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(B) Let $P_1, R_1$ and $P_2, R_2$ be the internal pressures and radii of the two soap bubbles, and $P_3, R_3$ be the internal pressure and radius of t

Vedclass · Accepted Answer

$3PV+4ST = 0$

Vedclass · Answer

(B) Let $P_1, R_1$ and $P_2, R_2$ be the internal pressures and radii of the two soap bubbles, and $P_3, R_3$ be the internal pressure and radius of the resulting single bubble.The internal pressure of a soap bubble is given by $P_{in} = P + \frac{4T}{R}$.Assuming the process is isothermal, the total amount of air (in terms of $PV$) remains constant: $P_1V_1 + P_2V_2 = P_3V_3$.Substituting the expressions for pressure and volume: $(P + \frac{4T}{R_1})(\frac{4}{3}\pi R_1^3) + (P + \frac{4T}{R_2})(\frac{4}{3}\pi R_2^3) = (P + \frac{4T}{R_3})(\frac{4}{3}\pi R_3^3)$.Expanding this, we get: $P(\frac{4}{3}\pi R_1^3 + \frac{4}{3}\pi R_2^3 - \frac{4}{3}\pi R_3^3) + \frac{16\pi T}{3}(R_1^2 + R_2^2 - R_3^2) = 0$.Here, $V = V_3 - (V_1 + V_2)$ is the change in volume, so $V_1 + V_2 - V_3 = -V$. Also, $S = S_3 - (S_1 + S_2)$ is the change in surface area, where $S_i = 4\pi R_i^2$, so $S_1 + S_2 - S_3 = -S$.Substituting these into the equation: $P(-V) + \frac{4T}{3}(-S) = 0$.Multiplying by $-3$, we get $3PV + 4ST = 0$.

Two soap bubbles coalesce to form a single bubble. If $V$ is the subsequent change in volume of contained air and $S$ is the change in total surface area, $T$ is the surface tension and $P$ is atmospheric pressure, then which of the following relations is correct?

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