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Because the energy available to them is less than that required to get them to rotate or vibrate.
Is translational energy also quantized?
For what it's worth, the translational KE is also quantised
If not, what makes the difference?
Temperature (in kelvin, since 2019) is NOT telling you something about the actual kinetic energy of a particle you can observe in reality. It is much more abstract than that, it is the kinetic energy a particle would have in the model.
A single particle cannot have a temperature.Temperature is only defined as a mesoscopic parameter, or, if you prefer, an emergent property of an ensemble.
I think that the experimental plot above plays important role in the development of equipartition theory, also the concept of degree of freedom. But the difference in the gradient of the curve shows that at least at some points, the energy distribution among different degrees of freedoms are not equal.
Or it shows that the number of degrees of freedom isn't an integer.
https://en.m.wikipedia.org/wiki/Degrees_of_freedom_(physics_and_chemistry)Both the rotational and vibrational modes are quantized, requiring a minimum temperature to be activated. The "rotational temperature" to activate the rotational degrees of freedom is less than 100 K for many gases. For N2 and O2, it is less than 3 K. The "vibrational temperature" necessary for substantial vibration is between 103 K and 104 K, 3521 K for N2 and 2156 K for O2. Typical atmospheric temperatures are not high enough to activate vibration in N2 and O2, which comprise most of the atmosphere.
Do you have supporting evidence for your hypothesis?
But the previous plot shows that it's already activated at 0 K.
Yes; your graph.Since the heat capacity is 1/2 R per degree of freedom, that graph is effectively a graph of "number of degrees of freedom per molecule" vs temperature.The sloping bits correspond to non integer values of the DoF.
The heat capacity at 0 K is zero."So the heat capacity must go to zero at absolute zero"fromhttps://en.wikipedia.org/wiki/Third_law_of_thermodynamicsSo, we know that it isn't.We know that the graph is wrong.
What's the slope of the change?
What's your sources?
Where's the switching point from 0 to 3R/2 ?
What's the minimum temperature where the specific heat of gas is 3R/2?
Quote from: hamdani yusuf on 12/05/2022 03:16:09What's the slope of the change?Upwards to the right.Obviously the gradient changes with temperature.Last time I saw it modeled, they fitted the hyperbolic tangent function to the data.I'm not sure if that was on theoretical grounds, or just because it's the right shape.Quote from: hamdani yusuf on 12/05/2022 03:09:14What's your sources?56 years of acquired experience.
Quote from: hamdani yusuf on 12/05/2022 15:20:50What's the minimum temperature where the specific heat of gas is 3R/2?It depends.What would you actually do with the information if I told you?Why not study science?Then you would be able to work it out rather than asking pointless questions.
Don't you think that science should be objective?
What does it mean when a system has 3.4 degrees of freedom?
Also "mesoscopic"? Did you really mean middle-sized?
Temperature is telling you about the average kinetic energy of a particle among a myriad of otherwise identical particles
They don't need to be identical. Porridge is inhomogeneous, but Goldilocks was able to measure its temperature. Temperature is the mean kinetic energy of all the particles in a sample.
I never suggested otherwise.
It depends.What would you actually do with the information if I told you?
Incidentally, you forgot to answer my questions.Please do so.
That you are a little less than half way between the flat bits of the graph at 3 and 4.