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3. Given 325 ml of saturated gas at 760 mmHg and 25°C, what would be its volume if dry at the same pressure and temperature?

I got confused when I read 'if dry' Does that mean I have to subtract something out of 760mmHg?
And would I used the combined gas law and solved for V2?

a teacher answered it but I had a question about their answer.
They said:" There will be less gas when the saturated gas is dried. Subtract the vapor pressure of water at 25 C (23.8 mm Hg) from the 760 mmHg pressure to get the initial dry gas partial pressure. Since the pressure is maintained at 760 mm, that means the volume must be reduced by the fraction (760- initial water vapor pressure)/760 "

How did they get 23.8mmHg?

And I wanted to know if I could subtract 47mmHg from 760mmHg to get the new number component?

thanks

Answer 1

Yes, when the problem says "dry" you subtract the vapor pressure of water AT THAT TEMPERATURE. How do we get that vapor pressure? We look it up in a table. The vapor pressure of water at 25 degrees C is 23.8 mm Hg. So 760-23.8 is the pressure of the dry gas (the other 23.8 is the pressure of the water in the gas). I don't know where you obtained the 47 or what is is. You may use the combined gas laws OR you may use PV = nRT, find the number of mols at the pressure of 760-23.8, then use that n and plug in the new conditions of T and P to find the new volume.