# If all molecules have basically the same KE at a given temperature, why do lighter isotopes evaporate more readily?

One of the proxies used for paleoclimate is the ratio of oxygen-18 to oxygen-16 in ice cores and in sedimentary rocks. The idea is that water molecules with oxygen-18 generally evaporate less readily than oxygen-16, and condense more readily; and this proclivity varies with temperature. One can then use the deviations of oxygen-18 concentration from a standard ratio to say something about Earth's temperature in the distant past.

Why do light water molecules evaporate more readily? It makes some intuitive sense that lighter molecules would evaporate more easily. But how does this square with the equipartition theorem and the idea that all molecules have the same amount of kinetic energy at a given temperature?

It would seem to me that because the intermolecular forces between two water molecules are independent of their mass, it should take a certain amount of energy to remove a water molecule from its neighbors in the liquid phase and move it far away. This energy requirement doesn't depend on the mass of the molecules, only on how the force on it varies with distance ($$W = \int \vec{F} \cdot d\vec{r}$$.) So by this argument, molecules with the same amount of kinetic energy would have the same likelihood of evaporation. And by the equipartition theorem, the translational kinetic energy of a water molecule is only dependent on the temperature, not on the mass of the molecule.

EDIT: Just to pre-empt the "it's velocity that matters, not energy" argument: I agree that lighter molecules would, on average, have a higher velocity. But from classical physics, we know that applying a fixed force to a smaller mass leads to a larger change in velocity, while a larger mass gets a smaller change in velocity. That would imply that the "escape velocity" required for a heavier molecule would be smaller; since it's heavier, its motion would not be affected as much by the force it feels from the other molecules. This would imply that while heavier molecules are moving more slowly on average, they also don't need to be moving as quickly to escape. In other words, the velocity threshold for evaporation should be lower for heavier molecules. These two effects (lower speeds and lower threshold for evaporation) should cancel each other out.

Where does my logic fail?

• Energy is not quite the same. The vibrational levels of the heavier molecule are deeper. Apr 21 at 21:22
• Apr 21 at 21:26
• One of the lessons by statistical thermodynamics is that there is no fix, uniform kinetic energy for the molecules of a compound. Instead, there is a distribution of molecules with higher, intermediate, and lower energy. The shape of this distribution is temperature dependent, too (e.g., slide 32/36 here). Depending on what you want to describe (and level of detail), this branch influenced by the work by Boltzmann offers a better description than the by classical thermodynamics. Apr 22 at 13:48
• @Buttonwood: Yeah, I've taught "thermodynamics and statistical mechanics" in my physics department a couple of times, and I always enjoy teaching the "Statistical mechanics" part more than the "thermodynamics" part. Apr 22 at 13:50
• @MichaelSeifert I then assume you present a question like the one by you «about water and water» explicitly comparing molecules differing only by their isotopic composition (e.g., in kinetics when it comes to the deuterium isotope effect). Because, within reason, intermolecular forces (e.g. cohesion) must stay the same for such a comparison; otherwise, acetic acid (60 g/mol, bp 118 C @ 1 atm) should evaporate at higher rate than benzene (78 g/mol, bp 80 C @ 1 atm). Apr 22 at 14:16

## 3 Answers

Evaporation depends on molecular velocity not on the energy

The simple mistake in your assumption is that evaporation is driven by molecules having enough energy to escape the liquid phase. This is wrong. Evaporation depends on whether the molecule is travelling fast enough to escape the liquid.

The molecules in the liquid will have, on average, the same kinetic energy. But this means that, in practice, the molecules will have a range of different kinetic energies across a wide distribution. But, since kinetic energy is a function of both molecular mass and molecular velocity, there will also be a distribution of velocities. And, crucially, the more massive molecules (in this case those with 18O) will have slightly lower velocities at the same average kinetic energy. So the lighter isotopes will be slightly more likely to have a high enough velocity to escape the liquid phase.

You can work through the mathematical details (taking into account the statistical distribution of kinetic energy and converting that to the distribution of velocities and then calculating the proportion of molecules with a velocity above the threshold for evaporation) and check that the proportion of light molecules is higher than the proportion of heavy molecules.

Clarification: why is the force-based view wrong

The argument that applying a fixed force leads to a smaller effect on a heavier molecule so heavier molecules don't need to moves as quickly to escape the liquid is flawed for several reasons.

It sounds relevant but isn't because the vast majority of all collisions don't lead to escape from the liquid. There are so many momentum (and KE) redistributing collisions that all we have to consider is the distribution of energy/momentum across the ensemble of molecules in the liquid. And what we need to know is the average proportion of molecules having enough velocity to escape after their last collision.

