Question

There is an air bubble of radius 1.0 mm in a liquid of surface tension 0.072 N/m and density 10³ kg/m³ . The bubble is at a depth of 10 cm below the free surface of the liquid. By what amount is the pressure inside the bubble greater than the’ atmospheric pressure?

Answer 1

Data : R = 10^-3 m, T = 0.072 N/m, ρ = 10³ kg/m³ , h = 0.1 m

Let the atmospheric pressure be p0. Then, the absolute pressure within the liquid at a depth h is

p = p₀ + ρgh

Hence, the pressure inside the bubble is

p_in = p₀ + \(\frac{2T}R\) = p₀ + pgh + \(\frac{2T}R\)

The excess pressure inside the bubble over the atmospheric pressure is

p_in = p_{0 = pgh +} \(\frac{2T}R\)

= (10³) (9.8) (0.1) + \(\frac{2(0.072)}{10^{-3}}\)

= 980 + 144 = 1124 Pa