# Entropy of a perfect gas at 0 kelvin

## Main Question or Discussion Point

Hello,

I'm still trying getting familiar with the concepts of statistical mechanics and thermodynamics , and there's this equation for S of an ideal system, which somehow perplexes me. Suppose the ideal gas is composed of N indistinguishable atoms , then :

$$S = \frac{U - U_{0} }{T} + k Ln \frac{q^N}{N!}$$

At T = 0 this expression becomes :

$$S = \frac{3}{2} Nk + k Ln \frac{{g_0}^N}{N!}$$

where g0 is level of degeneracy of the ground state.

Is there anything wrong with this ? If not , why doesn't the entropy equal to zero at T = 0 kelvin ? Doesn't that contradict the 3rd law of thermodynamics ?

S = kLnW = 0 because at T = 0 , W = 1 .

What am I doing wrong ?

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You aren't doing anything wrong And it certainly doesn't contradict the third law of thermodynamics- it IS the third law of thermodynamics!
You might be getting confused with the second law, which states that the entropy of any isoloated system increases? It doesn't contradict this because whatever is being cooled down can't be considered as "isolated", so you have to consider wherever the energy of the object being cooled is going to as part of the system, and the total entropy of that system increases.
Also, you can't actually attain absolute zero!

Hello muppet , thanks for replying.

Honestly I do not fully understand the 3rd law , so yes I might be confusing this. However, as I understand, at T = 0 all the atoms are at the ground state , so W = 1. Then S = 0 at T = 0. This should be true for any system , correct ? Why then is it not true for the ideal gas if I use the equation above.

Concerning absolute zero, why can this temperature not be attained ? I have been sudying statistical mechanics on my own and lately have concentrated on a simple 2-level system. I plotted S , U and q as a function of T . For U and q , there seems to be a discontinuity at T = 0. However , the functions can be plotted for negative temperatures in kelvin. Why then can we not go below 0 kelvin ?

One last thing for anyone who can help , why do write ( according to the equipartition principle ) that U = U0 + C RT ( C depends on the type of molecule ).

Why is the U0 included ? The book I use says that at T = 0 , all thermal motion ceases the and only contribution to the internal energy is the internal energy of the molecules. Does that mean it's the energy of bonding in the molecules ? i.e. it's related to quantum mechanics ? If so , then would atoms have U0 = 0 ?

Thanks,

Sorry Andrew, I didn't read your original post sufficiently carefully. I'll try and give a more intelligent reply tomorrow when it's not 5am local time. If it isn't going to zero at t=0 then yes it is contradicting the third law of TD. Where did you get your equations from?
Btw, you're a brave man for teaching yourself statistical mechanics. Most expend the minimum of effort under suffrance olgranpappy
Homework Helper
Hello,

I'm still trying getting familiar with the concepts of statistical mechanics and thermodynamics , and there's this equation for S of an ideal system, which somehow perplexes me. Suppose the ideal gas is composed of N indistinguishable atoms , then :

$$S = \frac{U - U_{0} }{T} + k Ln \frac{q^N}{N!}$$

At T = 0 this expression becomes :

$$S = \frac{3}{2} Nk + k Ln \frac{{g_0}^N}{N!}$$

where g0 is level of degeneracy of the ground state.

Is there anything wrong with this ? If not , why doesn't the entropy equal to zero at T = 0 kelvin ? Doesn't that contradict the 3rd law of thermodynamics ?
Yes. Your equation for S must not be valid at very low temperature.

Sorry Andrew, I didn't read your original post sufficiently carefully. I'll try and give a more intelligent reply tomorrow when it's not 5am local time. If it isn't going to zero at t=0 then yes it is contradicting the third law of TD. Where did you get your equations from?
Btw, you're a brave man for teaching yourself statistical mechanics. Most expend the minimum of effort under suffrance Thanks for the encouragement. I got my equations directly from the book. I'll write the the derivation below but for a system composed of distinguishable atoms. There's no real difference except the N! added to the partition function in the latter case , which should not make any difference at T = 0.

olgranpappy said:
Yes. Your equation for S must not be valid at very low temperature.
Hello olgranpappy,

Indeed there must be something wrong with the equation in the limit of T = 0 , but what could be the reason ? The book gives the derivation but mentions nothing about absolute temperatures. I am very confused about this.

From the boltzman formula and then using stirling's approximation :

$$S = k Ln \left ( \frac{N!}{n_1!n_2!...} \right ) = k \left ( NLnN - N - \sum_{i=1}^N n_i Ln ( n_i ) + \sum_{i=1}^N n_i \right )$$

The system is closed so :

$$\sum_{i=1}^N n_i = N$$

$$S = k \left ( NLnN - \sum_{i=1}^N n_i Ln ( n_i ) \right ) = k \sum_{i=1}^N \left ( n_i Ln \frac{N}{n_i} \right )$$

From the boltzman distribution :

$$q n_i = N e^{ -E_i / kT }$$

$$S = Nk Lnq + \frac{\sum_{i=1}^N n_i E_i}{T}$$

The internal energy of an ideal gas is :

$$U = U_0 + \sum_{i=1}^N n_i E_i$$

Then the equation obtained is :

$$S = k Lnq^N + \frac{U - U_0}{T}$$

There many other ways to derive this equation and it is probably covered in all statistical mechanics books, the only limitation is that the particles are not interacting which is why the internal energy of the system is the direct summation of the energies of the particles, and ofcourse the number of particles should be high.

