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What would the curve that describes the change of internal energy ($U$) with the volume ($V$) be like if the change in the temperature is negative?

My try:

For a ideal monoatomic gas, we have $U=(3/2)NkT$, where $N$=number of particles and $k$=boltzmann constant. So $$U=(3/2)pV$$

M question is: If the pressure changes, how is it possible to describe $U$ as a function of $V$?

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    $\begingroup$ Gases will never have negative temperature. Try reading about spin systems $\endgroup$ – Hydro Guy Jan 7 '15 at 16:43
  • $\begingroup$ well so I think that is the change of U with T is negative...I think! $\endgroup$ – pipita Jan 7 '15 at 16:57
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Negative Temperature can arise only in systems which have strict upper limits on the highest energy state that the system's constituent particles can be in. What a negative temperature is telling you is that they only way to take on more energy is to force more and more constituent particles in their highest energy state. That is, the probability of any particle's being in its highest energy state approaches unity, so the number of possible arrangements of the system (system microcomplexions) decreases as the system "tries more and more in vain" to increase its internal energy by making it more and more certain of finding a particle in the highest energy state.

Ideal gasses are not this kind of system: the energy of possible states is unbounded. So the temperature cannot be negative.

As an exploratory exercise, try comparing a system of thermalised quantum harmonic oscillators, whose energy states are unbounded in energy, with a system of two state systems.

For the ensemble of quantum harmonic oscillators, the mean oscillator energy is:

$$\left<E\right> = \frac{\hbar\,\omega}{2}\,\coth\left(\frac{1}{2}\,\beta\,\hbar\,\omega \right)$$

The Shannon entropy (per oscillator) is then:

$$S = -\sum\limits_{n = 0}^\infty p(n) \log p(n) = \frac{\beta\,\hbar\,\omega\,e^{\beta \,\hbar\,\omega}}{e^{\beta\,\hbar\,\omega}-1} - \log \left(e^{\beta\,\hbar\,\omega}-1\right)$$

so the thermodynamic temperature is then given by (noting that the only way we change this system's energy is by varying $\beta$):

$$T^{-1} = \partial_{\left<E\right>} S = \frac{\mathrm{d}_\beta S}{\mathrm{d}_\beta \left<E\right>} = \beta$$

Mull over and play with these equations a bit and you see that $\beta$ is positive and can be any positive real. So the temperature is unbounded: by heating the system up, you're letting the particles access higher and higher energy states and there is no limit to this process (at least from the above theory alone: pace, for example, the Bekenstein bound). Informally, the energy states range over unboundedly bigger "alphabets" as heat is added, so that the bigger alphabets encode more and more information and there is no limit to this process.

Now let's do the ensemble of two state particles: let the energy eigenstates be of energy $E_0$ (ground state) and $E_1$ (raised state), with a probability $p$ that a given particle is in the raised state. The mean energy per particle is then:

$$\langle E\rangle = (1-p)\,E_0 + p\, E_1$$

which can be re-arranged to:

$$p = \frac{\langle E\rangle -E_0}{E_1-E_0}$$

and the Shannon entropy per particle:

$$\begin{array}{lcl}S&=&-p\,\log p-(1-p)\,\log(1-p)\\&=&-\frac{\langle E\rangle -E_0}{E_1-E_0}\,\log\left(\frac{\langle E\rangle -E_0}{E_1-E_0}\right)-\frac{E_1-\langle E\rangle}{E_1-E_0}\,\log\left(\frac{E_1-\langle E\rangle}{E_1-E_0}\right)\end{array}$$

which you can plot to see that the entropy is maximal when $\langle E\rangle=\frac{1}{2}(E_1+E_0)$ and decreases with internal energy when $\langle E\rangle>\frac{1}{2}(E_1+E_0)$. The temperature given by $T^{-1} = \partial_{\left<E\right>} S$ is large and negative when $\langle E\rangle$ is just whisker above the mean $\frac{1}{2}(E_1+E_0)$, asymptoting to nought through negative values as $\left<E\right>$ approaches $E_1$, as there is less and less scope to add energy to the system.

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  • $\begingroup$ Hi! The question is really "how is the curve of U(internal energy) with volume if the temperature is negative"? i cnsidered T=(partial U/partial S) at constant V and N (number of particules, but now i cant say the change of U with V if T<0...can yoou hep? $\endgroup$ – pipita Jan 9 '15 at 13:58
  • $\begingroup$ @pipita Yes, sorry, I was a bit hasty - and I also answered before one of your edits. I'm going to delete this answer when I have saved it away and hopefully give you a real answer, but I'm a little busy right now. BTW nice user name: reminds me of a childhood friend (Philipa) who called herself "Pippa" and also of a friend's dog Pip who is a rogue but he's highly smart and a great deal of fun just the same! $\endgroup$ – WetSavannaAnimal Jan 12 '15 at 0:52

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