2022.03.07 — kelvin equation
let's consider a (spherical) drop of some homogeneous fluid is floating around in some closed container that is initially evacuated.
the drop will begin to evaporate, filling the initially empty space around the drop with a gas composed of the same molecules that make up the drop.
this evaporation continues until the rate at which molecules leaving the drop and entering the gas equals the rate of molecules from the gas phase entering the drop.
in thermodynamic terms, this occurs when the chemical potentials $\mu_g$ and $\mu_l$ of the gas and liquid phases become equal:
$$
\mu_g(p_g) = \mu_l(p_l)
$$
$p_g$ and $p_l$ represent the pressure in the gas and liquid, respectively.
for a flat surface these pressures must be equal in thermal equilibrium, and the pressure $p_s$ at which this occurs is known as the "vapor pressure" of the fluid (at a given temperature).
for a closed surface, however, these pressures differ by the "laplace pressure" $\Delta p_{LP}$ of the gas-fluid interface:
$$
p_l - p_g = \Delta p_{LP}
$$
where
$$
\Delta p_{LP} = \frac{2 \gamma}{r}
$$
where $\gamma$ is the surface tension of the interface and $r$ is the radius of the drop.
taking into account the laplace pressure, our equilibrium condition can be stated as
$$
\mu_g(p_g) = \mu_l(p_g+p_{LP}) \tag{1}
$$
in other words, as a result of the curvature of the gas-liquid interface, the vapor around the drop will not settle at $p_s$ but instead at some different pressure p_g, which we can express as the sum of $p_s$ and some difference $\Delta p_g$.
by what amount $\Delta p_g$ then does the vapor pressure change due to the effect of the laplace pressure?
we tackle this by analyzing separately the left- and right-hand sides of equation 1.
to begin, we note that, at a fixed temperature, the change $\Delta \mu_g$ in the chemical potential of an ideal gas going from a pressure $p_o$ to $p' \equiv p_o + \Delta p$ is given by
$$
\Delta \mu_g = RT \ln \frac{p'}{p_o}
$$
if we assume $\Delta p \ll p_o$ then, using $\ln \left( 1 + x \right) \approx x$ we obtain
$$
\Delta \mu_g \approx \frac{RT}{p_o} \Delta p = v_o \Delta p
$$
where $v_o$ is the molar volume of a gas at temperature $T$ and pressure $p_o$.
we can therefore tentatively approximate the left-hand side of our equilibrium condition (equation 1) as
$$
\mu_g(p_g) = \mu_g(p_s + \Delta p_g) = \mu_g(p_s) + v_g \Delta p_g \tag{2}
$$
where $v_g$ is the molar volume of the gas at temperature $T$ and pressure $p_s$.
at the end of the calculation, when we arrive at an expression for $\Delta p_g$, we can analyze the conditions for which the assumption $\Delta p_g \ll p_s$ is valid.
next, we use the fundamental equation of calculus:
$$
\int_a^b \frac{d}{dx} f(x) dx = f(b) - f(a)
$$
to rewrite the right-hand side of equation 1 as:
$$
\mu_l(p_g+p_{LP}) = \mu_l(p_s + \Delta p_g + \Delta p_{LP}) = \mu_l(p_s) + \int_{p_s}^{p_s + \Delta p_g + \Delta p_{LP}} \left( \frac{\partial \mu_l}{\partial p} \right)_T dp
$$
then, from the gibbs-duhem relation for a single component system:
$$
d\mu = s dT + v dp
$$
where $s$ and $v$ are respectively the molar entropy and molar volume, we note that $v = \left( \frac{\partial \mu}{\partial p} \right)_T$, so that
$$
\mu_l(p_g+p_{LP}) = \mu_l(p_s) + \int_{p_s}^{p_s + \Delta p_g + \Delta p_{LP}} v_l dp
$$
where $v_l$ is the molar volume of the liquid droplet.
assuming that $v_l$ is roughly constant with pressure (i.e. the fluid is incompressible), we arrive at
$$
\mu_l(p_g+p_{LP}) = \mu_l(p_s) + v_l \left(\Delta p_g + \Delta p_{LP}\right) \tag{3}
$$
ok, now with equations 2 and 3 we have our new forms of the left- and right-hand sides of equations 1, respectively.
making these substitutions we obtain
$$
\mu_g(p_s) + v_g \Delta p_g = \mu_l(p_s) + v_l \left(\Delta p_g + \Delta p_{LP}\right)
$$
since $p_s$ is defined to be the vapor pressure of the (flat!) gas-fluid interface, we have
$$
\mu_g(p_s) = \mu_l(p_s)
$$
so that our equilibrium condition simplifies to
$$
v_g \Delta p_g = v_l \left(\Delta p_g + \Delta p_{LP}\right)
$$
solving for $\Delta p_g$ we obtain
$$
\Delta p_g = \frac{v_l \Delta p_{LP}}{v_g - v_l}
$$
since liquids are typically much denser than gases (i.e. $v_g \gg v_l$), we can simplify this to
$$
\Delta p_g = \frac{v_l}{v_g} \Delta p_{LP} \tag{4}
$$
so we arrive at the simple (and I suppose sensible) result that the increase in vapor pressure due to the laplace pressure is smaller than the laplace pressure by a factor equal to the ratio $\frac{v_g}{v_l}$.
to give a sense of scale, for water at room temperature we have a vapor pressure of $\approx \frac{1}{30}$ atmospheres, so that $v_g \approx 30 \times 22 \text{L} \approx 600 \text{L}$ while $v_l \approx 0.02 \text{L}$, so that the change in vapor pressure is about $\frac{600}{0.02} \approx 3 \times 10^4$ times smaller than the laplace pressure.
if earlier we had not made the simplification permitted by the assumption $\Delta p_g \ll p_s$, we would have at the end arrived not at equation 4, but rather at the kelvin equation:
$$
p_g = p_s + \Delta p_g = p_s e^{\frac{2 v_l \gamma}{R T r}}
$$
you can check that this equation reduces to equation 4 when the exponent is small, i.e. when
$$
\frac{2 v_l \gamma}{R T r} \ll 1
$$
so for which droplet radii $r$ is equation 4 valid?
for water at room temperature ($v_l \approx 20 \text{mL}$, $\gamma \approx 70 \frac{\text{mN}}{\text{m}}$), you can show that the exponent in the kelvin equation becomes unity at $r \approx 1 \text{nm}$, so that the approximate equation is applicable down to microscopic drop sizes.
finally, how can we understand this increase in the vapor pressure in a microscopic sense?
as the radius of the drop decreases, the curvature increases.
with increasing curvature, the bonding of molecules at the surface is weaker, since the average molecule at the interface has fewer neighbors to bond with.
the reduced binding of molecules increases the evaporation rate.
since at equilibrium the rate of molecules entering and exiting the interface is equal, the pressure in the gas phase must be higher around a surface to compensate the increased rate of evaporation.