Angular velocity of an oxygen molecule.

[j_namtirach]

Homework Statement

The atoms in the oxygen molecule [O][/2] may be considered to be point masses separated by a distance of 1.2 x 10^-10 m. The molecular speed of an oxygen molecule at s.t.p. is 460 m/s. Given that the rotational kinetic energy of the molecule is two-thirds the of its translational kinetic energy, calculate its angular velocity at s.t.p. assuming that molecular rotation takes place about an axis through the centre of, and perpendicular to, the line joining the atoms.

Homework Equations

K.E. = 1/2 mv^2
K.E. = 1/2 Iω^2
I = ∑mr^2
n = N/Na
m = nμ

The Attempt at a Solution

no. moles of oxygen = 1/6.02 x 10^23 = 1.66 x 10^-24 mol

mass of oxygen molecule = (1.66 x 10^-24) x 32 = 5.32 x 10^-23 g = 5.32 x 10^-26 kg

translational K.E. of oxygen molecule = 0.5 x (5.32 x 10^-26) x 460 = 1.22 x 10^-23 J

rotational K.E. of oxygen molecule = 2/3 x 1.22 x 10^-23 J = 8.15 x 10^-24 J

I = 2(mr^2) = 2((5.32 x 10^-32) x (6 x 10^-11)^2) = 3.83 x 10^-46 kg m^2

angular velocity of oxygen molecule, ω = sqrt(((8.15 x 10^-24) x 2)/3.83 x 10^-46) = 2.06 x 10^11 rad/s


I think the correct answer is 6.3 x 10^12 rad/s. I'm not sure where I have gone wrong here. I'm not sure if I used the correct method for working out the moment of inertia. Any guidance would be much appreciated.

Thanks.

 

[CAF123]

You do not have to compute the mass of the oxygen atoms to obtain your answer. You can treat the two oxygen atoms as point masses rotating about an axis through the centre of rotation. So, if you carry the calculation through symbolically, the mass term should cancel.

Should you not have $E_{\text{rot}} = \frac{2}{3} E_{\text{trans}}$?

 

[j_namtirach]

OK, I've got it now, thanks!

 

[Prabin Acharya]

I don't know if this is the way...but I kind of got the close answer....

we have d = 1.2 × 10^-10
r = d/2 = 0.6 × 10 ^-10
v = 460

3 × rotational KE = 2 × translational KE
(3 . I . w^2 ) / 2 = (2 . m . v^2) / 2
3 . m . r^2 . w^2 = 2 . m . v^2
w^2 = (2/3) . v^2/r^2
w = ✓(2/3) . v/r
w = ✓(2/3) . 460/0.6×10^-10
w = ✓(2/3) . (7.67 × 10^12)
w = 6.26 × 10^12 rad/s

 

[berkeman]

I don't know if this is the way...but I kind of got the close answer....

Welcome to PF.

We usually don't allow full solutions to be posted in Homework threads (except by the Original Poster, of course), but since this question thread is 8 years old, it is okay in this case.

In active schoolwork threads, please provide hints, ask probing questions, find mistakes, etc. But please do not post solutions in active schoolwork threads -- we require that the student do the bulk of the work. Thanks.

 

[haruspex]

I don't know if this is the way...but I kind of got the close answer....

we have d = 1.2 × 10^-10
r = d/2 = 0.6 × 10 ^-10
v = 460

3 × rotational KE = 2 × translational KE
(3 . I . w^2 ) / 2 = (2 . m . v^2) / 2
3 . m . r^2 . w^2 = 2 . m . v^2
w^2 = (2/3) . v^2/r^2
w = ✓(2/3) . v/r
w = ✓(2/3) . 460/0.6×10^-10
w = ✓(2/3) . (7.67 × 10^12)
w = 6.26 × 10^12 rad/s

Your only mistake is in regards to precision. The diameter is only given to two sig figs, so your answer should not have more. Rounding your 6.26 to 6.3 completes it.

 

[mjc123]

There are two errors in the OP:

translational K.E. of oxygen molecule = 0.5 x (5.32 x 10^-26) x 460 = 1.22 x 10^-23 J

You have forgotten to square the speed.

I = 2(mr^2) = 2((5.32 x 10^-32) x (6 x 10^-11)^2) = 3.83 x 10^-46 kg m^2

You need to use the mass of an oxygen atom, not an oxygen molecule. (Or else, like @Prabin Acharya, use the molecular mass and the formula I = mr².)

 

[haruspex]

There are two errors in the OP

Which is eight years old.