The surface tension of the soap water solutio | vedclass

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(D) Given:Diameter of the soap bubble $d = 5 	ext{ mm} = 5 	imes 10^{-3} 	ext{ m}$.Radius $R = \frac{d}{2} = 2.5 	imes 10^{-3} 	ext{ m}$.Surface

Vedclass · Accepted Answer

$5 	imes 10^{-6} 	ext{ J}$

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(D) Given:Diameter of the soap bubble $d = 5 	ext{ mm} = 5 	imes 10^{-3} 	ext{ m}$.Radius $R = \frac{d}{2} = 2.5 	imes 10^{-3} 	ext{ m}$.Surface tension $T = \frac{1}{10 \pi} 	ext{ N m}^{-1}$.$A$ soap bubble has two surfaces (inner and outer), so the total surface area $A_{total} = 2 	imes (4 \pi R^2) = 8 \pi R^2$.The free energy $E$ is given by $E = T 	imes A_{total}$.Substituting the values:$E = \left( \frac{1}{10 \pi} ight) 	imes 8 \pi 	imes (2.5 	imes 10^{-3})^2$$E = \frac{8}{10} 	imes 6.25 	imes 10^{-6}$$E = 0.8 	imes 6.25 	imes 10^{-6} = 5 	imes 10^{-6} 	ext{ J}$.

The surface tension of the soap water solution is $\frac{1}{10 \pi} \text{ N m}^{-1}$. The free energy of the surface layer of a soap bubble of diameter $5 \text{ mm}$ will be:

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