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Topic: vaporization of methyl salicylate  (Read 5169 times)

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Offline rleung

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vaporization of methyl salicylate
« on: May 23, 2006, 07:54:53 PM »
Hi,

At 25 degrees C, the vapor pressure of methyl salicylate is 0.131 torr.  I have to find how many liters of air must pass over the compound at 25 degrees C to vaporize 6.57x10^-6 mol.  The answer I have in the solutions book is to use PV=nRT:

(0.131/760 atm)(V)=(6.57x10^-6 mol)(0.0821)(298-K)

and solve for V, which will be the answer.  Apparently, this seems to mean that the volume of air needed to vaporize that amount of methyl salicylate will equal the volume of methyl salicylate present as vapor in air at that temperature and pressure.  So, I guess my question is whether or not this generalizes to almost every situation (does the amount of air needed to vaporize something always equal the volume occupied by the gaseous molecules of the substance you wish to vaporize)?  Thanks.

Ryan

Offline Borek

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Re: vaporization of methyl salicylate
« Reply #1 on: May 24, 2006, 03:31:39 AM »
my question is whether or not this generalizes to almost every situation (does the amount of air needed to vaporize something always equal the volume occupied by the gaseous molecules of the substance you wish to vaporize)?

That's good approximation.
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