Question

The atoms in a gas (gas constant R=8.31 J/mol-K) can be treated as classical particles if their de Broglie wavelength ƒÉ is much smaller than the average separation between the particles d. Consider monatomic Helium gas (mHec2 = 3727MeV), molar mass 4g/mol) at 1.0 atmosphere of pressure (1.0~105) and at room temperature (T=293K).

a. Estimate d for this gas
b. Find the average de Broglie wavelength of the atoms in the gas ( KE = (3/2) KB T, KB = 8.617~10-E5 eV/K
c. Find the pressure that would make d equal to ă from part b

Answer 1

a) The number density of He atoms under those conditions is
n = 6*10^23 atoms/24.4 liters = 2.4*10^19 atom/cm^3
The volume per atom is 1/n = 4*10^-20 cm^2 and the average distance between molecules is the cube root of that, or
3.5*10^-7 cm.

b. The average momentum of the He atoms is
p = sqrt(2mE) where E is the average kinetic energy,
E = (3/2) kT = 6.1*10^-14 erg. So,
p = sqrt(2*6.1*10^-14*4*1.67*10^-24)
= 9.0*10^-19 g cm/s
The de Broglie wavelength is
L = h/p = (6.6*10^-27 g cm^2/s)/(9*10^-19 g cm/s) = 7.3*10^-9 cm

This is about 50 times less than the spacing between atoms.

c) To make d equal to L, you have to decrease d by a factor of 50.
This means the density (and pressure) would have to increase by a factor of 50^3