Question

The tidal lung volume of human breathing, representing the amount of air inhaled and exhaled in a normal breath, is 500 cm3. (Assume atmospheric pressure.)(a) What is the number of molecules of air inhaled with each human breath when the air temperature is 18.0°C?________molecules(b) If the molar mass of air is 28.96 g/mol, what is the mass of air molecules inhaled with each breath? (Assume the air temperature is 18.0°C.)(c) It has been calculated that all of the air in Earth's atmosphere could be collected into a sphere of diameter 1,999 km at a pressure of 1.00 atm. What is the mass of the air in Earth's atmosphere? (Assume the density of air used in this calculation was 1.225 kg/m3.)(d) If all 7 billion humans on Earth inhaled simultaneously, what percentage of the atmosphere would be inhaled during this process? (Assume the air temperature is 18.0°C everywhere on Earth.)

Answer 1

Step-by-step explanation:

Temperature of air = 18°C = 273 + 18 = 291 K .

volume = 500 cc = 0 .5 litre .

pressure = one atmosphere ( atm) .

From gas equation , we can calculate this volume at NTP as follows.

volume = .5 x ( 273 / 291 ) litre

= 0.469 litre .

In any gas at NTP , 22.4 litre contains 6.02 x 10²³ molecules

.469 litre will contain 6.02 x 10²³ x .469 / 22.4 molecules

= 126 x 10²⁰ molecules .

b )

one mole = 6.02 x 10²³ molecules

6.02 x 10²³ molecules has weight of 28.96 grams

126 x 10²⁰ molecules has weight of 28.96 x 126 x 10²⁰ / 6.02 x 10²³ grams

= .606 gram .

c )

volume of all the air in the atmosphere = volume of sphere

= 4 / 3 x π x R³

= ( 4 / 3) x 3.14 x (999.5 x 10³ )³ m³

= 4.18 x 10¹⁸ m³

density of air = 1.225 kg / m³

mass of air = 1.225 x 4.18 x 10¹⁸ kg

= 5.12 x 10¹⁸ kg

d )

volume of air inhaled by 7 billion people

= . 5 x 7 x 10⁹ litre

= 3.5 x 10⁶ m³ .

Total volume of air in atmosphere = 4.18 x 10¹⁸ m³

required percentage

= 3.5 x 10⁶ x 100 / 4.18 x 10¹⁸

= .8373 x 10⁻¹⁰ % .