I want to ask a question about working out the energy of a hydrogen bond between two water molecules, $w_{AA}$ using the chemical potentials of vapour and condensed phases.

I was reading K. Dill, Molecular Driving Forces, 2nd Ed, 2011, p. 257 which lines out the framework for the Clapeyron equation and I was confused on the following:

The book states that you can calculate $w_{AA}$ using the following formula, which is derived by calculating the chemical potentials of the vapour $\mu_v$ and $\mu_c$ respectively:

$$\mu_v = kT\ln\left(\frac{p}{p_{int}^{*}}\right) $$

where $\mu_v$ is the chemical potential of a a pure substance in the liquid phase, $p$ is the partial pressure of the vapour phase and $p_{int}^*$ is the standard vapour pressure

The chemical potential of the condensed phase, $\mu_c$ is achieved by differentiation of the Helmholtz equation, with the knowledge that the $TS$ term of $U-TS$ is set to $0$ since a pure liquid's molecules are indistinguishable upon interchange.

$$\mu_c = \left(\frac{\partial F}{\partial N}\right)_{T,V} = \frac{zw_{AA}}{2}$$

where $z$ is the number of neighbours of each molecule in the liquid phase, $w_{AA}$ is the bonding interaction between two molecules, and divided by 2 to account for any double counting.

Using $\mu_A = \mu_B$, both terms can be equated:

$$kT\ln\left(\frac{p}{p_{int}^{*}}\right) = \frac{zw_{AA}}{2}$$

and more simply written as

$$\ln\left(\frac{p}{p_{int}^{*}}\right) = \frac{zw_{AA}}{2kT}$$

and for those interested, this can be expressed as:


Now, I was presented with the following question from the book:

Given that water’s vapor pressure is $1 \ atm$ at $T = 100^{\circ}$ and $0.03 \ atm$ at $T = 25^{\circ}C$, find the value of $zw_{AA}$.

This all makes sense.

However, Dill states the following

Take the logarithm of the equation to get the boiling pressure $p_1$ at temperature $T_1$ as $$\ln\left(\frac{p_1}{p_{int}^{*}}\right) = \frac{zw_{AA}}{RT_1}$$ and the boiling pressure $p_2$ at temperature $T_2$ as $$\ln\left(\frac{p_2}{p_{int}^{*}}\right) = \frac{zw_{AA}}{RT_2}$$
Subtract the second equation from the first to get $$\ln\left(\frac{p_2}{p_1}\right) = \frac{zw_{AA}}{2R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$$ neglecting the small temperature dependence on $p_{int}^*$.

However, I'm not sure whether this derivation is correct. Below shows my working out from the second equation stated from the start of the body:

First of all, they transformed

$$\ln\left(\frac{p}{p_{int}^{*}}\right) = \frac{zw_{AA}}{2kT}$$


$$\ln\left(\frac{p_1}{p_{int}^{*}}\right) = \frac{zw_{AA}}{RT_1}$$

and somehow stated the relationship between the gas constant $R$ and $k$ is $R = 2k$ which doesn't make sense.

Also, the denominator term containing the $2$ disappears in both of their initial statements for $T_1$ and $T_2$ e.g. $\ln\left(\frac{p_1}{p_{int}^{*}}\right) = \frac{zw_{AA}}{RT_1}$, yet it returns in the final equation $\ln\left(\frac{p_2}{p_1}\right) = \frac{zw_{AA}}{2R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$ and I can't see that from the following working that I did to prove this equation does not work (taking $R = 2k$ to be true):

$$\ln\left(\frac{p_1}{p_{int}^{*}}\right) = \frac{zw_{AA}}{RT_1}, \ln\left(\frac{p_2}{p_{int}^{*}}\right) = \frac{zw_{AA}}{RT_2}$$

Subtracting the second equation from the first:

$$\ln\frac{P_1}{P_2} = \frac{zw_{AA}}{RT_1} - \frac{zw_{AA}}{RT_2} = \frac{zw_{AA}}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$

which does not compare with the derived result.

My question is therefore two-fold:

  1. How has Dill interchanged $R$ and $k$ in his solution?
  2. Is there an error in his derivation for his final expression.

The solution (as transcribed above) is also attached as a screenshot for illustration (for illustrative purposes only):

Model solution for the problem delivered by Dill, used only for illustrative purposes to support the question

  • 1
    $\begingroup$ Switching between k and R is just a matter of changing units from J to J/mol (or energy to molar energy units). The factor of two should be retained and its disappearance from some of the expressions is was probably a typo. $\endgroup$ – Buck Thorn Oct 9 '20 at 16:33
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    $\begingroup$ Going between R and k is more than just a dimensional conversion. You also need to divide or multiply by $N_A$, since R is J/K/mol, while k is J/K/particle: $R = N_A k$. $\endgroup$ – theorist Oct 10 '20 at 2:02
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    $\begingroup$ @theorist you are strictly right of course. You encounter this type of manipulation so often that you ignore the finer details. Forgot to note: upon conversion from k to R, $w_{AA}$ must be expressed as a molar quantity. $\endgroup$ – Buck Thorn Oct 10 '20 at 7:49
  • $\begingroup$ @BuckThorn yeah that's what I thought - all quantites relevant would then have to be expressed explicitly as molar quantites - shame that this book doesn't choose to mention it properly. Many thanks for your suggestions! I think I've solved the matter! $\endgroup$ – vik1245 Oct 12 '20 at 1:12

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