Estimate how many molecules of air are in each 2.1 -L breath youinhale that were

Question

Estimate how many molecules of air are in each 2.1 -L breath youinhale that were also in the last breath Galileo took. Assume theatmosphere is about 10 km high and of constant density.

I tried using the PV=nRT equation and then also using the bozemanconstant but I got an answer of 5.123E22 which is incorrect, so i'mnot sure of what to do. Here's what i did:

PV=nRT
PV=(N/Na)RT
N= # of molecules in volume V
Na= Avogadro's number

R/Na=k=1.38E-23

PV=NkT
N=PV/kT
P= 1.0atm
V=2.1L so 2.1E-3 m^3
k=1.38 E-23
T=300K

what am I doing wrong?

Explanation / Answer

The volume of air when we inhale is V = 2.1 L = 2.1 *10-3 m3 The height of the atmosphere is 10 km and has a constantdensity. We know from the relation P * V = n * R * T or P * V = (N/Na) * R * T ------------(1) Here,P = * g * h Here, = 0.001225 g/cm3 = 1.225kg/m3,g = 9.8 m/s2 and h = 10 km = 10 *103 m or P = 1.225 * 9.8 * 10 * 103 = 12.005 *104 Pa and N is the number of molecules,Na is theavogadro's number and R = 8.314 J/mol/K and T = 25 oC =(25 + 273) K = 298 K From equation (1),we get N = (P * V/R * T) * Na Here,Na= 6.022 * 1023mol-1 Here,Na= 6.022 * 1023mol-1 or N = (12.005 * 104 * 2.1 * 10-3/8.314* 298) * 6.022 * 1023 = 0.6128 * 1023 = 6.128* 1022 Therefore,the number of molecules of air in each 2.1-L ofbreath we inhale is N = 6.128 * 1022. Therefore,the number of molecules of air in each 2.1-L ofbreath we inhale is N = 6.128 * 1022.

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