Problem Statement
An oil-in-water emulsion has droplets of radius $R=1\ \mu\text{m}$ and $\sigma_{oil-water}=0.05\ \text{N/m}$. Find the excess pressure inside a droplet and the free energy per droplet relative to bulk separation.
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: Laplace pressure inside a spherical droplet:
Step 2 — Apply the relevant physical law or equation: $$\Delta p = \frac{2\sigma}{R} = \frac{2\times0.05}{10^{-6}} = 10^5\ \text{Pa} = 1\ \text{atm}$$
Step 3 — Solve algebraically for the unknown: Surface energy per droplet:
Step 4 — Substitute numerical values with units: $$E_{surface} = 4\pi R^2\sigma = 4\pi\times(10^{-6})^2\times0.05 = 6.28\times10^{-13}\ \text{J}$$
Step 5 — Compute and check the result: In $k_BT$ units ($k_BT\approx4.1\times10^{-21}\ \text{J}$ at 300 K):
Step 6: $$\frac{E_{surface}}{k_BT} \approx \frac{6.28\times10^{-13}}{4.1\times10^{-21}} \approx 1.5\times10^8$$
Worked Calculation
$$\Delta p = \frac{2\sigma}{R} = \frac{2\times0.05}{10^{-6}} = 10^5\ \text{Pa} = 1\ \text{atm}$$
$$E_{surface} = 4\pi R^2\sigma = 4\pi\times(10^{-6})^2\times0.05 = 6.28\times10^{-13}\ \text{J}$$
$$\frac{E_{surface}}{k_BT} \approx \frac{6.28\times10^{-13}}{4.1\times10^{-21}} \approx 1.5\times10^8$$
Laplace pressure inside a spherical droplet:
$$\Delta p = \frac{2\sigma}{R} = \frac{2\times0.05}{10^{-6}} = 10^5\ \text{Pa} = 1\ \text{atm}$$
Surface energy per droplet:
$$E_{surface} = 4\pi R^2\sigma = 4\pi\times(10^{-6})^2\times0.05 = 6.28\times10^{-13}\ \text{J}$$
In $k_BT$ units ($k_BT\approx4.1\times10^{-21}\ \text{J}$ at 300 K):
$$\frac{E_{surface}}{k_BT} \approx \frac{6.28\times10^{-13}}{4.1\times10^{-21}} \approx 1.5\times10^8$$
This enormous surface energy ($\gg k_BT$) means emulsions are thermodynamically unstable — they will eventually coalesce. Kinetic stability (from surfactants preventing film drainage) is what keeps emulsions like milk usable.
Answer
$$\boxed{\frac{E_{surface}}{k_BT} \approx \frac{6.28\times10^{-13}}{4.1\times10^{-21}} \approx 1.5\times10^8}$$
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.
Leave a Reply