Question

If for two gases of molar masses ${M_A}$ and ${M_B}$ at temperature ${T_A}$ and ${T_B}$ , ${T_A}{M_B} = {T_B}{M_A}$ , then which property has the same magnitude for both the gases?
A. Density
B. Pressure
C. KE per mole
D. RMS speed

Answer 1

Hint: Start using the module for finding the constant parameters already given in the question. Look for the physical property which best suits with the constant criteria observed.Here the condition given is ${T_A}{M_B} = {T_B}{M_A}$ for two specific gases. Since two specific gases are chosen at first it needs to be found out which are the similar characters which are observed in gases.

Complete step by step answer:
It becomes easier to find a constant physical property if the similarity in character or constant factor can be determined.
From there we can get,
$\dfrac{{{T_A}}}{{{M_A}}} = \dfrac{{{T_B}}}{{{M_B}}}$
This proves that under the specific condition $\dfrac{T}{M}$is constant for all the gases.
Now to understand the KE of the specific gases the equation for KE needs to be chosen, which is
$KE = \dfrac{1}{2}m{v^2}$
Therefore, the KE required the velocity but the molar mass changes in both the gases and hence it differs in the gases. This is why the RMS velocity is chosen here, with the given equation
${V_{rms}} = \dfrac{{\sqrt {3RT} }}{{\sqrt M }}$
Hence in this equation, there are few parts that are constant, such as the 3R here is constant. This is because R is the gas constant with a constant value. Therefore, the variable left here is $\dfrac{{\sqrt T }}{{\sqrt M }}$ which is seen to be constant under the given condition for the two gases. Hence the RMS speed or the root mean square speed of the gases is the same.

So, the correct answer is Option D.

Note: The similarity and dissimilar characters are to be observed based on the physical properties or parameters already mentioned in the question. If the equations for all the given characters in the options are known all the characters need to be tried for finding if the constant character is maintained.