The lower end of a capillary tube of diameter 2.0 mm is dipped 8.00cm below the surface of water in a beaker. What is the pressure required in the tube in order to blow a hemispherical bubble at its end in water? The surface tension of water at temperature of the experiments is `7.30 xx 10^(-2) Nm^(-1)`. 1 atmospheric pressure =` 1.01 xx 10^(5) Pa`, density of water `= 1000 kg//m^(3), g=9.80 ms^(-2)`. also calculate the excess pressure.

Explanation

The excess pressure in a bubble of gas in a liquid is given by `2S//r` where S is the surface tension of the liquid gas interface. You should note there is only one liquid surface in this case. (For a bubble of liquid in a gas, there are two liquid surfaces, so the formula for excess pressure in that case is `4S//r`) THe radius of the bubble is r. Now the pressure outside the bubble `P_0` equals atmospheric pressure plus the pressure due to 8.00 cm of water column. That is
`P_0=(1.01 times 10^5 Pa+0.08m times 1000 kg m^-3 times 9.80 ms^-2)`
`=1.01784 times 10^5 Pa`
Therefore, the pressure inside the bubble is
`P_1=P_0+2S//r`
`=1.01784 times 10^5 Pa+(2 times 7.3 times 10^-2 Pa m//10^-3m)`
`=(1.01784+0.00146)times 10^5 Pa`
`=1.02 times 10^5 Pa `
Where the radius of the bubble is taken to be equal to the radius of the capillary tube. since the bubble is hemispherical (The answer has been rounded off to three significant figures.) The excess pressure in the bubble is 146 Pa.