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What is the rotational kinetic energy of one molecule of Cl2

  • Thread starter Kawrae
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In a crude model of a rotating diatomic molecule of chlorine (Cl2), the two Cl atoms are 2.00e-10 m apart and rotate about their center of mass with angular speed w = 1.40e12 rad/s. What is the rotational kinetic energy of one molecule of Cl2, which has a molar mass of 70.0 g/mol?

>> I wasn't exactly sure how to do this, so I thought I'd try with the rotational kinetic formula from the first half of General Physics. That formula was 1/2(m)(r^2)(w^2). I don't think it gave me the answer they were looking for... it wants an answer in Joules and I'm not sure where to start or how to finish it. Or if this is even along the right steps to take...

:grumpy:
 

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  • #2
Dr Transport
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show your work so far......
 
  • #3
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Okay... I have:

r=2.00e-10 m
w=1.40e12 rad/s
m=70.0 g/mol

1/2 m(r^2)(w^2)
= 1/2 (70)(2.00e-10)^2(1.40e12)^2
= 2.74e6 J

Hmmm... is r wrong? Or should I change w to degrees? I'm not sure what to do... :yuck:
 
  • #4
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Kawrae said:
Okay... I have:

r=2.00e-10 m
w=1.40e12 rad/s
m=70.0 g/mol

1/2 m(r^2)(w^2)
= 1/2 (70)(2.00e-10)^2(1.40e12)^2
= 2.74e6 J

Hmmm... is r wrong? Or should I change w to degrees? I'm not sure what to do... :yuck:
It's good practice to include the units when showing your working. Correcting your working,

K = 1/2 (70.0 g/mol)(2.00e-10 m)^2 (1.4e12 rad/s)
= 2.74e6 g m^2/mol/s.

Now, I'm not sure what the significance of that derived unit is, but it's certainly not energy :wink:.

Your equation [itex]K = \frac{1}{2}Mr^2\omega^2[/itex] is clearly wrong as it is dimensionally inconsistent --- one side has units of kg m^2/s^2 and the other has g m^2/mol/s. The equation you're looking for is [itex]K = \frac{1}{2}I\omega^2[/itex] where [itex]I[/itex] is the moment of inertia of the rotating body. The equation for the moment of inertia of a continuous body is [itex]I= \int r^2\, dm[/itex], which, for the Cl[itex]_2[/itex] molecule is given by the discrete sum [itex]I = (1.00E-10 {\rm m})^2 m + (1.00E-10 {\rm m})^2 m[/itex]. To find the value of [itex]m[/itex], use the relationships[itex]n = \frac{m}{M}[/itex] and [itex]n = \frac{N}{N_\mathrm{A}}[/itex]. That should be all the info you need to complete the question.

P.S. Avoid using degrees at all in physics, except in particular disciplines such as astronomy.
 
  • #5
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Hi Kawrae,

In my opinion you are not so wrong with your approach. E_c=(1/2)*m*v^2 must work for all physical systems. When you have two particles in system you must add the two corresponding energies. But, m must be the mass of the particle (in Kg) and v=r*w where r is the radius of the trajectory (and not the relative distance between particles). So, you have

E_c=(1/2)*m*(r/2)^2*w^2+(1/2)*m*(r/2)^2*w^2=m*(r/2)^2*w^2=(1/4)*r^2*w^2

(here r is the distance between the two particles...then r/2 is the radius of the trajectory)

Now you have to calculate the mass m of a single atom (see the post of jdstokes).

Good luck!
 
Last edited:

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