Find(), the rms angular speed of the dumbbell about a single axis (taken to be t

Question

Find(), the rms angular speed of the dumbbell about a single axis (taken to be the x axis), assuming that the dumbbell is lined up on the z axis and is in equililbrium at temperature T Express the rms angular speed in terms of T, kB, m, d, and other given quantities. View Available Hint(s) Me SubmitP Previous Answers X Incorrect; Try Again; 6 attempts remaining; no points deducted Variables are case sensitive. Check to make sure you are using the correct case Part E What is the typical rotational frequency Jrot for a molecule like N2 at room temperature (25 CAssume that d for this molecule is 1 A -10-10 m. Take the total mass of an N2 molecule to be m 4.65 x 10-26 kg. You will need to account for rotations around two axes (not just one) to find the correct frequency. Express frot numerically in hertz, to three significant figures. Hz Submit uest Answer

Explanation / Answer

Ans:-

A] Rotational energy = 1/2KB T

By the Equipartition Therom

1/2KB T = ½ Ix*Wx^2

Moment of inertia of a dumbbell is Ix = 2md^2/4

KB T = 1/2md^2Wx^2

Sqrt (Wx^2) =sqrt ( 2 KB T/md^2)

B] A bimolecular gas particle like the nitrogen molecule has two degrees of freedom for rotational motion.
E = (2/2)kT = kT

The rotational kinetic energy of object of moment of inertia J rotating at angular velocity is:
E = (1/2)J²

Consider the nitrogen molecule as two point masses connected by thin massless rod. Take the middle of the rod as axis of rotation. Moment of inertia equals mass times squared distance to the axis of rotation
J = 2 (m/2)(d/2)² = (1/4)md²

Angular velocity and frequency are related as:
= 2f

Therefore
E = (1/2) (1/4)md² (2f )² = (1/2)²md²f²
and
kT = (1/2)²md²f²
=>
f = [ 2kT/(²md²) ]
= (2 1.38×10²³JK¹ 298.15K / (² 4.65×10²kg (10¹m)²) ]
= 1.34×10¹² Hz

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.