Two soap bubbles of radius R1 and R2 coalesce isothermally in vacuum to form a single soap bubble. The radius of the bigger bubble formed is

Question

Two soap bubbles of radius R1 and R2 coalesce isothermally in vacuum to form a single soap bubble. The radius of the bigger bubble formed is (A) (1/R)=(1/R1)+(1/R2) (B) (1/R)=(1/R1)-(1/R2) (C) √R12 + R22 (D) (1/R)=(1/R12)+(1/R22)

Tardigrade Admin · Accepted Answer

P1V1+P2V2=PV (no. of moles constant) (4 T/R1)⋅ (4/3)π R13+(4 T/R2)⋅ (4/3)π R23=(4 T/R)(4/3)π R3 R=√R12 + R22

Q. Two soap bubbles of radius $R_{1}$ and $R_{2}$ coalesce isothermally in vacuum to form a single soap bubble. The radius of the bigger bubble formed is

$\frac{1}{R} = \frac{1}{R _{1}} + \frac{1}{R _{2}}$

$\frac{1}{R} = \frac{1}{R _{1}} - \frac{1}{R _{2}}$

$R_{1}^{2} + R_{2}^{2}$

$\frac{1}{R} = \frac{1}{R _{1}^{2}} + \frac{1}{R _{2}^{2}}$

$P_{1} V_{1} + P_{2} V_{2} = P V$ (no. of moles constant)
$\frac{4 T}{R _{1}} \cdot \frac{4}{3} π R_{1}^{3} + \frac{4 T}{R _{2}} \cdot \frac{4}{3} π R_{2}^{3} = \frac{4 T}{R} \frac{4}{3} π R^{3}$
$R = R_{1}^{2} + R_{2}^{2}$

Q. Two soap bubbles of radius R1​ and R2​ coalesce isothermally in vacuum to form a single soap bubble. The radius of the bigger bubble formed is

R1​=R1​1​+R2​1​

R1​=R1​1​−R2​1​

R12​+R22​​

R1​=R12​1​+R22​1​