Question

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 "Caesar's last breath," which estimates that you, right now, are breathing some of the molecules exhaled by Julius Caesar just before he died. The assumptions made are: The gas molecules from Caesar's last breath are now evenly dispersed in the atmosphere. The atmosphere is 50 km thick, has an average temperature of 15 °C , and an average pressure of 0.20 atm . The radius of the Earth is about 6400 km . The volume of a single human breath is roughly 500 mL . Perform the calculations, reporting all answers to two significant figures.

Answer 1

Answer:

Explanation:

(a)

Since the earth is assumed to be a sphere.

Volume of atmosphere = volume of (earth +atm osphere) — volume of earth

$= \frac{4}{3}\pi(6400+ 50)^3 - \frac{4}{3}\pi (6400) ^3\\\\= \frac{4}{3}\pi(6192125000) km’^3\\= 2.6\times 10^{19} m^3$

Hence the volume of atmosphere is $2.6\times 10^{19} m^3$

(b)

Write the ideal gas equation as foll ows:

$PV = nRT\\\\n\frac{0.20atm\times 2.6\times10^{19} m^3}{0.08206L\, atm/mok\, K \times (15+273+15)K}\times \frac{1L}{10^{-3}m^3}\\\\= 2.20\times 10^{20} moles$

$no.\, of\, molecules = 2.20\times 10^{20} moles \times \frac{6.022\times10^{23}\,molecules}{1mole}= 13.3\times10^{43} molecules$

Hence the required molecules is $13.3\times10^{43} molecules$

(c)

Write the ideal gas equation as follows:

$PV =nRT \\\\n=\frac{1.0 atm \times 0.5L }{0.08206 L\, atm/mol\,K \times (37 +273.1 5)K} = 0.0196 moles$

$no.\, of\, molecules = 0.0196 moles \times\frac{6.022\times10^{23} molecules} {Imole}= 1.2\times 10^{23} molecules$

Hence the required molecules in Caesar breath is $1.2\times 10^{23} molecules$

(d)

Volume fraction in Caesar last breath is as follows:  

$Fraction,\, X =\frac{12\times 10 molecules}{13.3\times 10^{43} \,molecules}= 9.0\times 10\, molecule/air\, molecule}$

(e)

Since the volume capacity of the human body is 500 mL.

$Volume\, of\, Caesar\, nreath\, inhale\, is =\frac{ 12\times 10^{22}\, molecules}{breath}\times \frac{9.0\times10^{-23} molecule}{air\, molecule}\\\\= 1.08 molecule/breath$