6.51 Bioengineering Application When air (21 mole% O 2 , 79% N 2 ) is placed in

Question

6.51  

Bioengineering Application

When air (21 mole% O2, 79% N2) is placed in contact with 1000 cm3 of liquid water at body temperature, 36.9°C, and 1 atm absolute, approximately 14.1 standard cubic centimeters [cm3(STP)] of gas are absorbed in the water at equilibrium. Subsequent analysis of the liquid reveals that 33.4 mole% of the dissolved gas is oxygen and the balance is nitrogen.

(a)  

Estimate the Henry's law coefficients (atm/mole fraction) of oxygen and nitrogen at 36.9°C.

(b)  

An adult absorbs approximately 0.4 g O2/min in the blood flowing though the lungs. Assuming that blood behaves like water and that it enters the lungs free of oxygen, estimate the flow rate of blood into the lungs in L/min.

(c)  

The actual flow rate of blood into the lungs is roughly 5 L/min. Identify the assumptions made in the calculation of Part b that are likely causes of the discrepancy between the calculated and actual blood flows.

Explanation / Answer

100 moles air oxygen = 21 moles; mole fraction of oxygen = 0.21

nitrogen =79 moles ;mole fraction of nitrogen = 0.79

water volume = 1000 cm3 , temperature = 36.9 oc = t , pressure =P = 1 atm

mole fraction of oxygen = 0.334

mole fraction of nitrogen = 0.664

volume of water at T1 = 0 +273 = 273 K and 1 atm V1=?

V2 = 1000 cm3 at T2 = 36.9 +273 = 309.9 K P2 =1 ATM

P1.V1/ T1 = P2V2/T2 ;V1 = V2/T2 * T1 = 1000 * 273 /309.9 =880.93 cm3

mole fraction of unabsorbed oxgen = 0.336 -0.21 = 0.126 moles

mole fraction of unabsorbed nitrogen = 0.79 - 0.664 = 0.126 moles

PART A- Gas mixture mole fraction = 14.1 /880. 93 =0.0160

oxygen mol fraction = 0.0160 *0.334=0.005354 =XO2

nitogen mol fraction = 0.0160 *0.666 = 0.010656 =XN2

according to Henry law-- P = H.XO2 For ,

oxygen, H = 0.334 / 0.005354 = =62.39 atm/ mol fraction

for nitrogen, H = 0.664 /0.010656 = 62.32 atm/ mol fraction

Part B

q 1 = density of gas (air ) = 1.29 kg /m3 at 273Kand 1 atm pressure

q2 = ? at t = 36.9+273=309.9K

q2=q1*309.9/273=1.464Kg/m3

flow rate=59L/min

mole frac =0.0160 oxygen mol fraction = 0.0160 *0.334=0.005354 =XO2 nitogen mol fraction = 0.0160 *0.666 = 0.010656 =XN2 according to Henry law-- P = H.XO2 For , oxygen, H = 0.334 / 0.005354 = =62.39 atm/ mol fraction for nitrogen, H = 1* 0.664 /0.010656 = 62.32 atm/ mol fraction

Part - b --- q 1 = density of gas (air ) = 1.29 kg /m3 at 273 ok and 1 atm pressure q2 = ? at t = 36

orbed oxygen = 0.336 -0.21 = 0.126 moles

mole fraction of unabsorbed nitrogen = 0.79 - 0.664 = 0.126 moles

PART A- Gas mixture mole fraction = 14.1 /880. 93 =0.0160

oxygen mol fraction = 0.0160 *0.334=0.005354 =XO2

nitrogen mol fraction = 0.0160 *0.666 = 0.010656 =XN2

according to Henry law-- P = H.XO2 Where H i

oxygen, H = 0.334 / 0.005354 = =62.39 atm/ mol fraction for nitrogen, H = 1* 0.664 /0.010656 = 62.32 atm/ mol fraction Part - b --- q 1 = density of gas (air ) = 1.29 kg /m3 at 273 ok and 1 atm pressure q2 = ? at t = 36

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