Chemical Forums
Chemistry Forums for Students => Inorganic Chemistry Forum => Topic started by: g36jordan on May 06, 2020, 06:25:15 PM
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My apologies, I am not sure if this is the right sub for this question but I figured it would be applicable.
I am looking for an explanation of vapor pressure as it relates to how to make a determination on how "volatile" a chemical is. I always understood the higher the vapor pressure, the more a chemical will naturally volatilize even at STP. How then, if the pressure exerted by air is 760 mmHg at STP, can something like Methylene Chloride (considered to be a pretty volatile chemical) volatilize into the air with a VP of about 353 mmHg? Wouldn't its vapor pressure need to increase to beyond 760 mmHg before the molecules jump into the air? How does this work? Something like formaldehyde, with a VP > 1 atm, is something I can understand, but in what way are chemicals, at STP, with vapor pressures less than 760 mmHg, still volatile?
I realize I must sound like an idiot, but I figured I, as a non-chemist, would ask the chemists.
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Whatever evaporates pushes out the air. Total pressure stays the same, just the (local) composition changes.
Unless you work with a closed container, then the pressure will build up.
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I am still confused, if evaporation can occur with a vapor pressure less than 760 mmHg then why does water need to boil (aka get up to a vapor pressure of 760 mmHg) to evaporate? Or can it evaporate at less VP?
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Evaporation occurs before boiling. This happens after a rainfall for instance.
If the vapour pressure exceeds the pressure of the surrounding gas (for instance the atmosphere), then the vapour can form bubbles in the liquid. The liquid boils.