What is the Typical Rotational Frequency of N₂ at Room Temperature?

In summary: E12 HzIn summary, the typical rotational frequency for a molecule like N2 at room temperature (25 C) is 1.34E12 Hz. The equation used to calculate this frequency is ω= sqrt((2*k_B*T)/(M*d^2)), where k_B is the Boltzman constant, T is the temperature in Kelvin, M is the mass of the molecule, and d is the atomic distance.
  • #1
jaded18
150
0
What is the typical rotational frequency f_rot for a molecule like N_2 (nitrogen) at room temperature (25 C)? Assume that d for this molecule is 1 angstrom = 10^{-10} m. Take the atomic mass of N_2 to be 4.65 * 10^{-26} kg.
_________________________________

I know that the rms angular speed which is the root mean square (rms) of the x component of velocity of the gas particles is = sqrt(2k_B(T)/(m(d^2))) where k_B is the Boltzman constant 1.38*10^-23 J/K. Ok so I have all the info I need I believe to solve for velocity but what about the wavelength component that I need to use this equation velocity=Freq*wavelength and to get my answer for frequency. What is this wavelength that I have to calculate for?
 
Physics news on Phys.org
  • #2
i found this equation relating frequency and veolcity --> angular velocity=2(pi)f
but i DONT UNDERSTAND why I'm not getting the right answer. i get velocity to be 4.2057*10^12 m/s and i SHOULD be able to get frequency by dividing it by 2pi, but f=6.69*10^11 is not right!

whats wrOong?!
 
  • #3
Hold on, you're getting confused. You don't want the de Broglie wavelength because this isn't a wave, it's a true classical rotation. The x component of velocity is just that--translational speed, not angular rotation. Also your title mentions the equipartition theorem so you know that will figure in somewhere.

Take a look at the equipartion theorem in your text. How many degrees of freedom does N2 have? What is the energy of each one? Then set it equal to the energy of a rotating object (what is that?)

Come back if you need more help.
 
  • #4
Jaded18, I just saw your other post in Introductory Physics. Please don't double post in the future. (And that forum is the appropriate one for homework problems.)
 
  • #5
marcusl said:
Hold on, you're getting confused. You don't want the de Broglie wavelength because this isn't a wave, it's a true classical rotation. The x component of velocity is just that--translational speed, not angular rotation. Also your title mentions the equipartition theorem so you know that will figure in somewhere.

Take a look at the equipartion theorem in your text. How many degrees of freedom does N2 have? What is the energy of each one? Then set it equal to the energy of a rotating object (what is that?)

Come back if you need more help.

I don't understand how the energy relates to frequency! Degrees of freedom for diatomic molecules: 5. energy of each one = 0.5k_B(T)= 0.5(1.38*10^-23 J/K)(25+273K)=2.056*10^-21= 0.5I(angular velocity)^2... what is I?. How come what i did in the beginning with the equation i found didn't work out to get the correct answer?
 
  • #6
Good, now you're on the right track; you properly found the energy of one degree of freedom and you set it equal to the rotational energy of that mode. I is the moment of inertia, which is the sum of (each mass times that mass's distance squared from the center). You can finish the problem now and solve for the angular velocity.

The reason you didn't get the right answer before is because you used equations that made no sense.
 
  • #7
first of all, i don't need the angular velocity. i need the rotational frequency. and also, the equations are all very real. it turns out that my ans was right and it was a mistake in the ans key... So yeah won't double post in the future, it's pretty much useless anyways.
 
  • #8
Well, I'm glad it worked out for you.
 
  • #9
Actually - For others who have to finish this problem there was an error in the above calculations. When dealing with N_2 there are 2 molecules of nitrogen when your solving for just 1. This means you'll need to cut your Mass in half:
(4.65E-26)/2 = M
sqrt((2*k_B*T(K)/M*d^2))= ω <---(Remember Kelvins)
f=(ω/2pi)
 

FAQ: What is the Typical Rotational Frequency of N₂ at Room Temperature?

What is the Equipartition theorem?

The Equipartition theorem is a fundamental principle in thermodynamics that states that in thermal equilibrium, the total energy of a system is equally divided among all of its degrees of freedom. This means that each degree of freedom, such as translational, rotational, or vibrational motion, will have an average energy of kT/2, where k is the Boltzmann constant and T is the temperature.

How does the Equipartition theorem relate to motion?

The Equipartition theorem is closely related to motion because it helps us understand how energy is distributed among the different types of motion of particles in a system. This includes the motion of molecules in a gas, the vibrations of atoms in a solid, and the rotations of molecules in a liquid. The theorem states that each type of motion will have an average energy of kT/2, which allows us to make predictions about the behavior of these systems.

What is the significance of the Equipartition theorem in physics?

The Equipartition theorem is significant in physics because it provides a simple and general way to calculate the average energy of a system in thermal equilibrium. This allows us to make predictions about the behavior of various systems, such as gases, solids, and liquids, and to understand their properties, such as specific heat capacity and thermal conductivity.

Are there any exceptions to the Equipartition theorem?

While the Equipartition theorem is generally applicable to many systems, there are some exceptions where it does not hold. One major exception is when a system is at very low temperatures, close to absolute zero. In this case, quantum effects become important and the distribution of energy among degrees of freedom is no longer equal.

How does the Equipartition theorem contribute to our understanding of thermodynamics?

The Equipartition theorem is a key concept in thermodynamics because it helps us understand how energy is distributed among the different types of motion in a system. This is important in understanding the behavior of various materials and systems, and is crucial in the development of technologies such as engines, refrigeration, and power plants.

Back
Top