14
$\begingroup$

Octane has a boiling point of 120 °C. Water has a boiling point of 100 °C. The definition of boiling point is, "the temperature which the liquid substance's saturated vapor pressure equals the atmospheric pressure". Volatile substances have higher saturated vapor pressure at a given temperature, than the lesser volatile substances. When temperature increased (heated), the saturated vapor pressure increases rapidly and non-linearly. So, the volatile substance's saturated vapor pressure reaches atmospheric pressure quicker than the lesser volatile substance, hence, reaching boiling pint at a lower temperature.

However, water which is a less volatile substance than octane (if you put two separate liquid spots from water and octane, octane evaporates faster) has a lower boiling point than octane at atmospheric pressure. Is there something I am missing? If my reasoning is wrong please correct me.

$\endgroup$
1

5 Answers 5

21
$\begingroup$

One thing you are missing is that air contains water, but usually does not contain octane. So for water, the process is:

$$\ce{H2O(l) <=> H2O(g)}$$

and for octane, it is

$$\ce{C8H18(l) <=> C8H18(g)}$$

For octane, the partial pressure of octane is virtually zero except near the liquid surface. For water, it depends on the humidity in the room.

If you look up the vapor pressure at room temperature, you will find that water has a higher vapor pressure. At equilibrium, the partial pressure of water will be higher than that of octane. However, the kinetics determine how fast a drop will evaporate (and the water drop will not evaporate at all if the humidity is higher than 100%).

$\endgroup$
6
  • $\begingroup$ Does that mean we applied Le Chatelier's principle here in terms of concentration of water and concentration of octane? $\endgroup$ Commented Oct 5, 2023 at 17:09
  • 2
    $\begingroup$ No, the system remains out of equilibrium. The forward and reverse rates depend on the rate constant (which might be very different for the two systems) and the partial pressure at the interface (which depend on bulk partial pressure and transport kinetics, which in turn depend on the shape of the drop and the object the drop is on). Overall, it is plenty complicated. $\endgroup$
    – Karsten
    Commented Oct 5, 2023 at 17:15
  • $\begingroup$ @Karsten Is there something to do with steam distillation? Octane that we talk here is petrol ( which is a mixture). As far as I know, octane is also mixed with alcohols to increase petrol's efficiency as a fuel. Alcohol being a polar substance, might be less soluble in octane, which is non polar ( I tried to find solubility of alcohol in octane, but couldn't find. so this is just an assumption.) So as in steam distillation, they boil at low temperature. (therefore, evaporate more at room temperature.) $\endgroup$ Commented Oct 5, 2023 at 19:30
  • $\begingroup$ Good point with mentioning existing humidity, I have forgotten to mention it. $\endgroup$
    – Poutnik
    Commented Oct 5, 2023 at 20:41
  • 1
    $\begingroup$ @donthababakka "Octane that we talk here is petrol" – No. Typical gasoline contains less than 1 % octane. $\endgroup$
    – Loong
    Commented Oct 8, 2023 at 9:04
14
$\begingroup$

The are three major factors, affecting observed relative volatility, ordered by assumed importance:

  • The molar mass. The liquid with higher molar mass evaporates faster than a liquid with the same vapor pressure but lower molar mass. This has two reasons:

    • That is because for the given vapor pressure, the mass concentration of vapor is proportional (as ideal gas approximation) to its molar mass: $\rho = \frac{pM}{RT}$
    • The higher molar mass leads to lower specific enthalpy of evaporation, so it need lower heat transfer for the same evaporation rate.: $ΔH_\text{evap,spec}=\frac{ΔH_\text{evap,molar}}{m}$
  • Another factor Karsten mentioned (and I as a former enlisted military meteorologist should remember to write) is water vapor presence, which slows down evaporation. Air already contains typically $\pu{20 to 100\%}$ of possible water vapor amount, what slows down water evaporation. That is not the case for other liquids.

  • The third factor is the molar evaporation enthalpy. The higher is the molar evaporation enthalpy of the liquid, the faster is the vapor pressure decrease with temperature.


The typical other case is toluene with the boiling point $\pu{111 ^\circ C}$, but significantly more volatile than water.


Note the van't Hoff equation about the dependence of liquid vapor pressure on absolute temperature in the integral form:

