Under isothermal conditions, two soap bubbles | vedclass

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Question

(B) For isothermal conditions, the total number of moles of air remains constant. The pressure inside a soap bubble of radius $r$ is given by $P_{in}

Vedclass · Accepted Answer

$\frac{P(c^{3}-a^{3}-b^{3})}{4(a^{2}+b^{2}-c^{2})}$

Vedclass · Answer

(B) For isothermal conditions, the total number of moles of air remains constant. The pressure inside a soap bubble of radius $r$ is given by $P_{in} = P + \frac{4T}{r}$, where $P$ is the external pressure and $T$ is the surface tension. Using the ideal gas law $PV = nRT$, since $T$ (temperature) is constant, $PV$ is proportional to the number of moles. Thus, $P_1V_1 + P_2V_2 = P_cV_c$. Substituting the values: $(P + \frac{4T}{a}) \cdot \frac{4}{3}\pi a^3 + (P + \frac{4T}{b}) \cdot \frac{4}{3}\pi b^3 = (P + \frac{4T}{c}) \cdot \frac{4}{3}\pi c^3$. Dividing by $\frac{4}{3}\pi$: $P(a^3 + b^3 - c^3) = 4T(c^2 - a^2 - b^2)$. Rearranging for $T$: $T = \frac{P(a^3 + b^3 - c^3)}{4(c^2 - a^2 - b^2)} = \frac{P(c^3 - a^3 - b^3)}{4(a^2 + b^2 - c^2)}$.

Under isothermal conditions, two soap bubbles of radii $a$ and $b$ coalesce to form a single bubble of radius $c$. If the external pressure is $P$, then the surface tension of the bubbles is:

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