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 inspired gas is at 24 degree C and 1 atm, and the expired gas is at body temperature and pressure (37 degree C and 1 atm). Nitrogen is not transported into or out of the blood in the lungs, so that (N_2)_in = (N_2)_out. Calculate the masses of O_2, CO_2, and H_2O transferred from the pulmonary gases to the blood or vice versa (specify which) per minute. Calculate the volume of air exhaled per milliliter inhaled. At what rate (g/min) is this individual losing weight by merely breathing? The rate at which oxygen is transferred from the air in the lungs to the blood is roughly proportional to [(p_O_2)_air - (p_O_2)_blood], where (p_O_2)_blood is a quantity related to the concentration of oxygen in the blood. Compared to regions where atmospheric pressure is 14.7 psia, what effect does the atmospheric pressure in Denver, which is approximately 12.1 psi, have on the transport rate and breathing rate? How does the body adjust to address this condition?

Explanation / Answer

(a). Volume of air inhaled with each breathe = 500 mL

Total volume of air inhaled per minute = 12 * 0.5

= 6.0 L

Moles of air inhaled = PV / RT

= 1.0 * 6.0 / 0.0821 * 297

= 0.246 moles / min

Let, moles of air exhaled be x.

Since (N2)in = (N2)out

0.246 * 0.774 = x * 0.75

x = 0.2534 moles /min

Moles of O2 transferred = Moles inspired - Moles expired

= 0.206 * 0.246 - 0.151 * 0.254

= 0.0123 moles /min

Mass of O2 transferred = 32 * 0.0123

= 0.393 g/min (transferred from the pulmonary gases to the blood)

Moles of CO2 transferred = 0.037 * 0.254 - 0 * 0.246

= 0.0094 moles /min

Mass of CO2 transferred = 44.01 * 0.0094

= 0.414 g/min (transferred from the blood to the pulmonary gases)

Moles of H2O transferred = 0.062 * 0.254 - 0.02 * 0.246

= 0.0108 moles /min

Mass of H2O transferred = 18.02 * 0.0108

= 0.194 g/min (transferred from the blood to the pulmonary gases)

(b). For the air inhaled, using PV = nRT

1 * Vin = 0.246 * R * 297   .........(1)

For air exhaled:

1 * Vout = 0.2534 * R * 310   ......(2)

Dividing equation (1) by (2):

Vout / Vin = (0.2534 * 310) / (0.246 * 297)

Vout / Vin = 1.075

Hence, Volume of air exhaled = 1.075 mL per milliliter inhaled.

(c). The rate (g/min) at which individual is losing weight by merely breathing = Mass transferred from the blood to the pulmonary gases - Mass transferred from the pulmonary gases to the blood

= (0.414 + 0.194) - 0.393

= 0.215 g/min

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