Under isothermal conditions, two soap bubbles | vedclass

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(C) For a soap bubble, the excess pressure is $\Delta P = \frac{4T}{r}$. The total pressure inside is $P_{in} = P + \frac{4T}{r}$.Under isothermal con

Vedclass · Accepted Answer

$\frac{P(R^3-r_1^3-r_2^3)}{4(r_1^2+r_2^2-R^2)}$

Vedclass · Answer

(C) For a soap bubble, the excess pressure is $\Delta P = \frac{4T}{r}$. The total pressure inside is $P_{in} = P + \frac{4T}{r}$.Under isothermal conditions, the number of moles of air $n$ is conserved, and $PV = nR_g	heta$ (where $R_g$ is the gas constant).For bubble $1$: $(P + \frac{4T}{r_1}) \cdot \frac{4}{3}\pi r_1^3 = n_1 R_g 	heta$.For bubble $2$: $(P + \frac{4T}{r_2}) \cdot \frac{4}{3}\pi r_2^3 = n_2 R_g 	heta$.For the combined bubble: $(P + \frac{4T}{R}) \cdot \frac{4}{3}\pi R^3 = (n_1 + n_2) R_g 	heta$.Substituting the expressions for $n_1$ and $n_2$:$(P + \frac{4T}{R}) R^3 = (P + \frac{4T}{r_1}) r_1^3 + (P + \frac{4T}{r_2}) r_2^3$.$PR^3 + 4TR^2 = Pr_1^3 + 4Tr_1^2 + Pr_2^3 + 4Tr_2^2$.$P(R^3 - r_1^3 - r_2^3) = 4T(r_1^2 + r_2^2 - R^2)$.$T = \frac{P(R^3 - r_1^3 - r_2^3)}{4(r_1^2 + r_2^2 - R^2)}$.

Under isothermal conditions, two soap bubbles of radii $r_1$ and $r_2$ combine to form a single soap bubble of radius $R$. If $P$ is the outside pressure, find the surface tension $T$ of the soap solution.

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