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Question: The pressure of air inside a soap bubble of diameter \[0.7{\rm{ cm}}\] is \[8{\rm{ mm}}\] of water a...

The pressure of air inside a soap bubble of diameter 0.7cm0.7{\rm{ cm}} is 8mm8{\rm{ mm}} of water above atmospheric pressure. The surface tension of soap solution is:
A. 9.8×10−1N/Nmm9.8 \times {10^{ - 1}}{\rm{ }}{{\rm{N}} {\left/ {\vphantom {{\rm{N}} {\rm{m}}}} \right. } {\rm{m}}}
B. 6.8×10−2N/Nmm6.8 \times {10^{ - 2}}{\rm{ }}{{\rm{N}} {\left/ {\vphantom {{\rm{N}} {\rm{m}}}} \right.} {\rm{m}}}
C. 7.2×10−2N/Nmm7.2 \times {10^{ - 2}}{\rm{ }}{{\rm{N}} {\left/ {\vphantom {{\rm{N}} {\rm{m}}}} \right. } {\rm{m}}}
D. 1.37×10−1N/Nmm1.37 \times {10^{ - 1}}{\rm{ }}{{\rm{N}} {\left/ {\vphantom {{\rm{N}} {\rm{m}}}} \right. } {\rm{m}}}

Explanation

Solution

We know that the pressure above the atmospheric pressure in a soap bubble can be written in the form of surface tension of the soap bubble and its diameter. We will convert the pressure from millimetres of water into Pascal.

Complete step by step answer:
Given:
The diameter of the soap bubble is D=0.7cm=0.7cm×(10mmcm)=7mmD = 0.7{\rm{ cm}} = 0.7{\rm{ cm}} \times \left( {\dfrac{{10{\rm{ mm}}}}{{{\rm{cm}}}}} \right) = 7{\rm{ mm}}.
The amount of excess pressure inside the bubble is ΔP=8mm\Delta P = 8{\rm{ mm}} of water.

Let us write the expression to convert pressure given in millimetres of water into Pascal. We know that one millimetre of water is given as:
1mm1 mm of water = 9.806Pa9.806{\rm{ Pa}}

We will use the above expression to convert 8mm8{\rm{ mm}} of water into the S.I. unit of pressure (Pascal). To do so, we will multiply eight on both sides of the above equation.
8mm8 mm of water = 8 \times 9.806{\rm{ Pa}}\\\
= 78.448Pa78.448{\rm{ Pa}}

We know that the formula for excess pressure inside a soap bubble is given as:

\Delta P = \dfrac{{4\sigma }}{R}\\\ \sigma = \dfrac{{R\Delta P}}{4} \end{array}$$……(1) Here $$\sigma $$ is the surface tension and R is the radius of the bubble. We know that the radius of the bubble is twice its diameter. $$\begin{array}{c} D = 2R\\\ R = \dfrac{D}{2} \end{array}$$ We will substitute $$7{\rm{ mm}}$$ for D in the above expression to get the value of radius R. $$\begin{array}{c} R = \dfrac{{7{\rm{ mm}}}}{2}\\\ = 3.5{\rm{ mm}} \end{array}$$ We will substitute $$78.448{\rm{ Pa}}$$ for $$\Delta P$$ and $$3.5{\rm{ mm}}$$ for R in equation (1) to get the value of surface tension of the bubble. $$\begin{array}{c} \sigma = \dfrac{{\left( {3.5{\rm{ mm}} \times \dfrac{{\rm{m}}}{{1000{\rm{ mm}}}}} \right)\left( {78.448{\rm{ Pa}} \times \dfrac{{{{{{\rm{N}} {\left/ {\vphantom {{\rm{N}} {\rm{m}}}} \right. } {\rm{m}}}}^2}}}{{{\rm{Pa}}}}} \right)}}{4}\\\ = 6.8 \times {10^{ - 2}}{\rm{ }}{{\rm{N}} {\left/ {\vphantom {{\rm{N}} {\rm{m}}}} \right. } {\rm{m}}} \end{array}$$ Therefore, the surface tension of the given soap solution is $$6.8 \times {10^{ - 2}}{\rm{ }}{{\rm{N}} {\left/ {\vphantom {{\rm{N}} {\rm{m}}}} \right.} {\rm{m}}}$$. **So, the correct answer is “Option B”.** **Note:** The formula for excess pressure inside the soap bubble is obtained using equations of equilibrium for the force acting on the bubble. Surface tension force and force due to pressure are the acting forces on the soap bubble.