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From my rudimentary understanding of chemistry, pressure should be directly proportional to temperature for a constant mass of gas under the same volume (Charles’s law).

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However, in the diagram above, the vapour pressure seems to follow a non-linear relationship with the boiling point temperature of water.

Why is it the case that they are non-linear, and is there a mathematical relationship between the two quantities?

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    $\begingroup$ Evaporation of liquid is very different phenomena, compared to heating up a gas at constant volume. $\endgroup$
    – Poutnik
    Commented Dec 26, 2023 at 20:48
  • $\begingroup$ See en.m.wikipedia.org/wiki/Vapour_pressure_of_water $\endgroup$
    – Jon Custer
    Commented Dec 26, 2023 at 20:54
  • $\begingroup$ It is linear if looked at differently. Check out the Clapeyron Equation in any P Chem textbook. As usual it involves the energy and entropy differences [ AKA Gibbs free energy] between the liquid and vapor. $\endgroup$
    – jimchmst
    Commented Dec 26, 2023 at 21:11
  • $\begingroup$ By other words, there is not a constant mass for the curves above. $\endgroup$
    – Poutnik
    Commented Dec 26, 2023 at 21:55

1 Answer 1

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If you took out vapor at some moment and if you put it in a container of a fixed volume then its pressure would linearly grow with temperature, as in the Charles' law.

The point is, amount of vapor above liquid is not constant. It grows roughly exponentially with temperature and so does the pressure.

For a narrow enough temperature interval to consider the evaporation enthalpy approximately constant, the vapor pressure dependency on temperature is:

$$p_2=p_1 \cdot \exp {\left(\frac{\Delta H_\text{evap}}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)\right)}$$

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