Rotational Kinetic Energy and distribution of diatomic molecules

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SUMMARY

The discussion focuses on calculating the rotational kinetic energy and moment of inertia for diatomic molecules, specifically oxygen (O2), at a temperature of 300 K. The correct approach for calculating the total rotational kinetic energy involves using the equation K(rot) = (f/2)kT, where f represents the degrees of freedom, and k is the Boltzmann constant. The moment of inertia for the oxygen molecule is calculated using I = mR^2, where m is the mass of the molecule derived from the molar mass divided by Avogadro's number, and R is the distance between the two oxygen atoms, given as 1.21×10-10 m.

PREREQUISITES
  • Understanding of rotational kinetic energy and its equations
  • Familiarity with the equipartition theorem
  • Knowledge of moment of inertia calculations for point masses
  • Basic principles of thermodynamics, including the Boltzmann constant
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  • Study the equipartition theorem in detail to understand energy distribution in diatomic molecules
  • Learn about the calculation of moment of inertia for various molecular geometries
  • Explore the relationship between temperature and kinetic energy in gases
  • Investigate the application of rotational motion equations in molecular physics
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Students and educators in physics, particularly those focusing on thermodynamics and molecular dynamics, as well as researchers studying the properties of diatomic gases.

merbear
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[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



Homework Equations



kT=2/3*<k>

where, k is Boltzmann 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


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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merbear said:
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 1/2 I \omega^2. I would assume the temperature is that given in part 1. (These questions all relate to the same situation.)
 
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, Boltzmann 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
 
For one thing, you are calculating the rotational energy per molecule, but the question asks for it per mole.
 

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