Radius of new soap bubble when two soap bubbles which coalesce ‖ Excess Pressure ‖ Surface Tension
⦿ Chapter : Surface Tension ⦿ Topic : Find the radius of new soap bubble when two soap bubbles coalesce ....
As the new bubble is formed under isothermal condition, so Boyle’s law holds good and
PV = P1 V1@P2V2
or, 4S/r . (4𝝅/3) r^3 = 4S/r 1 . (4𝝅/3) r 1 ^3 + 4S/r 2 . (4𝝅/3) r 2 ^3
∴ r = √ ( r 1 ^ 2 + r2 ^ 2 )
This is the radius of new soap bubble when two soap bubbles coalesce .
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Can you please explain to me why you used for P1, P2 and P expression for excess pressure inside the bubble and not P(atmospheric) - 4S/R ? Are these pressures equivalent?