Excess pressure is the pressure difference between the concave and convex sides of a curved liquid interface caused by surface tension. It acts inward toward the center of curvature, maintaining the mechanical equilibrium of liquid drops, bubbles, and menisci.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Calculation of excess pressure for spherical droplets, soap bubbles, and cylindrical liquid surfaces; basic work-energy principles involving surface tension.
PGT extension
Derivation of Young-Laplace equation for general curved surfaces with two principal radii of curvature, contact angles, capillary rise under pressure, and advanced energy methods.
Prerequisites
Surface tension and surface energy fundamentals
Forces of cohesion and adhesion
Hydrostatic pressure and Pascal's law
Origin of Excess Pressure
Molecules on a curved liquid surface experience a net inward resultant cohesive force due to asymmetry in molecular neighbors.
This inward pull generates a continuous inward normal force per unit area, making the pressure immediately inside a concave surface greater than that outside.
The magnitude of this pressure difference is independent of the absolute external pressure and depends strictly on surface tension and curvature.
Excess Pressure in a Spherical Liquid Droplet
A liquid droplet possesses only one free surface exposed to air or gas.
The excess pressure inside a droplet acts uniformly across the spherical interface of radius R.
It balances the inward force exerted by the surface tension along the circumference of any diametral plane.
Excess Pressure in a Soap Bubble
A soap bubble in air has two distinct free surfaces: an inner surface and an outer surface.
Because both surfaces contribute to the restoring force due to surface tension, the total excess pressure is double that of a solid droplet of the same radius.
The thin film of liquid encloses air on both sides, requiring both interfaces to be considered.
General Curved Surface and Young-Laplace Equation
For a general non-spherical liquid-gas interface, the curvature is defined by two principal radii of curvature, R1 and R2.
The Young-Laplace equation relates the excess pressure to the surface tension and the sum of the reciprocal principal radii.
When R1 = R2 = R, the equation reduces to the standard spherical interface formula.
Derivations
Excess Pressure in a Spherical Liquid Droplet
Conditions: Isothermal expansion, spherical geometry, uniform surface tension T and radius R.
Let the initial radius of a spherical liquid droplet be R and surface tension be T.
Suppose a small virtual increase in radius by dR occurs due to an internal pressure increment ΔP.
The outward mechanical work done by the excess pressure is d$W = (Excess Pressure) × (Change in Volume) = ΔP × (4\piR^{2} dR)$.
The increase in surface potential energy is d$U = T × (Increase in Surface Area) = T × d(4\piR^{2}) = T × (8\piR dR)$.
Equating the mechanical work done to the increase in surface energy: ΔP (4πR2 dR) = T (8πR dR).
Solving for ΔP yields the required expression.
Result: ΔP = \frac{2T}{R}
Excess Pressure in a Soap Bubble
Conditions: Soap bubble with two free surfaces (inner and outer), radius R, surface tension T.
A soap bubble has two surfaces in contact with air: an inner surface and an outer surface, both having approximately radius R.
When the radius increases by dR, the increase in total surface area is 2 × (8πR dR) = 16πR dR.
The increase in surface energy is d$U = T × (16\piR dR)$.
The work done by the excess pressure ΔP for a volume change d$V = 4\piR^{2} dR$ is d$W = ΔP (4\piR^{2} dR)$.
Equating work done to the change in surface energy: ΔP (4πR2 dR) = T (16πR dR).
Solving for ΔP gives the final relation.
