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Does it make sense that entropy goes to infinity as temperature goes to infinity? For an ideal monoatomic 1D gas I have an expression for entropy \begin{equation} S=Nk_{B}Ln(\frac{Ve^{\frac{5}{2}}\sqrt{2\pi m k_{B}T}}{N h}) \end{equation} Where $N$ is the number of particles, $h$ the Planck's constant. $V$ the volume of the box, $e$ Euler's number, $m$ the mass of each particle. This expression tends to infinity when temperature goes to infinity... why does this have sense? I was used to finding asymptotic behaviours for entropy as a function of temperature, so this result is confusing me

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    $\begingroup$ Is there a particular reason you would expect the entropy to be bounded as $T\rightarrow \infty$? $\endgroup$ – J. Murray Jun 13 at 21:27
  • $\begingroup$ Well, mainly that I don't understand what would mean physically an infinite entropy, and scondly, that the exercises I have been doing have always bounded entropies $\endgroup$ – Juan Pablo Arcila Jun 13 at 21:32
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You may get answers based on statistical mechanics that are more helpful in resolving your confusion (higher temperature means a steadily increasing number of accessible states), but let me just mention a couple of relations from classical thermodynamics.

Firstly, for the ideal gas, the heat capacity at constant volume $C_V$ is just a constant ($3R/2$ for a monatomic gas in 3D, for instance). It follows, if you integrate with respect to temperature at constant volume, that the internal energy $U$ is proportional to temperature. So, if $T\rightarrow\infty$, then $U\rightarrow\infty$. Is this something to worry about? I don't see why.

But also, it is an identity that $$ C_V = T \left( \frac{\partial S}{\partial T}\right)_V $$ So to get the entropy difference of an ideal gas for two different temperatures, at the same volume, just integrate $C_V/T$ with respect to $T$. Since $C_V$ is constant, this just gives $C_V(\ln T_2 - \ln T_1)$. This seems to be what you are concerned about. Some people worry about the limit $T\rightarrow 0$, and the answer (given elsewhere on Physics SE) is that the ideal gas model is simply inapplicable in this limit. At extremely high temperatures, one will start to see the model become inapplicable as well (e.g. relativistic effects). But over its range of applicability, which is quite wide for low density gases, there is nothing intrinsically unphysical about the logarithmic increase in entropy with temperature.

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Does it make sense that entropy goes to infinity as temperature goes to infinity?

At constant volume it makes sense for the entropy to increase with temperature since, at constant volume: $$ dS = \frac{dU}{T} $$ and $T$ is positive and we expect $U$ to increase with increasing $T$. However, this does not mean the entropy goes to infinity.

The exact details of how the entropy increases and whether it increases to infinity or just increases asymptotically to some limit will depends on the details of the system under study.

However, for many "normal" systems one might expect the entropy to increase to infinity. Roughly, one could argue that this is true because the Hamiltonian is bounded below but not above for "most" systems. That is, we expect there is some lowest energy "ground state" but no "highest energy state." Because there is no highest energy state, there is no limit to how many more states are readily accessible as the energy increases and no necessary limit to the entropy.

To put this into more concrete terms, what we really want to know is how the entropy changes with temperature at constant volume... but this can be written in terms of how the energy changes at constant volume $C_V$ as: $$ \left(\frac{\partial S}{\partial T}\right)_V = \frac{1}{T}\left(\frac{\partial U}{\partial T}\right)_V = \frac{C_V}{T}\;. $$

In general, $C_V$ can depend on temperature, but for some systems (like the "ideal gas") $C_V$ is independent of temperature. In the "ideal gas" case, the entropy increases as the log of the temperature (as shown in the other answer).

So, it does seem to make sense that the entropy can go to infinity as the temperature goes to infinity, but it is not necessarily the case for all systems.

In fact, we know of simple examples where the entropy does not go to infinity at infinite temperature. One example of this is the so-called "two-state" system (See: http://home.thep.lu.se/~larsg/Site/SM2.pdf).

For the "two-state" system the entropy continues to increase as the temperature approaches infinity, but it does not increase to inifinity. Rather, the entropy approaches approaches a maximum entropy of $S=Nk\ln(2)$ asymptotically. But, as indicated at in the above discussion, this is due to the single particle Hamiltonian being bounded both above and below for this example.

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  • $\begingroup$ In general, $C_V$ depends on any two of the three state parameters ($p, T, V$). In ideal fluids (gases or perhaps even liquids), $C_V$ depends only on temperature. Finally, the assumption that $C_V$ is constant must be stated explicitly for the system. It is not a characteristic of something called an "ideal" system as this explanation implies. $\endgroup$ – Jeffrey J Weimer Jun 14 at 0:09
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It's because kinetic energy, unlike most other forms of energy, is unbounded from above. It can increase without limit. This implies that the heat capacity need not fall to zero as the temperature goes up, and this is confirmed when we look at phase space and the available momentum states. There are more and more states available at any given energy as the energy increases.

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