Latent heat and binding energy

[v_pino]

Homework Statement

Show that the potential height (energy required to separate 2 nitrogen molecules from each other) for N2 is ~ 0.01eV, given that latent heat of evaporation is 210 Jg^-1 and molecular weight is 28 for N2.

Homework Equations

I'm thinking that this equation might be of use:

Latent heat = potential height x 0.5 x N x n

N = number of atoms in sample
n = coordination number (number of nearest neighbors surrounding a given atom)

The Attempt at a Solution

28 / 6.02 x 10^23 = 4.65 x 10^-23 g per nitrogen atom

How should I progress from here? Using dimensional analysis, it appears that I might have to multiply (4.65 x 10^-23) by 210 .

Thanks x

 

[anathema]

To solve this problem, we will use the equation given in the forum post:

Latent heat = potential height x 0.5 x N x n

First, we need to find the number of nitrogen molecules in 1 gram (g) of nitrogen gas. This can be found using the molecular weight (MW) of nitrogen (28 g/mol) and Avogadro's number (6.02 x 10^23 molecules/mol):

Number of molecules = 1 g / (28 g/mol) x (6.02 x 10^23 molecules/mol) = 2.15 x 10^22 molecules

Next, we need to determine the coordination number (n) for nitrogen molecules. Since nitrogen gas is made up of diatomic molecules (N2), the coordination number is 2.

Now, we can plug in the values into the equation:

210 J/g = potential height x 0.5 x (2.15 x 10^22) x 2

Solving for the potential height, we get:

Potential height = (210 J/g) / (0.5 x 2.15 x 10^22 x 2) = 0.0098 J/molecule

Finally, we convert this value to electron volts (eV) by dividing by the conversion factor of 1.602 x 10^-19 J/eV:

Potential height = (0.0098 J/molecule) / (1.602 x 10^-19 J/eV) = 6.1 x 10^-20 eV/molecule

Therefore, the potential height for separating two nitrogen molecules from each other is approximately 0.01 eV, which is consistent with the given value in the forum post.