Notes : Derivation Excess Pressure Inside Liquid Drop and Soap Bubble
Drops and Bubbles
Small liquid drops and soap bubbles are generally spherical because of surface tension. A sphere has the minimum surface area for a given volume, so it possesses the minimum surface energy. Surface tension always tries to reduce the surface area of a liquid, causing the drop to become spherical.
A liquid drop has only one liquid-air surface, whereas a soap bubble has two liquid-air surfaces (inner and outer). Due to surface tension, the pressure inside the drop or bubble is always greater than the pressure outside. This pressure difference is called Excess Pressure.
Principle of Derivation
The derivation of excess pressure is based on the principle of conservation of energy.
When a liquid drop expands slightly:
- The excess pressure inside the drop performs work to increase its size.
- The surface area of the drop increases.
- As the surface area increases, the surface energy also increases.
Therefore,
Work Done by Excess Pressure=Increase in Surface Energy
This principle is used to derive the expression for excess pressure inside a liquid drop and a soap bubble.
Derivation of Excess Pressure Inside a Liquid Drop
Step 1: Consider a Spherical Liquid Drop
Consider a spherical liquid drop of radius r.
Let,
- Radius of the drop = r
- Surface tension = T
- Pressure inside the drop = Pi
- Pressure outside the drop = P0
Since surface tension pulls the liquid molecules inward, the pressure inside the drop must be greater than the pressure outside.
Hence, the excess pressure is
$$ \Delta P=P_i-P_0 $$
Suppose the radius of the drop increases by a very small distance dr.
The new radius becomes
$$ r+dr $$
Step 2: Calculate the Outward Force
Pressure is defined as force acting per unit area.
$$ \text{Pressure}=\frac{\text{Force}}{\text{Area}} $$
Therefore,
$$ \boxed{\text{Force}=\text{Pressure}\times\text{Area}} $$
The excess pressure acts uniformly over the entire surface of the spherical drop.
The surface area of a sphere is
$$ 4\pi r^2 $$
Hence, the outward force acting on the drop is
$$ F=(P_i-P_0)\times4\pi r^2 $$
This force tends to increase the size of the drop, while surface tension resists this expansion.
Step 3: Calculate the Work Done by Excess Pressure
Assume that the radius of the drop increases by a very small distance dr.
Work done is given by
$$ \boxed{\text{Work Done}=\text{Force}\times\text{Displacement}} $$
Substituting the value of force,
$$ W=(P_i-P_0)\times4\pi r^2\times dr $$
Thus, the work done by the excess pressure during the small expansion of the liquid drop is
$$ \boxed{W=(P_i-P_0)\,4\pi r^2dr} $$
This work is stored as surface energy because the surface area of the drop increases.
What's Next?
In the next part, we will calculate the increase in surface area and surface energy, apply the relation
$$ \text{Work Done}=\text{Increase in Surface Energy} $$
and derive the final expression
$$ \boxed{P_i-P_0=\frac{2T}{r}} $$
Step 4: Calculate the Increase in Surface Area
As the radius of the liquid drop increases from r to r + dr, its surface area also increases.
Initial surface area of the drop is
$$ A_1=4\pi r^2 $$
Final surface area after expansion is
$$ A_2=4\pi(r+dr)^2 $$
Therefore, the increase in surface area is
$$ \Delta A=A_2-A_1 $$
$$ \Delta A=4\pi(r+dr)^2-4\pi r^2 $$
Expanding the Square
Using the identity
$$ (a+b)^2=a^2+2ab+b^2 $$
we get
$$ (r+dr)^2=r^2+2rdr+dr^2 $$
Substituting into the equation,
$$ \Delta A = 4\pi(r^2+2rdr+dr^2)-4\pi r^2 $$
Cancelling the common term,
$$ \Delta A = 8\pi rdr+4\pi dr^2 $$
Since dr is extremely small,
$$ dr^2\approx0 $$
Therefore,
$$ \boxed{\Delta A=8\pi rdr} $$
Step 5: Calculate the Increase in Surface Energy
Every liquid surface possesses surface energy.
