Question

Let `barv,v_(rms) and v_p` respectively denote the mean speed. Root mean square speed, and most probable speed of the molecules in an ideal monatomic gas at absolute temperature T. The mass of a molecule is m. Then
A. no molecules can have a speed greater than `sqrt2v_(rms)`
B. no molecule can have a speed less than `v_p//sqrt2`
C. `v_p lt barv lt v_(rms)`
D. the average kinetic energy of a molecules is `3/4mv_p^2`.

Answer 1

Correct Answer - C::D
(c,d)
We know that
`barv=sqrt((8RT)/(piM)),v_(rms)=sqrt((3RT)/M) and `v_p=sqrt((2RT)/M)`
From these expressions, we can conclude that
`v_pltbarvltv_(rms)`
Also the average kinetic energy of gaseous molecule is
`barE=1/2mv_(rms)^2=1/2m(3/2v_p^2)=3/4mv_p^2`