How to find gauge pressure inside a soapy bubble?
You are outside on a beautiful summer day with your dog, Rusty. One of Rusty's favorite pastimes is chasing bubbles around the yard. You use a mixture of dish soap and water, which has a surface tension of 0.034 N/m. You note that the largest bubble you can blow with this is 1.02 cm in diameter, and the ambient air pressure that day is 97.2 kPa. What is the gauge pressure inside the bubble? The only equation I know is change in Pressure=4*gamma(surface tension)/r The pressure in the bubble should be higher. What I've worked out so far: radius=0.0051 m P(bubble) - (97200 Pa) = 4*(0.034 N/m^2)/(0.0051 m) and I got 97226.67 Pa, but that's wrong. Can anyone help me with how to solve this? Am I using the wrong equation? Or did I just calculate something wrong? Thanks
Why does an air bubble increase in size at the surface of the liquid?
Boyle's law says PV=RTHere P is the pressure, V is the volume and T is the remperature. R is a constant. When T is constant we getPV=Constant.When bubble is deep down it has water pressure and atmosphere pressure. As it raises, water pressure which is proportional to the depth, reduces. This will give V more. Volume increases.