[hamdani yusuf (OP)]

Because the energy available to them is less than that required to get them  to rotate or vibrate.

Is translational energy also quantized?
If not,  what makes the difference?

[Bored chemist]

Is translational energy also quantized?

For what it's worth, the translational KE is also quantised

If not,  what makes the difference?

Good question.
I'm saying that you consider all the ways in which the molecule can carry energy when you consider temperature.
So I'm not saying that translational energy is "special" and it's the only one which counts.

So there isn't a difference (apart from a practical one- unless the box is very small, the translational energy levels are very close together and the effect is not measurable).

You need to ask the people who think that vibrational energy doesn't count why they think it's different.

[hamdani yusuf (OP)]

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.

[alancalverd]

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.

Pedant mode engaged.
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.

[Eternal Student]

Hi.

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.

   Yes, absolutely true.   Did I undermine that?  Sorry if I did.
   Also "mesoscopic"?  Did you really mean middle-sized?   I'd be alright with a macroscopic property myself.

    I think the flow of the original statement(s) were already becoming excessively cumbersome.  It certainly wasn't that a particle had a temperature, it doesn't but it does have a kinetic energy.   Or at the very least, it does in the model.

  Temperature is telling you about the average kinetic energy of a particle among a myriad of otherwise identical particles that make up the whole macroscopic object  (and in the case of a theoretical temperature scale, it's still only information for the average k.e. of a particle in the model ).

Best Wishes.

[Bored chemist]

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.

[hamdani yusuf (OP)]

Or it shows that the number of degrees of freedom isn't an integer.

That's new to me. Do you have supporting evidence for your hypothesis?

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.[7] 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.[1] The "vibrational temperature" necessary for substantial vibration is between 103 K and 104 K, 3521 K for N2 and 2156 K for O2.[1] Typical atmospheric temperatures are not high enough to activate vibration in N2 and O2, which comprise most of the atmosphere.

The article doesn't mention about quantization of translational energy. But the previous plot shows that it's already activated at 0 K.

[Bored chemist]

Do you have supporting evidence for your hypothesis?

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.

But the previous plot shows that it's already activated at 0 K.

The heat capacity at 0 K is zero.

"So the heat capacity must go to zero at absolute zero"
from
https://en.wikipedia.org/wiki/Third_law_of_thermodynamics

So, we know that it isn't.
We know that the graph is wrong.

It would be quicker if you started by learning science from the bottom up, rather than trying to understand the complicated bits before you understand the underlying principles.

[hamdani yusuf (OP)]

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.

What's your sources?

[hamdani yusuf (OP)]

The heat capacity at 0 K is zero.

"So the heat capacity must go to zero at absolute zero"
from
https://en.wikipedia.org/wiki/Third_law_of_thermodynamics

So, we know that it isn't.
We know that the graph is wrong.

Where's the switching point from 0 to 3R/2 ?
What's the slope of the change?

[Bored chemist]

What'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.

What's your sources?

56 years of acquired experience.

[hamdani yusuf (OP)]

Where's the switching point from 0 to 3R/2 ?

I'll rephrase it to make it more precise.
What's the minimum temperature where the specific heat of gas is 3R/2?

[Bored chemist]

What'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.

[hamdani yusuf (OP)]

What'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.

What's your sources?

56 years of acquired experience.

What does it mean when a system has 3.4 degrees of freedom?

[hamdani yusuf (OP)]

What'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?

[Bored chemist]

Don't you think that science should be objective?

I never suggested otherwise.

Incidentally, you forgot to answer my questions.
Please do so.

What does it mean when a system has 3.4 degrees of freedom?

That you are a little less than half way between the flat bits of the graph at 3 and 4.

[alancalverd]

 Also "mesoscopic"?  Did you really mean middle-sized? 

somewhere between microscopic (where quantum effects add significant noise to your measurement) and astronomical (where simultaneity cannot be assumed). Usually refers to manageable things like bricks and cows. 

 

  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.

[Eternal Student]

Hi.

    Thanks for your comments @alancalverd .   

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.

    The original statement(s) I made (pages back, for example post #278) concerns the modern definition of temperature in kelvin since May 2019.  In the models they use for that theoretical temperature scale,  all the particles of one species are assumed to be identical to each other (apart from having their own velocity which follows a Boltzmann distribution).   There's just no way I could have all the i dotted and all the t crossed.

      Kinetic theory provides a microscopic account of temperature for some bodies of material, especially gases, based on macroscopic systems' being composed of many microscopic particles, such as molecules and ions of various species, the particles of a species being all alike.

https://en.wikipedia.org/wiki/Temperature#Kinetic_theory_of_gases

     Anyway, the particles of a species in the porridge are identical (in the model) but the porridge can have more than one species of particle.   
- - - - - - - - - -

The first bowl was <unknown>,  the second bowl was <unknown>  but the last was just <unknown>

   As it happens I don't think there is a good theoretical model for porridge.  As such Goldilocks probably can't determine the temperature of her porridge on the modern (post 2019) kelvin scale.   That temperature is simply "unknown" or "undetermined" at this time.
     ....In an ideal gas, and in other theoretically understood bodies, the Kelvin temperature is defined to be proportional to the average kinetic energy of non-interactively moving microscopic particles, which can be measured by suitable techniques....     [ https://en.wikipedia.org/wiki/Temperature#Theoretical_scales ]
      ... making the clear implication that if the porridge is not theoretically understood,  specifically that we don't have a good model for it right now, then  .... who knows,  best not give it a temperature in kelvin (kelvin post 2019)....
   
    It would be worth me saying this again:  The very latest entirely theoretical approach to defining temperature came in from May 2019 and Wikipedia articles on temperature are getting edited every week trying to bring them inline with these changes.   I'm not claiming to be an expert on the new approach.   

Best Wishes.

[hamdani yusuf (OP)]

I never suggested otherwise.

It seems you did.

It depends.
What would you actually do with the information if I told you?

Incidentally, you forgot to answer my questions.
Please do so.

Which one?

[hamdani yusuf (OP)]

That you are a little less than half way between the flat bits of the graph at 3 and 4.

What's the physical interpretation of that?
What makes it better than interpreting that rotational energy hasn't been fully activated?