# I Entropy and temperature question

1. Jan 31, 2017

### Lacplesis

Hi folks, I have a question, I will first write down some old truths and then ask what is unclear to me.
Now we know that Boltzmann's constant is the average kinetic energy of one molecule of ideal gas, related to its temperature T.
The German scientist Clausius defined entropy change of some substance as the amount of energy dispersed reversibly at a specific temperature T:
dS=dQ/T

What is weird to me is that according to this formula and the definitions of entropy in thermodynamics it turns out that small changes in temperature at or near absolute zero correspond to large or infinite change in entropy , in other words as we get closer to T zero the entropy skyrockets, but this doesn't sound logical.
Also ideal gas has greater heat capacity at higher temperatures than at absolute zero as does most other liquids , gasses and even solids , so where is the misunderstanding here? Or maybe it's the problem with what we define absolute zero - either a true zero temperature or simply the lowest possible energy/temperature state?

I hope you can understand my question thank you.

2. Jan 31, 2017

### jbriggs444

As we get closer to T zero, the incremental change in entropy per incremental change in energy skyrockets. Solve $dS=\frac{dQ}{T}$ for T and see what that implies for small T.

3. Jan 31, 2017

### Staff: Mentor

You have to be a little careful here. This is at a fixed temperature, so it doesn't say anything about changes in temperature. It is about changes in entropy and energy at a fixed temperature.

If T is small then you have to decrease the entropy a lot to remove even a small amount of heat. Conversely, even a small amount of additional heat will dramatically increase the entropy. Since nature wants to increase entropy, you get energy flowing from hot to cold since the increase in the entropy of the cold reservoir is more than the decrease in the entropy of the hot reservoir

4. Feb 1, 2017

### Lacplesis

Ok I understand that a close zero temp as compared to even ordinary room temp is a huge difference and if such two systems would somehow interact thermodynamically then the entropy increase in the cold system would by far steeper and bigger than the entropy decrease in the room temp. system.So in this example it seems logical to think that a very "cold" system's entropy changes are huge, but isn't then also the change affected by the other temp. of the other system , if we take the universe to be a closed system and having multiple systems within it like stars and galaxies then as the ultimate heat death comes some time along in the future and the universe turns into a big near zero temp region , and say somehow an external system interacts with such a universe and the temp of that system is only a few degrees higher then the entropy change would be very small for the same zero temp system that had a previously much bigger entropy change due to the system that interacted with it being much higher temp correct?

So all in all would it be correct to say that the rise and fall line angle of entropy at or near zero temp is tied to the heat capacity/temp of the other system or whatever is interacting with the at or near zero system ?

Or another example if i have a liquid at 0K and a small piece of copper at room temp. I throw the copper piece into the liquid, now due to the much smaller heat capacity due to the size of the copper , the copper would now be the one undergoing the biggest change in entropy in this case it's entropy would decrease , now would this decrease be comparable with the steepness of an increase line for the small temp changes at or near 0K for entropy ?
Also in this case since the copper was very small and the 0K liquid large with much bigger capacity , what is the entropy increase for the liquid is it still high no matter with what kind of system it interacts , it only matters that it's T is raised a few degrees ?

I think i'm having problems understanding why entropy for the same gas or liquid for example can't be linear all the way through the temperature starting from 0K up to infinity where entropy is lowest at 0K and then gradually increases , if the entropy increases dramatically at or near 0K then does the same rapid change happens to molecules and the movement of atoms too? But that can't be true because then it would mean that while adding just a few degrees we would have added alot more energy since the movement of atoms in an ideal gas for example is directly proportional to it's energy/temperature, isn't that so ?

5. Feb 1, 2017

### Khashishi

0K is a idealization. It is a limit case that can't be reached. That's because temperature is defined as a differential:
$\frac{1}{T} \equiv \frac{dS}{dE}$
But, in quantum mechanics, S and E are quantized, so the differential only can be calculated if you make some coarse grained approximation.
It is easier to work and think in terms of "inverse temperature" $1/T$ than temperature. 0K is equivalent to a division by zero of inverse temperature, so that's another reason it cannot be reached.

There's no reason to expect temperature to scale linearly with entropy (aka inverse temperature scales inversely with entropy). It generally doesn't do so. You need to calculate the entropy at a given energy, and then take the derivative. The entropy is $S = k \ln \Omega$, so it scales logarithmically with the number of states. The number of states increases as you increase the energy, since you have more ways of allocating energy among the available degrees of freedom. For example, for a single atom of monoatomic gas, you have three directions to move in, so you can allocate one unit of energy in three ways, or n units of energy in (n choose 3) ways. If you have many atoms in the gas, there are many more ways to allocate the energy, since you can divvy up the energy between various particles (but you need to consider indistinguishability). Anyways, you have logarithm of combinatoric functions. It's unlikely to get a linear function out of a derivative of logarithm of combinatoric functions.