Question

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

Answer 1

The rms speed of the molecules of gas A is twice that of gas B. The molecular mass of A is one fourth to that of B.

Answer: Option B

Explanation:

Measuring the speed of particles at a given point in time results in a large distribution of values. Some molecules can move very slowly, others very fast, and because they are still moving in different directions, the speeds may be zero. (Velocity, vector quantity that corresponds to the speed and direction of the molecule.)

To correctly estimate the average velocity, you must take the squares of the mean velocity and take the square root of this value. This is known as the root mean square (rms) velocity and is shown as follows:

                 $V_{r m s}=\sqrt{\frac{3 R T}{M}}$

Where,

M – Gas’s molar mass

R – Molar mass constant

T – Temperature (in Kelvin)

Given data is rms speed for gas molecule A is twice that of gas molecule B. So,

                 $\left(V_{r m s}\right)_{A}=2\left(V_{r m s}\right)_{B}$

Therefore, equating the molecule’s rms speed formula for both A and B,

                  $\sqrt{\frac{3 R T}{M_{A}}}=2(\sqrt{\frac{3 R T}{M_{B}}})$

On squaring both sides, we get,

                 $\frac{3 R T}{M_{A}}=4\left(\frac{3 R T}{M_{B}}\right)$

By solving the above equations, we get,

                 $M_{A}=\frac{M_{B}}{4}$