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Feb4-13, 12:52 PM
P: 645
Quote Quote by Philip Wood View Post
Boiling is a very special case of evaporation. It involves bubbles of vapour forming below the surface of the water. The bubbles usually form on minute concavities on the inside of the vessel below the liquid level, or on lime-scale or other solids below the water level. [The water is evaporating into minute air pockets in these concavities.] The bubbles are able to grow, and then break away and rise to the surface, when the saturated vapour pressure inside them is equal to the external pressure on them. This pressure is that of the atmosphere (about 101000 Pa) plus the extra pressures (usually negligible) of the liquid column above them and of surface tension at the bubble surface. So at what temperature is the svp of water equal to 101000 Pa? 100°C !

Ordinary evaporation from the exposed surface of a liquid is the unco-ordinated escape of individual molecules. They do not have to fight against atmospheric pressure, except inasmuch as they will collide with the molecules of air once they have escaped. So ordinary evaporation does not require the svp of the liquid to equal atmospheric pressure, as it does not involve the escape of bubbles of vapour having to grow and escape through the liquid against atmospheric pressure.
Oh but could you explain how the particles gain enough energy to evaporate? Since heat transfers from a region of high temperature to a region of low temperature how would it apply to the heat transfer to allow evaporation? I'm having trouble understanding how the heat energy can accumulate at the surface of the liquid..

Thanks for the help :)