Question 16.1: vessel containing gaseous argon is maintained at 300 K and 1...
vessel containing gaseous argon is maintained at 300 K and 1 atm. Presuming a hard-sphere diameter of 3.42 Å, calculate the volumetric collision rate between argon atoms within the vessel. Discuss the implications of your result.
From Appendices A and C, the mass of an argon atom is
m = (39.948)(1.6605 × 10^{−24} g) = 6.63 × 10^{−23} g,
while its hard-sphere diameter is given as
σ = 3.42 × 10^{−8} cm.
From Eq. (15.33), the number density within the vessel is
n=\frac{N}{V}=\frac{P}{k T}, (15.33)
n=\frac{P}{k T}=\frac{\left(1.013 \times 10^{6} erg / cm ^{3}\right)}{\left(1.3807 \times 10^{-16} erg / K \right)(300 K )}=2.45 \times 10^{19} cm ^{-3}.
Substituting into Eq. (16.18),
Z=2 n^{2} \sigma^{2}\left(\frac{\pi k T}{m}\right)^{1 / 2};
we find that the volumetric collision rate becomes
Z=2\left(2.45 \times 10^{19} cm ^{-3}\right)^{2}\left(3.42 \times 10^{-8} cm\right)^{2}\left[\frac{\pi(300)\left(1.38 \times 10^{-16} g \cdot cm ^{2} / s ^{2}\right)}{\left(6.63 \times 10^{-23} g \right)}\right]^{1 / 2}
= 6.22 × 10^{28} collisions/cm³ · s.
In other words, we have determined that, at room temperature and pressure, over 10^{28} collisions occur between argon atoms every second in a volume of only one cubic centimeter. This unfathomable number still represents over 10^{16} collisions per picosecond! Obviously, a binary collision rate of this magnitude will be very effective in maintaining local thermodynamic equilibrium.