Result: ΔP = \frac{4T}{R}
Formula sheet
Formula / law
Equation
Variables and units
Conditions
Excess Pressure in Liquid Droplet
$$\Delta P = \frac{2T}{R}$$
T = Surface tension, R = Radius of droplet, \Delta P = Excess pressure N/m2 or Pa
Spherical liquid droplet with single surface
Excess Pressure in Soap Bubble
$$\Delta P = \frac{4T}{R}$$
T = Surface tension, R = Radius of bubble, \Delta P = Excess pressure N/m2 or Pa
Spherical bubble with inner and outer surfaces
Excess Pressure in Cylindrical Liquid Jet or Column
$$\Delta P = \frac{T}{R}$$
T = Surface tension, R = Radius of cylinder, \Delta P = Excess pressure N/m2 or Pa
Cylindrical liquid surface (one principal radius is infinite)
Young-Laplace Equation (General Surface)
$$\Delta P = T \left( \frac{1}{R_1} + \frac{1}{R_2} \right)$$
T = Surface tension, R_1, $R_2 = Principal radii of curvature$ N/m2 or Pa
General curved liquid-gas or liquid-liquid interface
Work Done in Blowing a Droplet
$$W = T \times 4\pi R^2$$
T = Surface tension, R = Final radius Joule (J)
Isothermal formation of a single-surface droplet from zero initial radius
Work Done in Blowing a Soap Bubble
$$W = T \times 8\pi R^2$$
T = Surface tension, R = Final radius Joule (J)
Formation of a soap bubble with two surfaces
Radius of Combined Soap Bubbles
$$R = \frac{R_1 R_2}{\sqrt{R_1^2 + R_2^2}}$$
R_1, $R_2 = Radii of two isothermal soap bubbles forming a common interface$ m
Isothermal isothermal coalescence or contact between two soap bubbles
Capillary Pressure Height Relation
$$h = \frac{2T \cos\theta}{r \rho g}$$
T = Surface tension, \theta = Contact angle, r = Capillary radius, \rho = Density, g = Acceleration due to gravity m
Cylindrical capillary tube dipping in liquid
Graphs and experiments
Excess Pressure vs. Inverse Radius Graph
Draw: A linear plot of excess pressure (ΔP) on the y-axis against the reciprocal radius (1/R) on the x-axis, showing a straight line passing through the origin.
Interpret: The slope of the line equals 2T for a droplet and 4T for a bubble, directly verifying proportionality between excess pressure and curvature.
Exam use: Tested in objective questions requiring identification of slope parameters or comparative analysis of droplet versus bubble scaling.
Interconnected Soap Bubbles Experiment
Draw: Two soap bubbles of unequal radii (R1 < R2) connected by a stopcock valve.
Interpret: Since ΔP is inversely proportional to R, the smaller bubble has a higher internal pressure and consequently deflates while inflating the larger bubble until equilibrium or film rupture occurs.
Exam use: Conceptual TGT/PGT questions regarding the direction of air flow between connected bubbles.
Jaeger's Method for Surface Tension
Draw: Schematic setup of a manometer connected to a capillary tube immersed in liquid, showing bubble formation at the tip.
Interpret: At the moment of bubble detachment, the radius of the bubble equals the radius of the capillary tube, and the maximum excess pressure is recorded via the manometer head.
Exam use: Experimental physics background questions on measuring surface tension and maximum bubble pressure.
Solved numericals
TGT core
Calculate the excess pressure inside a soap bubble of radius 2 cm. Given the surface tension of soap solution is 0.03 N/m.
Identify given parameters: Radius R = 2 cm = 0.02 m, Surface tension $T = 0.03 N/m$.
Recall the formula for excess pressure in a soap bubble: Δ$P = 4T / R$.
Substitute the values: Δ$P = (4 × 0.03) / 0.02$.
Calculate the final value: Δ$P = 0.12 / 0.02 = 6 N/m^{2}$.
Answer: 6 Pa
TGT core
If the excess pressure inside a liquid droplet is 800 N/m2 and its radius is 0.5 mm, find the surface tension of the liquid.
Identify given parameters: Δ$P = 800 N/m^{2}$, $R = 0.5 mm = 5 × 10^{-4} m$.
Recall the formula for a liquid droplet: Δ$P = 2T / R$, hence $T = (\DeltaP × R) / 2$.
Two soap bubbles of radii 4 cm and 6 cm coalesce isothermally in vacuum to form a single soap bubble. Calculate the radius of the new bubble.
Since the process is isothermal, the total amount of air (PV product) or surface area conservation applies. For soap bubbles, total surface area before coalescence equals surface area after coalescence.
Initial surface area of bubble 1: A1 = 8πR12 = 8π(0.04)2.
Initial surface area of bubble 2: A2 = 8πR22 = 8π(0.06)2.
Final surface area of combined bubble: $A = 8\piR^{2}$.
The liquid-gas interface is assumed to be in static or quasi-static equilibrium.
Surface tension T is assumed to be uniform across the entire interface, neglecting local thermal or compositional gradients (Marangoni effect).
The droplet or bubble is assumed to be perfectly spherical, which holds true only when gravitational forces are negligible compared to surface tension forces (i.e., small Bond number).
Common misconceptions
Common error: A liquid droplet has two free surfaces similar to a soap bubble.
Correct: A liquid droplet has only one free surface, whereas a soap bubble has two free surfaces (inner and outer).
