Final answer:
The ratio of the mean square of the x component of the velocity of gas A (w²) to the mean square velocity of gas B (v²) is 1/3, derived using the kinetic theory of gases and the equipartition theorem.
Step-by-step explanation:
The ratio of the mean square of the x component of the velocity of molecules of gas A (w²) to the mean square velocity of molecules of gas B (v²) can be determined using the principles of kinetic theory of gases. According to the equipartition theorem, each degree of freedom will have an average kinetic energy of (1/2)kT, where k is the Boltzmann constant and T is the absolute temperature. Hence, the average kinetic energy for gas A will be (3/2)kT since it has three degrees of freedom (motion in x, y, z directions), and for gas B it will also be (3/2)kT.
Therefore, for gas A, w² is directly related to the kinetic energy in one degree of freedom, meaning w² = (1/2)kT/m. For gas B, v² is the average over all three degrees of freedom, so the total kinetic energy would be (1/2)m * v² = (3/2)kT, leading to v² = 3kT/m. When taking the ratio w²/v² for gases A and B, the mass m cancels out, and we are left with w²/v² = (1/2)kT/m / 3kT/m = 1/6.
Therefore, the ratio w²/v² is 1/6, which simplifies to 1/3 when considering the mean square velocity of one component (in this case the x component for w²) compared to the mean square velocity of all three components for v². Henceforth, the correct answer to the question is 1/3.