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Question

I calculated the classical heat capacity of a diatomic gas as $C_V = (9/2)Nk_B$, however the accepted value is $C_V = (7/2)Nk_B$.

I assumed the classical Hamiltonian of two identical atoms bound together as $$ H = \dfrac{1}{2m}( |\bar{p}_2|^2 + |\bar{p}_2|^2)+ \dfrac{\alpha}{2} |\bar{q}_1-\bar{q}_2|^2. $$ I calculated the partition function of $N$ particles as $$ Z = \left( \iiint_{-\infty}^{\infty} \iiint_{-\infty}^{\infty} \iiint_{-\infty}^{\infty} \iiint_{-\infty}^{\infty} e^{-\beta H} ~d^3q_1~d^3p_1~d^3q_2~d^3p_2 \right)^N \propto V^N T^{(9/2)N}. $$ I calcuated the heat capacity as $$ C_V = \dfrac{\partial }{\partial T} \left( k_B T^2 \dfrac{\partial \ln(Z)}{\partial T} \right) = \dfrac{9}{2}k_BN. $$

Why does the classical argument fail?

Classical Derivation

The partition function is \begin{align} Z &=& \left( \frac{1}{h^6} \int \mathrm{e}^{- \beta H(\bar{q}_1,\bar{q}_2,\bar{p}_1,\bar{p}_2)} ~d^{3}q_1 ~d^{3}q_2 ~d^{3}p_1 ~d^{3}p_2 \right)^N \\&=& \left( \frac{1}{h^6} \int \mathrm{e}^{- \beta ((|\bar{p}_1|^2+|\bar{p}_2|^2)/(2m)+\alpha |\bar{q}_1-\bar{q}_2|^2/2)} ~d^{3}q_1 ~d^{3}q_2 ~d^{3}p_1 ~d^{3}p_2 \right)^N \end{align} A useful gaussian integral \begin{align} \int_{-\infty}^{\infty} e^{-\gamma (x-x_0)^2}dx = \sqrt{\dfrac{\pi}{\gamma}} \end{align} The partition function can be evaluated using separated integrals \begin{align} \iiint_{-\infty}^{\infty} \mathrm{e}^{- \beta |\bar{p}_1|^2} ~d^{3}p_1 = \iiint_{-\infty}^{\infty} \mathrm{e}^{- \beta |\bar{p}_2|^2} ~d^{3}p_2 = \left(\sqrt{\dfrac{\pi}{\beta}}\right)^3 \end{align} and \begin{align} \iiint_{-\infty}^{\infty} \iiint_{-\infty}^{\infty} \mathrm{e}^{- \beta \alpha |\bar{q}_1-\bar{q}_2|^2/2 } ~d^{3}q_1 ~d^{3}q_2 = \left( \sqrt{\dfrac{\pi}{\beta \alpha/2}} \right)^3 \iiint_{-\infty}^{\infty} ~d^{3}q_1 = \left( \sqrt{\dfrac{\pi}{\beta \alpha/2}} \right)^3 V \end{align} The last set of integrals are improper integrals. One has to take the limit as the space approaches infinite containment. In that limit, integrating one set of variables $d^3q_2$ approaches the limit of a finite Gaussian term, while the other $d^3q_1$ approaches the diverging value of the total volume of the gas.

The partition function is \begin{align} Z &=& \left( h^{-6} \left(\sqrt{\dfrac{\pi}{\beta}}\right)^3 \left(\sqrt{\dfrac{\pi}{\beta}}\right)^3 \left( \sqrt{\dfrac{\pi}{\beta \alpha/2}} \right)^3 V \right)^N \\&=& \left( h^{-6} \left(k_B T \pi\right)^{9/2} \left( \dfrac{2}{\alpha} \right)^{3/2} V \right)^N \\&=& \left( h^{-6} \left(k_B \pi\right)^{9/2} \left( \dfrac{2}{\alpha} \right)^{3/2} \right)^N V^N T^{9N/2} \end{align}

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    $\begingroup$ You're probably going to have to show us more details of your calculation. It is known that the classical approximation should fail at low temperatures (when the spacing of the quantized energy levels becomes comparable to the thermal energy scale). Moreover, "low temperature" can include room temperature for many diatomic gases. However, the classical value (ignoring the quantum effects) is $C_v = (7/2) k N$, not $(9/2) k N$, so you've probably also made a mistake somewhere. $\endgroup$ – Michael Seifert Jul 14 '15 at 20:49
  • $\begingroup$ @MichaelSeifert Sorry I made a typo. 5/2 was supposed to be 7/2. $\endgroup$ – linuxfreebird Jul 14 '15 at 21:08
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    $\begingroup$ Fair enough, but 9/2 is still incorrect for the integral above. I suspect you'll have to add more details to your derivation before your real question above can be addressed. $\endgroup$ – Michael Seifert Jul 15 '15 at 2:20
  • $\begingroup$ @MichaelSeifert I uploaded the classical formulation. $\endgroup$ – linuxfreebird Jul 15 '15 at 7:57
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The potential energy for a diatomic molecule is not $$ U(\vec{q}_1, \vec{q}_2) = \frac{\alpha}{2} |\vec{q}_1 - \vec{q}_2|^2 $$ but is instead $$ U(\vec{q}_1, \vec{q}_2) = \frac{\alpha}{2} (|\vec{q}_1 - \vec{q}_2| - r_0)^2, $$ where $r_0$ is the equilibrium bond distance. The important difference here is that in your version, any displacement of the vector $\vec{q}_1 - \vec{q}_2$ will result in a quadratic change in the potential energy; whereas in the correct version, there will be two directions in "configuration space" that correspond to no change in the potential energy. Remember that the equipartition theorem basically says that every degree of freedom that contributes quadratically to the energy will then contribute $\frac{1}{2} k$ to $C_V$. These two spurious energetic degrees of freedom are what are giving you $C_V = \frac{9}{2} k N$ instead of $C_V = \frac{7}{2} k N$.