A droplet is solid liquid with air outside, while a soap bubble is a thin film of liquid enclosing air inside and surrounded by air outside.
Common error: Excess pressure acts outward from the center of the bubble.
Correct: Excess pressure acts inward from the concave side toward the convex side (i.e., inward toward the center of curvature).
The concave side of a curved interface always experiences a higher pressure than the convex side.
Common error: When two soap bubbles of different sizes touch, air flows from the larger bubble to the smaller bubble.
Correct: Air flows from the smaller bubble into the larger bubble.
The smaller bubble has a higher excess pressure (ΔP ∝ 1/R), driving gas toward the lower-pressure larger bubble.
Quick revision
Excess pressure in a liquid droplet: 2T/R; in a soap bubble: 4T/R.
Young-Laplace equation: Δ$P = T(1/R_1 + 1/R_2)$.
Work done in blowing a droplet is 4πR2T and for a soap bubble is 8πR2T.
Smaller bubbles have higher excess pressure and tend to deflate into larger bubbles.
Must remember
The excess pressure inside a spherical liquid droplet is given by $\Delta P = \frac{2T}{R}$ due to a single free surface.
A soap bubble has two free surfaces (inner and outer), making the excess pressure $\Delta P = \frac{4T}{R}$.
For a cylindrical liquid column or jet, the excess pressure is $\Delta P = \frac{T}{R}$ since one principal radius of curvature is infinite.
The general Young-Laplace equation for any curved liquid-gas interface is $\Delta P = T \left( \frac{1}{R_1} + \frac{1}{R_2} \right)$.
When two soap bubbles of radii $R_1$ and $R_2$ coalesce isothermally, the radius of the resulting single bubble or common interface radius is given by $R = \frac{R_1 R_2}{\sqrt{R_1^2 + R_2^2}}$.
Air always flows from a smaller soap bubble into a larger bubble because excess pressure is inversely proportional to radius ($\Delta P \propto \frac{1}{R}$).
Glossary
Surface Tension ($T$)The property of a liquid surface that causes it to behave as a stretched elastic membrane, measured in $\text{N/m}$.
Excess Pressure ($\Delta P$)The pressure difference between the inside and outside of a curved liquid surface, expressed in $\text{N/m}^2$ or $\text{Pa}$.
Young-Laplace EquationA nonlinear partial differential equation that describes the capillary pressure difference sustained across the interface between two static fluids.
Principal Radii of Curvature ($R_1, R_2$)The radii of the circular arcs representing the maximum and minimum curvatures at a point on a general surface.
Comparison of Liquid Droplet, Soap Bubble, and Cylindrical Jet
Parameter
Liquid Droplet
Soap Bubble
Cylindrical Liquid Jet
Number of Free Surfaces
One
Two (Inner and Outer)
One
Excess Pressure Formula
$\Delta P = \frac{2T}{R}$
$\Delta P = \frac{4T}{R}$
$\Delta P = \frac{T}{R}$
Work Done in Formation
$W = T \times 4\pi R^2$
$W = T \times 8\pi R^2$
$W = T \times 2\pi R L$
Practice Quiz
Choose an option to reveal the explanation.
1. What is the excess pressure inside a spherical liquid droplet of radius $R$ and surface tension $T$?
Explanation: A liquid droplet has only one free surface exposed to air. Balancing the inward force due to surface tension along a diametral plane gives $\Delta P \times \pi R^2 = 2\pi R T$, leading to $\Delta P = \frac{2T}{R}$.
2. Why is the excess pressure inside a soap bubble double that of a liquid droplet of the same radius?
Explanation: A soap bubble encloses air inside and is surrounded by air outside, creating both an inner and an outer liquid-air interface. Both surfaces contribute to the restoring force, resulting in $\Delta P = \frac{4T}{R}$.
3. What is the excess pressure inside a cylindrical liquid column or jet of radius $R$?
Explanation: For a cylindrical surface, one principal radius of curvature is $R$ and the other is infinite ($R_2 = \infty$). Substituting into the Young-Laplace equation gives $\Delta P = T \left( \frac{1}{R} + \frac{1}{\infty} \right) = \frac{T}{R}$.
4. The general Young-Laplace equation for a curved interface with principal radii $R_1$ and $R_2$ and surface tension $T$ is given by:
Explanation: The Young-Laplace equation relates the excess pressure to surface tension and the sum of the reciprocal principal radii of curvature: $\Delta P = T \left( \frac{1}{R_1} + \frac{1}{R_2} \right)$.
