The kinetic molecular theory attributes an average translational kinetic energy of 32RTN to each particle. What rms speed woul a mist particle of mass 10−12g have a root temperature (27∘C) according to kinetic theory of gases.
Answers
Answered by
5
Answer:Let mass of mist particle be m , then K.E of this particle = 12mu2
Where u is the its rms velocity
Also K.E per molecule = 3RT2×N
∴12mu2=3RT2×N
u=[3RN×Tm]−−−−−−−−−√
m=10−12gR=8.314×107ergT=300K
u=3×8.314×107×3006.023×1023×10−12−−−−−−−−−−−−−√
=0.35cmsec−1
Explanation:
Similar questions