Question

a soap bubble of radius R is surrounded by another bubble of radius 2R . take surface tension =S then, the pressure inside the smaller bubble in excess of atmospheric pressure will be.?

Answer 1

A soap bubble of radius R is surrounded by another bubble of radius 2R. Take surface tension S. What will be the pressure inside the smaller bubble in excess of atmospheric pressure?

Solution: According to the Laplace equation, excess pressure, created with surface tension of the spherical surface of the liquid is equal to $\Delta p = \frac{2S}{R}$ . In case of soap bubbles, the excess pressure of air inside them is doubled due to the presence of two interfaces, one inside and one outside: $\Delta p_{b} = \frac{4S}{R}$ .

Excess pressure of the air inside the bigger bubble will be: $\Delta p_{B} = \frac{4S}{2R} = \frac{2S}{R}$ ;

Excess pressure of the air inside the smaller bubble will be: $\Delta p_{S} = \frac{4S}{R}$ ;

Air pressure difference between the smaller bubble and the atmosphere will be equal to the sum of excess pressures inside the bigger and the smaller bubbles: $\Delta p = \Delta p_{B} + \Delta p_{S} = \frac{2S}{R} + \frac{4S}{R} = \frac{6S}{R}$ .

Answer: $\frac{6S}{R}$ .