5. How much work is required to blow a spherical soap bubble of final radius $R$ with surface tension $T$, assuming isothermal conditions?
Explanation: A soap bubble has two free surfaces. The total surface area is $2 \times (4\pi R^2) = 8\pi R^2$. The work done is equal to the surface tension multiplied by the total change in surface area: $W = T \times 8\pi R^2$.
6. When two soap bubbles of radii $R_1$ and $R_2$ coalesce isothermally into a single soap bubble of radius $R$, what is the expression for $R$?
Explanation: Due to isothermal conservation of surface area for soap bubbles ($8\pi R_1^2 + 8\pi R_2^2 = 8\pi R^2$), we obtain $R = \sqrt{R_1^2 + R_2^2}$.
7. Two soap bubbles of unequal radii $R_1$ and $R_2$ ($R_1 < R_2$) are in contact with a common interface. What is the radius of curvature $R$ of this common interface?
Explanation: The pressure difference across the common interface balances the difference in excess pressures of the two bubbles: $\frac{4T}{R} = \frac{4T}{R_1} - \frac{4T}{R_2}$. Solving for $R$ yields $R = \frac{R_1 R_2}{R_2 - R_1}$.
8. When two soap bubbles of different radii touch each other, air flows:
Explanation: Since excess pressure is inversely proportional to radius ($\Delta P \propto \frac{1}{R}$), the smaller bubble has a higher internal pressure, causing air to flow from the smaller bubble into the larger bubble.
9. What is the work done in forming a single-surface liquid droplet of radius $R$ from zero initial radius?
Explanation: A liquid droplet has only one free surface with area $4\pi R^2$. The work done is $W = T \times 4\pi R^2$.
10. Which of the following statements regarding the origin of excess pressure is correct?
Explanation: Molecules on a curved liquid surface experience an asymmetry in neighbors, creating a net inward cohesive force. This generates an inward normal force per unit area, making the pressure immediately inside greater than outside.
11. If the radius of a soap bubble is doubled while keeping the temperature constant, the excess pressure inside the bubble becomes:
Explanation: Excess pressure in a soap bubble is $\Delta P = \frac{4T}{R}$. If radius $R$ is doubled, $\Delta P$ becomes half of its original value.
12. What happens to the total surface energy when two identical soap bubbles coalesce into a single soap bubble under isothermal conditions?
Explanation: Since total surface area decreases upon coalescence ($2 \times 4\pi R^2 > 4\pi (\sqrt{2}R)^2$), the total surface energy (which is $T \times \text{Area}$) decreases.
13. The excess pressure inside a bubble of radius $R$ submerged completely at a depth $h$ in a liquid of density $\rho$ is:
Explanation: The excess pressure $\Delta P$ across the bubble film depends only on surface tension and radius ($\frac{4T}{R}$ for a gas bubble in liquid, which has two surfaces or one depending on context; standard gas bubble in liquid has one interface, so $\frac{2T}{R}$). Wait, for a gas bubble inside a liquid, there is only one surface, so excess pressure is $\frac{2T}{R}$. Total absolute pressure inside would be hydrostatic pressure plus excess pressure.
14. For a soap bubble, if the radius decreases from $R_1$ to $R_2$ due to isothermal compression, the work done by the gas inside is:
Explanation: Work done by the gas equals the decrease in total surface energy: $W = T \times \Delta A = T \times [8\pi R_1^2 - 8\pi R_2^2] = 8\pi T (R_1^2 - R_2^2)$.
15. In capillary rise, the height $h$ to which liquid rises in a tube of radius $r$ is related to surface tension $T$ and contact angle $\theta$ by:
Explanation: Equating the upward force due to surface tension ($2\pi r T \cos\theta$) to the weight of the liquid column ($\pi r^2 h \rho g$) gives $h = \frac{2T \cos\theta}{r \rho g}$.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. The excess pressure inside a soap bubble of radius $R$ is $P_1$ and that inside a liquid droplet of radius $R$ is $P_2$. The ratio $\frac{P_1}{P_2}$ is:
Explanation: Excess pressure for a soap bubble is $P_1 = \frac{4T}{R}$ and for a liquid droplet is $P_2 = \frac{2T}{R}$. Therefore, $\frac{P_1}{P_2} = \frac{4T/R}{2T/R} = 2:1$.
