21
$\begingroup$

enter image description here

The graph of rotational heat capacity above shows a small maximum before approaching the equipartition value. What is the origin/physical explanation of this maximum?

$\endgroup$
15
+400
$\begingroup$

Most heat capacities go through a maximum as the temperature increases. $C_V = \left( \frac{dU}{dT} \right)_V$ so a maximum in $C_V$ corresponds to a minimum in $\left( \frac{dT}{dU} \right)_V$, i.e. the point where the temperature changes very little as energy is being supplied to the system. At this point (most of) the energy is being used to excite molecules from the ground state to the excited state, rather than increasing the kinetic energy (i.e. the temperature) of the molecules.

I have made video about it here.

For a more quantitative explanation, consider a two-state system with a singly degenerate ground-state and a excited state with degeneracy $g$, that is $\varepsilon$ high in energy than the ground state. The heat capacity can be written as $$C_V = \frac{\varepsilon^2}{RT^2}p_0p_1 $$ where $p_0$ and $p_1$ is the probability of being in the ground and excited state, respectively $$ p_0 = \frac{1}{1+ge^{-\varepsilon/kT}}$$ $$ p_1 = \frac{ge^{-\varepsilon/kT}}{1+ge^{-\varepsilon/kT}}$$

Now some plots for $N_A \varepsilon$ = 2.5 kJ/mol and $g$ = 100

enter image description here

The first plot shows $p_0p_1$ (green), $C_V$ (red) and $\frac{\varepsilon^2}{RT^2}$ (blue, right $y$-axis). This shows that the general shape of $C_V$ as a function of temperature comes from $p_0p_1$

enter image description here

The second plot shows $p_0$ (red), $p_1$ (green), $p_0p_1$ (yellow) and $\frac{\varepsilon^2}{RT^2}$ (blue, right $y$-axis) - right $y$-axis. This shows that the shape of $p_0p_1$ come from the shape of $p_0$ and $p_1$, and that the peak occurs where the ground and exited states are equally likely ($p_0 = p_1$). The peak in $C_V$ occurs at slightly lower temperature due to the $\frac{\varepsilon^2}{RT^2}$ term, but ultimately the peak in $C_V$ derives from the peak in $p_0p_1$.

$\endgroup$
  • 1
    $\begingroup$ With a two state model, there is always a maximum. However, when Hicks (J. Amer. Chem. Soc. 48, 1520) tried to calculate Cv for HCl using degeneracies, 2, 4, 6... there was no maximum. Only when Hutchisson (J. Amer. Chem. Soc. 50, 1895) used the correct degeneracies 1, 3, 5... was a graph with a maximum obtained. Similarly, there is no maximum for orthohydrogen. So while the two state model always predicts a maximum, it really depends upon the actual degeneracies. $\endgroup$ – DavePhD Jun 9 '15 at 13:09
12
$\begingroup$

It is not a general truth that there is a maximum before approaching the equipartition value.

For example, for $\ce{H2}$ as it exists under ordinary conditions, there is no such maximum, because orthohydrogen predominates (in a 3:1 statistical ratio over parahydrogen) and orthohydrogen must always rotate.

enter image description here

enter image description here

So, as seen above, the existence of the greater-than-equipartition-value maximum is tied to the existence of a state where the molecule does not rotate. Parahydrogen has such a state, while orthohydrogen does not. In this sense, the maximum could be thought of as due to depopulation of the non-rotating state.

Mathematically, it is very difficult to express the form of the heat capacity curve over the full temperature range. As seen in other answers and page 9 here, in the limit of temperature approaching zero or infinity, it is not extraordinarily difficult to calculate, but at intermediate values it is.

Historically, a curve with the maximum was developed by Fritz Reiche in 1919 from Planks theory. See the presentation The Rotational Specific Heat of Hydrogen From Old Quantum Theory especially around slide 42 "The calculations become horrendous!". And of course he had to do them by hand, only to see that the result was an apparent failure, in the sense that it did not correctly reproduce the heat capacity curve for hydrogen. See also the 1919 "A new formula for the Temperature Variation of the Specific Heat of Hydrogen" Phys. Rev. vol. 11, pages 156-158. Only later when the significance of nuclear spin was understood was the problem fully explained.