Well... on a line by line basis the part I can't follow is where
$$S = Nk Lnq + \frac{\sum_{i=1}^N n_i E_i}{T}$$
comes from, but I don't think that's our problem. The reason I suspect that it's not valid at low temperatures is that Stirling's theorem here doesn't just require that the number of particles is large, but that the number of arrangements of particles is large. As you get closer to absolute zero in a system of particles with negligible energy of interaction, the number of accessible energy states will decrease, so the term
$$\frac{N!}{n_1!n_2!...}$$
gets smaller as each factorial term in the denominator will get larger. That's my best guess!

atyy
I believe the problem is that S=kBln(w) is the formula for the entropy in the microcanonical ensemble. In this ensemble, the energy is fixed, and w=(number of microstates consistent with that energy). So you cannot use as input the Boltzmann formula which relates probabilities for states of different energy (but you can derive it from this approach). For the ideal gas in the microcanonical ensemble, the microstates are usually labelled using the position and momentum of each particle.

Try Eric Poisson's Chapter 2 of his stat mech notes:
http://www.physics.uoguelph.ca/~poisson/research/notes.html

Or Kardar's lecture 13:
http://ocw.mit.edu/OcwWeb/Physics/8-333Fall-2005/CourseHome/index.htm [Broken]

BTW, one thing to be careful of in statistical mechanics is that there are all sorts of different ensembles which are not conceptually equivalent and give different answers - but for most purposes, the answers from different ensembles is (magically) close enough to live with.

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atyy
I have just realised that although what I wrote above seems to be technically right, a more important point is that the classical ideal gas does not obey the third law, which from the microcanonical viewpoint, is the assumption that the system has a unique ground state.

http://hyperphysics.phy-astr.gsu.edu/Hbase/therm/entropgas.html

Kardar's lecture 4:
http://ocw.mit.edu/OcwWeb/Physics/8-...Home/index.htm [Broken]

Poisson's Statistical Mechanics notes Eq 2.9:
http://www.physics.uoguelph.ca/~poisson/research/notes.html

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You made a mistake with the N! factor somewhere. At T = 0, U tends to U0 faster than T goes to zero, because the heat capacity tends to zero. To see this, you must also quantize the translational degrees of motion, the formula U = 3/2 k T is not valid at low temperatures, even if the gas is ideal. At T = 0, the entropy is given by

$$S = N k \log(g_{0})$$

Which is what you would expect.

atyy thanks a lot for the links , lots of useful stuff there ... I'm still reading them.
However, I do not follow when you say that the ideal gas does not obey the 3rd law. Isn't the 3rd law fundamental ? That is , all systems - real or deal - should obey that law no matter what. And what do you mean by a "unique ground state " ? Does that mean go = 1 ? If that's the case then the equation might be correct. Anyway I will keep reading the interesting lecture notes you've provided and reply later. It might take me some time though.

Count Iblis , the N! is absent because I assumed the particles to be distinguishable. Anyway your idea is interesting, but then the formula obtained doesnt signify the entropy is zero unless go =1 , is this true for an ideal gas? I fail to see why.

Btw Concerning Uo , I've been trying to understand the quantum mechanical treatment of a particle in a box model ( 1-D ) . The minimum energy that the particle has in not zero , infact it can never be zero in contrast to the classical treatment ( KMT ) which says that at T = 0 K the energy of the particle is zero. Is this why the term Uo is added to U? i.e. it's a correction factor ? Can I say then that for a perfect gas Uo = h2 / 8mL2

atyy
However, I do not follow when you say that the ideal gas does not obey the 3rd law. Isn't the 3rd law fundamental ? That is , all systems - real or deal - should obey that law no matter what. And what do you mean by a "unique ground state " ? Does that mean go = 1 ?
Yes, unique ground state means go=1.

Classical systems do not obey the third law of thermodynamics. As long as it is valid to describe a system classically, only the first two laws apply.

The third law is also thought not to hold for some phases of matter such as "glass"

At very low temperatures, ideal classical gases do not exist experimentally. Instead, there are ideal quantum gases of identical particles which obey either Fermi or Bose-Einstein statistics.

Actually, there is no "pure" classical statistical mechanics, because classical "identical particles" can be distinguished by their different positions at all times. Quantum mechanics has truly identical particles since each particle no longer has a well-defined position at all times - this justifies the N! for identical particles to avoid the Gibbs mixing paradox.