Problem Statement
Derive the critical radius $R^*$ for a droplet nucleus in a supersaturated vapour.
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.
- See the step-by-step solution for the specific equations applied.
- All quantities are in SI units unless otherwise stated.
Step-by-Step Solution
Step 1 — Identify given quantities and set up the problem: The free energy change to form a droplet of radius $R$ in a supersaturated vapour (supersaturation ratio $S = p/p_\infty > 1$):
Step 2 — Apply the relevant physical law or equation: $$\Delta G = 4\pi R^2\sigma – \frac{4}{3}\pi R^3\frac{k_BT\ln S}{v_l}$$
Step 3 — Solve algebraically for the unknown: where $v_l$ is the molecular volume in the liquid. $\Delta G$ has a maximum (barrier) at the critical radius:
Step 4 — Substitute numerical values with units: $$\frac{d\Delta G}{dR} = 0 \implies R^* = \frac{2\sigma v_l}{k_BT\ln S}$$
Step 5 — Compute and check the result: Or using the Kelvin equation: $R^* = \frac{2\sigma V_m}{RT\ln S}$.
Step 6: Droplets smaller than $R^*$ re-evaporate; larger ones grow spontaneously. The nucleation rate is $\propto e^{-\Delta G^*/k_BT}$ where $\Delta G^* = \frac{16\pi\sigma^3 v_l^2}{3(k_BT\ln S)^2}$.
Worked Calculation
$$\Delta G = 4\pi R^2\sigma – \frac{4}{3}\pi R^3\frac{k_BT\ln S}{v_l}$$
$$\frac{d\Delta G}{dR} = 0 \implies R^* = \frac{2\sigma v_l}{k_BT\ln S}$$
$$\text{Numerical result} = \text{given expression substituted with values}$$
The free energy change to form a droplet of radius $R$ in a supersaturated vapour (supersaturation ratio $S = p/p_\infty > 1$):
$$\Delta G = 4\pi R^2\sigma – \frac{4}{3}\pi R^3\frac{k_BT\ln S}{v_l}$$
where $v_l$ is the molecular volume in the liquid. $\Delta G$ has a maximum (barrier) at the critical radius:
$$\frac{d\Delta G}{dR} = 0 \implies R^* = \frac{2\sigma v_l}{k_BT\ln S}$$
Or using the Kelvin equation: $R^* = \frac{2\sigma V_m}{RT\ln S}$.
Droplets smaller than $R^*$ re-evaporate; larger ones grow spontaneously. The nucleation rate is $\propto e^{-\Delta G^*/k_BT}$ where $\Delta G^* = \frac{16\pi\sigma^3 v_l^2}{3(k_BT\ln S)^2}$.
Answer
$$\boxed{\frac{d\Delta G}{dR} = 0 \implies R^* = \frac{2\sigma v_l}{k_BT\ln S}}$$
Physical Interpretation
The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.
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