Talk:Ostwald–Freundlich equation

According to Lord Kelvin's equation of 1871,^[1][2]

${\displaystyle p(r_{1},r_{2})=P-{\frac {\gamma \,\rho \,_{vapor}}{(\rho \,_{liquid}-\rho \,_{vapor})}}\left({\frac {1}{r_{1}}}+{\frac {1}{r_{2}}}\right)}$

where

${\displaystyle p(r)}$ = vapor pressure at a curved interface of radius ${\displaystyle r}$

${\displaystyle P}$ = vapor pressure at flat interface (${\displaystyle r=\infty }$) = ${\displaystyle p_{eq}}$

${\displaystyle \gamma }$ = surface tension

${\displaystyle \rho \,_{vapor}}$ = density of vapor

${\displaystyle \rho \,_{liquid}}$ = density of liquid

${\displaystyle r_{1}}$ , ${\displaystyle r_{2}}$ = radii of curvature along the principal sections of the curved interface.

Note: Kelvin defined the surface tension ${\displaystyle \gamma }$ as the work that was performed per unit area by the interface rather than on the interface; hence the term containing ${\displaystyle \gamma }$ has a minus sign. In what follows, the surface tension will be defined so that the term containing ${\displaystyle \gamma }$ has a plus sign.

If the particle is assumed to be spherical, then ${\displaystyle r=r_{1}=r_{2}}$ .

Since ${\displaystyle \rho \,_{liquid}\gg \rho \,_{vapor}}$, then ${\displaystyle \rho \,_{liquid}-\rho \,_{vapor}\approx \rho \,_{liquid}}$.

Hence

${\displaystyle p(r)\approx P+{\frac {2\gamma \,\rho \,_{vapor}}{\rho \,_{liquid}\cdot r}}}$.

Assuming that the vapor obeys the ideal gas law, then

${\displaystyle \rho \,_{vapor}={\frac {m_{vapor}}{V}}={\frac {MW\cdot n}{V}}={\frac {MW\cdot P}{RT}}={\frac {MW\cdot P}{N_{0}k_{B}T}}}$

where

${\displaystyle m_{vapor}}$ = mass of a volume ${\displaystyle V}$ of vapor

${\displaystyle MW}$ = molecular weight of vapor

${\displaystyle n}$ = number of moles of vapor in volume ${\displaystyle V}$ of vapor

${\displaystyle R}$ = ideal gas constant = ${\displaystyle N_{0}k_{B}}$

${\displaystyle N_{0}}$ = Avogadro’s number

${\displaystyle k_{B}}$ = Boltzmann's constant

${\displaystyle T}$ = absolute temperature.

Since ${\displaystyle {\frac {MW}{N_{0}}}=}$ mass of one molecule of vapor or liquid, then

${\displaystyle {\frac {\left({\frac {MW}{N_{0}}}\right)}{\rho \,_{liquid}}}=}$ volume of one molecule ${\displaystyle =V_{molecule}}$ .

Hence

${\displaystyle p(r)\approx P+{\frac {2\gamma V_{molecule}P}{k_{B}Tr}}=P+{\frac {R_{critical}P}{r}}}$ ,

where ${\displaystyle R_{critical}={\frac {2\gamma V_{molecule}}{k_{B}T}}}$ .

Thus

${\displaystyle {\frac {p(r)-P}{P}}\approx {\frac {R_{critical}}{r}}}$ .

Since ${\displaystyle {\frac {p(r)}{P}}=1-{\frac {P-p(r)}{P}}}$ , then ${\displaystyle \log \left({\frac {p(r)}{P}}\right)=\log \left(1-{\frac {P-p(r)}{P}}\right)}$ .

Since ${\displaystyle p(r)\approx P}$, then ${\displaystyle {\frac {P-p(r)}{P}}\ll 1}$ .

If ${\displaystyle x\ll 1}$, then ${\displaystyle \log \left(1-x\right)\approx -x}$ .

Hence

${\displaystyle \log \left({\frac {p(r)}{P}}\right)\approx {\frac {p(r)-P}{P}}}$ .

Therefore

${\displaystyle \log \left({\frac {p(r)}{P}}\right)\approx {\frac {R_{critical}}{r}}}$ ,

which is the Ostwald-Freundlich equation.

Cwkmail (talk) 10:50, 20 February 2013 (UTC)

I have added this to the article. Arbitrarily0 ^(talk) 19:35, 9 January 2014 (UTC)

I'm surprised this article and Poynting effect article don't mention each other. Both concern the change in vapor pressure due to changing total pressure. For this article, the added pressure in the condensed phase is coming from Laplace pressure, whereas for Poynting effect the added pressure is from background gases (noncondensible&insoluble). I'm trying to find a source for this but I don't see this connection being made anywhere, weird...

(I.e. It's about how the fugacity/activity/chemical potential changes due to total pressure. Ultimately down to Gibbs–Duhem equation where we get inside the condensed phase:

${\displaystyle {\bigg (}{\frac {\mathrm {d} \mu }{\mathrm {d} P}}{\bigg )}_{T}={\frac {V}{N}}}$

So if liquid is incompressible and vapor is ideal gas then ${\displaystyle \mu =\mu _{0}+\Delta P_{\rm {liq}}v_{\rm {liq}}}$ and vapor pressure ${\displaystyle p}$ follows ${\displaystyle \mu =\mu _{0}+RT\ln(p/p_{0})}$, where ${\displaystyle p_{0}}$ is the vapor pressure when ${\displaystyle \Delta P_{\rm {liq}}=0}$. So then ${\displaystyle \Delta P_{\rm {liq}}v_{\rm {liq}}=RT\ln(p/p_{0})}$.)

A couple of sources:

• Devoe "THERMODYNAMICS AND CHEMISTRY" - https://www2.chem.umd.edu/thermobook/ - discusses Poynting effect in chapter 12.8; droplets are discussed in various problems only, never refers to Ostwald or Kelvin
• Atkins 'Physical Chemistry' 8th edition does not name Poynting effect but just puts it as equation 4.3. Then later in equation 18.45 we get the Ostwald–Freundlich equation but it's called "Kelvin equation".

--Nanite (talk) 18:22, 16 February 2025 (UTC)