In a mixture of the gases oxygen and helium, which statement is valid: (a) the helium molecules will be moving faster than the oxygen molecules, on average; (b) both kinds of molecules will be moving at the same speed; (c) the oxygen molecules will, on average, be moving more rapidly than the helium molecules; (d) the kinetic energy of the helium will exceed that of the oxygen; (e) none of the above.

a

Explanation:

Given:

In a mixture of the gases oxygen and helium, which statement is valid:

(a) the helium molecules will be moving faster than the oxygen molecules, on average

(b) both kinds of molecules will be moving at the same speed

(c) the oxygen molecules will, on average, be moving more rapidly than the helium molecules

(d) the kinetic energy of the helium will exceed that of the oxygen

(e) none of the above

Solution:

- We will use Boltzmann distribution to answer this question. The root mean square speed of molecules of a gas gives the average speed as follows:

V_rms = sqrt (3 k T / m)

- Where, k is the Boltzmann constant, T is the temperature and m is the mass of a single molecule of a gas.

- In general, a mixture has a constant equilibrium temperature T_eq.

- So the v_rms is governed by the mass of a single molecule.

- We know that mass of single molecule of Oxygen is higher than that of Helium molecule. Hence, the relation of mass is inversely proportional to square of root mean speed. So the helium molecules will be moving faster than the oxygen molecules.

- Note: The kinetic energy of the mixture remains constant because it is due to the interaction of the molecules within i. e oxygen and helium. Which makes the kinetic energy independent of mass.

E_k = 0.5*m*v_rms^2

E_k = 0.5*m * (3*k*T / m)

E_k = 0.5*3*k*T

Hence, E_k is only the function of Temperature which we already established to remain constant at equilibrium.