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(D) Let the radii of the two soap bubbles be $a$ and $b$ respectively,and the radius of the resulting single larger bubble be $c$.The excess pressure

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$3 P V+4 T A=0$

Vedclass · Answer

(D) Let the radii of the two soap bubbles be $a$ and $b$ respectively,and the radius of the resulting single larger bubble be $c$.The excess pressure inside a soap bubble is $\Delta P = \frac{4 T}{r}$,where $T$ is surface tension and $r$ is the radius.The internal pressure of the bubbles are $P_a = P + \frac{4 T}{a}$,$P_b = P + \frac{4 T}{b}$,and $P_c = P + \frac{4 T}{c}$.Assuming isothermal conditions and conservation of the number of moles of air,$P_a V_a + P_b V_b = P_c V_c$.Substituting the volumes $V_a = \frac{4}{3} \pi a^3$,$V_b = \frac{4}{3} \pi b^3$,and $V_c = \frac{4}{3} \pi c^3$,we get:$(P + \frac{4 T}{a})(\frac{4}{3} \pi a^3) + (P + \frac{4 T}{b})(\frac{4}{3} \pi b^3) = (P + \frac{4 T}{c})(\frac{4}{3} \pi c^3)$.Simplifying,$P(a^3 + b^3 - c^3) + 4 T(a^2 + b^2 - c^2) = 0$.The change in volume is $V = V_c - (V_a + V_b) = \frac{4}{3} \pi (c^3 - a^3 - b^3)$,so $a^3 + b^3 - c^3 = -\frac{3 V}{4 \pi}$.The change in surface area is $A = 4 \pi c^2 - (4 \pi a^2 + 4 \pi b^2) = 4 \pi (c^2 - a^2 - b^2)$,so $a^2 + b^2 - c^2 = -\frac{A}{4 \pi}$.Substituting these into the equation: $P(-\frac{3 V}{4 \pi}) + 4 T(-\frac{A}{4 \pi}) = 0$.Multiplying by $-4 \pi$,we get $3 P V + 4 T A = 0$.

Two soap bubbles combine to form a single bubble. In this process,the change in volume and surface area are respectively $V$ and $A$. If $P$ is the atmospheric pressure,and $T$ is the surface tension of the soap solution,the following relation is true :

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