An adult takes about 12 breaths per minute, inhaling roughly 500 mL of air with

Question

An adult takes about 12 breaths per minute, inhaling roughly 500 mL of air with each breath. The molar compositions of the inspired and expired gases are as follows: The inured gas is at 24 degree C and 1 atm, and the expired gas it at body temperature and pressure 37 degree C 1 atm Nitrogen is not transported into or out of the blood in the lungs, so that (N_2)_a = (N_2)_out. (a) Calculate the masses O_2, CO_2, and H_2O transferred from the pulmonary gases to the blood or vice versa (specify which) per minute. (b) Calculate the volume of air exhaled per milliliter inhaled. (c) At what rate (g/min) is this individual sling weight by merely breathing?

Explanation / Answer

Let us have an idea of the whole process

The adult can take 12 breaths / minute

In each breath the air inhaled = 500mL

so in each minute the air inhaled = 500 X 12 = 6000 mL

Now the mole concept for inhaled and exhaled air is

(i) Let the moles of air inhaled = n1 moles / minute

the composition = O2 = 20.6%

                             CO2 = 0%

                            N2 = 77.4%

                           H2O = 2%

(ii) Let the moles of air exhaled = n2 mole / minute

he composition = O2 = 15.1%

                             CO2 = 3.7%

                            N2 = 75%

                           H2O = 6.2%

a) from ideal gas equation we can calculate the moles of air inhaled

PV = nRT

T = 24C = 297.15 K

n = ?

R = 0.0821 Latm / mol K

P = 1 atm

V = 6000mL = 6 L

n = PV / RT = 1 X 6 / 0.0821 X 297.15 = 0.246 moles

So the moles of air inhaled per minute = 0.246 moles / minute = n1

Moles of N2 inhaled = 0.774 X 0.246 = 0.1904 moles / minute = moles of N2 exhaled

So moles of air exhaled = 0.1904 / 0.75 = 0.254 moles / minute = n2

1 ) mass of O2 inhaled = Moles of O2 inhaled X molecular weight = 0.206 X 0.246 X 32 = 1.62 g

Mass of O2 exhaled = Moles of O2 exhaled X molecular weight = 0.151 X 0.254 X 32 = 1.23 g

so mass of O2 consumed = Mass of O2 inhaled - mass of O2 exhaled = 1.62 - 1.23 = 0.39 g O2 / min

2) Mass of CO2 transfered = Moles of CO2 exhaled X molecular weight = 0.037 X 0.254 X 44 = 0.414 g CO2 /min

3) Mass of H2O inhaled = Moles of H2O inhaled X molecular weight = 0.02 X 0.246 x 18 = 0.0886 grams

Mass of H2O exhaled = Moles of H2O exhaled X molecular weight = 0.062 X 0.254 X 18 = 0.283 grams

Mass of H2O transfered = Mass of H2O exhaled - Mass of H2O inhaled = 0.283 - 0.0886 = 0.1944 grams H2O / min

b) volume of air exhaled = nRT / Pressure = 0.254 X 0.0821 X 310 K / 1 atm = 6.46 Litres = 6460 mL

The volume of air inhaled = 6.00 Litres = 6000mL

Air exhaled / air inhaled = 6460 / 6000 = 1.078 mL / mL

The total mass lost = (0.1944 grams H2O / min   0.414 g CO2 /min) - ( 0.39 g O2 / min   )

Total mass lost = 0.2184 g / minute

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