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:
$$p=p_{int}^*e^{\frac{zw_{AA}}{2kT}}$$
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
$$\ln\left(\frac{p}{p_{int}^{*}}\right) = \frac{zw_{AA}}{2kT}$$
to
$$\ln\left(\frac{p_1}{p_{int}^{*}}\right) = \frac{zw_{AA}}{RT_1}$$
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:
- How has Dill interchanged $R$ and $k$ in his solution?
- 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):
