Question:

Two soap bubbles each with radius r1r_1 and r2r_2 coalesce in vacuum under isothermal conditions to form a bigger bubble of radius RR. Then RR is equal to

Updated On: Apr 5, 2024
  • r12+r22\sqrt{r_1^2 + r_2^2}
  • r12r22\sqrt{r_1^2 - r_2^2}
  • r1+r2r_1 + r_2
  • r12r222\frac{\sqrt{r_1^2 - r_2^2}}{2}
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The Correct Option is A

Solution and Explanation

By Boyle's law
pV=p V = constant
So, p1V1+p2V2=pV p_{1} V_{1}+p_{2} V_{2} =p V
where p1=Tr1p_{1}=\frac{T}{r_{1}}
V1=43πr13V_{1}=\frac{4}{3} \pi r_{1}^{3}
p2=2Tr2p_{2} =\frac{2 T}{r_{2}}
V2=43πr23V_{2} =\frac{4}{3} \pi r_{2}^{3}
p=2TR p =\frac{2 T}{R}
V=43πR3V =\frac{4}{3} \pi R^{3}
2Tr1×43πr13+2Tr2×43πr23=2TR×43πR3\frac{2 T}{r_{1}} \times \frac{4}{3} \pi r_{1}^{3}+\frac{2 T}{r_{2}} \times \frac{4}{3} \pi r_{2}^{3}=\frac{2 T}{R} \times \frac{4}{3} \pi R^{3}
2T×43π(1r1×r3+1r2×r23)=2T×43π(1R×R3)2 T \times \frac{4}{3} \pi\left(\frac{1}{r_{1}} \times r^{3}+\frac{1}{r_{2}} \times r_{2}^{3}\right)=2 T \times \frac{4}{3} \pi\left(\frac{1}{R} \times R^{3}\right)
r12+r22=R2r_{1}^{2}+r_{2}^{2}=R^{2}
R=r12+r22R=\sqrt{r_{1}^{2}+r_{2}^{2}}
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Concepts Used:

Gas Laws

The gas laws were developed at the end of the 18th century, when scientists began to realize that relationships between pressure, volume and temperature of a sample of gas could be obtained which would hold to approximation for all gases.

The five gas laws are:

  • Boyle’s Law, which provides a relationship between the pressure and the volume of a gas.
  • Charles’s Law, which provides a relationship between the volume occupied by a gas and the absolute temperature.
  • Gay-Lussac’s Law, which provides a relationship between the pressure exerted by a gas on the walls of its container and the absolute temperature associated with the gas.
  • Avogadro’s Law, which provides a relationship between the volume occupied by a gas and the amount of gaseous substance.
  • The Combined Gas Law (or the Ideal Gas Law), which can be obtained by combining the four laws listed above.