What is the Form of the Lennard-Jones Potential Between Two Water Molecules?

[ProPatto16]

Homework Statement

Deduce the form of the Lennard-Jones potential between two water molecules.

given/known:
molar enthalpy of evaporation of water is 4.1x10⁴J/mol
density is 10³kg/m³
Assume water molecule is surrounded by approximately ten nearest-neighbours.
Take M_r = 18

Homework Equations

from the sources i have:
V(r)=4ε[-(σ/r)⁶+(σ/r)¹²]

The Attempt at a Solution

ε is a measure of how strongly the molecules attract each other.
σ is the distance at which the intermolecular potential is zero.
r is the distance of separation between both particles.

its a bit unclear what the question means by "deduce the form" but I am assuming it means find the value r.

to find r i first find the atomic diameter d= [(M_r(X)x10^-3)/(pN_A)]^1/3

where X is number of molecules
M_r = 18
p given at 10³kg/m³
N_A is avogadros constant.

and X I am unsure if it is 2 (because the question says two molecules) or if its 11 (because of the molecule plus its 10 "neighbours")

if X is 2 i get a diameter of 1.815x10^-10
giving a radius of 9.07x10^-11

if X is 11 i get d = 3.20x10^-10
making r = 1.60x10^-10

now to get the "form" i just sub an r into the potential equation??

however, i don't entirely understand the "ten nearst-neighbours" and what its used for.
i also don't use the molar enthalpy.
molar enthalpy isn't mentioned anywhere in my notes but online it says enthlpy is a measure of energy of the system.

some clarification would be much appreciated
thanks

 

[ProPatto16]

kept at it and found

molecular enthalpy of sublimation H_m.s = 1/2(nN_AΔE)
where n is the number of "nearest neighbours"
solving for ΔE i got 1.36x10_-20 J which seems appropriate.

and ΔE = ε when r is at equilibrium distance.

so subbing that back into the lennard-jones and simplifying i get

5.44x10^-20[-(σ/0.72x10^-10)⁶+(σ/0.72x10^-10)¹²]

is that meant to be "the form of the Lennard-Jones 6-12 potential" that the question says deduce?