Let us consider a diatomic gas whose molecules have the shape of a dumbell. In this model, the center of mass of the molecule can translate in the `x,y` and `z` directions. In addition, the molecule can rotate about three mutually perpendicular axes. We can neglect the rotation about the `y` -axis because the moment of inertia `I_(y)` and the rotational energy `1/2 I_(y)omega^(2)` about this axis are negligible compared with those associated with the `x` and `z` axes. (If the two atoms are taken to the point masses, then `I_(y)` is identically zero). Thus, there are five degrees of freedom: three associated with the translation motion and two associated with the rotational motion. Because each degree of freedom contributes, on average, `1/2k_(B)T` of energy per molecule, the total internal energy for a system of `N` molecules is:
`E_("int")=3N(1/2k_(B)T)+2N(1/2k_(B)T)=5/2NK_(B)T=5/2nRT`...........(i)
We can use this result to find the molar specific heat at constant volume:
`C_(v)=1/n (dE_("int"))/(dT)=1/n d/(dT)(5/2nRT)=5/2R`................(ii)
From equation (i) and (ii) we find that
`C_(P)=C_(V)+R=7/2R`
`gamma=(C_(P))/(C_(V))=(7//2R)/(5//2r)=7/5=1.40`
In the vibratory model, the two atoms are joined by an imaginary spring. The vibrational motion adds two more degrees of freedom, which correspond to the kinetic energy and the potential associated with vibrations along the length of the molecule about its centre of mass.

The root mean square angular velocity of a diatomic molecule (with each atom of mass `m` andn interatomic distance `a`) is given by:

Options

A. `sqrt((mk_(B)T)/(ma^(2)))`
B. `sqrt((2k_(B)T)/(ma^(2)))`
C. `sqrt((k_(B)T)/(ma^(2)))`
D. `sqrt((k_(B)T)/(2ma^(2)))`

Answer

B

Explanation

`1/2 I_(x) omega^(2) +1/2 I_(y) omega^(2)=f/2 kT{f_("fractional")=2}`
`I_(x)=I_(y)impliesmur^(2)=m/2 a^(2) {mu=(m_(1) m_(2))/(m_(1)+m_(2))}`
`(ma^(2))/2 omega^(2)=2/2 klT mu= m/2`
`omega= sqrt((2KT)/(ma^(2)))`