Emulsification of 10 , mL of oil in water pro | vedclass

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(C) The energy spent in the formation of droplets is given by the product of the total surface area created and the surface tension at the interface.G

Vedclass · Accepted Answer

$900$

Vedclass · Answer

(C) The energy spent in the formation of droplets is given by the product of the total surface area created and the surface tension at the interface.Given:Surface tension $(\gamma) = 0.03 \, J \, m^{-2}$Number of droplets $(n) = 24 	imes 10^{18}$Area of each droplet $(A_{single}) = 12.5 	imes 10^{-16} \, m^2$Total surface area $(A_{total}) = n 	imes A_{single}$$A_{total} = (24 	imes 10^{18}) 	imes (12.5 	imes 10^{-16}) \, m^2$$A_{total} = 24 	imes 12.5 	imes 10^2 \, m^2 = 300 	imes 10^2 \, m^2 = 30000 \, m^2$Energy spent $(W) = A_{total} 	imes \gamma$$W = 30000 \, m^2 	imes 0.03 \, J \, m^{-2} = 900 \, J$Wait,re-calculating: $24 	imes 12.5 = 300$. $300 	imes 10^2 = 30000$. $30000 	imes 0.03 = 900 \, J$. The correct answer is $900 \, J$.

Emulsification of $10 \, mL$ of oil in water produces $24 \times 10^{18}$ droplets. If the surface tension at the oil-water interface is $0.03 \, J \, m^{-2}$ and the area of each droplet is $12.5 \times 10^{-16} \, m^2$,the energy spent in the formation of oil droplets is $.... \, J$.

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