# Evaporation and Intermolecular Forces

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So we did this lab in class in which we inserted a temperature probe into various samples liquids, removed the probes, and then measured the temperature drop over time of the evaporating liquids. Two of the liquids we had to test and compare were n-heptane and n-hexane. According to my predictions, the heptane, due to its longer chain structure, would have a greater instantaneous dipole, which would mean greater intermolecular attraction and higher boiling point. This would have suggested that there would have been a greater temperature drop in the heptane than the hexane because more energy would be required to separate the heptane molecules. The experiment proved me wrong; the heptane had a lower temperature drop than the hexane but took longer to evaporate. I can understand why it would take heptane longer to evaporate, but why was the temperature drop in the hexane greater then the heptane? The same results occurred in the testing of the other liquids where I predicted that one compound would have a higher temperature drop but was wrong. I asked my teacher about this, and she was baffled because her predictions were the same as mine... Can anybody please explain?

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You should not look at the temperature drop after a certain time, but at the temperature drop, after a certain amount has evaporated. So, if you wait, until e.g. 5 grams are evaporated, then I expect that with heptane the temperature drop is larger than with hexane.

What you did, is measure temperatures every few seconds (minutes?) and compare these. That is wrong. Less heptane evaporates per unit of time than hexane. Apparently the effect of slower evaporation outbalances (and in fact overrides) the effect of more energy consumption per gram of compound.

Repeat the experiment with two balances (accurate) and start with e.g. 10 gram of each liquid. Each time, when e.g. 0.1 gram is evaporated, measure the temperature. Now you'll have graphs of evaporated mass vs. temperature. If you make graphs for both compounds, then I expect that the heptane case shows lower temperatures. When performing this experiment, however, you have to take into account that the liquids, cooling down due to evaporation, also receive heat from the surrounding air. That may spoil the results somewhat, this effect will be stronger in the heptane case.

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You are right in assuming that heptane has the larger BP and your reasoning is pretty good. The larger the molecule the larger the surface area the greater the attraction due to induced dipoles. However, it is really the vapour pressure that is important. The vapour pressure is the pressure exerted by gaseous molecules when in contact with their liquid phase. The lower the boiling point the larger the vapour pressure. The larger the vapour pressure the greater the rate at which you can evapourate the liquid for a given temperature (p is proportional to n because PV = nRT). The faster it evapourates the faster the liquid left behind cools. Therefore hexane will cool faster than heptane

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