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Homework Help: Entropy calculation

  1. Mar 24, 2010 #1
    1. The problem statement, all variables and given/known data
    A thermally insulated cylinder, closed at both ends, is fitted with a frictionless heat-conducting piston that divides the cylinder into two parts. Initially, the piston is clamped in the center with 1 liter of air at 300 K and 2 atm pressure on one side and 1 liter of air at 300K at 1 atm pressure on the other side. The piston is released and reaches equilibrium in pressure and temperature at a new position. Compute the final pressure and temperature and increase of entropy if air is assumed to be the ideal gas. What irreversible process has taken place?


    2. Relevant equations
    [tex]\Delta[/tex]S = [tex]\Delta[/tex]Q/T
    3. The attempt at a solution
    This is an isothermal free expansion. So temperature remains constant. I know the final pressure will be 1.5 atm (intuitively, (1+2)/2), but how can I compute this out by ideal gas law?

    By PV=nRT,
    (101325)(1000/1000000)=n (8.31)(300)
    n1(the compartment with 1atm)=0.040643802
    n2(the compartment with 2atm) = 0.081828
    TOTAL number of moles = 0.12193

    When I search through the Internet, I find this equation
    [tex]\Delta[/tex]S=nRln(Vf/Vi) = 0.12193(8.31)ln2 = 0.702323.........

    but the answer is 0.0566, what's wrong with my calculation???
     
  2. jcsd
  3. Mar 25, 2010 #2

    Mapes

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    Why did you use [itex]V_f/V_i=2[/itex]? That ratio is supposed to represent the final and initial volumes.
     
  4. Mar 25, 2010 #3
    Actually, from the phase'1 liter of air', i am not quite sure about it is about the volume or number of moles. Moreover, when I think this question again, it is not free expansion as the sentence "The piston is released and reaches equilibrium in pressure and temperature at a new position." provides the fact that the piston is still in the cylinder. Then another question comes, if it is not free expansion, then, what irreversible process has taken place.

    Together with this question, i have one more question to ask...
    Answer: 1.5 atm. 300K, 0.0566 J/K
    For the first answer, the piston stops moving when the pressure of the two compartments are equal, that is 1.5 atm.
    From the second answer, it seems that the temperature is assumed to be unchanged. However, we know that PV=nRT, it may be the case that the temperature increase.
    Anybody can help me to clarify? It may be crucial in solving the increase in entropy.
    Thanks You.
     
  5. Mar 25, 2010 #4

    Mapes

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    The liter is a measure of volume. By applying an assumption (ideality), and knowing the pressure and temperature, one can calculate the amount of gas in moles.

    It is not a free expansion (because the gas does not expand into a vacuum), but it is a spontaneous expansion.

    Can the gases end up at different temperatures, considering the boundary between them?
    Can the gases end up at an average temperature different from 300K, considering the boundary between the gases and the outside environment?
     
  6. Mar 27, 2010 #5
    I can figure out the calculation of the increase in entropy now. However, I still cannot understand why the temperatures remain unchanged. Thank You.
     
  7. Mar 27, 2010 #6

    Mapes

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    I find it most useful to take a reductio ad absurdum approach here: assume the temperature does change in one or both of the chambers. What are the implications? Do these implications make sense, considering what we know about the barrier properties, the system boundary conditions, and the energy of an ideal gas as a function of temperature?
     
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