1.

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»

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

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

The excess pressure inside the bubble over the atmospheric pressure is

pin = p0 = pgh + \(\frac{2T}R\)

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

= 980 + 144 = 1124 Pa



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