Temperature of a diatomic gas is 300k. If the moment of inertia of its molecules is 8.28*10^-38 gm-cm^2. Calculate their root mean square angular velocity.
7 answers
Please give the complete solution
I=8.28•10⁻³⁸ g•cm²=8.28•10⁻⁴⁵kg•m²,
I=2mr²/3 =>
radius of diatomic molecule
r=sqrt{3•I/2m}
Root mean square speed v =sqrt{3kT/m}.
v = ωr.
Angular root mean square speed
ω=v/r= =sqrt{2kT/I} =
=sqrt{2•1.38•10⁻²³•300/8.28•10⁻⁴⁵}=
=1•10¹² rad/s
I=2mr²/3 =>
radius of diatomic molecule
r=sqrt{3•I/2m}
Root mean square speed v =sqrt{3kT/m}.
v = ωr.
Angular root mean square speed
ω=v/r= =sqrt{2kT/I} =
=sqrt{2•1.38•10⁻²³•300/8.28•10⁻⁴⁵}=
=1•10¹² rad/s
Thanks, Elena...
Thank you so much
R can either be radius or the gas constant.It is not both simultaneously.Dont think R as an electron.Please.
wrong she has taken 10^-45 it should be 10^-35
radius of an molecule is negligible bro so it cannot be uder consideration makes not valid