= Cloud-chamber expansion model
{title2=$S=p_v/p_{\rm sat}(T)$}
Consider an illustrative <water>–air <mixture>, initially saturated at $T_0=293.15\,\mathrm K$. Let $\lambda=V/V_0$ be the expansion ratio and approximate the carrier <gas> by a uniform <ideal gas> with constant <heat capacity ratio> $\gamma=1.40$. Before significant <condensation>, assume fixed numbers of carrier and <vapor> <molecules>, negligible incoming <heat>, and the reversible <adiabatic expansion> relation. The <first law of thermodynamics>, $nC_V\,dT=-p\,dV$, with the <ideal gas law>, gives
$$
T(\lambda)=T_0\lambda^{1-\gamma},\qquad p_v(\lambda)=p_0\lambda^{-\gamma}.
$$
The <vapor> follows the same <temperature>, so its <partial pressure> scales as $T/V$. A rapid real <piston> motion need not be reversible; the formula is an idealized bulk estimate after neglecting flow and losses.
Use a physically motivated <Integrated Clausius-Clapeyron relation> for the equilibrium <water> <vapor pressure>, with stipulated rounded constants $p_0=2339\,\mathrm{Pa}$ at $T_0$, molar <latent heat> $L_m=45000\,\mathrm{J\,mol^{-1}}$ and <molar gas constant> $R=8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}$:
$$
p_{\rm sat}(T)=p_0\exp\left[\frac{L_m}{R}\left(\frac1{T_0}-\frac1T\right)\right],\qquad
\ln S=-\gamma\ln\lambda+\frac{L_m}{RT_0}\left(\lambda^{\gamma-1}-1\right).
$$
These constants define the example, without fitting any historical measurements. Throughout the plotted interval $1\leq\lambda\leq1.17$, the <temperature> stays above the freezing point of <water>. Initially
$$
\left.\frac{d\ln S}{d\ln\lambda}\right|_{\lambda=1}
=-\gamma+(\gamma-1)\frac{L_m}{RT_0}\simeq5.99>0.
$$
Thus the decrease in <vapor pressure> initially beats dilution. At $\lambda=1.15$, the model gives $T=277.21\,\mathrm K$, $p_v=1923.33\,\mathrm{Pa}$, $p_{\rm sat}=809.07\,\mathrm{Pa}$ and $S=2.3772$.
\Image[/cloud-chamber-expansion-and-nucleation-model.png]
{title=Cooling by <adiabatic expansion>, <vapor pressure> and the neutral-droplet <nucleation> barrier in a stipulated <cloud chamber> model}
{description=Original calculations, not measured traces. The neutral-droplet model illustrates a barrier; it omits the charged clusters responsible for the detector's ion-assisted response.}
Why does $S>1$ not instantly make every <molecule> a visible drop? In the capillarity approximation of <classical nucleation theory>, a neutral spherical drop of radius $r$ has a positive <surface energy> cost and a negative bulk <Gibbs free energy> contribution. For molecular <liquid> volume $v_l$, <surface tension> $\sigma$ and <Boltzmann constant> $k_B$,
$$
\Delta G(r)=4\pi\sigma r^2-\frac{4\pi r^3}{3v_l}k_BT\ln S,\qquad
r_* =\frac{2\sigma v_l}{k_BT\ln S},\qquad
\Delta G_* =\frac{16\pi\sigma^3v_l^2}{3(k_BT\ln S)^2}.
$$
Differentiating $\Delta G$ shows that the <critical nucleus> at $r_*$ is a maximum: smaller neutral drops tend to shrink, whereas larger ones can grow. With stipulated $\sigma=0.075\,\mathrm{J\,m^{-2}}$, $v_l=2.99\times10^{-29}\,\mathrm{m^3}$ and $k_B=1.380649\times10^{-23}\,\mathrm{J\,K^{-1}}$, the $\lambda=1.15$ example gives $r_*=1.353\,\mathrm{nm}$ and $\Delta G_*/(k_BT)=150.3$. A substantial neutral-drop barrier can coexist with <supersaturated vapor>.
This is deliberately a neutral-drop comparison. An <ion> changes the cluster <free energy> through electrical interactions and molecular hydration; the formula does not calculate the ion-assisted <nucleation> threshold. At nanometre radii, using bulk <surface tension> is itself approximate. There is no predicted droplet count, growth time or detection efficiency here. Once many drops grow, <latent heat> release, <vapor> depletion and <heat transfer> from the chamber walls invalidate the dry <adiabatic expansion> trajectory. The example therefore explains the direction of the effect and the need for <nucleation>, without extrapolating its curves to Wilson's cloud limits.
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