the object+environment:

    [itex]\Delta S_{o+e} = \Delta S_o + \Delta S_e = Q_{tot} \left( \frac{1}{T_o} - \frac{1}{T_e} \right)[/itex]

    And there is another equation:

    [itex]\Delta S_{o+e} = C_o \ln \left( \frac{T_f}{T_o} \right) + C_e \ln \left( \frac{T_f}{T_e} \right)[/itex]

    I want to use the second one. But again how do I find the Ce and Co? The only info I am given is the number of moles. Any guidance is appreciated.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Mar 17, 2012 #2
    My question really is what formulas need to be used for finding these particular heat capacities?
     
    Last edited: Mar 17, 2012
  4. Mar 17, 2012 #3

    Andrew Mason

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    Where does the number of moles factor into your equation?
    Will the final temperature be closer to 300 or 500 K? Why?

    If volume constant? Is pressure constant?

    AM
     
  5. Mar 18, 2012 #4
    How can I use the number of moles for finding C? Because that equation doesn't directly take the number of moles into account...

    How can I find that out? :confused: Intuitively, I think the final temprature would be somewhere between 300 and 500 K.

    If volume is constant then the pressure would be constant as long as the number of moles don't change (we can see that from the equation P=nRT/V).
     
  6. Mar 19, 2012 #5
    Since the hotter gas will lose some heat and the colder gas would gain that much heat I could use the equation [itex]Q=mc\Delta T[/itex] for finding Tf:

    [itex]m_1 C_1 (T_f - 300) = m_2 C_2 (T_f - 500)[/itex]

    Since it is the same type of gas in both partitions, is it okay to cancel out the C's on both sides (i.e C1=C2)?

    To find the masses using the number of moles we can use use m=nM... but we don't know the molar mass of the gas. But would it be okay to also cancel out the M's on both sides?
     
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