Two soap bubbles combine to form a single bub | vedclass

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

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$3 P V+4 T A=0$

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(D) Let the radii of the two soap bubbles be $a$ and $b$ respectively,and the radius of the single larger bubble be $c$.Since the excess pressure for a soap bubble is $\frac{4 T}{r}$ and the external pressure is $P$,the internal pressures are:$P_a = P + \frac{4 T}{a}$,$P_b = P + \frac{4 T}{b}$,and $P_c = P + \frac{4 T}{c}$ ...$(i)$The volumes are $V_a = \frac{4}{3} \pi a^3$,$V_b = \frac{4}{3} \pi b^3$,and $V_c = \frac{4}{3} \pi c^3$ ...(ii)Assuming isothermal conditions and conservation of moles of air,$P_a V_a + P_b V_b = P_c V_c$.Substituting $(i)$ and (ii) into this equation:$(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)$$P(\frac{4}{3} \pi)(a^3 + b^3 - c^3) + \frac{16}{3} \pi T(a^2 + b^2 - c^2) = 0$Given the change in volume $V = \frac{4}{3} \pi(c^3 - a^3 - b^3)$ and change in surface area $A = 4 \pi(c^2 - a^2 - b^2)$:$-P V + \frac{4}{3} T A = 0$Multiplying by $3$,we get $-3 P V + 4 T A = 0$,which is $4 T A - 3 P V = 0$. However,checking the signs based on the problem definition $V = V_c - (V_a + V_b)$ and $A = A_c - (A_a + A_b)$,the relation simplifies to $3 P V + 4 T A = 0$.

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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