Question

The collision diameter, d, of an H2 molecule is about 0.25 nm = 2.5 * 10^-10 m. For H2 gas at 0 degrees celsius and 1 bar, calculate the following:

a. Root-mean-square velocity

b. The translational kinetic energy of 1 mol of H2 molecules

c. The number of H2 molecules in 1 mL of the gas

d. The mean free path

e. The number of collisions each H2 molecule undergoes in 1 s

f. The total number of intermolecular collisions in 1 s in 1 mL of the gas

Answer 1

For H2, M = 2.0 g/mol = 2.0x10^-3 Kg/mol

T = 0 C = 273 K

(a):  Root-mean-square velocity, Vrms = underroot(3RT/M)

= underroot(3 x 8.314 JK-1mol-1 x 273 K / 2.0x10^-3 Kg/mol)

=> Vrms = 1845 m/s (answer)

(b): translational kinetic energy of 1 mol of H2 molecules, KE = (3/2)nRT = 1.5 x 1 x 8.314 JK-1mol-1 x 273 K

= 3405 J/mol or 3.405 KJ/mol (answer)

(c): n = PV/RT = 0.987 atm x 0.001 L / 0.0821 L.atm.mol-1K-1 x 273 K = 4.4036x10^-5 mol

Hence number of H2 molecules = 4.4036x10^-5 mol x (6.023x10²³ molecule / 1 mol)

= 2.65x10¹⁹ molecules (answer)

(d): d = 2.5x10^-10 m , T = 273 K

P = 1 bar = 10⁵ Pa

k_b = 1.38x10^-23

Formulae for mean free path, l = k_bx T / (2)xd²P

Putting values in the above equation we get, l =

=> l = 1.3615x10^-7 m = 136.15 nm (answer)