Chemical potential in Thermodynamics to find latent heat of vaporization

[musicure]

Homework Statement

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?

Homework Equations

L = J/mol

W = -ΔE = Nμ_o + $\frac{NvkΔT}{2}$

k = 1.381x10^-23

Avogrado's Constant = 6.022x10²³

The Attempt at a Solution

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

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

Now for the $\frac{NvkΔT}{2}$ term:

$\frac{NvkΔT}{2}$ = $\frac{N(3k)(700)}{2}$ = 8732 for liquid

$\frac{NvkΔT}{2}$ = $\frac{N(5k)(700)}{2}$ = 14.6x10⁶ for gas


How do I combine these to solve for L, the latent heat of vaporization?

 

[TSny]

Hello musicure. Welcome to the forum.

W = -ΔE = Nμ_o + $\frac{NvkΔT}{2}$

The Attempt at a Solution

μ_o = -0.12eV x 1.602x10^-20 = -1.9224x10^-20 J/atomNow for the $\frac{NvkΔT}{2}$ term:

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

In the formula $\frac{NvkΔT}{2}$ 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?

 

[musicure]

Hello musicure. Welcome to the forum.

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 + $\frac{NvkT}{2}$

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

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

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


Gas state v=5

$\frac{NvkT}{2}$ = $\frac{5(1.381e-23)(700)}{2}$ = 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, E_liquid = E_gas

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

Nμo = -17393 J/mol
μo = $\frac{-17393}{(6.022e23)(1.602e-19)}$ = -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?

 

[TSny]

-W = E = Nμo + $\frac{NvkT}{2}$

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|μ_o| + (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 |μ_o| is the energy required to pull one molecule out of the liquid. Thus, N|μ_o| 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.

 

[musicure]

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."

 

[TSny]

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

E = Nu_o + NvkT/2

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

ΔE = N(Δu_o) + N(Δv)kT/2

The question only specifies u_o for the liquid, but just below equation (4.2) the text states u_o ≈ 0 for the gas phase. Note that both Δu_o 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.

 

[musicure]

Oh I see.. I didn't realize μ_o 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!

 

[TSny]

I think that's right.