Consider two ideal gases, A & B, at the same temperature. The rms speed of the molecules of gas A is twice that of gas B. How does the molecular mass of A compare to that of B? (a) It is twice that of B. (b) It is one half that of B (c) It is four times that of B (d) It is one fourth that of B. (e) It is 1.4 times that of B.

 To compare the molecular mass of gases A and B based on their root mean square (rms) speeds, we can use the formula for the root-mean-square speed of gas molecules, which is given by:

v_rms = sqrt(3kT/m)

where v_rms is the root mean square speed, k is the Boltzmann constant, T is the temperature, and m is the molecular mass of the gas.

Given that both gases are at the same temperature, we can ignore the factors of k and T, as they are constants and will be the same for both gases. Thus, the formula simplifies the relationship between rms speed and molecular mass to:

v_rms ∝ sqrt(1/m)

Rearranging this proportionality gives us:

m ∝ 1/v_rms^2

If the rms speed of gas A is twice that of gas B (v_rms,A = 2 * v_rms,B), then plugging this into the proportionality yields:

m_A ∝ 1/(2*v_rms,B)^2 = 1/(4*v_rms,B^2)

Comparing this to the molecular mass of gas B (m_B ∝ 1/v_rms,B^2), we have:

m_A / m_B = (1/(4*v_rms,B^2)) / (1/v_rms,B^2) = 1/4

This shows that the molecular mass of gas A is one fourth that of gas B. Hence, the correct answer is:

(d) It is one fourth that of B.