As adiabatic vessel contains `n_(1) = 3` mole of diatomic gas. Moment of inertia of each molecule is `I = 2.76 xx 10^(-46) kg m^(2)` and root-mean-square angular velocity is `omega_(0) = 5 xx 10^(12) rad//s`. Another adiabatic vessel contains `n_(2) = 5` mole of a monatomic gas at a temperature `470 K`. Assume gases to be ideal, calculate root-mean-square angular velocity of diatomic molecules when the two vessels are connected by a thin tube of negligible volume. Boltzmann constant `k = 1.38 xx 10^(-23) J//:"molecule"`.

Explanation

We know according to law of equipartition of energy, each gas molecule has `(1)/(2)kT` energy associated with each of its degrees of freedom. As a diatomic gas molecule has two rotational degrees of freedom, its total rotational energy must be `2xx(1)/(2)` kT = kT.
If initial temperature of diatomic gas is `T_(1)`, we have
`(1)/(2)I omega_("rms")^(2)=kT_(1)`
or `" "T_(1)=(I omega_("rms")^(2))/(2k)`
`=((2.76xx10^(-46))(5xx10^(12))^(2))/(2xx1.38xx10^(-23))`
`T_(1)=250K`
When the two vessels containing diatomic and monoatomic gases are connected, these gases exchange this thermal energy but no energy is lost to surrounding as vessels are adiabatic. Thus this mixing of gases tables place at constant volume, total internal energy of system remains constant. If `T_(f)` be the final temperature of the sustem, we have
`T_(f)=(f_(1)n_(1)T_(1)+f_(2)n_(2)T_(2))/(f_(1)n_(1)+f_(2)n_(2))" "...(2.78)`
Here for diatomic gas
`f_(1)=5,n=3" and "T_(1)=250" K"`
For monoatomic gas
`f_(2)=3,n_(2)=5" and "T_(2)=470" K"`
Thus from equation-(2.78)
or `" "T_(f)=(5xx3xx250+3xx5xx470)/(5xx3+3xx5)`
`T_(f)=360" K"`
Thus final mixture of the two gases is at temperature 360 K. If final rms angular velocity of diatomic gas molecules is `omega_("rms f")`, according to law of equipartition of energy, we have
`(1)/(2)I omega_("rms f")^(2)=kT`
`omega_("rms f")=sqrt((2kT)/(I))`
`=sqrt((2xx1.38xx10^(-23)xx360)/(2.76xx10^(-46)))`
`omega_("rms f")=6xx10^(12)" rad"//"s"`