The final paragraph of Dissenter's question here is worthy of standing alone:

[H]ow does one square a high heat of vaporization with a low boiling point? If it takes a lot of energy to vaporize something, then how can that something have a low boiling point?

(It's similar to this old question by Friend of Kim, but different in that it emphasizes an apparent contradiction.)


1 Answer 1


This seeming contradiction can be reconciled by examining the thermodynamic quantities involved. First, per Wikipedia, the enthalpy of vaporization is "the enthalpy change required to transform a given quantity of a substance from a liquid into a gas at a given pressure." Written symbolically:

$$ \Delta H_{\mathrm{vap}} = H_{\mathrm{gas}} - H_{\mathrm{liq}} $$

For this situation, it is most helpful to consider $H_i$ as the enthalpy per mole ("molar enthalpy") of the species of interest in phase $i$.

Formally, again per Wikipedia, the boiling point for a substance is "the temperature at which the vapor pressure of the liquid equals the pressure surrounding the liquid and the liquid changes into a vapor." This is not so helpful for answering this question, however.

Another way of defining the boiling point is the temperature at which the liquid and the vapor of the substance are in equilibrium, for a closed system (no mass can get in or out) containing only the pure substance of interest at a given pressure$^*$. The equilibrium condition here can be expressed in terms of molar Gibbs free energies (free energy per mole$^{**}$) as:

$$ G_{\mathrm{liq}} = G_{\mathrm{gas}} $$

By definition, the molar free energy of a phase is equal to the molar enthalpy minus the temperature times the molar entropy:

$$ G = H - TS $$

So, the boiling point equilibrium condition can be rewritten as ($T_b$ is the boiling point):

$$ H_{\mathrm{liq}} - T_bS_{\mathrm{liq}} = H_{\mathrm{gas}} - T_bS_{\mathrm{gas}} $$

Subtracting $H_{\mathrm{liq}}$ and $-T_bS_{\mathrm{gas}}$ from both sides gives:

$$ T_b\Delta S_{\mathrm{vap}} = T_b\left(S_{\mathrm{gas}} - S_{\mathrm{liq}}\right) = H_{\mathrm{gas}} - H_{\mathrm{liq}} = \Delta H_{\mathrm{vap}} $$

Here, $\Delta S_{\mathrm{vap}}$ is the molar entropy of vaporization, which is analogous to $\Delta H_{\mathrm{vap}}$. The boiling point is thus:

$$ T_b = \frac{\Delta H_{\mathrm{vap}}}{\Delta S_{\mathrm{vap}}} $$

So, what does this result mean?

Qualitatively, as noted in answers to the linked questions, $\Delta H_{\mathrm{vap}}$ is a measure of how much energy must be added to one mole of molecules in the liquid to free them from their associations with the other liquid molecules. The physical meaning of $\Delta S_{\mathrm{vap}}$ is somewhat harder to describe, but corresponds roughly to the increase in the 'freedom to wiggle' afforded by the transition to the gas phase from the liquid.

So, in order to have a large $\Delta H_{\mathrm{vap}}$ (lots of energy required to drive molecules into the gas phase) but also a low $T_b$ (low temperature at which molecules "like" to transition into the gas phase) a chemical must have a large $\Delta S_{\mathrm{vap}}$, which means that each molecule gains a significant amount of 'freedom of movement' from entering the gas phase. This makes sense in the case of ethanol versus water in Dissenter's question: ethanol is bulkier than water, so even though they both have similar values of $\Delta H_{\mathrm{vap}}$, ethanol gains more 'freedom to wiggle' in the gas phase and thus has a lower $T_b$.

This also helps to provide a qualitative explanation for why $T_b$ is proportional to pressure: as the pressure of the gas phase increases, the 'freedom to wiggle' gained by entry to the gas phase decreases, because higher $P$ means more collisions with gas-phase molecules. Thus, one can expect $\Delta S_{\mathrm{vap}}$ to decrease with increasing pressure ($S_{\mathrm{liq}}$ isn't affected nearly as much by pressure changes as is $S_{\mathrm{gas}}$), which translates directly to the experimentally observed increase in $T_b$.

More details on the relationship between boiling point and pressure can be found in the Wikipedia articles on the saturation temperature and the Clausius-Clapeyron relation. More information on equilibrium and Gibbs free energy can be found on www.chem1.com.

$^*$ In this model system, the "knob" one has available (see Gibbs' phase rule) to change the equilibrium pressure and temperature is to add to or remove from the system some number of moles of the species of interest.

$^{**}$ As orthocresol rightly notes in a comment, the equilibrium must be expressed in terms of the chemical potentials of each species present. However, in this case there is only one species present and thus examining the total molar Gibbs free energy is sufficient.

  • 3
    $\begingroup$ Brian, I disagree with your definition of boiling point as "the temperature at which the liquid and the vapor of the substance are in equilibrium". In fact, in those conditions the liquid and vapor of a substance will always reach an equilibrium at any temperature: this is governed by the vapor pressure of that substance and has nothing to do with being at the boiling point. Fortunately this does not affect the rest of the answer, i.e. your "boiling point equilibrium condition" happens to be correct for any temperature, as long as both phases are present, including at the boiling point. $\endgroup$
    – MarcoB
    Jun 6, 2015 at 18:13
  • $\begingroup$ To be pedantic, the condition for equilibrium is that the chemical potentials (which is, in this case, equal to the molar Gibbs free energies) of both phases are equal. Most of the equations should be written in terms of molar quantities, actually. $\endgroup$ Jun 6, 2015 at 19:44
  • $\begingroup$ @orthocresol Pedantry appreciated; edit made. While $\Delta H_{\mathrm{vap}}$ was described obliquely as being intensive, it was definitely a point worth making more explicit throughout the explanation. $\endgroup$
    – hBy2Py
    Jun 6, 2015 at 21:57
  • $\begingroup$ @MarcoB This was something that bugged me when initially writing up the answer, actually. I have revised the answer somewhat to indicate the system considered in the analysis to be closed and composed solely of the chemical species of interest. I believe this makes the argument accurate, though I welcome critique as errors may well remain -- thermo is not my strongest subject. $\endgroup$
    – hBy2Py
    Jun 6, 2015 at 21:59
  • $\begingroup$ Am accepting as answer since dissent tapered off. Will gladly revisit if critique resumes in the future. $\endgroup$
    – hBy2Py
    Jun 10, 2015 at 14:50

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