Question

If the diameter of a soap bubble is 10 mm and its surface tension is 0.04 N m⁻¹, find the excess pressure inside the bubble.

Answer 1

excess pressure inside the soap bubble is given by, $P=\frac{4S}{r}$

where S is surface tension of soap bubble ,
r is the radius of the soap.

given, surface tension of soap bubble, S = 0.04 N/m
diameter of soap bubble, d = 10mm
radius of soap bubble , r = 5mm = 5 × 10^-3 m

now, excess pressure inside the soap bubble , P = 4 × 0.04/5 × 10^-3

P = 4 × 0.04/0.005 = 160/5 = 32 N/m²

hence, excess pressure inside bubble is 32 N/m².

Answer 2

The pressure inside the Soap bubble is more than the pressure outside, This is because it is to balance the surface tension too.

The Excess pressure inside a bubble is given by,

$\frac{2T}{r}$

Soap bubbles contains 2 surfaces. Hence, The Excess pressure inside a Soap bubble is given by,

$\frac{2T}{r} + \frac{2T}{r} = \frac{4T}{r}$

Here,

$\frac{4T}{r} \: \begin{cases} \sf T \: - \: Surface \: Tension \\ \sf \: r \: - \: radius \: of \: the \: bubble \: \\ \end{cases}$

Given

1. Diameter of soap bubble (2r) = 10mm
2. Surface tension (T) = 0.04 N/m

⇒ 2 * Radius = Diameter

Radius of soap bubble = 10 ÷ 2 = 5mm

Excess pressure inside the Soap bubble is,

$\: \: \: \: \: \: \: \: \: \: \: = \frac{4T}{r} \\ \\ \: \: \: \: \: \: \: \: \: \: \: = \frac{4(0.04)N/m}{5 \times {10}^{ - 3} m} \\ \\ \: \: \: \: \: \: \: \: \: \: \: = \frac{160}{5} N/m^{2} \\ \\ \: \: \: \: \: \: \: \: \: \: \: = 32N/m^{2}$

Therefore, The Excess pressure inside the given soap bubble is 32N/m²