Suppose we have a diatomic molecule. Its center of mass has an average kinetic energy given by

$$ \frac{1}{2} (m_1 +m_2) v_{cm}^2 = \frac{3}{2}k_BT $$

and using this we can derive the ideal gas law

$$ PV = n k_BT $$ where $n$ is the number of molecules. We do this by thinking about how much momentum is transferred per unit time and the equipartition theorem.

At temperatures high enough that the atoms break apart, they do not move together and the same computation has to be repeated for each of component atom. The ideal gas law is still the same but if we still denote the number of molecules as $n$ then the number of atoms is $2n$ and we get

$$ PV=2nk_BT $$

My question is what happens during the transition? What happens when temperatures are high enough for the diatomic molecules to not completely break apart but the constituent atoms to move away far from each other?

  • $\begingroup$ I realize that this actually may be a sharp transition in the $n \to \infty$ limit and somewhere in the details also lies the issue of the time taken over which pressure is being measured. I am asking to see if someone already has an answer before diving into the detailed computation. $\endgroup$ Commented Jul 3, 2019 at 8:04

2 Answers 2


What happens when temperatures are high enough for the diatomic molecules to not completely break apart but the constituent atoms to move away far from each other?

Atoms in a molecule can't move away from each other without the molecule breaking.

The chemical bond acts like a spring and once we add enough energy to the molecule to overcome the bond energy, the molecule breaks.

As a very rough approximation:

The equipartition theorem says that each degree of freedom of a molecule has equal average energy. A molecule has $3$ translational degrees of freedom.

The diatomic molecule has $2$ rotational and

$$f_{vib} = 3n_{at} - 5 = 1$$

vibrational degree of freedom (where $n_{at}$ is the number of atoms in the molecule). Since the harmonic oscillator has potential and kinetic energy, the vibrational degree of freedom is for the purpose of equipartition theorem double-counted, which gives us number $2$. Together with rotational degrees of freedom, we have $4$ degrees of freedom that contribute to stressing the bond, and $3$ translational degrees of freedom.

From that follows that $\frac47$ of energy given to the molecule will be expended on stressing the bond.

The energy given to the molecule before it breaks apart is equal to bond energy. Since the bond acts like a mechanical spring with certain stiffness which breaks when the molecule gets energy greater or equal to the bond energy, we can see that we can't stretch the average bond length to an arbitrarily high extent without the molecule breaking (even though we can stretch it to some extent).

So at any moment, we'll have two or three mixed gasses - the diatomic molecules, the monoatomic molecules of one kind and the monoatomic molecules of the second kind. For each component, the ideal gas law holds separately, so

$$P_{\text{molecules}}V = N_{\text{molecules}}kT$$ $$P_{\text{atoms kind 1}}V = N_{\text{atoms kind 1}}kT$$ $$P_{\text{atoms kind 2}}V = N_{\text{atoms kind 2}}kT$$

  • $\begingroup$ Of course atoms can "move away" from each other without breaking the bond. Take your model of the spring. Let the constant be $k$ then the typical length of the spring is $x \sim \sqrt{k_BT/k}$. This increases with $T$ and note that in the spring model the atoms keep moving apart without breaking. In a more realistic model the potential becomes shallow when the atoms are far apart. But nevertheless there is a region where the atoms move far apart without the molecule breaking. It is this region I am asking about. In the spring model it will be when $x$ is the size of the box. $\endgroup$ Commented Jul 5, 2019 at 12:57
  • $\begingroup$ Your degrees of freedom counting seems incorrect. For a diatomic molecule it is 6 coming from the translation of the two atoms. You can massage it into vibrational and rotational but it will remain 6 degrees of freedom. Not that it is relevant for the discussion. $\endgroup$ Commented Jul 6, 2019 at 6:30
  • $\begingroup$ @BorunChowdhury (Part 1) Sorry, I meant to type "can't move far away," but the word "far" slipped somewhere. When the atoms are far apart, the molecule doesn't exist anymore, because atoms have electric orbitals repelling each other, so if you move atoms far apart, they stay far apart forever (unless you crash them into each other with at least activation energy. There is no region where atoms move far apart without the molecule breaking. $\endgroup$ Commented Jul 6, 2019 at 13:05
  • $\begingroup$ @BorunChowdhury (Part 2) The diatomic molecule does have $6$ degrees of freedom, but since vibrational degrees of freedom need to be counted twice for the purpose of the equipartition theorem, this gives you $7$. They don't come only from the translation of the two atoms, they also come from their vibrations and the rotation of the molecule (and from their electrons, but I ignored the electron degrees of freedom for simplicity). $\endgroup$ Commented Jul 6, 2019 at 13:16
  • $\begingroup$ I am not really interested in a realistic molecule. Ideal gas law anyways does not apply to realistic molecules. $\endgroup$ Commented Jul 6, 2019 at 19:30

I went ahead and solved it. The details of the calculation have been posted on my blog but the main point is the following.

I assume a diatomic molecule made of the same atoms connected by a spring with spring constant $k$ to make things simpler. Working in the com coordinates $x$ and the relative distance $y$ the partition function of a single molecule in one dimension is

$$ Z_1 = 4 \int_0^{L/2-y/2} dx \int_0^L dy e^{-y^2/X^2} \\ =X^2 \left( \eta \sqrt{\pi} ~erf( \eta) - (1-e^{-\eta^2}) \right) $$

where $X=\frac{2k_B T}{k}$ is a length scale corresponding to the typical distance between the atoms and $\eta= \frac{L}{X}$.

We have put a subscript to denote that this is the partition function in 1 dimension for 1 molecule. Thus the full partition function is

$$ Z={Z_1}^{3n} $$

and we will henceforth write $L$ as $V^\frac{1}{3}$. The pressure is given by

$$ \frac{P}{k_BT} = \frac{1}{Z} \frac{\partial Z}{\partial V} \\ = 3n \frac{1}{Z_1} \frac{\partial Z_1}{\partial V} \\ = \frac{n}{V} \frac{ 1 }{1 -\frac{(1-e^{-\eta^2})}{\eta \sqrt{\pi} ~erf(\eta)} } $$

We plot the correction term

enter image description here

This confirms the intuition behind the question.


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