At non-boiling point, I understand that when you add heat, temperature increases and also the amount of gas increases, so by using $PV=nRT$ at constant volume, the increase in $n$ and $T$ would increase $P$.

Thus, vapour pressure increases with temperature.

However, at boiling point, if you add heat, temperature does not increase and the energy is used to generate more gas molecules.

So if you are at a equilibrium where there is $\pu{1 mol}$ of $\ce{A(l)}$ and $\pu{1 mol}$ of $\ce{A(g)}$, after you add heat and shift equilibrium to $\pu{0.5 mol}$ of $\ce{A(l)}$ and $\pu{1.5 mol}$ of $\ce{A(g)}$ wouldn’t the vapour pressure increase even though temperature is the same by the ideal gas law– $PV=nRT$, where increase in $n$ increases $P$?

So why is vapour pressure is constant at boiling point?

  • $\begingroup$ Of course it would increase, if you keep the volume constant (as would T, by the way). Who said it wouldn't? The pressure is constant in a constant-pressure setting, but that's kinda trivial. $\endgroup$ – Ivan Neretin Sep 20 '20 at 8:36
  • $\begingroup$ At the boiling point under constant pressure, the temperature is constant. Seems you are trying to make sense of this by looking at (likely sloppily written) definitions in a book. Keep the experiment in your mind! $\endgroup$ – Karl Sep 20 '20 at 8:41
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    $\begingroup$ By the way, "vapour pressure" is only defined for $T < T_{Bp}$. At $T >= T_{Bp}$, vapour pressure makes no sense. $\endgroup$ – Karl Sep 20 '20 at 8:43
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    $\begingroup$ It is fixed if you keep the pressure fixed. Otherwise it isn't. $\endgroup$ – Ivan Neretin Sep 20 '20 at 9:06
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    $\begingroup$ Imagine you heat water. When reaching 100°C, the water boils at ordinary pressure. If you carry out this operation in a closed syringe, the volume of the gas will steadily increase, the number of moles of gas will also increase. You may continue heating, the volume of gas will increase, but the pressure remains at 1 atm. and the temperature remains at 100°C. At the end, when all liquid water has been vaporized, heating the vapor even more will increase its temperature and its volume. But the presence of two phases prevents the temperature from increasing by heating. $\endgroup$ – Maurice Sep 20 '20 at 10:51

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