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Question

1. 1.) the gas molecules from ceaser/s last breath are now evenly dispersed in the atmosphere. 2. The atmosphere is 50 km thick, has an average temperature of 15 \deg C, and an average pressure of 0.20 atm. 3. The radius of the Earth is about 6400 km.4. The volume of a single human breath is roughly 500 mL.Perform the following calculations, reporting all answers to two significant figures.
2. Calculate the total volume of the atmosphere.
3. Calculate the total number of gas molecules in the atmosphere. _____ m3Calculate the number of gas molecules in Caesar\'s last breath (37 \deg C and 1.0 atm)._____ molecules
4. What fraction of all air molecules came from Caesar\'s last breath?_____ molecules
5. About how many molecules from Caesar\'s last breath do you inhale each time you breathe

Answer 1

The volume of the atmospheric layer is the volume of a sphere with radius from the earth center to the outer fringes of atmospheric gases, minus the the volume of the earth.

Let V_e = volume of the earth, R_e = radius of the earth

V_a = volume of the air layer, D_a = thickness of the air layer

V_t = volume to the outer fringes R_t = radius of the outer fringes

Then Va = Vt - Ve (1)

Note that the volume of a sphere is (4/3)*pi* R³

Also note that Rt = Re + Da (2)

Thhen from equ(1), V_a = (4/3)*pi*R_t³ - (4/3)*pi*R_e³

Substituting equ(2) we get V_a = (4/3)*pi* (R_t³ - R_e³) = (4/3)*pi* [ (R_e+D_a)³ - R_e³ ]

Rather than expanding the sum-cubed, let's put in the numbers.

V_a = (4/3)*pi* [ (6400+50)³ - (6400)³ ]

= 1.33*3.14* [ 268.336E9 - 262.144E9 ]

= 1.33*3.14* 6.19E9

V_a = 25.9E9 = 25.9 billion km³ (about 6.3 billion cubic miles!)

For the total number of gas molecules in the atmosphere, we start with the Ideal Gas Law:

PV = nRT

n = PV/RT = (0.20 atm)*(25.9E9 km³)*(10³ m/km)³ *(10² cm/m)³ *(10^-3 L/cm³)

(0.08205 L*atm/degK*mole * 288.15 degK)

(note the change to Kelvin temperature: 15 C = 288.15K)

Doing the arithmetic gives n = 2.19E20 moles

Converting the moles to molecules requires Avogadro's Number:

N = n * N_A = 2.19E20 moles * 6.023E23 molecules/mole

N = 13.2 E43 = 1.32 E44 molecules

Calculating the number of gas molecules in Caesar\'s last breath

Again using PV=nRT ,

n(C_lb) = PV/RT = (1 atm)*(500 ml)*(10^-3 L/ml) / [ (0.08205 L*atm/degK/mole)*(300.15 degK) ]

n(C_lb) = 0.0203 moles

N(C_lb) = 0.0203 * 6.023E23 = 0.122E23 = 1.22 E22 molecules

The fraction of all air molecules from Caesar's last breath is

F = N(C_lb) / N(atm) = 1.22E22 / 1.32E44

F = 0.924 E-22 = 9.24 E-23

About how many molecules from Caesar\'s last breath do you inhale?

Assuming the 500ml breath volume from the Caesar calculation,

N(breath) = 1.22 E22 molecules

N(Caesar in modern breath) = N(C) * F = 1.22 E22 * 9.24 E-23 = 11.3 E-1 = 1.13

= about 1 molecule per breath (weird, huh?)