Surface Energy is given by
Surface Energy = Surface Tension × Surface Area
Hence, the increase in surface energy becomes
$$ \Delta E=T\times\Delta A $$
Substituting the value of ΔA,
$$ \Delta E=T(8\pi rdr) $$
Therefore,
$$ \boxed{\Delta E=8\pi Trdr} $$
Step 6: Apply the Principle of Conservation of Energy
According to the principle,
$$ \boxed{\text{Work Done}=\text{Increase in Surface Energy}} $$
Substituting the values,
$$ (P_i-P_0)\,4\pi r^2dr = 8\pi Trdr $$
Cancel the common terms 4π and dr.
$$ (P_i-P_0)r^2=2Tr $$
Dividing both sides by r²,
$$ \boxed{P_i-P_0=\frac{2T}{r}} $$
Final Result for a Liquid Drop
The excess pressure inside a liquid drop is
$$ \boxed{\Delta P=\frac{2T}{r}} $$
where,
- T = Surface tension
- r = Radius of the drop
Derivation of Excess Pressure Inside a Soap Bubble
A soap bubble differs from a liquid drop because it has two liquid-air surfaces:
- Outer surface
- Inner surface
Therefore, the increase in surface energy is twice that of a liquid drop.
Step 1: Work Done by Excess Pressure
The outward force acting on the bubble is
$$ F=(P_i-P_0)\times4\pi r^2 $$
If the radius increases by a small distance dr,
$$ W=(P_i-P_0)\times4\pi r^2dr $$
Step 2: Increase in Surface Energy
Increase in surface area of one surface is
$$ 8\pi rdr $$
Since a soap bubble has two surfaces,
$$ \Delta A=2\times8\pi rdr $$
$$ \boxed{\Delta A=16\pi rdr} $$
Hence,
$$ \Delta E=T\times16\pi rdr $$
$$ \boxed{\Delta E=16\pi Trdr} $$
Step 3: Apply the Energy Principle
According to the principle,
$$ \text{Work Done}=\text{Increase in Surface Energy} $$
Therefore,
$$ (P_i-P_0)\,4\pi r^2dr = 16\pi Trdr $$
Cancelling common terms,
$$ (P_i-P_0)r^2=4Tr $$
Dividing by r²,
$$ \boxed{P_i-P_0=\frac{4T}{r}} $$
Final Result for a Soap Bubble
The excess pressure inside a soap bubble is
$$ \boxed{\Delta P=\frac{4T}{r}} $$
Summary of Excess Pressure
| Object | Number of Surfaces | Excess Pressure |
|---|---|---|
| Liquid Drop | One | $\Delta P=\frac{2T}{r}$ |
| Air Bubble (Cavity) in Liquid | One | $\Delta P=\frac{2T}{r}$ |
| Soap Bubble | Two | $\Delta P=\frac{4T}{r}$ |
Difference Between Liquid Drop and Soap Bubble
| Liquid Drop | Soap Bubble |
|---|---|
| Has only one liquid-air surface. | Has two liquid-air surfaces (inner and outer). |
| Increase in surface area = $8\pi r\,dr$ | Increase in surface area = $16\pi r\,dr$ |
| Surface energy increases only once. | Surface energy increases twice. |
| $\Delta P=\frac{2T}{r}$ | $\Delta P=\frac{4T}{r}$ |
Important Formulae
Surface Area of Sphere
$$ A=4\pi r^2 $$
Increase in Surface Area
$$ \Delta A=8\pi r\,dr $$
Surface Energy
$$ E=T\times A $$
Increase in Surface Energy
$$ \Delta E=T\Delta A $$
Work Done
$$ W=F\times d $$
Force Due to Excess Pressure
$$ F=(P_i-P_0)\times4\pi r^2 $$
Key Points
- Surface tension always tries to reduce the surface area of a liquid.
- A sphere has the minimum surface area for a given volume.
- Therefore, liquid drops become spherical.
- Pressure inside a liquid drop is greater than outside.
- Pressure inside a soap bubble is greater than outside.
- A soap bubble has two liquid-air surfaces.
- Excess pressure is inversely proportional to radius.
- Smaller the radius, greater is the excess pressure.
- The derivation is based on the principle: Work Done = Increase in Surface Energy
Solved Concept
Why is the pressure inside a liquid drop greater than outside?
Surface tension pulls the liquid surface inward. To balance this inward pull, the pressure inside the drop becomes greater than the outside pressure.
Why is the excess pressure inside a soap bubble twice that of a liquid drop?
A soap bubble has two liquid-air surfaces. Hence, the increase in surface energy is twice that of a liquid drop, making the excess pressure twice as large.
Frequently Asked Questions (FAQs)
Q1. What is excess pressure?
Excess pressure is the difference between the pressure inside and outside a liquid drop or soap bubble.
Q2. Why are liquid drops spherical?
Because a sphere has the minimum surface area for a given volume and therefore minimum surface energy.
Q3. What is the SI unit of surface tension?
Newton per metre (N m-1).
Q4. Which has greater excess pressure—a liquid drop or a soap bubble of the same radius?
A soap bubble has greater excess pressure because it has two liquid-air surfaces.
Q5. On which quantity is excess pressure directly proportional?
