Those solids behave more quantum mechanically, and haven't yet reached the equipartition limit at 298 K. That is the subject of the next video: ruclips.net/video/ZvwgqRmX4ks/видео.html
Thanks for the video! I have one question, for each of the particles there are only 3 translational + 3 vibrational doF, with no rotational doF as they are closely bonded together, which leads to the molar heat capacity being 6N*(1/2)R=3NR for N particles. What is wrong in my derivation?
You're taking a different point of view than presented in the video. In my explanation, the "particle" is the single "molecule" composed of N atoms, so it has a lot of vibrational degrees of freedom. In your point of view, you have N particles, each of which is a single atom. It is definitely possible to count the degrees of freedom from your point of view, although the identification of these degrees of freedom should be slightly different than you stated. Remember to pay attention to both KE and PE degrees of freedom. Each of your N particles (atoms) does have 3 translational dof. These contribute to both KE and to PE (because PE increases ~quadratically as they collide with their neighbors). This gets us to 6N contributions. They do not contribute to the rotational energy, as you said, because they are spherical. As single particles, they don't have vibrational degrees of freedom of their own. So there is no contribution from vibrations. The total is still the same: 6N ( 1/2 R) = 3R.
Thanks the video really helps! Though one question, isn't it (3N-(6/2))R instead of (9/2)? I know you can just get rid of that fraction anyways cause N >> 1, but I'm just asking for the sake of understanding the calculation before :)
Yes! You're absolutely right. I made an arithmetic error, and didn't notice it because that term got dropped. Thanks for pointing out the error. Hopefully your comment will prevent others from being confused.
Great video!! I wanted to know what was going on with the cases of exceptions with heat capacity lower than 24 J/mol K?
Those solids behave more quantum mechanically, and haven't yet reached the equipartition limit at 298 K. That is the subject of the next video: ruclips.net/video/ZvwgqRmX4ks/видео.html
Thanks for the video!
I have one question, for each of the particles there are only 3 translational + 3 vibrational doF, with no rotational doF as they are closely bonded together, which leads to the molar heat capacity being 6N*(1/2)R=3NR for N particles. What is wrong in my derivation?
You're taking a different point of view than presented in the video. In my explanation, the "particle" is the single "molecule" composed of N atoms, so it has a lot of vibrational degrees of freedom. In your point of view, you have N particles, each of which is a single atom.
It is definitely possible to count the degrees of freedom from your point of view, although the identification of these degrees of freedom should be slightly different than you stated. Remember to pay attention to both KE and PE degrees of freedom.
Each of your N particles (atoms) does have 3 translational dof. These contribute to both KE and to PE (because PE increases ~quadratically as they collide with their neighbors). This gets us to 6N contributions.
They do not contribute to the rotational energy, as you said, because they are spherical.
As single particles, they don't have vibrational degrees of freedom of their own. So there is no contribution from vibrations.
The total is still the same: 6N ( 1/2 R) = 3R.
Thanks the video really helps!
Though one question, isn't it (3N-(6/2))R instead of (9/2)? I know you can just get rid of that fraction anyways cause N >> 1, but I'm just asking for the sake of understanding the calculation before :)
Yes! You're absolutely right. I made an arithmetic error, and didn't notice it because that term got dropped.
Thanks for pointing out the error. Hopefully your comment will prevent others from being confused.
Amazing Videos! Please Keep Up The Great Work, Truly Appreciated
Thanks, glad you like them
Thank you for this video!
You're welcome!