BIO. There is a maximum depth at which a diver can breathe through a snorkel tube (Fig. E12.17) because as the depth increases, so does the pressure difference, which tends to collapse the diver's lungs. Since the snorkel connects the air in the lungs to the atmosphere at the surface, the pressure inside the lungs is atmospheric pressure. What is the external– internal pressure difference when the diver's lungs are at a depth of 6.1 m (about 20 ft)? Assume that the diver is in fresh-water. (A scuba diver breathing from compressed air tanks can operate at greater depths than can a snorkeler, since the pressure of the air inside the scuba diver's lungs increases to match the external pressure of the water.)

Video transcript

Hi everyone. Today we are going to determine the pressure difference between the inside and the outside of the lungs when the lifesaver is 5m below the surface of the water. So a person breathes with eight of the pressure difference between the lungs and the atmosphere. And this lifesaver on this mission decides to use historical tube that links lungs to the atmosphere at the surface of the water. The increasing water depth collapses the chest cavity and consequently decreases the pressure difference experience within the lungs. So in this particular example we want to assume that the inside inside pressure is equal to the atmospheric pressure. And in this case we know that the lifesaver is five m below the surface of the water, which means that the age is going to be five m. That's pretty much all the information given. And we want to look into this diagram right here which represent our current situation. So we call that to calculate the pressure difference the pressure formula to determine what the actual pressure is under water can be calculated using this formula Soapy Will equals to p A. T. M. Which is the pressure pressure at atmospheric plus H. Multiplied by ro multiplied by G. H. S. Gonna be this depth or the height of the water row is just the role of the water and G. Is the gravitational acceleration. So we are required to calculate the pressure difference in this particular example and recall that in the problem statement. It is given that the pay the pressure at the inside of the lungs is equals to the pressure at the atmosphere, so P. N. Equals p atmosphere. Well, pressure outside is going to be the pressure calculated using this formula right here. Right? So pressure outside is going to be P at the atmosphere plus H. Rho G. So the delta pressure, which is what is being asked. The pressure difference is going to be P out minus P. N. So that will then be P A. T. M plus H rho G minus P. N. Which is also going to be P A. T. M. So we can cancel the P A. T. M. So the delta pressure is essentially just a church road G. We know the Hs five m, we know what the role of the water is and we know what the G. S. So we can plug everything in. So this is going to be five m multiplied by, I'm just gonna write down 1000 kg per meter cube, multiplied by 9.8 m per second squared. And this will come out to be 49, pascal, 49,000 pascal. So that will corresponds to option a and that will essentially be the answer of our problem. So if you guys have any more question or any sort of confusion on this particular example, please look into our other lesson videos regarding to stop it and that will be all for this particular practice problem. Thank you