Question 16.1: vessel containing gaseous argon is maintained at 300 K and 1...

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.