So, we all know that there are two ways to evaporate a liquid, either by increasing its temperature through heating, which could be explained microscopically, as we're providing the molecules of the liquid the necessary energy to free themselves from cohesion intermolecular forces, plus the energy necessary to push air molecules out of their way as the molecules of the substance are rising, due to a transition to a gaseous state. So far so good. Now, let's talk about the second method. Liquids could be evaporated by lowering the pressure of the air above their surface, which makes it easier for fluid molecules to rise, and the more pressure is reduced, the more the liquid evaporates until its container is empty. My question is what role do cohesion forces play in this method?? Why does reducing pressure cause molecules to rise? I mean, aren't they attracting each other, or is there any dependence between cohesion forces and pressure?


1 Answer 1


There are both thermodynamic and kinetic aspects of the evaporation rate.

Stronger cohesion suppresses evaporation, but at a nonzero temperature—which is any temperature—thermal fluctuations always result in some nonzero likelihood that a molecule will detach from the surface of the condensed phase, entering the gas phase. Put another way, every material, no matter how strongly bonded, has a nonzero equilibrium vapor pressure. That's the thermodynamic aspect.

Net evaporation corresponds to more molecules departing than (re)attaching. Reducing the ambient pressure kinetically speeds up evaporation to reach the equilibrium vapor pressure because there are fewer gas molecules that could redirect the evaporating vapor back into the condensed phase. That's the kinetic aspect.

To discuss the relation between cohesion and the equilibrium vapor pressure quantitatively, first, a quick review:

  • The matter around us is doing all it can to reduce its Gibbs free energy $G$ (because this maximizes total entropy). The stable equilibrium state at some temperature and pressure is the one with the lowest Gibbs free energy at those conditions.

  • The Gibbs free energy can be interpreted as the internal energy $U$ of a system, plus the work needed to push the surroundings out of the way ($PV$), minus the free heating the system gets from the surroundings to bring it up to ambient temperature ($TS$). Putting this all together, we have $G\equiv U+PV-TS$, which is minimized for systems in thermal and mechanical contact with their surroundings.

  • The Gibbs free energy can also be interpreted as Nature’s opposing preferences for strong, dense bonding (low internal energy $U$ and small volume $V$, i.e., a low enthalpy $H\equiv U+PV$) but also many possible arrangements (high entropy $S$). The minus sign reflects this opposition. The extent of the internal energy below that of the gas state corresponds to the cohesive energy you mention.

  • The tradeoff between enthalpy minimization and entropy maximization of a parcel of matter of interest is mediated through the ambient pressure $P$ and temperature $T$, the coefficients in the Gibbs free energy definition. A larger pressure favors a denser state (condensed matter). A higher temperature favors a higher-entropy state (gas). You may find this framework useful because it presents pressure and temperature as parallel driving forces, with analogous results. Importantly, shifting to a lower enthalpy—as with condensation or freezing—is an exothermic process that heats the rest of the universe. So one way or another, total entropy is maximized.

  • Because phase equilibrium corresponds to minimization of the total Gibbs free energy, it follows that the chemical potentials of the phases present are then equal. (The chemical potential $\mu\equiv\left(\frac{\partial G}{\partial N_i}\right)_{T,P,N_{j\neq i}}$ is just the partial molar Gibbs free energy at constant temperature and pressure—its minimization governs which direction matter shifts. The Gibbs free energy in differential form is the fundamental relation $dG=-S\,dT+V\,dP+\sum_i\mu_i\,dN_i$, from which one can verify the above definition. The enthalpy in differential form is $dH=d(U+PV)=d(G+TS)=T\,dS+V\,dP+\sum_i\mu_i\,dN$.)

  • Net evaporation from a liquid is thus spontaneous until $$\mu_{\mathrm{liquid}}=\mu_{\mathrm{gas}};$$ $$\Delta \mu=0,$$ where $\Delta$ indicates a difference between phases. Now, the chemical potential $\mu$ of some phase at some concentration can be expressed as $$\mu=\mu_0+RT\ln a,$$ where $\mu_0$ is the chemical potential of the pure phase at a reference state, $R$ is the gas constant, and $a$ is the thermodynamic activity. The activity of a pure condensed phase is 1 (essentially, its concentration is 100%), and the activity of an ideal gas is $\frac{p}{p_0}$, where $p$ is the partial pressure and $p_0$ is a reference pressure, which we'll take as atmospheric pressure. So at equilibrium between a liquid and a rarefied vapor above it, we have $$\mu_{0\mathrm{,~liquid}}=\mu_{0\mathrm{,~gas}}+RT\ln\frac{p}{p_0}\tag{1}.$$ At the boiling point, the vapor pressure equals the surrounding pressure; essentially, nucleated bubbles have a sufficiently high pressure to push liquid aside. From $p=p_0$ and from $$G=H-TS;$$ $$\mu=h-Ts;$$ $$\Delta\mu=\Delta h-T\Delta s\tag{2}$$ at constant temperature and pressure (with the per-mole enthalpy $h$ and entropy $s$), we have $$\Delta h=T_\mathrm{boiling}\Delta s;$$ $$\Delta s=\frac{\Delta h}{T_\mathrm{boiling}}=\frac{L}{T_\mathrm{boiling}}.\tag{3}$$ with latent heat of vaporization $$L=\Delta u+P\Delta v\approx\Delta u+RT,\tag{4}$$ where $\Delta u$ represents the cohesive energy of the condensed phase per mole and where I've assumed that the condensed phase has a negligible molar volume compared that of the gas phase, which is $v_\mathrm{gas}=\frac{RT}{P}$.

Given all this, how does stronger cohesion quantitatively affect evaporation? From Eqs. 1-4, we have an equilibrium vapor pressure dependence of

$$p\approx p_0\exp\left[-\left(\frac{\Delta u}{RT}+1\right)\left(1-\frac{T}{T_\mathrm{boiling}}\right)\right]\tag{5},$$

indicating an exponential dependence on the cohesive energy $\Delta u$. That is, the better the bonding, the exponentially lower the vapor pressure. Considering matter in general, from a vapor pressure table of the elements, we do indeed see that more-refractory elements have a far lower vapor pressure at any temperature.

Is the model suitably predictive? For water, $\Delta u\approx 37.6-41.7\,\mathrm{kJ/mol}$ between 20°C and 100°C. The equilibrium vapor pressure at 20°C is then predicted from Eq. 5 to be $p\approx 0.020-0.030\,\mathrm{atm}$. The experimental value is $p=0.023\,\mathrm{atm}$. The agreement arguably justifies the deep dive into the thermodynamic concepts reviewed above.

  • $\begingroup$ okay, but that didn't answer my question, why aren't the intermolecular forces hindering the process of vaporization, I mean if we set vacuum in the whole space above the liquid, it will evaporate (at least that's what I think) till its last molecule, right? $\endgroup$ Commented Apr 24 at 19:15
  • $\begingroup$ Please see my edits focusing on this point, and let me know if anything's unclear. $\endgroup$ Commented Apr 24 at 23:18

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