Respuesta :
Answer:
option (d)
Explanation:
The relation between the rms velocity and the molecular mass is given by
v proportional to \frac{1}{\sqrt{M}} keeping the temperature constant
So for two gases
[tex]\frac{v_{A}}{v_{B}}=\sqrt{\frac{M_{B}}{M_{A}}}[/tex]
[tex]\frac{2v_{B}}{v_{B}}=\sqrt{\frac{M_{B}}{M_{A}}}[/tex]
[tex]{\frac{M_{B}}{M_{A}}} = 4[/tex]
[tex]{\frac{M_{B}}{4}} = M_{A}[/tex]
If the RMS speed of the molecules of gas A is twice that of gas B then the molecular mass of A compared to that of B is one-fourth that of B.
How does the molecular mass of A compare to that of B?
We know that the RMS velocity of the molecule is given as,
[tex]V = \dfrac{1}{\sqrt{M}}[/tex]
Given to us
RMS speed of the molecules of gas A is twice that of gas B, therefore, [tex]V_A = 2 V_B[/tex]
Substitute the value of RMS in the equation [tex]V_A = 2 V_B[/tex],
[tex]\dfrac{1}{\sqrt{M_A}} = \dfrac{2}{\sqrt{M_B}}\\\\\\\sqrt{\dfrac{M_B}{M_A}} = 2\\\\\\\dfrac{M_B}{M_A} = 4[/tex]
Hence, If the RMS speed of the molecules of gas A is twice that of gas B then the molecular mass of A compared to that of B is one-fourth that of B.
Learn more about RMS speed:
https://brainly.com/question/15514391
