InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
The temperature of a gas consisting of rigid diatomic moleculoes is T = 300 K. Calculate the angular root-mean square velocity of a rotating molecules if its moment of inertia is ` I = 2.0 xx 10^(-40) kg m^(2)`. |
|
Answer» The degrees of freedom of rotatioal motion = 2 `:. lt lt epsilon gt gt_("rot") = 2 xx (1)/(2) kT` (by equipartition principle) `= lt lt 2 xx (1)/(2) I omega^(2) gt gt` `kT = I lt lt omega^(2) gt gt implies omega_(rms) = sqrt((kT)/(I))` Therefore, `omega_(rms) = sqrt((1.38 xx 10^(-23) xx 300)/(2 xx 10^(-40))) = 4.5 xx 10^(9) "rad"//s` |
|