$\endgroup$
5
$\begingroup$

Since now this question has a bounty of 400 points, I think it now deserves some well thought answer. As a part of that endeavor, I am adding another elaborate answer.
The overall contribution to $C_v=C_{v,\text{trans}}+C_{v,\text{rot}}+C_{v,\text{vib}}$ and translational and vibrational partition function don’t show that maxima and reaches equipartition value monotonically and overall this three contribution results in $C_v=3R$, which is an expected result.
To get the rotational partition function, we use rigid motor approximation. For a simple diatomic molecule, the moment of inertia is $I=\mu \times r_0^2$ , where ${\mu}$ is reduced mass and is calculated as $\mu=\frac{m_1 \times m_2}{m_1+m_2}$ and ${r_0}$ is equilibrium distance between two atoms. Now, from quantum mechanics we know that energy level of rotational partition function can be written as
$$\epsilon_l=l(l+1)\frac{\hbar^2}{2I}$$ and defining characteristic temperature as $\theta_r=\frac{\hbar^2}{2Ik}$, we can rewrite that equation as $$\epsilon_l=l(l+1)k\theta_r$$ Also from quantum mechanics, we know degeneracy of energy levels are ${g_l=(2l+1)}$
Now the partition function $$q_r=\sum g_l\operatorname{e}^{-\epsilon_l/kT}=\sum (2l+1)\operatorname{e}^{-l(l+1)\theta_r/T}$$ Now at high temperature closely spaced energy states can be considered as continuous variable. So, now we can replace the sum by integral and to make our life easier, we can perform a variable change here. Let, $x=l(l+1)$ so $\mathrm{d}x=(2l+1)$
So, $$q_r=\int_0^\infty\operatorname{e}^{-x\theta_r/T}\mathrm{d}x=\left[-(\theta_r/T)\operatorname{e}^{-x\theta_r/T}\right]_0^\infty = \theta_r/T$$ So, at high temperature: $$E_r=NkT^2 \frac{\delta \ln q_r}{\delta T}=NkT$$ So, $$C_v=\frac{\delta E_r}{\delta T}=Nk$$ Now at low temperature: $$q_r=\sum\limits_{l=0}^\infty(2l+1)\operatorname{e}^{-l(l+1)\theta_r/T}$$ So, $$q_r=1+3\operatorname{e}^{-2\theta_r/T}+5\operatorname{e}^{-6\theta_r/T}+\ldots$$
So, $$\ln q_r=\ln\left(1+3\operatorname{e}^{-2\theta_r/T}\right)$$ Neglecting higher order terms.
Now, $\ln(1+x)=x-0.5x^2+\ldots\approx x$
Here, at low temperature, the second term of $\ln(1+3\operatorname{e}^{-2\theta_r/T})$ is small, so $$\ln q_r=3\operatorname{e}^{-2\theta_r/T}$$ So, $$E_r=NkT^2(\frac{\delta \ln q_r}{\delta T})=6Nk\theta_r\operatorname{e}^{-2\theta_r/T}$$ So, $$C_v=\frac{\delta E_r}{\delta T}=3Nk(\frac{2\theta_r}{T})^2\operatorname{e}^{-2\theta_r/T}$$ So, like my previous answer the explanation is the same from mathematical point of view: It’s a trade off between square term and exponential term. Moreover I tried to relate the degeneracy of level with heat capacity.
Probability of finding a molecule in $l$th state
$$P_l=(2l+1)\operatorname{e}^{-l(l+1)\theta_r/T}/q_r$$ So, that means the degeneracy of the highest probable energy level is the maxima of $P_l$ with respect to $l$. So, if we solve the equation $\frac{\delta P_l}{\delta l}=0$ for $l$, we will find that $$l_{\text{max}}=(T/2\theta_r)^{1/2}-1/2\approx (T/2\theta_r)^{1/2}$$ So, that means at low temperature the degeneracy of highest probable energy level is increasing with increasing temperature (since at high temperature all energy levels are equally probable, that’s why it doesn’t matter at high temperature). So, that could have a effect on the steepness of the heat capacity curve.

I am still trying to find the physical reason for this behavior. This is what i got if i plot my equation Scale is arbitrary
Internal energy is increasing with temp. but the rate of internal energy change is fast at the beginning but after a certain temp. it is being slowed down.
And that's the reason for that maxima. Now, heat capacity is actually the fluctuation of energy or it can be written as
$C_v=<E^2>-<E>^2/kT^2$
At high temp. since all energy levels are almost equally probable and their numerical value difference is less, that's why the apparent change or fluctuation is almost constant. But at low temp. we have a certain energy level with high occupancy, that's why the change in fluctuation is very noticeable.
Now if possible we could do a Monte Carlo simulation to see this behavior but that would be time consuming. Still i am thinking about it and hoping that we would be able to find a satisfactory answer from physical perspective.

$\endgroup$
  • 1
    $\begingroup$ Rather than adding a new answer, consider expanding your first answer. I can hardly follow this answer, as it contains a lot of formulas with not well defined quantities. Please also consider using a display maths environment. Please have a look here and here. $\endgroup$ – Martin - マーチン Jun 4 '15 at 6:28
  • $\begingroup$ @mamun a few questions: in the last equation, if P is probability, and probability is less than 1, how can 1/2 be approximated as zero? Is the answer for homonuclear or heteronuclear? Is it valid to base the answer on assuming low temperature? solar.njit.edu/~leej/lecture/ph641/ch06_summary.pdf $\endgroup$ – DavePhD Jun 4 '15 at 14:41
  • $\begingroup$ Sorry, my mistake. It's not probability, it's the degeneracy of the highest probable energy level or $l_{max}$. This is for heteronuclear case. For homonuclear case, there is a symmetry requirement of the total wave function. $\endgroup$ – mamun Jun 4 '15 at 15:58
  • $\begingroup$ @mamun ok, but aren't you simultaneously assuming T is small ("now at low temperature") and large (in the last line, where $T/2\theta$ is approximated as large compare to 1/2)? $\endgroup$ – DavePhD Jun 4 '15 at 16:14
  • $\begingroup$ Your point is right. For realistic case low temp. limit is ${T<\theta _r}$, So, $(T/\Theta_r)^0.5$ should be on the same order of 0.5, and hence it can't be neglected. But my point was to prove that ${l_{max}}$ is proportional to ${T^{0.5}}$ or ${l_{max}}$ is increasing with temperature. $\endgroup$ – mamun Jun 4 '15 at 17:04
4
$\begingroup$

OK, i tried to refresh my stat mech class materials. If you derive the formula for ${C_v}$, then you will find that at high temp. ${C_v=Nk=R}$ and at low temp. $C_v=3Nk( \frac{2\theta_r}{T})^2 \exp(\frac{-2\theta_r}{T})$. So, At low temp. limit there is a trade off between the square term and exponential term and results in the above mentioned maxima. This is the explanation from mathematical POV but from physical POV, i think the answer lies in (I am not quite sure) the separation of rotational spectra and degeneracy number.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.