ThermodynamicsHard

Question

Let and vp respectvely denote the mean speed, root mean squre speed and most monoatomic gas at absolute temperture t. The mass of the molecule is m. Then

Options

A.no molecule can have a speed greater than (√2vrms)
B.no molecule can have a speed less than
C. < vp < Vrms 
D.The average kinetic A and B with frictionaless is (mvp2)

Solution

vrms =

and vp =
From these expressions, we can see that vp < < vrms
Again, vrms = vp
and average kinetic energy of a gas molecule Ek = mv2rms

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