[SOLVED] Rotational Kinetic Energy and distribution of diatomic molecules


by merbear
Tags: diatomic, distribution, energy, kinetic, molecules, rotational, solved
merbear
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#1
Feb17-08, 04:14 PM
P: 12
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 1.2110-10m. 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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Doc Al
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Feb17-08, 04:49 PM
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Quote Quote by merbear View Post
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.
Look up the equipartion theorem. How many rotational degrees of freedom are there for a diatomic molecule?

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.
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.

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.
Again, you are dealing with rotational motion here, so consider the answer to part 1. And the rotational KE is [itex]1/2 I \omega^2[/itex]. I would assume the temperature is that given in part 1. (These questions all relate to the same situation.)
merbear
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#3
Feb19-08, 03:29 PM
P: 12
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

Doc Al
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Feb19-08, 03:48 PM
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[SOLVED] Rotational Kinetic Energy and distribution of diatomic molecules


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


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