# Rotational Kinetic Energy and distribution of diatomic molecules

1. Feb 17, 2008

### merbear

[SOLVED] Rotational Kinetic Energy and distribution of diatomic molecules

a) Calculate the total rotational kinetic energy of the molecules in 1.00 mol of a diatomic gas at 300 K.

b) Calculate the moment of inertia of an oxygen molecule (O2) for rotation about either the x- or y-axis shown in the figure. Treat the molecule as two massive points (representing the oxygen atoms) separated by a distance 1.21×10-10m. The molar mass of oxygen atoms is 16 g/mol.

c) Find the rms angular velocity of rotation of an oxygen molecule about either an x- or y-axis

2. Relevant equations

kT=2/3*<k>

where, k is boltzman constant and K is kinetic energy

I=mR^2

E(rot)= 1/2*I*angular velocity(for x)^2 + 1/2*I*angular velocity(for y)^2

3. The attempt at a solution

To find the answer to the first part of the problem I used the first equation listed and got 6.21E-21 J, but that answer is incorrect. I do not know how else to approach the problem. I think it went wrong because K in that equation is the average kinetic energy and not the rotational kinetic energy, but I couldn't find another equation that would work.

For the second part I used I=mR^2. To find m, I took the molar mass and divided by avagadros number and I used the distance given for R.

my answer was: 3.89E-46 kg*m^2, but I don't think that was correct.

For the third part, I would think that you would use Vrms equals the squareroot of (3kT/m). But we are not given the temperature. So I don't know how to go about this part either.

I would really appreciate help on these three parts. Thank you!

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2. Feb 17, 2008

### Staff: Mentor

Look up the equipartion theorem. How many rotational degrees of freedom are there for a diatomic molecule?

What's the rotational inertia of a point mass? mR^2. But m must equal the mass of each atom and R the distance to the center. And you have two atoms, of course.

Again, you are dealing with rotational motion here, so consider the answer to part 1. And the rotational KE is $1/2 I \omega^2$. I would assume the temperature is that given in part 1. (These questions all relate to the same situation.)

3. Feb 19, 2008

### merbear

Still confused about Rotational Kinetic Energy and angular velocity

I was able to figure our the moment of inertia by using the equation I= 2mL^2

However, I tried Part A and C and I still can't figure it out.

For finding the rotational kinetic energy I used:

K(rot)= (f/2)*kT

When solving using f= 2 for the degrees of freedom, boltzman constant for k, and 300 K for temperature I get 4.14E-21 J. But when I input that into the program it comes up incorrect.

For part C, I think I am getting it wrong still because I am using an incorrect value for K(rot). I am using the equation: K(rot) = 1/2 I w^2.

Please let me know why my approach to part A was incorrect.

Thank you

4. Feb 19, 2008

### Staff: Mentor

For one thing, you are calculating the rotational energy per molecule, but the question asks for it per mole.