Nitrogen molecules (N2) are moving all around you right now at room temperature (~300 K). Take the radius of a single nitrogen atom to be 0.1 nm (1 nm = 10^-9 m) and the density of air in this room to be 1.2 kg/m^3. A nitrogen molecule contains a total of 28 nucleons (protons and neutrons). You can approximate the air as being 100% N2 for this problem. a) What is the effective cross-sectional area of a nitrogen molecule) b) Determine the number density of nitrogen molecules in the room. c) Calculate the mean free path of N2 in this environment. Explain in words what this value means.

Chemistry & Chemical Reactivity
10th Edition
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Chapter3: Chemical Reactions
Section3.9: Classifying Reactions In Aqueous Solution
Problem 2.2ACP
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Nitrogen molecules (N2) are moving all around
you right now at room temperature (~300 K).
Take the radius of a single nitrogen atom to be
0.1 nm (1 nm = 10^-9 m) and the density of air in
this room to be 1.2 kg/m^3. A nitrogen molecule
contains a total of 28 nucleons (protons and
neutrons). You can approximate the air as being
100% N2 for this problem.
a) What is the effective cross-sectional area of a
nitrogen molecule)
b) Determine the number density of nitrogen
molecules in the room.
c) Calculate the mean free path of N2 in this
environment. Explain in words what this value
means.
Transcribed Image Text:Nitrogen molecules (N2) are moving all around you right now at room temperature (~300 K). Take the radius of a single nitrogen atom to be 0.1 nm (1 nm = 10^-9 m) and the density of air in this room to be 1.2 kg/m^3. A nitrogen molecule contains a total of 28 nucleons (protons and neutrons). You can approximate the air as being 100% N2 for this problem. a) What is the effective cross-sectional area of a nitrogen molecule) b) Determine the number density of nitrogen molecules in the room. c) Calculate the mean free path of N2 in this environment. Explain in words what this value means.
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