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It is said that hot air holds more water vapour.

  1. I guess this means that the saturation vapour pressure of water in air increases with temperature. Is this correct?

  2. How can one derive this result? This seems related https://en.wikipedia.org/wiki/Clausius%E2%80%93Clapeyron_relation but it assumes quite a bit of prior knowledge. Also, the final result seems to imply that the derivative of pressure with respect to temperature actually decreases quadratically with temperature. But wouldn't that imply that hot air can hold less water vapour (pressure decreases)?

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It is correct, though the idea that air has a 'holding' rôle is, I think unhelpful. The air simply gives a background partial pressure.

Consider first a liquid evaporating into the space above it in a closed container. The vapour is 'saturated' when as many molecules of vapour return per second to the liquid by random bombardment of its surface as leave the liquid by evaporation. If the liquid temperature is raised the evaporation rate increases, because more molecules will have enough energy to escape the relatively short-range intermolecular forces. This will increase the concentration, $\nu_{vap}$, of molecules in the vapour. So the return rate of molecules, $\tfrac 14 \nu_{vap} \bar c$ per unit area of liquid surface increases until it equals the number leaving per unit area per second. But since $\nu_{vap}$ has increased, the vapour pressure, $\frac 13 \nu_{vap} m \bar{c^2}$ has increased.

We can obtain an expression for the saturated vapour pressure, $p_{sat vap}$, by a quite different approach: integrating the Clausius-Clapeyron equation, treating the molar volume, $V_{liq}$, of the liquid as negligible, so $$\frac{dp}{dT}=\frac {\Delta H_{vap}}{TV_{satvap}}.$$ Here, $V_{satvap}$ is the molar volume at the 'wholly vapour' end (point X, say) of the horizontal part of the isotherm on a p_V diagram. We shall assume, as an approximation, that the saturated vapour behaves as an ideal gas, so $V_{satvap}=RT/p$ can be applied along the curve through the points, X, for the different isotherms.

So, eliminating $V_{sat vap}$ we now have $$\frac{dp}{dT}=\frac {\Delta H_{vap}\ p}{RT^2}.$$ Separating variables, integrating (using the approximation that $\Delta H_{vap}$ is independent of temperature) and tidying we get $$p_{sat vap}=p_0\exp\left(\tfrac{\Delta H_{vap}}{RT_0}\right)\exp\left(-\tfrac{\Delta H_{vap}}{RT}\right)$$ in which $p_0$ is the saturated vapour pressure at some reference temperature, $T_0$. If we put $P_0=p_0\exp\left(\tfrac{\Delta H_{vap}}{RT_0}\right)$, then our equation becomes simply $$p_{sat vap}=P_0\exp\left(-\tfrac{\Delta H_{vap}}{RT}\right).$$ The equation confirms the increase of $p_{sat\ vap}$ with temperature, $T$, though, because of the approximations we've made, great accuracy is not expected.

It is pleasing, though, that our equation features the Boltzmann factor, $\exp \left(-\frac {\Delta H_{vap}}{RT}\right)$ as this is what we'd expect from statistical mechanics for molecules needing a threshold energy to escape from a liquid.

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The higher the temperature, the greater the tendency of a liquid to become a vapor. The physical reason is that the kinetic energy of molecules increases with the temperature, which makes it easier to escape to the vapor phase. This explains qualitatively the increase of the saturation pressure with temperature.

Clausius Clapeyron Equation

The Clausius-Clapeyron equation says the same thing, quantitatively. This equation gives the derivative of the saturation pressure wrt temperature: $$ \frac{dP^\text{sat}}{dT} = \frac{\Delta H_\text{vap}}{T (V_V-V_L)} $$ where $\Delta H_\text{vap}$ is the enthalpy of vaporization, $V_V$ is the molar volume of the saturated vapor and $V_L$ is the volume of the saturated liquid. Since $\Delta H_\text{vap}\geq 0$ and $V_V>V_L$, the derivative $dP^\text{sat}/dT$ is always greater than zero. This shows that $P^\text{sat}$ increases with temperature.

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  • $\begingroup$ I fixed the obvious typos, feel free to edit yourself if there are more. $\endgroup$
    – Themis
    Commented May 24, 2023 at 22:15
  • $\begingroup$ Mostly right, but the widely claimed "physical reason is that kinetic energy increases ..." is not so clear. For example, if you increase the pressure on a vapour at constant $T$ (below critical temperature), it will condense. The preferred phase is really a question of entropy not energy (or to be precise, Gibbs function). $\endgroup$ Commented May 25, 2023 at 14:48
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    $\begingroup$ @AndrewSteane Yes – and above the critical point there is no phase transition at all. There is no denying that the explanation of phase transitions is thermodynamic (Gubbs energy, chemical potential), not mechanical (kinetic/potential energy). But on a qualitative level the mechanical picture is simple enough and conveys the main idea: increasing $T$ increases the escape tendency from the liquid by overcoming intermooecular attraction, increasing $P$ increases the escape tendency from the vapor. $\endgroup$
    – Themis
    Commented May 26, 2023 at 12:10
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A simple bit of physical reasoning that applies, is the minimization of Gibbs free energy at equilibrium. Water vapor has greater entropy than liquid water, and to minimize $$G = E - T \times S$$ where G is Gibbs free energy, E is thermal energy, T is absolute temperature, and S is entropy...

well, the entropy of that vapor becomes a more important part of the minimization when T rises.

There are other consequences; you cool by evaporation on your skin, but if salt builds up... salt water has more entropy than fresh, so evaporation cooling is more effective if you rinse off with fresh water after a dip in the ocean.

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  1. Correct.
  2. If you want to avoid using the Clausius-Clapeyron relation, you can quantify the saturation pressure on the coexistence diagram. Another thing, you mention that the derivative of saturation pressure decreases with temperature, which is true, but it remains positive, which means saturation pressure still increases as temperature increases. You can also look up other approximations of saturation pressure as a function of temperature
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