Excess pressure is directly proportional to surface tension.
Multiple Choice Questions (MCQs)
1. The excess pressure inside a liquid drop is
- A. $\frac{T}{r}$
- B. $\frac{2T}{r}$ ✓
- C. $\frac{3T}{r}$
- D. $\frac{4T}{r}$
2. The excess pressure inside a soap bubble is
- A. $\frac{2T}{r}$
- B. $\frac{4T}{r}$ ✓
- C.$\frac{6T}{r}$
- D. $\frac{8T}{r}$
3. A soap bubble has
- A. One surface
- B. Two surfaces ✓
- C. Three surfaces
- D. Four surfaces
True or False
- ✔ Surface tension tries to minimize surface area. (True)
- ✔ Pressure inside a liquid drop is greater than outside. (True)
- ✔ A soap bubble has only one surface. (False)
- ✔ Excess pressure decreases as radius increases. (True)
- ✔ A sphere has minimum surface area for a given volume. (True)
Fill in the Blanks
- The shape of a liquid drop is Spherical.
- The SI unit of surface tension is N m-1.
- Excess pressure inside a liquid drop is $\frac{2T}{r}$.
- Excess pressure inside a soap bubble is $\frac{4T}{r}$.
- A soap bubble has Two liquid-air surfaces.
Very Short Answer Questions with Answers
1. What is Excess Pressure?
Excess pressure is the difference between the pressure inside and outside a liquid drop or soap bubble.
$$ \Delta P=P_i-P_0 $$
2. What is Surface Tension?
Surface tension is the property of a liquid by which its free surface behaves like a stretched elastic membrane and tends to occupy the minimum possible surface area.
3. Why are liquid drops spherical?
A sphere has the minimum surface area for a given volume. Since surface tension always tries to minimize the surface area, liquid drops become spherical.
4. Write the formula for excess pressure inside a liquid drop.
$$ \boxed{\Delta P=\frac{2T}{r}} $$
5. Write the formula for excess pressure inside a soap bubble.
$$ \boxed{\Delta P=\frac{4T}{r}} $$
Short Answer Questions with Answers
1. Why is the pressure inside a liquid drop greater than outside?
Surface tension pulls the surface molecules inward, tending to reduce the size of the drop. To balance this inward pull, the pressure inside the drop becomes greater than the outside pressure. This difference is called excess pressure.
2. Differentiate between a liquid drop and a soap bubble.
| Liquid Drop | Soap Bubble |
|---|---|
| Has one liquid-air surface. | Has two liquid-air surfaces. |
| Excess pressure = $\frac{2T}{r}$ | Excess pressure = $\frac{4T}{r}$ |
| Surface energy increases once. | Surface energy increases twice. |
3. State the principle used to derive excess pressure.
The derivation is based on the principle of conservation of energy:
Work Done by Excess Pressure = Increase in Surface Energy
4. Why is excess pressure inversely proportional to the radius?
From the relation
$$ \Delta P=\frac{2T}{r} $$
it is clear that excess pressure varies inversely with radius. Therefore, smaller drops have greater excess pressure than larger drops.
Long Answer Questions with Answers
1. Derive the expression for excess pressure inside a liquid drop.
Consider a spherical liquid drop of radius r. Let the inside pressure be Pi and the outside pressure be P0. If the radius increases by a small amount dr, the work done by the excess pressure is
$$ W=(P_i-P_0)\times4\pi r^2dr $$
The increase in surface area is
$$ \Delta A=8\pi rdr $$
Therefore, the increase in surface energy is
$$ \Delta E=T\times8\pi rdr $$
Applying the relation
$$ \text{Work Done}=\text{Increase in Surface Energy} $$
$$ (P_i-P_0)4\pi r^2dr=8\pi Trdr $$
Simplifying,
$$ \boxed{P_i-P_0=\frac{2T}{r}} $$
2. Derive the expression for excess pressure inside a soap bubble.
A soap bubble has two liquid-air surfaces. Therefore, the increase in surface area becomes
$$ 16\pi rdr $$
The increase in surface energy is
$$ 16\pi Trdr $$
Using
$$ \text{Work Done}=\text{Increase in Surface Energy} $$
$$ (P_i-P_0)4\pi r^2dr=16\pi Trdr $$
On simplification,
$$ \boxed{P_i-P_0=\frac{4T}{r}} $$
3. Why do small drops have greater excess pressure than large drops?
The excess pressure inside a liquid drop is given by
$$ \Delta P=\frac{2T}{r} $$
Since excess pressure is inversely proportional to the radius, a smaller drop has a larger excess pressure, whereas a larger drop has a smaller excess pressure.
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