At ordinary body temperature (37 oC), the solubili...

Question

At ordinary body temperature (37 oC), the solubility of N2 in water at ordinary atmospheric pressure (1.0 atm) is 0.015 g/L. Air is approximately 78 mol % N2.At a depth of 100 ft in water, the external pressure is 4.0 atm. If a scuba diver suddenly surfaces from this depth, how many milliliters of N2 gas, in the form of tiny bubbles, are released into the bloodstream from each liter of blood?

Jules Bruno · Accepted Answer

We're being asked to calculate the milliliters of N2 gas released into the blood stream.For this problem, we will do the following steps:Step 1: Calculate Henry’s Law constant for the gasRecall that the solubility of a gas is given by Henry’s law:Sgas = kH·Pgaswhere:Sgas = solubility of the gas (in mol/L or M)kH = Henry’s law constant for the gasPgas = partial pressure of the gas[readmore]The Solubility of the gas is in mol/L or Molarity. Recall that molarity is given by:Molarity = mol of soluteliter of solventThe partial pressure of the gas can be calculated from the given total pressure using Dalton’s law for partial pressure:Pgas= xgas·PtotalwherePgas is the partial pressure of the gasxgas is the mole fraction of the gasPtotal is the total pressure of the gasesThe mole fraction of the gas is given byxgas=ngasntotalwherexgas is the mole fraction of the gasngas is the mole of the gasntotal is the total mole of the gasesStep 2: Calculate Solubility of N2 in mol/L at depth of 100 ft in water using the calculated Henry’s law constant for the gasStep 3: Calculate the amount of N2 released from each liter of bloodWe can use the Ideal gas law to calculate the volume of N2. Recall that the Ideal gas law is given by:Recall that the Ideal gas law is given by: PV = nRTwhere:P = pressure in atmV = volume in litersR = ideal gas law constant = 0.0821 L•atm/mol•KT = temperature in KelvinWe can isolate the volume PVP = nRTP V = nRTPLet's perform the steps enumerated.Step 1: We determine Henry's law constant for N2● We calculate the solubility of N2 in mol/L or MWe have SN2 = 0.015 g/L. We need to convert the solubility from g/L to mol/L or M. We will need the molar mass of N2.The molar mass of N2N2     2 N x 14.00 g/mol N = 28 g/mol                                      Sum = 28 g/molWe convert g/L to mol/LSN2= 0.015 g N21 L N2 ×1 mol N228 g N2SN2 = 5.357 x 10-4 mol/L or 5.357 x 10-4 M● We calculate the partial pressure of N2 The partial pressure of N2, is not given but %N2 and total pressure are given. We calculate the mole fraction of N2 . We are given N2 = 78% which means ngas = 78 mol and ntotal = 100 molxN2=nN2ntotalxN2=78 mol100 molxN2 = 0.78We have Ptotal = 1 atm. Let’s calculate PN2 by using Dalton’s law of partial pressurePN2= xN2·PtotalPN2=(0.78)(1 atm)PN2 = 0.78 atm●We calculate kH. Substitute the values into Henry’s Law.SN2= kH · PN25.357×10-4molL=kH ×0.78 atmSolve for kH5.357×10-4molL0.78 atm=kH ×0.78 atm0.78 atmkH = 6.868 x 10-4 mol/L•atmStep 2: Now, we use kH to calculate solubility of N2 at depth of 100 ft.● We calculate the partial pressure of N2The partial pressure of N2, is not given but %N2 and total pressure are given. To determine the partial pressure of N2, we use Dalton’s law for partial pressure.We have x2 = 0.78 and PTotal = 4 atm. We plug these values into the equation.PN2 =xN2 · PTotalPN2 =(0.78) (4 atm)PN2 = 3.12 atm● We solve for the solubility of the gas using Henry's Law. Substitute the values into the equationSN2 = kH·PN2SN2 = 6.868×10-4molL·atm×3.12 atmSN2 = 2.143 x 10-3 mol/LStep 3: Calculate the amount of N2 released from each liter of blood● We calculate the difference in solubility: The difference between the mol/L at sea level and mol/L at depth of 100 ft is the amount of released N2 gas into the bloodstream. So,SN2 = SN2 at depth of 100 ft -  SN2 at sea levelSN2 = 2.143×10-3molL-5.357×10-4molLSN2 = 1.6073 x 10-3 mol/L● We calculate the amount of N2 in moles: Since we are to calculate the amount released from each liter of blood, we will multiply the solubility by 1 L and get the number of moles.molN2 = 1.6073×10-3molL×1 LmolN2 = 1.6073 x 10-3 mol● We calculate the amount of N2 in mL: We will use the Ideal gas law to calculate the volume of the gas given.n(number of moles) = 1.6073 x 10-3 molT (body temperature) = 37°C + 273 = 310 KP (at the surface) = 1 atmWe plug the values into the equationV = nRTPV = 1.6073×10-3 mol ×0.0821 L·atmmol·K×310 K1 atmV = 0.04091 LWe are asked for the volume in milliliters. So we convert L to mLV= 0.04091 L × 1 mL10-3 LV = 40.91 mLThere are 40.91 mL of N2 gas in the form of tiny bubbles, that are released into the bloodstream from each liter of blood.

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