Why is vaporisation rate increased by a decrease in pressure

• Roroy
In summary, the rate of vaporization can be increased by increasing the temperature or reducing the pressure. An increase in temperature leads to more energy, making it more likely for bonds to break and molecules to become vapor. A reduction in pressure reduces the force needed for molecules to overcome and vaporize, which can be observed by different boiling points at different heights. The vaporization rate depends on the vapor pressure, which is affected by the ratio between mass transfer from the liquid into the gas phase. The system works by moving farther from equilibrium, making evaporation faster. In real-life systems, a pressure cooker uses high pressure to increase the boiling point of water, while Mt. Everest's low pressure means a lower temperature is needed to reach boiling point.
Roroy
In my textbook, it says that the rate of vaporisation can be increased by (a) increasing the temperature or (b) reducing the pressure.

I understand why an increase in temp can lead to more molecules breaking away from the liquid and entering the vapor state. More temp = more energy = more likely for the bonds to break and for molecules to become vapor.

However, I don't understand why a reduction in pressure would have this effect. An increase in pressure is similar to an increase in temperature, isn't it?

My logic is that an increase in pressure leads to more molecular collisions, which results in an increase in energy = more likely for bonds to break. Therefore, why does a reduction in pressure lead to an increase in vaporisation rate?

Appreciate any help.

The air acts like a cap on the fluid, its weight is the pressure. Less pressure means less force to overcome by the molecules to vaporize (and displace air molecules). This effect can also be observed by different boiling points of water in different heights.

ProfuselyQuarky and Drakkith
fresh_42 said:
The air acts like a cap on the fluid, its weight is the pressure. Less pressure means less force to overcome by the molecules to vaporize (and displace air molecules). This effect can also be observed by different boiling points of water in different heights.

What pressure are we talking about? The vapour pressure or the total pressure? The vaporization rate depends on the vapour pressure because it affects the ratio between the mass transfer from the liquid into the gas phase and vice versa. But I do not think that changing the total pressure at constant vapour pressure has a significant effect on vaporization as long as it is above the equilibrium vapour pressure.

More air pressure means the vaporization pressure must be higher to reach the equilibrium and vice versa.

fresh_42 said:
More air pressure means the vaporization pressure must be higher to reach the equilibrium and vice versa.

Not at all. Equilibrium pressure of the saturated vapor depends only on the temperature.

So how does the system work?
Cause I've been googling and all I can find is that above explanation that a reduction in pressure means that the equilibrium is reached faster.

Part of the problem is, the statement

Roroy said:
rate of vaporisation can be increased by (...) reducing the pressure

is a bit ambiguous, as it doesn't clearly state - reducing pressure of what. And there are two pressures involved here - the total pressure (including all the inert gases) and the partial pressure of vapor.

I take it it was intended to mean "removing the gases present above the liquid" - in which case both the total and the partial pressure are getting lower. That means moving system farther from the equilibrium, so the evaporation gets faster. The same effect would be achieved by reducing pressure of the vapor only (without changing the pressure of the inert gases).

That's all assuming temperature is low enough that the liquid doesn't boil and that the temperature remains constant (which means enough heat is delivered to the liquid to not let the liquid cool down, evaporation requires energy).

So, if Saturated Vapor depends on temp, then wouldn't it be right in saying:

In a system where the atmospheric pressure (all inert gases, as you mentioned) is high, then a higher temp would be required to increase vapour pressure enough to equal the atmospheric pressure (at which point the liquid reaches its boiling point).

Similarly, in a system with low atmospheric pressure (again, only counting the pressure exerted by air molecules), then a lower temp is fine to increase vapour pressure enough to equal the atmospheric pressure (liquid reaches its boiling point).

Applying this on two real life systems: a pressure cooker vs Mt. Everest.

In a pressure cooker, the high pressure causes the liquid to reach boiling point at a much higher temp (let's say that water has a boiling point of 200C in a pressure cooker). This means that the food cooks in 200C water, which means it cooks faster.

On Mt. Everest, the low pressure means that the liquid needs a lower temp to reach boiling point (let's say water has a boiling point of 90C on Mt. Everest). This means that food cooks in 90C water, hence cooks slower.

Is that all correct? Cause it seems to match all of the answers given, and all of the research I've done.

Actually much less than 90°C at Mt Everest (something around 71) and it doesn't go up to 200°C in a pressure cooker (around 120, depending on the valve). But yes, you are about right.

Surya97
Thanks for the help

fresh_42 said:
More air pressure means the vaporization pressure must be higher to reach the equilibrium and vice versa.
I agree with Borek. It's only the partial pressure of the water vapor in the air above the liquid (I.e., the relative humidity) that matters in terms of pressure (for straight evaporation below the boiling point). Once the boiling point is reached, the surface area for evaporation increases.

Last edited:
fresh_42
Borek said:
Not at all. Equilibrium pressure of the saturated vapor depends only on the temperature.
I remember vaguely that vapour pressure over a liquid increases with increasing pressure.

DrDu said:
I remember vaguely that vapour pressure over a liquid increases with increasing pressure.
Within the context of the OP's question, how important do you feel it was to mention this typically minor non-ideality (related to the so-called "Poynting effect" of total pressure on the fugacity of the liquid) at the risk of confusing the OP? Please explore this for us with some sample calculations to illustrate the magnitude of the correction. For example, at 1 atm total pressure, how much does this correction affect the equilibrium vapor pressure of water?

Bystander
So this is called Poynting effect. Just looked it up on wikipedia. Does not seem to be due to nonideality and they cite even an interesting application of it. But you are right, it is apparently too small to be relevant for the vaporisation of water at ambient pressure.
A more relevant aspect of pressure dependence of vaporisation is the increased free pathlength for particles leaving the surface or higher diffusivity at lower pressure. I.e. the air on a water surface acts like a lid and the better so the higher its pressure.

1. Why does a decrease in pressure increase the rate of vaporisation?

According to the kinetic theory of gases, vaporisation occurs when molecules at the surface of a liquid gain enough energy to overcome the intermolecular forces holding them together. When pressure is decreased, the number of molecules above the liquid's surface decreases, reducing the number of collisions between molecules and making it easier for molecules to escape into the gas phase. This results in a higher vaporisation rate.

2. How does the boiling point of a liquid change with decreasing pressure?

As pressure decreases, the boiling point of a liquid also decreases. This is because at lower pressures, the vaporisation rate increases and the liquid molecules escape more easily into the gas phase. As a result, the temperature required for the liquid to reach its vapor pressure (boiling point) is lower.

3. Can the vaporisation rate be increased indefinitely by decreasing pressure?

No, the vaporisation rate cannot be increased indefinitely by decreasing pressure. As the pressure approaches a very low value, the number of molecules above the liquid's surface becomes so small that the vaporisation rate reaches a maximum and cannot be increased any further.

4. How does altitude affect the vaporisation rate of liquids?

Altitude affects the vaporisation rate of liquids in a similar way to decreasing pressure. As altitude increases, the air pressure decreases, resulting in a lower number of molecules above the liquid's surface. This reduces the number of collisions between molecules and leads to a higher vaporisation rate.

5. Is the effect of pressure on vaporisation rate the same for all liquids?

No, the effect of pressure on vaporisation rate can vary depending on the properties of the liquid. Some liquids may have weaker intermolecular forces, making it easier for molecules to escape into the gas phase at lower pressures. Others may have stronger intermolecular forces, resulting in a smaller change in vaporisation rate with decreasing pressure.

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