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We know vapour pressure is dependent on temperature. But why it is not dependent on atmospheric pressure or pressure applied on the gas surrounding vapour-water equilibrium?

Pressure does influence vaporization/evaporation process. If amount of vapor decreases due to uncrease in pressure shouldn't the vapour pressure decrease too?

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  • $\begingroup$ The vapor pressure is a property of the water as is the temperature. The pressure is a property of the environment the water is within. When the two pressures are equal it's called the boiling point. It's the difference in these pressures that influences evaporation. $\endgroup$ – R. Rankin Apr 15 at 19:29
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But why is [the vapour pressure] not dependent on atmospheric pressure or pressure applied on the gas surrounding vapour-water equilibrium?

Does this beg the question? In other words, is the vapor pressure really independent of the surrounding pressure? Pressurizing a system tends to increase its energy, which we'd expect to affect the vapor pressure as an indication of the state of the condensed matter beneath it. Let's see.

The vapor pressure can be modeled as satisfying $$\mu=\mu_0+RT\ln\left(\frac{p_v}{p_0}\right),$$

where $\mu$ is the chemical potential (i.e., the molar Gibbs free energy, with $d\mu=-s\,dT+v\,dP$ for a closed system, where $s$ is the molar entropy, $T$ is the temperature, $v$ is the molar volume, and $P$ is the system pressure), $\mu_0$ is a reference value, $R$ is the gas constant, $p_v$ is the equilibrium vapor pressure in atm, and $p_0=1\,\mathrm{atm}$.

Differentiating at constant temperature, we obtain

$$d\mu=\frac{RT}{p_v}dp_v=v\,dP;$$

$$\frac{dp_v}{dP}=\frac{vp_v}{RT}.$$

Therefore, for small changes in $p_v$, $$\Delta p_v\approx\frac{vp_v}{RT}\Delta P.$$

But $\frac{vp_v}{RT}$ is generally a very small number ($10^{-8}$ for water at room temperature, for example). Thus, it may seem that the vapor pressure is not affected by the surrounding pressure—but it is.

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Let's clarify the context. In a system of water-vapor equilibrium, with no other gases present, the pressure of the vapor is the same as the pressure that is applied by the surroundings on the system (e.g. by a piston on top of the vessel).

Now imagine we have a mixture of vapor and another gas (say nitrogen). The pressure on the piston will be the sum of vapor pressure and nitrogen pressure. the gas particles are so far apart, that they rarely interact with each other, and simply coexist in the same space, each pushing on the piston on their own.

Nitrogen pressure doesn't affect the vaporization/condensation process. This means the net pressure can have different values without affecting the water/vapor behavior. So, the net pressure isn't a relevant parameter for determining the vapor pressure.

The temperature, on the other hand, is the same for every substance in equilibrium. So increasing temperature will increase the vapor pressure and vaporization and condensation rates of water.

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  • $\begingroup$ Could it be like this.. That the air particles did obstruct surface molecules to vaporize but the energy that triggered those molecules to vaporize was released to the surrounding which eventually resulted in rise of pressure on the water surface, therefore the outcome was same? $\endgroup$ – MSKB Apr 15 at 19:40

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