Yes, the force argument matters for individual collisions (if they are not purely elastic), but, if you simulated a drop of liquid down to the molecule level, you would find that the distribution of energy/momentum would match the statistical one expected from theory and you would have gained no new knowledge. Follow an individual molecule and, yes, the thinking about force will apply. But follow the distribution of velocities across the ensemble of molecules and what matters is the proportion of molecules having a high enough escape velocity and that is governed by the statistical distribution. Think about the properties of the ensemble of molecules not the individual molecule.

And the force argument ("the escape velocity will also be lower") doesn't work at the ensemble level. All that matters is the velocity after the last collision and the only way to know that is to look at the statistics of the whole system not the individual forces on individual molecules.

• To add a comment to your clear answer I would mention that as evaporation occurs the total mass of liquid diminishes and at the same time as only the quick molecules escape the kinetic energy of the molecules remaining in the liquid is continually being lowered. Thus cooling accompanies evaporation which explains the action of most refrigerators. Apr 22 at 12:33
• From physics classes, I know that applying a fixed force to a smaller mass leads to a larger change in velocity, while a larger mass gets a smaller change in velocity. That would imply that the "escape velocity" required for a heavier molecule would be smaller; since it's heavier, its motion is not affected as much by the force it feels from the other molecules. This would imply that while heavier molecules are moving more slowly on average, they also don't need to be moving as quickly to escape, and the effects should cancel each other out. What am I neglecting? Apr 22 at 12:36
• @MichaelSeifert Your argument about force and mass is irrelevant as the average speeds are determined by the statistical distribution (guaranteed by equipartition). There is no "fixed force" governing evaporation or the speeds of individual molecules: there is only a vast number of momentum-redistributing molecular collisions which ensure the KE of all molecules is distributed in a particular statistical distribution. Apr 22 at 12:46
• @MichaelSeifert see my additions to the answer. The key point is that–even if you did a complete molecule-level simulation including the detail of attractive forces–all that matters is the overall distribution of velocity after the last collision. equipartition in a large collection of molecules guarantees that the apparent logic of that individual molecule's view is irrelevant. Apr 22 at 13:19
• @matt_black Probably, I did not think through these lines of thought until reading the OP. But, if the reasoning may be simplified (maybe I go too far here) to $$E_{kin} = \frac{1}{2} mv^2 \Leftrightarrow v = \sqrt{ \frac{2E_{kin}}{m}}$$ then, assuming the same $E_{kin}$ per molecule of $\ce{H2^{16}O}$ and $\ce{H2^{18}O}$ would mean the ratio of their velocities were about 1.05 : 1.00. Which in turn reminds me about heavy water, $\ce{D2O}$, (of molecular mass more similar to the one of $\ce{H2^{18}O}$ than $\ce{H2^{16}O}$) and its higher boiling point ($\pu{101 ^\circ{}C}$) than $\ce{H2O}$. Apr 22 at 13:28

As very simple model to find what properties are important in evaporation consider a molecule that when close to the surface moves around in a cell of side $$a$$ and collides with frequency $$f$$ into its neighbouring molecules. There is an area $$a^2$$ through which it can escape the surface and in a second it moves on average a distance $$v= 2af$$ back and forth in the cell, which is the average velocity.

The surface presents a barrier $$\Delta E_b$$ to escape due to intermolecular forces and the chance of exceeding this is proportional to $$\exp(-\Delta E_b/k_BT)$$ at temperature $$T$$. Thus the number leaving the surface per unit area per second is proportional to the frequency of hitting a side of the cell at the top of the liquid, which is proportional to its velocity, and the chance of escape over the potential barrier or $$\sim v\exp(-\Delta E_b/k_BT)$$.

At a given energy the more massive molecule will have the smaller velocity but the barrier remains the same as this is determined by intermolecular forces and so the more massive molecule evaporates less.

The simplest and most direct explanation is that isotopically heavier molecules have higher boiling points. Regular water (R) boils at 100.0$$^o$$ C. Heavy water (H) boils at 101.4$$^o$$ C. A mixture of the two (R + H) can be separated by a distillation column. The heavier molecules require more energy to evaporate, to push away one atmosphere.

Any similarity to (astronomical) orbital mechanics is misleading. The orbital metaphor is like throwing a stone, involving one total energy. Liquid evaporation (even at lower temperatures) is like wiggling thru a crowd: an equilibrium of molecules leaving and returning to the liquid, with a net loss of liquid (like diving thru a gap in traffic).

Evaporation of either R or H at lower temperatures relies on diffusion of the escaped molecule to a distance so far away that it is no longer in equilibrium with the liquid. Here the higher velocity of an escaped molecule gets it farther away quicker, and less likely to return to the liquid. Heavier molecules with the same energy move slower, don’t get as far away as fast - thus evaporate slower.

Looking at the macroscopic systems (like pure R and H) rather than individual molecules focuses on measurable properties (bp). It all boils down to differences in boiling point.