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 monoatomic gas at absolute temperature T. The mass of a molecule is m. Then

Options

A. no molecule can have speed greaer than `v_(rms)`
B. no molecule can have sped less than `(v_(P))/(sqrt(2))`
C. `v_(P)ltbarv lt v_(rms)`
D. the average kinetic energy of molecle is `3/4 mv_(p)^(2)`

Explanation

`v_(rms)=sqrt((3RT)/M)` and
`barv=sqrt((8RT)/(piM))=sqrt(2.55 (RT)/M)`
`v_(P)=sqrt((2RT)/M)`
Hence c and d are correct.