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I know that if you evaporate a molecule of water, that it must be 100C in order to be gaseous, so that it will take that heat with it if it wasn't already that temperature. This has a cooling effect if, for example, you're standing around wet.

I also know that if pressure is reduced then boiling temperature is likewise reduced, and the same for increasing pressure.

What I'm unsure of is whether reducing pressure changes the 100C that gaseous water must be.

In particular, in a vacuum water would basically instantly boil. Say due to the strength of the vacuum the boiling temperature is -50C. Does the gaseous water still need to be 100C, or has that temperature also been reduced? Would the water molecules that are boiling at -50C still need to be 100C to be gaseous, and so each one takes 150 degrees of energy with it when it evaporates in the vacuum? Or has the 100C been reduced to -50C, so the boiling water is already energetic enough to be gaseous, and does not extract any additional energy from other nearby matter?

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I know that if you evaporate a molecule of water, that it must be 100C in order to be gaseous

It's very difficult to assign a temperature to an individual molecule. The temperature we measure is an average property over a collection of particles.

But you could say that under normal circumstances (pressure of 1 atmosphere) if the water vapor is less than 100C, most of it will condense back into a liquid.

so that it will take that heat with it if it wasn't already that temperature.

Usually you think about heating the water as two separate steps. Some energy is added to the water to raise it to the boiling temperature (based on the heat capacity of the water), and then additional energy is necessary to vaporize the water (based on the latent heat of vaporization). So even without increasing temperature, vaporization still removes energy from the surroundings.

Would the water molecules that are boiling at -50C still need to be 100C to be gaseous, and so each one takes 150 degrees of energy with it when it evaporates in the vacuum?

No. When a gram of water boils (turns from liquid to vapor) it takes 2260 J of energy, but the vapor remains at the same temperature. If you create a container of vapor by lowering the pressure in room temperature water, the vapor doesn't suddenly become hot.

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In simple words, the liquid molecules turn into gaseous state after they've acquired a certain amount of energy. At normal pressure, raising the temperature to 100C helps the molecules in attaining that energy, thus moving away from each other (because of increased randomness) to such an extent that it eventually becomes a gas. As pressure is reduced, the spaces between the molecules increase anyway, so less energy is required to be given to them so that they can move away from each other and become a gas.
In vacuum, because there's no pressure at all, the water boils off instantly.

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