Two spherical soap bubbles formed in vacuum h | vedclass

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(A) Since the bubbles are in a vacuum,the pressure of air inside them is given by $P_1 = \frac{4\sigma}{r_1}$ and $P_2 = \frac{4\sigma}{r_2}$.Since th

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(A) Since the bubbles are in a vacuum,the pressure of air inside them is given by $P_1 = \frac{4\sigma}{r_1}$ and $P_2 = \frac{4\sigma}{r_2}$.Since the temperature remains unchanged,the number of moles of air is conserved,which implies $P_1V_1 + P_2V_2 = PV$ according to the ideal gas law.Substituting the expressions for pressure and volume $(V = \frac{4}{3}\pi r^3)$:$\left(\frac{4\sigma}{r_1}ight) \left(\frac{4}{3}\pi r_1^3ight) + \left(\frac{4\sigma}{r_2}ight) \left(\frac{4}{3}\pi r_2^3ight) = \left(\frac{4\sigma}{r}ight) \left(\frac{4}{3}\pi r^3ight)$This simplifies to $r_1^2 + r_2^2 = r^2$.Given diameters $D_1 = 3.0 \, mm$ and $D_2 = 4.0 \, mm$,the radii are $r_1 = 1.5 \, mm$ and $r_2 = 2.0 \, mm$.$r^2 = (1.5)^2 + (2.0)^2 = 2.25 + 4.0 = 6.25$.$r = \sqrt{6.25} = 2.5 \, mm$.The diameter of the new bubble is $D = 2r = 2 	imes 2.5 = 5.0 \, mm$.

Two spherical soap bubbles formed in vacuum have diameters $3.0 \, mm$ and $4.0 \, mm$. They coalesce to form a single spherical bubble. If the temperature remains unchanged,find the diameter of the bubble so formed in $mm$.

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