The excess pressure within a soap bubble is twice the excess pressure within a second soap bubble. If the volume of the first bubble is n times the volume of the second bubble, the value of n is

Question

Quantrex Academy · Accepted Answer

Correct answer: D. 0.125

The excess pressure inside a soap bubble of radius

R

and surface tension

T

is given by

P = \dfrac{4T}{R}

.

Given that

P_1 = 2P_2

, we can write:

\dfrac{4T}{R_1} = 2

\left(

\dfrac{4T}{R_2} \right)<br><br>

R_2 = 2R_1 \Rightarrow

\dfrac{R_1}{R_2} = \dfrac{1}{2}<br><br>The volume of a soap bubble is

V =

\dfrac{4}{3}\pi

R^3

.<br><br>Given that

V_1 = nV_2

, we have:<br><br>

n =

\dfrac{V_1}{V_2} = \left( \dfrac{R_1}{R_2} \right)^3

Substituting the ratio of radii:

n = \left(

\dfrac{1}{2} \right)^3 =

\dfrac{1}{8} =

0.125

The excess pressure within a soap bubble is twice the excess...

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