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PeterPoPS
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[SOLVED] Equipartition theorem
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)
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}
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??
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??