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Chemical potential in Thermodynamics to find latent heat of vaporization

  1. Feb 2, 2013 #1
    1. The problem statement, all variables and given/known data
    The latent heat of vaporization of a substance is defined as the amount of heat required to transform the substance from its liquid into its vapor phase. A certain material has a boiling point of 700K and its molecules have three degrees of freedom in the liquid phase but five in the gas phase. If the chemical potential, µ in the liquid phase is -0.12 eV, what is the latent heat of vaporization in J/mol?


    2. Relevant equations

    L = J/mol

    W = -ΔE = Nμo + [itex]\frac{NvkΔT}{2}[/itex]

    k = 1.381x10-23

    Avogrado's Constant = 6.022x1023

    3. The attempt at a solution

    μo = -0.12eV x 1.602x10-20 = -1.9224x10-20 J/atom

    multiply by avogadro's constant to get J/mol
    o = -1.9224x10-20 x 6.022x1023 = -11576.7 J/mol

    Now for the [itex]\frac{NvkΔT}{2}[/itex] term:

    [itex]\frac{NvkΔT}{2}[/itex] = [itex]\frac{N(3k)(700)}{2}[/itex] = 8732 for liquid

    [itex]\frac{NvkΔT}{2}[/itex] = [itex]\frac{N(5k)(700)}{2}[/itex] = 14.6x106 for gas


    How do I combine these to solve for L, the latent heat of vaporization?
     
  2. jcsd
  3. Feb 2, 2013 #2

    TSny

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    Hello musicure. Welcome to the forum.
    Check the conversion factor from eV to Joules. The power of 10 doesn't look correct.

    In the formula [itex]\frac{NvkΔT}{2}[/itex] you have a temperature change ΔT. But the temperature doesn't change during the phase change. Are you sure you wrote the equation for W correctly?
     
    Last edited: Feb 2, 2013
  4. Feb 2, 2013 #3
    Thanks.


    Yes, it should be -1.9224x10-21 J/mol

    And you're right, I should be using the other work equation...

    -W = E = Nμo + [itex]\frac{NvkT}{2}[/itex]

    I'll give this question another try now, because I've had some time to think about it.

    Liquid state v=3
    Nμo = (6.022e23)(-0.12eV)(1.602e-19) = -11577 J/mol

    [itex]\frac{NvkT}{2}[/itex] = [itex]\frac{3(1.381e-23)(700)}{2}[/itex] = 8731 J/mol

    Eliquid = -11577 J/mol + 8731 J/mol = -2844.7 J/mol


    Gas state v=5

    [itex]\frac{NvkT}{2}[/itex] = [itex]\frac{5(1.381e-23)(700)}{2}[/itex] = 14549 J/mol


    At this point I got stuck.. so I assumed that if ΔT=0, then ΔE=0 (is this true??). If that is the case, Eliquid = Egas

    Egas = -2844.7 J/mol = Nμo + 14549 J/mol

    Nμo = -17393 J/mol
    μo = [itex]\frac{-17393}{(6.022e23)(1.602e-19)}[/itex] = -0.18eV

    The solution for has Nμo has units J/mol (which the question asks for), so is this the correct answer for Latent heat of vaporization?
     
  5. Feb 2, 2013 #4

    TSny

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    This still doesn't look quite correct to me. Work, W, is always associated with a change in energy, ΔE, rather than energy itself. Can you give a reference for where you obtained this formula? I have not actually seen this formula before. I searched several physical chemistry texts as well as the internet but could not find latent heat expressed this way.

    As stated in the problem, the (molar) heat of vaporization is the energy that must be added to convert 1 mole of liquid to vapor. It looks like your formula is suggesting that this can be calculated as the sum of two terms:

    N|μo| + (1/2)N(Δv)kT

    where I introduced the Δ to get the change in the number of degrees of freedom of a molecule in going from liquid to gas.

    Roughly, the absolute value of the chemical potential |μo| is the energy required to pull one molecule out of the liquid. Thus, N|μo| is for N molecules.

    The second term appears to account for the additional energy needed by the extra degrees of freedom of motion of the gas molecules. Each degree of freedom of a molecule at temperature T will take up (on average) (1/2)kT of energy. The second term appears to take care of this for N molecules.

    So, perhaps the sum of the two terms (for N = Avogadro's number) gives the total energy required to vaporize 1 mole of liquid. That's my best guess. Again, it would be helpful if you could state a reference for the formula.
     
  6. Feb 3, 2013 #5
    My textbook is called Thermodynamics & Statistical Mechanics by Stowe

    Here's a link to it in my dropbox....

    https://www.dropbox.com/s/06ohioz83zdmwyc/Thermodynamics & Statistical Mechanics by Stowe.pdf

    We're in about chapter 4-5 and I thought to use the Work equation when I saw what was stated on the bottom of pg. 73:


    "At phase transitions we may add large amounts of heat to a system, without any change in temperature. Where does this added energy go? According to equation 4.13, we can see two possibilities: since N and T are constant, only u0 and ν can change. There is always a change in u0 during a phase transition, because molecular arrangements in the new phase are different, resulting in different potential energies. The number of degrees of freedom per molecule, ν might also change, because the change of phase could change the constraints on the motions of the individual molecules."
     
  7. Feb 3, 2013 #6

    TSny

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    OK. So, equation (4.13) gives the average energy of N molecules of a substance as

    E = Nuo + NvkT/2

    In a phase transition, only uo and v can change. So the energy needed to change N molecules from liquid to gas is

    ΔE = N(Δuo) + N(Δv)kT/2

    The question only specifies uo for the liquid, but just below equation (4.2) the text states uo ≈ 0 for the gas phase. Note that both Δuo and Δv are positive for your problem.

    Anyway ΔE should give you the molar latent heat of vaporization if you use the appropriate number of molecules N.
     
  8. Feb 3, 2013 #7
    Oh I see.. I didn't realize μo was 0 for gas phase.

    So the answer is simply

    ΔE = N(Δuo) + N(Δv)kT/2 = N(0.12eV) + N(2)k(700)/2 = 17398 J/mol?

    Thank you so much, that makes a lot more sense... I knew that ΔE couldn't possibly be zero lol

    Thanks again!
     
  9. Feb 3, 2013 #8

    TSny

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    I think that's right.
     
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