# A soap bubble is formed using a mixture of detergent and water. The surface tension of the mixture is 0.030 Nm^-1. If the bubble has a radius of 2 cm and atmospheric pressure is 101.2 kPa, what is the gauge pressure inside the bubble?

Mar 26, 2016

$\text{3 Pa}$

#### Explanation:

Your tool of choice for this problem will be the Laplace - Young equation, which allows you to calculate the Laplace pressure, i.e. the pressure difference that exists between the inside and the outside of a curved surface that acts as a boundary between a liquid region and a gas region

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \Delta p = \gamma \left(\frac{1}{R} _ 1 + \frac{1}{R} _ 2\right) \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, where

$\gamma$ - the surface tension of the liquid
${R}_{1}$ ,${R}_{2}$ - the principal radii of curvature

The idea here is that a difference in pressure between the two sides of a flat liquid surface gives rise to a force that can only be balanced by the surface tension of the liquid if the surface of the liquid is curved.

The great thing about a spherical shape such as a bubble is that the two principal radii of curvature are equal. This implies tha the Laplace pressure will be equal to

$\Delta p = \gamma \cdot \left(\frac{1}{R} + \frac{1}{R}\right)$

which is equivalent to

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \Delta p = \gamma \cdot \frac{2}{R} \textcolor{w h i t e}{\frac{a}{a}} |}}} \to$ the Laplace pressure for spherical shapes

Before plugging in your values to get the Laplace pressure for your soap bubble, make sure that you convert the radius of the bubble from centimeters to meters by using the conversion factor

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{1 m" = 10^2"cm}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

You will thus have

$\Delta p = {\text{0.030 N m"^(-1) * 2/(2 * 10^(-2)"m") = "3 N m}}^{- 2}$

As you know, ${\text{N" * m}}^{- 2}$ is equivalent to Pascal, $\text{Pa}$, which means that the Laplace pressure is equal to

$\Delta p = \text{3 Pa}$

Now, gauge pressure essentially tells you the difference that exists between the absolute pressure and ambient air pressure. In other words, gauge pressure is zero-referenced against normal pressure, i.e. air pressure at sea level, given to you as $\text{101.2 kPa}$.

In this context, the Laplace pressure will actually be equivalent to the gauge pressure. In other words, the pressure inside the soap bubble will be equal to

${P}_{\text{inside bubble" = "101,200 Pa" + "3 Pa" = "101,203 Pa}}$

The gauge pressure uses $\text{101,200 Pa}$ as the zero reference, so you will get a gauge reading of

${P}_{\text{gauge" = Deltap = "3 Pa}}$