2. If two soap bubbles of radii $R_1$ and $R_2$ combine under isothermal conditions to form a single soap bubble of radius $R$, the correct relation is:
Explanation: Under isothermal conditions, total surface area is conserved: $8\pi R_1^2 + 8\pi R_2^2 = 8\pi R^2$, which yields $R = \sqrt{R_1^2 + R_2^2}$.
3. The work done in blowing a soap bubble from a radius $r_1$ to $r_2$ with surface tension $T$ is:
Explanation: The work done is equal to the surface tension times the increase in total surface area of the two surfaces: $W = T \times \Delta A = T \times [8\pi r_2^2 - 8\pi r_1^2] = 8\pi T (r_2^2 - r_1^2)$.
4. The excess pressure inside a cylindrical liquid drop or jet of radius $R$ and surface tension $T$ is given by:
Explanation: For a cylindrical surface, one principal radius is $R$ and the other is $\infty$. Substituting into Young-Laplace equation gives $\Delta P = \frac{T}{R}$.
5. Two soap bubbles have radii in the ratio $2:1$. What is the ratio of excess pressures inside them?
Explanation: Excess pressure $\Delta P \propto \frac{1}{R}$. Thus, $\frac{\Delta P_1}{\Delta P_2} = \frac{R_2}{R_1} = \frac{1}{2}$ or $1:2$.
6. What is the radius of the common interface when two soap bubbles of radii $3\text{ cm}$ and $4\text{ cm}$ are in contact?
Attempt these questions under a ten-minute limit using a separate timer.
1. Which of the following equations represents the Young-Laplace equation for a general curved liquid surface?
Explanation: The Young-Laplace equation for any general curved surface is $\Delta P = T \left( \frac{1}{R_1} + \frac{1}{R_2} \right)$.
2. If a soap bubble of radius $R$ contracts to radius $r$ isothermally, the energy released in the process is:
Explanation: Energy released equals decrease in surface energy: $\Delta U = T \times \Delta A = T \times (8\pi R^2 - 8\pi r^2) = 8\pi T (R^2 - r^2)$.
3. The excess pressure inside a spherical bubble of radius $R$ formed in a liquid of surface tension $T$ is:
Explanation: A gas bubble inside a liquid has only one liquid-gas interface exposed to the liquid, so the excess pressure is $\frac{2T}{R}$.
4. What is the ratio of work done in blowing a soap bubble from radius $R$ to $2R$ versus blowing it from $0$ to $R$?
Explanation: Work from $0$ to $R$ is $W_1 = 8\pi T R^2$. Work from $R$ to $2R$ is $W_2 = 8\pi T ((2R)^2 - R^2) = 8\pi T (3R^2)$. Thus, $\frac{W_2}{W_1} = 3:1$.
5. If the surface tension of soap solution is $T$, the pressure inside a soap bubble of radius $R$ exceeds the atmospheric pressure by:
Explanation: The pressure difference between inside and outside is the excess pressure, which for a soap bubble is $\Delta P = \frac{4T}{R}$.
6. A drop of liquid of radius $R$ is broken into $1000$ identical droplets. If $T$ is surface tension, the work done in this process is:
Explanation: Volume conservation gives $r = \frac{R}{10} = 0.1 R$. Increase in surface area is $\Delta A = 1000 \times (4\pi r^2) - 4\pi R^2 = 1000 \times 4\pi (0.01 R^2) - 4\pi R^2 = 40\pi R^2 - 4\pi R^2 = 36\pi R^2$. Work done $W = T \times 36\pi R^2$.
7. Why does air flow from a smaller soap bubble into a larger one when connected?
Explanation: Excess pressure is inversely proportional to radius ($\Delta P \propto \frac{1}{R}$). The smaller bubble has higher pressure, causing air to flow toward the lower-pressure larger bubble.
8. The principal radii of curvature of an ellipsoidal liquid meniscus are $R_1 = 20\text{ cm}$ and $R_2 = 30\text{ cm}$. If surface tension is $0.075\text{ N/m}$, what is the excess pressure?
10. Which of the following physical quantities has the same dimensions as excess pressure multiplied by radius?
Explanation: Since $\Delta P = \frac{2T}{R}$ or $\frac{4T}{R}$, the product $(\Delta P \times R)$ has the dimensions of surface tension $T$ (force per unit length, $\text{MT}^{-2}$). Alternatively, $T$ is force per unit length.