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Thermodynamics - Two gases in a container

  1. Apr 14, 2016 #1
    1. The problem statement, all variables and given/known data
    Two kilograms of air at 5 bar, 350 K, and 5 kg of carbon monoxide, initially at 4 bar, 450 K, are confined to opposite sides of a rigid, well-insulated container by a partition. The partition is free to move and allows conduction from one gas to the other without energy storage in the partition itself. The air and carbon monoxide each behave as ideal gases with constant specific heat ratio, k = 1.395. Determine the equilibrium temperature.

    2. Relevant equations
    Q - W = delta U

    Cp - Cv = R

    Cp/Cv = k

    Rair = 8.314 / Mwtair

    Rco = 8.314 / Mwtco

    Q = mCv(T2-T1)

    3. The attempt at a solution

    Carbon Monoxide
    Cp - Cv = 0.297
    Cp / Cv = k

    Cv(k) - Cv = 0.297

    1.395Cv - Cv = 0.297
    0.395Cv = 0.297
    Cv = 0.75

    Air
    Cp - Cv = 0.287
    Cp / Cv = k

    1.395Cv - Cv = 0.287
    Cv = 0.73

    This is where i feel i am going wrong.

    Q = mCv(T2-T1)
    Q = 0 as insulated closed system


    Carbon Monoxide Air
    0 = 4 x 0.75 (T2 - 450) 0 = 2 x 0.73 (T2 - 350)
    0 = 3T2 - 1350 0 = 1.46T2 - 511

    I made the two equations equal each other (not sure if that is correct)

    1.46T2 - 511 = 3T2 - 1350
    839 = 1.54T2
    T2 = 544.8 K

    I know this is the wrong answer but I am not sure why.

     
  2. jcsd
  3. Apr 14, 2016 #2
    There are 5 kg of CO, not 4. Also, since the change in internal energy is zero, $${5}(0.75)(T_2-450)+{2}(0.73)(T_2-350)=0$$
     
  4. Apr 14, 2016 #3
    Ah I see! Thank you for your help.
     
  5. Apr 17, 2016 #4

    Pol

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    How would you also work out the volume and pressure at equilibrium ?
     
  6. Apr 17, 2016 #5
    Does the total volume change? What is the initial volume of each gas? Let V1 represent the final volume of air and V2 represent the final volume of CO. In terms of V1 and V2, what is the final pressure of each gas? At equilibrium, how do these pressures compare?
     
  7. Apr 17, 2016 #6

    Pol

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    The total volume does change , but I don't know what equation to use to work out the pressure ?
     
  8. Apr 17, 2016 #7
    Actually, the total volume doesn't change. The only thing that happens is that the partition moves, so one chamber gets larger and the other gets smaller. To get the final pressure, I need you to answer the other questions that I asked.

    Chet
     
  9. Apr 17, 2016 #8

    Pol

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    I'm not really sure , I don't know the initial volume of each gas
     
  10. Apr 17, 2016 #9
    Use the ideal gas law. You have the mass (i.e., the number of moles), the temperature, and the pressure of each gas.
     
  11. Apr 17, 2016 #10

    Pol

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    R the gas constant is 8.214 which I use aswell?
     
  12. Apr 17, 2016 #11
    If you use that (and the value is 8.314), you need to convert bars to Pa. You are aware of having to use consistent units, correct?
     
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