$$\ln{\frac{p_2}{p_1}=\frac{ΔH_\mathrm{evap}}{R}(\frac{1}{T_1}-\frac{1}{T_2})}$$

Substance Molar evaporation enthalpy ($\pu{kJ/mol}$) Molar mass ($\ce{g/mol}$)
Water 40.65 18.02
Toluene 36.08 92.14
n-Octane 39.1 114.23
$\endgroup$
12
  • 2
    $\begingroup$ I suspect the molar mass difference really is the biggest factor at work here. Assuming we're talking about the vapour phases as ideal gasses (which we are), the natural quantity to talk about is number of molecules. Two puddles of water and octane with the same volume (what "feels intuitively right" to compare) actually refers to amounts of molecules which differ by a factor of $\mathrm{\frac{114.23}{18.02}\frac{0.997}{0.703}=8.99}$ times - there is simply just much less octane to evaporate. $\endgroup$ Commented Oct 6, 2023 at 10:40
  • $\begingroup$ @NicolauSakerNeto I do agree, assuming the effect is on the order molar mass > vapor saturation > evaporation enthalpy. The mentioning order was not meant as order of importance. I sil reorder that. $\endgroup$
    – Poutnik
    Commented Oct 6, 2023 at 10:51
  • $\begingroup$ Since molar enthalpies are similar it is the Number o molecules that is pertinent $\endgroup$
    – jimchmst
    Commented Oct 6, 2023 at 15:58
  • 1
    $\begingroup$ @BuckThorn I mean, you see 2 comparable drops of water and n-octane and you do not say "Look, the n-octane drop contains 8 times less molecules.". You say "Look, the n-octane drop has similar volume and mass as the water one.". // I also assume that in pseudostatic scenario is not the determining factor the rate of bidirectional mass transfer between phases, but mass concentration of saturated vapor. $\endgroup$
    – Poutnik
    Commented Oct 10, 2023 at 9:32
  • 1
    $\begingroup$ That too, it is quite high for water. Like 2 effects of 1 cause ( pM/RT and ΔH_evap,spec) of the molecular mass. $\endgroup$
    – Poutnik
    Commented Oct 10, 2023 at 9:36
6
$\begingroup$

It is a case of careful observation and awareness of the construction of matter. Water and octane are molecular so a given volume with similar densities [1 vs 0.8] have very different # of molecules. The molar heats of vaporization are similar ~10Kcal/mole, [I leave it to you to look up the exact data]. Approximately similar masses with the same available energy and similar vapor pressures the fewer molecules will evaporate faster. As the MW difference increases the differences will change. Compare kerosene or cetane.

Also drop shape and surface area are important when comparing rates. Heat transfer is important.

$\endgroup$
6
$\begingroup$

Firstly, there are 18 different isomers of octane, each having a unique boiling point.

The gasoline standard is that "100 octane" corresponds to pure iso-octane (2,2,4-trimethylpentane).

Iso-octane has a boiling point of 99°C.

Secondly, the mass-rate of evaporation, according to the Hertz-Knudsen equation, given everything else is equal, is proportional to the square root of molecular weight. Therefore, if the vapor pressure of water and iso-octane are approximately the same, the mass-rate of evaporation of octane should be greater.

See also problem 1b here.

Thirdly, since the density of octane is about 0.7 relative to water, if equal-volume drops are being evaporated, the rate of evaporation will be even greater than the mass-rate.

$\endgroup$
3
  • 4
    $\begingroup$ I don't see why this is downvoted. There are indeed many isomers of octane. 2,2,4-trimethylpentane does have a boiling point of 99C, lower than water (while some other isomers, particularly n-octane, have a lot higher boiling point.) We should be sure we are comparing the same substance. Also, the per-mass latent heat of octane is indeed much lower than water. (They have similar per-mol latent heat, but the molecular weight of octane is much higher. ) $\endgroup$ Commented Oct 7, 2023 at 1:39
  • $\begingroup$ It looks like 4 ups despite You complicated issues. the same argument applies regardless of the isomer. also the last sentence is puzzling. Molecules determine what is happening. $\endgroup$
    – jimchmst
    Commented Oct 11, 2023 at 0:36
  • $\begingroup$ @jimchmst Each isomer is unique; lower boiling point than water is different than higher. The premise of the question is "if you put two separate liquid spots from water and octane, octane evaporates faster", which sounds like equal volume (as opposed to equal mass or equal moles), $\endgroup$
    – DavePhD
    Commented Oct 11, 2023 at 2:08
1
$\begingroup$

The boiling point is when it starts making bubbles inside.

"Liquids may change to a vapor at temperatures below their boiling points through the process of evaporation. Evaporation is a surface phenomenon in which molecules located near the liquid's edge, not contained by enough liquid pressure on that side, escape into the surroundings as vapor. On the other hand, boiling is a process in which molecules anywhere in the liquid escape, resulting in the formation of vapor bubbles within the liquid."


"a liquid will evaporate until the surrounding air is saturated"

Evaporation is a type of vaporization that occurs on the surface of a liquid as it changes into the gas phase. High concentration of the evaporating substance in the surrounding gas significantly slows down evaporation, such as when humidity affects rate of evaporation of water.


All things are not equal.

The boiling point is an important property because [all things being equal, e.g., in a vacuum] it determines the speed of evaporation. Small amounts of low-boiling-point solvents like diethyl ether, dichloromethane, or acetone will evaporate in seconds at room temperature, while high-boiling-point solvents like water or dimethyl sulfoxide need higher temperatures, an air flow, or the application of vacuum for fast evaporation.

$\endgroup$
1
  • 1
    $\begingroup$ "at atmospheric pressure" - at STP, which is 0% humidity and does not naturally occur on planet Earth? What you're "missing" is having conducted the experiment in a vacuum chamber. $\endgroup$
    – Mazura
    Commented Oct 6, 2023 at 5:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.