Question

If He(g) has an average kinetic energy of 4890 J/mol under certain conditions, what is the root mean square speed of F2(g) molecules under the same conditions?

Answer 1

According to Graham's law of diffusion, This law states that the relative velocity of one gas to another is the square root of the inverse ratio of their masses.
V(gas A) / V(gas B) = √ M(gas B) / M(gas A)

V(He) / V(F₂) = √ (38.0 / 4.00) = √(9.5) = 3.08 ==> this tells us that He diffuses at a rate of 3.08 times faster than does the heavier fluorine molecule.

Under the same conditions of temperature and pressure, all gases have the same average kinetic energy; this is because temperature is defined to be a measure of the average kinetic energy of the particles of the substance. We know that KE = 1/2 mv².

So for He: 4890-J/mol = 0.5 x (4.00-g/mol) x v².

Let's expand our J unit to kg•m²/s². ==> 4890-kg•m²/s²•mol = 0.5 x 4.00-g/mol x v² ==>
9780-kg•m²/s²•mol = (4.00-g/mol x 1kg/1000-g) x v²
v² = 9780-kg•m²/s²•mol / 0.004-kg/mol

both sides square root gives
v = 1564-m/s for He

F₂ = 1564-m/s / 3.08 = 508-m/s

The root mean square speed of F₂(g) molecules under the same conditions = 508 m/s

Thank You!