Question

1- A person performing ambient pressure diving descends to a depth of 30.0 m. a) Calculate the total pressure exerted on the diver by the atmosphere and the surrounding water. (2 pts) b) Assuming the pressure of the gas mixture inhaled by the diver is adjusted to equal the ambient pressure at that depth, calculate the concentration of oxygen dissolved in the diver's tissues (mol/L). Assume the gas mixture is 21% oxygen. Assume the tissue fluid can be treated as water and the solubility of oxygen in tissue fluid is governed by Henry's law at T = 298 K. (2 pts)

Answer 1

o alda) Total Press wre exerted on the diver at a depth of 30.om is Total Pocess wre - Almospheric press wre + water pressure =latm + depth (m) x density kg/m3 - latm + 30m x 1kg m 3 - lam + 30 kg { like lates m2 Total Pressure. - lahm +3 Daton - 31 atm. 6) we know herrys law 2x soluludity Kit Solubility = KH x P xe here are have to find the solubility of a Solubility - Ku* Poz we know total vanmbient pressure = 31 atm. we know the % O₂ in gas mixture - al mixliore - 21% So the partial pressure of O2 = 21x31 = 6.51 atm. Po = 6:51 atm. Henry's constant of oxygen = 1:3x103 moll an atm' - Solulaidi ly = 1-3x10"? mo lletm'x 8.51 atm - 8.463x10 mol/ - 8.463 mmol L