1.

Let vecv,v_(rms)andv_(p) respectively denote the mean speed, the root mean square speed and the most probable speed of the molecules in an ideal monatomic gas at absolute temperature T. The mass of a molecule is m. Then

Answer»

no molecule can have speed greater than `v_(RMS)`
no molecule can have speed less than `v_(p)//sqrt(2)`
`v_(p)ltvecvltv_(rms)`
the AVERAGE kinetic ENERGY of a molecule is `(3)/(4)mv(p)^(2)`

Solution :`v_(rms)=sqrt((3RT)/(M)),BARV=sqrt((8RT)/(piM))=sqrt(2.5(RT)/(M))`
`v_(p)=sqrt((2RT)/(M))`
`therefore v_(rms):barv:v_(p)=sqrt(3):sqrt(2.5):sqrt(2)`
So, `v_(p)ltbarvltv_(rms)and(v_(rms))/(v_(p))=sqrt((3)/(2))orv_(rms)^(2)=(3)/(2)v_(p)^(2)`
`therefore` Average kinetic energy
`=(1)/(2)mv_(rms)^(2)=(1)/(2)mxx(3)/(2)v_(p)^(2)=(3)/(4)mv_(p)^(2)`


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