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 surface, 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_(o)` equals atmospheric pressure plus the pressure due to 8.00 cm of water column. that is
`P_(o)=(1.01xx10^(5)Pa+0.08mxx1000kgm^(-3)xx9.80ms^(-2))`
`=1.01784xx10^(5)Pa`
Therefore,t he pressure inside the bubble is
`P_(i)=P_(o)+2S//r`
`=1.01784xx10^(5)Pa+(2xx7.3xx10^(-2)Pa" "m//10^(-3)m)`
`=(1.01784+0.00146)xx10^(5)Pa`
`=1.02xx10^(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 the three significant figures). the excess pressure in the bubble is 146 Pa.