At a given temperature , the most probable velocity of the molecules of gas A is the same as the average velocity of the molecules of gas B. which has larger molar mass?

Explanation

Suppose, the molar masses of gases A and B are `M_(A) and M_(B)`, and the temperature of both gases is T K.
Therefore, at this temperature, the most probable velocity of the molecules of gas A, `c_(m)=sqrt((2RT)/(M_(A)))` and the average velocity of the molecules of gas B, `overline(c)=sqrt((8RT)/(piM_(B)))`
As given, `c_(m)=overline(c)`
`therefore sqrt((2RT)/(M_(A)))=sqrt((8RT)/(piM_(B))) or, (1)/(M_(A))=(4)/(piM_(B)) " "therefore M_(B)=1.27M_(A)`
Therefore, gas B has higher molar mass.