Enrico Fermi (1901-1954) was a famous physicist who liked to pose what are now k
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Enrico Fermi (1901-1954) was a famous physicist who liked to pose what are now known as Fermi problems in which several assumptions are made in order to make a seemingly impossible estimate. Probably the most famous example is the estimate of the number of piano tuners in Chicago using the approximate population of the city and assumptions about how many households have pianos, how often pianos need tuning, and how many hours a given tuner works in a year Another famous example of a Fermi problem is "Caesars last breath" which estimates that you, right now, are breathing some of the molecules exhaled by Julius Caesar just before he died. Assumptions 1. The gas molecules from Caesar's last breath are now evenly dispersed in the atmosphere 2. The atmosphere is 50 km thick, has an average temperature of 15 oC, 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. Calculate the total volume of the atmosphere. Number Calculate the total number of gas molecules in the atmosphere Number molecules continued below... Calculate the number of gas molecules in Caesars last breath (37 °C and 1.0 atm) O Previous (8 Give Up & View Solution Check Answer Next Ext HintExplanation / Answer
Ans. #1. Volume of Earth = 4/3 (pi) r3 = (4/3) x 3.14159 x (6400km3) = 1.098 x 1012 km3
Radius of earth’s atmosphere = Radius of earth + Thickness of atmosphere
= 6400 km + 50 km = 6450 km
Volume of atmosphere + Earth = (6450km3) = 1.124 x 1012 km3
Volume of atmosphere = (Volume of earth + atmosphere) – volume of earth
= 1.124 x 1012 km3 - 1.098 x 1012 km3
= 2.6 x 1010 km3
= 2.6 x 1010 x 109 m3 ; [1 km3 = 109 m3]
= 2.6 x 1019 m3
#2. Volume of atmosphere, V = = 2.6 x 1019 m3 = 2.6 x 1023 L ; [1 m3= 1000 L]
Ideal gas equation: PV = nRT - equation 1
Where, P = pressure in atm
V = volume in L
n = number of moles
R = universal gas constant= 0.0821 atm L mol-1K-1
T = absolute temperature (in K) = (0C + 273.15) K
Putting the values in equation 1-
0.2 atm x (2.6 x 1023 L) = n x (0.0821 atm L mol-1K-1) x (288.15 K)
Or, n = (0.52 x 1023 atm L) / (23.657115 atm L mol-1) = 2.198 x 1021 mol
Therefore, number of moles of total gas = 2.198 x 1021 mol
Now, number of molecules = Moles x Avogadro number
= 2.198 x 1021 mol x (6.022 x 1023 molecule /mol)
= 1.324 x 1044 molecules
#3. Putting the value in equation 1-
1.0 atm x (0.500 L) = n x (0.0821 atm L mol-1K-1) x (300.15 K)
Or, n = 0.0202903 mol
Now, number of molecules = Moles x Avogadro number
= 0.0202903 mol x (6.022 x 1023 molecule /mol)
= 1.222 x 1022 molecules
#4. Fraction of all air molecules came from Caesar’s breath =
Number of air molecules in Caesar’s breath / Total number of air molecules
= 1.222 x 1022 molecule / 1.324 x 1044 molecules
= 9.23 x 10-23
That is, there is 9.23 x 10-23 molecule from Caesar’s breath per 1 molecule is the atmosphere.
#5. Molecules from Caesar’s breath inhaled per breathe =
Molecules inhaled per breath x Fraction of Caesar’s breathe molecule in it
= 1.222 x 1022 molecule x 9.23 x 10-23 ; [See #3 for 1.222 x 1022]
= 1.13 molecule
= 1 molecule ; (number of molecules must be a whole number)
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