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Change in internal energy

  1. Nov 19, 2011 #1
    1. The problem statement, all variables and given/known data
    In a closed vessel there was some water vapor. After the water vapor dissociated, the temperature decreased by 10%, and the volume by 20%. Find the change in internal energy.
    [itex](U{initial}-U{final})/U{initial}[/itex]

    2. Relevant equations



    3. The attempt at a solution
    Let's state, that not all of water vapor dissociated. Let's say that in the beggining there was [itex]n[/itex] moles of [itex]H_2O[/itex] , and [itex]x[/itex] moles of it dissociated.
    Since [itex]xH_2O -> xH_2 + 0,5xO_2[/itex] , then we get [itex]x+0,5x[/itex] moles in total of hydrogen and oxygen combined.The total moles of gas molecules after the dissociation would be [itex](n-x) + (x+0,5x)=n+0,5x[/itex]
    Then i can write the ideal gas law before and after the dissociation.
    [itex]pV=nRT_1[/itex]
    [itex]0.8pV=0,9(n+0,5x)RT_1[/itex]
    to continue, i have to find x in terms of n , but after solving the above, i get that x is a negative number. Please help !
     
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  3. Nov 19, 2011 #2

    Andrew Mason

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    Are you sure there is a decrease and not an increase in the volume? Are we to assume constant pressure?

    AM
     
  4. Nov 19, 2011 #3
    Thanks for your feedback.
    Yes, the volume did in fact decrease. Maybe you're right , the pressure isn't constant, i can't find another explanation. But in that case there isn't enough data to solve this problem.
    If i assume that all the water vapor dissociated, then why is the volume change given ? Since the internal energy only depends on the temperature.
     
  5. Nov 20, 2011 #4

    I like Serena

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    It seems to me that the change in volume is not relevant, nor is the change in pressure.

    From the temperatures and the molar heat capacities of water, hydrogen and air, you can find the relative change in internal energy.
     
  6. Nov 20, 2011 #5

    Andrew Mason

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    I think you have to assume that all the vapour dissociates. If there is no heat exchange with the surroundings and if the pressure is constant, you can apply the first law:

    ΔQ = ΔU + PΔV

    where ΔQ is the enthalpy of dissociation of water. So ΔU = ΔQ - PΔV. Does that help?

    AM
     
  7. Nov 20, 2011 #6

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    Suppose we have n moles of H2O(g), then we would get a total of (3/2)n moles of H2 and O2 combined.

    With the ideal gas law we would get:
    [tex]{nRT_1 \over V_1} = P_1[/tex]
    and so:
    [tex]P_2 = {3 \over 2} {n R T_2 \over V_2} = {3 \over 2} {n R \cdot 0.9 \ T_1 \over 0.8 \ V_1} = {27 \over 16} P_1[/tex]

    Doesn't that make it seem unlikely that the pressure would remain constant?
     
  8. Nov 20, 2011 #7

    Andrew Mason

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    If the system is closed - insulated from the surroundings, the energy required for dissociation must come from the internal energy. Although the number of moles increases, I expect from the question (I haven't worked it out) that the decrease in internal energy results in a net reduction in volume. The whole process takes place at constant temperture.

    AM
     
  9. Nov 21, 2011 #8

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    Can't the problem be resolved without taking insulation or dissociation energy into account?
    The state before and the state after seem to be uniquely defined and independent of the process path.
     
  10. Nov 23, 2011 #9

    Andrew Mason

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    I meant to say constant pressure. Sorry for any confusion.

    The problem is that water vapour cannot be treated as an ideal gas.

    To determine its internal energy, you would have to know its specific heat capacity at constant volume. Since Cv is a function of temperature for water vapour, how do you determine its Cv without knowing its temperature?

    AM
     
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