Enter values to see results
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Pa
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kPa
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atm
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m⁻¹
Enter values to see results
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Pa
—
kPa
—
atm
—
m⁻¹
The Young-Laplace Equation Calculator computes the pressure difference across a curved liquid interface due to surface tension. This fundamental equation in capillarity and fluid mechanics explains why curved liquid surfaces create a pressure jump — small bubbles have higher internal pressure than large ones, and liquid in narrow capillaries rises against gravity. The equation relates the pressure difference (ΔP) to the surface tension (γ) and the curvature of the interface, described by the principal radii of curvature. This calculator supports both spherical geometries (bubbles, droplets) and general curved surfaces with two different radii of curvature. The Young-Laplace equation is essential in microfluidics, emulsion science, foam physics, lung physiology (alveoli), inkjet printing, and the design of porous media and membrane filtration systems.
The Young-Laplace equation in its general form is:
$$\Delta P = \gamma \left(\frac{1}{R_1} + \frac{1}{R_2}\right) = 2\gamma H$$
where ΔP is the pressure difference across the interface (interior minus exterior), γ is the surface tension, R₁ and R₂ are the two principal radii of curvature, and H = (1/R₁ + 1/R₂)/2 is the mean curvature.
For a spherical interface (R₁ = R₂ = r):
$$\Delta P = \frac{2\gamma}{r}$$
For a soap bubble (two surfaces): ΔP = 4γ/r. For a cylindrical interface (one flat direction, R₂ → ∞): ΔP = γ/R₁.
The equation shows that smaller interfaces have higher pressure differences. A water droplet of radius 1 μm has ΔP ≈ 1.46 atm, while a 1 mm droplet has ΔP ≈ 146 Pa. This enormous dependence on size governs nucleation, Ostwald ripening, capillary penetration, and many other phenomena.
The Laplace pressure (ΔP) is always higher on the concave side of the interface. For a droplet, the interior pressure exceeds the exterior. For a bubble in liquid, the gas inside is at higher pressure. The magnitude of ΔP determines several important phenomena: capillary rise (h = 2γcosθ/(ρgr)), Ostwald ripening (small particles dissolve because their higher internal pressure raises the chemical potential), critical nucleation radius (embryos must exceed a minimum size for the bulk energy gain to overcome the Laplace pressure penalty), and foam drainage (pressure differences drive liquid flow between bubbles of different sizes). The mean curvature H is used in more advanced calculations involving non-spherical interfaces.
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A water droplet with 1 μm radius has an internal pressure 1.44 atm above atmospheric. This enormous Laplace pressure is why very small droplets evaporate faster than larger ones (Kelvin effect) and why cloud droplet formation requires supersaturation.
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A soap bubble with 1 cm radius and γ = 25 mN/m (soap solution) has ΔP = 5 Pa across each surface (total = 10 Pa for both surfaces). Note: for soap bubbles, double the calculator result because there are two air-liquid interfaces.
The Young-Laplace equation (ΔP = 2γ/r) shows that pressure difference is inversely proportional to radius. Smaller radius means higher curvature, which requires a greater pressure difference to maintain the interface shape. This is why when two bubbles of different sizes are connected, gas flows from the small (high pressure) to the large (low pressure) bubble.
Capillary action is the rise (or depression) of liquid in a narrow tube due to the Laplace pressure created by the curved meniscus. For a wetting liquid (θ < 90°) in a circular capillary: h = 2γcos(θ)/(ρgr), where ρ is the liquid density and r is the capillary radius. Water rises in glass capillaries; mercury is depressed (θ > 90°).
Lung alveoli are tiny air sacs (radius ~100 μm) lined with a thin liquid film. Without surfactant, the Laplace pressure would collapse small alveoli while expanding large ones (instability). Pulmonary surfactant reduces surface tension to ~1–25 mN/m and varies with compression, stabilizing alveoli of different sizes and preventing respiratory distress.
Ostwald ripening is the growth of large particles at the expense of small ones in a dispersion. The higher Laplace pressure in small particles increases their chemical potential and solubility (Kelvin/Thomson equation). Material dissolves from small particles and deposits on large ones, driven by the pressure difference. This causes coarsening of emulsions, crystals, and foams over time.
A soap bubble has two air-liquid interfaces (inner and outer surfaces of the soap film), each contributing a Laplace pressure of 2γ/r. The total pressure difference is ΔP = 4γ/r for a bubble. This is double that of a liquid droplet of the same radius. Enter γ for the soap solution and double the result.
A flat interface (R₁ = R₂ = ∞) has zero curvature and zero Laplace pressure (ΔP = 0). This corresponds to a free liquid surface far from walls. Any deviation from flatness creates a pressure difference that either drives the liquid to restore flatness (for gravity-dominated surfaces) or maintain the curved shape (for surface tension-dominated droplets).
The capillary number Ca = μv/γ compares viscous forces to surface tension forces, where μ is viscosity and v is velocity. At low Ca (< 10⁻⁵), surface tension dominates and interfaces maintain equilibrium shapes. At high Ca, viscous forces deform interfaces. Ca governs droplet breakup in flow, oil recovery, and printing processes.
In porous media, the Young-Laplace equation determines the capillary pressure: the pressure required to force a non-wetting fluid through a pore. Smaller pores require higher capillary pressure: Pc = 2γcos(θ)/r_pore. This is the basis of mercury intrusion porosimetry, membrane filtration, and oil reservoir engineering.
In nucleation theory, a new phase (bubble, crystal) must overcome the Laplace pressure to grow. The critical radius r* = 2γ/(ΔP_supersaturation) is the minimum radius above which growth is thermodynamically favorable. Below r*, the embryo spontaneously dissolves because the surface energy cost exceeds the bulk energy gain.
Yes, the general form uses two principal radii of curvature (R₁, R₂), which can differ for non-spherical shapes. For ellipsoidal droplets, pendant drops, and menisci in complex geometries, R₁ ≠ R₂. The equation becomes ΔP = γ(1/R₁ + 1/R₂). Analytical solutions exist for simple geometries; numerical methods are needed for complex shapes.
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