Why is 0 K unattainable?

[Kakashi]

At constant pressure we have:

$$ dq_{p}=dH=C_{p} dT $$ which implies

The enthalpy change dH reflects the increase in internal energy associated with molecular motions such as rotational and vibrational modes. The atoms, molecules or ions that compose a system can undergo several types of motion including translation, rotation, and vibration. The greater the motion (the higher the temperature), the greater the number of possible microstates and therefore the higher the entropy. A perfectly ordered system with only a single microstate would have 0 entropy. The only system that meets this criteria is a perfect crystal at 0 K in which each component is fixed in place within a crystal lattice and exhibits no motion.

Near absolute zero, the heat capacity approaches zero:

$$ Cp \to 0 $$ as $$ T→0 $$

$$ dT = \frac{dq_{p}}{C_{p}} $$

Because it is impossible to perfectly isolate a system from its surroundings (dq_{p} =/0) (for example due to radiation from the environment), there will always be some small heat transfer. Therefore it seems impossible to keep a system exactly at T=0. Is this reasoning correct for explaining why absolute zero is unattainable?

 

[sbrothy]

I'd guess Heisenberg's uncertainty principle would disallow it? At least in theory. As for in practice...?

 

[sbrothy]

Proof of the Nernst heat theorem

The Nernst heat theorem is probed from purely thermodynamic arguments connected with the second law of thermodynamics, and alien to the vanishing of the specific heats, or to the unattainability of the zeroth isotherm. If the proof is accepted the second law of thermodynamics would extend its applicability and the third postulate of thermodynamics would be narrowed to the fact that the entropy of a finite-density, chemically homogeneous body must not be negative.

EDIT: I may have over-estimated the amount of time this question sat unanswered. I'm glad some real competence is getting involved. Looking forward to more....

 

[.Scott]

I was hoping one of the PF QM experts would catch this one. But let me take a whack at it.

The actual refrigeration techniques are described in this wiki article on laser cooling.
Of course, once refrigerated, the atoms will tend to warm up for reasons explained in the OP.
Regarding QM effects, you can have a population of atoms at the lowest possible quantum state.
This is described in the Bose-Einstein condensate wiki article.

Your entire sample will not be at 0K, but an arbitrarily high portion of those atoms can be.
(See correction here.)

 

[PeterDonis]

Your entire sample will not be at 0K, but an arbitrarily high portion of those atoms can be.

This doesn't make sense; temperature is not a property of individual atoms. It's a property of the system. For example, see here (the NIST site is relevant because they set the SI standard definition of the Kelvin, which requires assessing the various techniques for measuring temperature and cooling things towards 0 K):

https://www.nist.gov/si-redefinition/kelvin/kelvin-thermodynamic-temperature

 

[PeterDonis]

I'd guess Heisenberg's uncertainty principle would disallow it?

I've seen that heuristic argument (though not in actual peer-reviewed literature), but I don' think it's correct. The uncertainty principle does not disallow a quantum system from being in its ground state. And since the ground state is an eigenstate of a Hermitian operator (it's the lowest energy eigenstate of the Hamiltonian), the uncertainty principle does not disallow an exact measurement of a system being in its ground state.

 

[PeterDonis]

Because it is impossible to perfectly isolate a system from its surroundings (dq_{p} =/0) (for example due to radiation from the environment), there will always be some small heat transfer. Therefore it seems impossible to keep a system exactly at T=0. Is this reasoning correct for explaining why absolute zero is unattainable?

It looks reasonable to me, but I'm not sure if this argument appears in the form you give it in the literature. For example, it doesn't look to me like this argument appears in the paper on the Nernst Theorem that was referenced earlier in the thread.

 

[PeterDonis]

Your entire sample will not be at 0K, but an arbitrarily high portion of those atoms can be.

Per my previous comment on this, I don't think it makes sense to describe this in terms of temperature.

I think a fair description in quantum terms would be that not all atoms in the BEC will be in the actual ground state; there will still be some distribution of states. (Perhaps a more accurate description would be that the quantum state of the condensate as a whole is a weighted superposition of multiple energy eigenstates.) But that very fact means that the condensate, as a thermodynamic system, is not at 0 K; it's at some finite positive temperature, because its average energy per atom is higher than the exact ground state energy.