Under isothermal condition, two soap bubbles | vedclass

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(A) For a soap bubble, the excess pressure inside is given by $\Delta P = \frac{4T}{r}$. The total pressure inside is $P_{in} = P + \frac{4T}{r}$.Sinc

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) For a soap bubble, the excess pressure inside is given by $\Delta P = \frac{4T}{r}$. The total pressure inside is $P_{in} = P + \frac{4T}{r}$.Since the process is isothermal, the number of moles of air remains constant, so $PV = 	ext{constant}$.For the two initial bubbles: $P_1 V_1 = (P + \frac{4T}{r_1}) \cdot \frac{4}{3} \pi r_1^3$ and $P_2 V_2 = (P + \frac{4T}{r_2}) \cdot \frac{4}{3} \pi r_2^3$.For the final bubble: $P_f V_f = (P + \frac{4T}{R}) \cdot \frac{4}{3} \pi R^3$.Since the total amount of air is conserved, $P_1 V_1 + P_2 V_2 = P_f V_f$.Substituting the expressions: $(P + \frac{4T}{r_1}) \frac{4}{3} \pi r_1^3 + (P + \frac{4T}{r_2}) \frac{4}{3} \pi r_2^3 = (P + \frac{4T}{R}) \frac{4}{3} \pi R^3$.$P r_1^3 + 4T r_1^2 + P r_2^3 + 4T r_2^2 = P R^3 + 4T R^2$.$P(r_1^3 + r_2^3 - R^3) = 4T(R^2 - r_1^2 - r_2^2)$.$T = \frac{P(r_1^3 + r_2^3 - R^3)}{4(R^2 - r_1^2 - r_2^2)} = \frac{P(R^3 - r_1^3 - r_2^3)}{4(r_1^2 + r_2^2 - R^2)}$.

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

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