Dust particles in the suspended state in a mono atomic gas are in thermal aquilibrium with the gas.If the temperature of the gas is 300K,find the mean KE of translation of dust particle,If the mass of a certain particle is 10^(-17) kg,calculate its rms speed (k=1.38xx10^(-23)J/K).

Dust particles in the suspended state in a mono atomic gas are in ther

1.	Dust particles in the suspended state in a mono atomic gas are in thermal aquilibrium with the gas.If the temperature of the gas is 300K,find the mean KE of translation of dust particle,If the mass of a certain particle is 10^(-17) kg,calculate its rms speed (k=1.38xx10^(-23)J/K).
Answer» Solution :According to kinetic theory of gases,KE=(3/2)kT, so for a particle in thermal equilibrium `T_(P)=T_(G)=T` so `(KE)_(PA)=(3)/(2)kT=(3)/(2)xx1.38xx10^(-23)xx300=6.21xx10^(-21)J` Now if .m, is the mass of particle and `v_(rms)` its rms SPEED, KE=`(1)/(2)mv_(rms)^(2)`, i.e, `V_(rms)=sqrt((2KE)/(m))` `therefore v_(rms)=sqrt((2xx6.21xx10^(-21))/(10^(-178)))=3.524xx10^(-2)m//s`