Just to show that I'm not making this up, let's do the integral. Define $\vec{Q} = \frac{1}{2}(\vec{q}_1 + \vec{q}_2)$ and $\vec{r} = \vec{q}_1 - \vec{q}_2$. $$ \begin{align} I = \iiint_{-\infty}^{\infty} \iiint_{-\infty}^{\infty} \mathrm{e}^{- \beta \alpha (|\bar{q}_1-\bar{q}_2|-r_0)^2/2 } ~d^{3}q_1 ~d^{3}q_2 &= \iiint_{-\infty}^{\infty} \iiint_{-\infty}^{\infty} \mathrm{e}^{- \beta \alpha (r-r_0)^2/2 } ~d^{3}Q ~d^{3}r \\ &= \left[ \iiint_{-\infty}^{\infty}~d^{3}Q \right] \left[ \iiint_{-\infty}^{\infty} \mathrm{e}^{- \beta \alpha (r-r_0)^2/2 } ~d^{3}r \right] \end{align} $$ The first integral gives a factor of $V$ as before. The second one is a little more complicated. The angular contribution is obviously $4\pi$, leaving $$ I = 4 \pi V \int_0^\infty r^2 \mathrm{e}^{- \beta \alpha (r-r_0)^2/2 } ~dr $$

This last integral isn't of the standard "useful Gaussian integral" form, and will not give a result that is exactly proportional to $\beta^{-1/2}$. However, in the limit of low temperature, it does approach this limit. Define $\tilde{r} = \sqrt{\beta \alpha} (r - r_0)$; then the integral becomes $$ I = \frac{4 \pi V}{\sqrt{\beta \alpha}} \int_{-\sqrt{\beta \alpha} r_0}^\infty \left( \frac{\tilde{r}}{\sqrt{\beta \alpha}} + r_0 \right)^2 e^{-\tilde{r}^2/2} \, d \tilde{r}. $$ In the low-temperature limit, we have $\beta \to \infty$, meaning that the lower limit of integration becomes $- \infty$ and the first term in the parentheses vanishes; thus, in this limit, $$ I \approx \frac{4 \sqrt{2} \pi^{3/2} V r_0^2 }{\sqrt{\beta \alpha}} \propto V T^{1/2} $$ as desired.

EDIT: The exact integral above can't actually be evaluated in closed form, but it can be expressed in terms of the normalized error function erf(x): $$ I = \frac{4 \pi^{3/2} V}{\sqrt{2}} \left[ \left(\frac{r_0^2}{\sqrt{\alpha \beta}} + \frac{1}{(\alpha \beta)^{3/2}} \right)\left( 1 + \text{erf} \left( \frac{r_0 \sqrt{\alpha \beta}}{\sqrt{2}} \right) \right) + \sqrt{\frac{2}{\pi}} \frac{r_0}{\alpha \beta} e^{-\alpha \beta r_0^2/2}\right]. $$ Note that if we set $r_0 \to 0$, we recover your result above (with $I \propto T^{3/2}$.) However, for non-zero $r_0$, we get a leading-order result proportional to $\sqrt{T}$, and a leading-order correction proportional to $T^{3/2}$ (as well as even smaller corrections proportional to $e^{-\alpha \beta r_0^2/2}$ times various powers of $T$, arising from the exponential term and the asymptotic expansion of the erf function.)

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  • $\begingroup$ What does the incorrect physical potential represent? Is it truly a non-physical potential, or does it represent an approximation of a more complicated potential? $\endgroup$ – linuxfreebird Jul 15 '15 at 15:31
  • $\begingroup$ It would represent a situation in which the atoms' energy was minimized when they were at the same location ($\vec{q}_1 = \vec{q}_2$, or when $r_0 = 0$.) You might be able to cook up a situation where this was the case, but it wouldn't correspond to a diatomic molecule; the point is that there should be some changes in the displacement vector $\vec{r}$ that don't change the potential energy at all. $\endgroup$ – Michael Seifert Jul 15 '15 at 16:25

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