Equipartition theorem and chlorine

[PeterPoPS]

[SOLVED] Equipartition theorem

Homework Statement

If the chlorine molecule at 290 K were to rotate at the angular frequency predicted by the equipartition theorem what would be the average centripetal force?
(The atmos of Cl are 2*10^-10 m apart and the mass of the chlorine atom 35.45 a.m.u)

(Correct answer is 1.6*10^-10 N)

Homework Equations

Kinetic energy
$$E_{kin} = \frac{mv^2}{2}$$

Centripetal force
$$F_c = \frac{mv^2}{r}$$

Boltzmann's constant
$$k_B = 1.3807*10^{-23} JK^{-1}$$

The Attempt at a Solution

Hi!

I tried to solve it this way.
I think of my Chlorine molecule as two points. One stationary and the other one circulating around it.
From the equipartition theorem i get that the average energy of the molecule is

$$E = \frac{7k_BT}{2}$$

I think that this is equal to the kinetic energy

$$E = \frac{7k_BT}{2} = \frac{mv^2}{2}$$

so

$$mv^2 = 7k_BT$$

Using this in the equation for the centriopetal force I get

$$F_c = \frac{7k_BT}{r} = 1.4014^{-10}N$$

which is not the right answer and I havn't used the mass at all in the problem.
What's wrong??

 

[Shooting Star]

For diatomic molecules, Eavg = 5/2 KbT.

 

[ogb p]

The equipartition theorem states that in thermal equilibrium, each degree of freedom of a system has an average energy of k_B*T/2, where k_B is Boltzmann's constant and T is the temperature. In the case of a rotating molecule, there are three degrees of freedom - two for translation and one for rotation. Therefore, the average energy of the molecule is 3*k_B*T/2.

To calculate the average centripetal force, we need to consider the kinetic energy of the rotating molecule. This can be calculated using the formula E_kin = 1/2*I*ω^2, where I is the moment of inertia and ω is the angular frequency. In this case, the moment of inertia for a diatomic molecule is given by I = m*r^2, where m is the mass of the molecule and r is the distance between the atoms.

Substituting these values into the equation, we get E_kin = 1/2*m*r^2*ω^2. Equating this to the average energy from the equipartition theorem, we get 3*k_B*T/2 = 1/2*m*r^2*ω^2. Solving for ω, we get ω = √(3*k_B*T/m*r^2).

Finally, we can calculate the average centripetal force using the formula F_c = m*ω^2*r. Substituting the value of ω we just calculated, we get F_c = m*3*k_B*T/m*r^2*r = 3*k_B*T/r. Plugging in the values given in the problem (m = 35.45 a.m.u, r = 2*10^-10 m, T = 290 K), we get F_c = 1.6*10^-10 N, which is the correct answer.

It is important to note that the equipartition theorem only gives us the average energy of a system and does not take into account the specific details of the system, such as the mass of the particles. This is why we need to use additional equations and consider the specific system to calculate the average centripetal force accurately.