Question

When air (=21 mole% O2, 79 mole% N2) is placed in contact with 1000 cm3 of liquid water at body temperature, 36.9C, 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 balanceis nitrogen. (a) Estimate the Henry's law coefficients (atm/mole fraction) of oxygen and nitrogen at 36.9C. (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.

Answer 1

100 moles air contain 21 moles of oxygen

Hence mole fraction of oxygen = 0.21

Nitrogen =79 moles ; mole fraction of nitrogen = 0.79

water volume = 1000 cm3 , temperature = t =36.9 oc =309.9 K , pressure =P = 1 atm and 14.1 cm3 of gas are absorbed in water at equilibrium .

Absorbed gas analysis - oxygen = 33.4 moles ; nitrogen = 100 - 33.4 =66.6

moles mole fraction of oxygen = 0.334

mole fraction of nitrogen = 0.664

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

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

V1 = P2.V2/ T2 * T1 / P1 = V2/T2 * T1 = 1000 * 273 /309.9 =880.93 cm3

mole fraction of unabsorbed oxgen = 0.336 -0.21 = 0.126

mole fraction of unabsorbed nitrogen = 0.79 - 0.664 = 0.126

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 = K_H.XO2 For , oxygen, K_H = 0.334 / 0.005354 = =62.39 atm/ mol fraction

For nitrogen, K_H = 1* 0.664 /0.010656 = 62.32 atm/ mol fraction

Part - b

An adult inhale approximately 12 times per minute.

So total air volume = 12*500 =6000 ML Per minute = 6000/1000 =6 L/min

In the air on the basis of - Oxygen - 21% Nitrogen -- 79%

Amount of Nitrogen exhaled =Amount of air inhaled

MOLE FRACTION OF NITROGEN IN THE EXHALED AIR = 0.75

MOLECULAR WT OF OXYGEN = 32

MOLECULAR WT OF NITROGEN = 28

AVERAGE MOLECULAR WT OF AIR = 32*21/100 + 79/100 *28=6.72 +22.12 =28.84 KG/KMOL

DENSITY OF AIR = 28.84/22.4 1,2875 =1.29 KG/M3

SO DENSITY OF NITROGEN = 1.29 * 79/100 =1.019 KG/M3

DENSITY OF OXYGEN = 1.29 *21/100 = 0,2709 =0.271 KG/M3

T1 = t1 +273

Here density of air =p1= 1.29 kg/m3 at t1 = 0 oc , at T1 = 273 +0 =273 K density of air = p2 =? at t2 = 37 oc = T2 = 273 +37 =310 ok p1/p2 = T2/T1 1.29 / p2 = 310/273 =1.136 p2 = 1.29/1.136 =1.136 kg/m3

Amount of air inhaled = volume *density = 6 /1000 *1.136 = 6.82*10^-3 kg/min

amount of air inhaled = amount of nitrogen exhaled mol fraction of nitrogen exhaled = 0.75 mol fraction of oxygen exhaled =1- 0.75 = 0.25 mol.fraction of N2 = 0.75 amount of nitrogen = 0.75 * 28 =21 kg /min

amount of oxygen = 0.25*32 =8 kg /min