Question

What volume will `1.50*mol` of an Ideal Gas occupy at `225*K`, at a pressure of `231*mm*Hg`?

Answer 1

`V=(nRT)/P~=90*L`

Explanation:

`V=(1.50*molxx0.0821*L*atm*K^-1*mol^-1xx225*K)/((231*mm*Hg)/(760*mm*Hg*atm^-1)`

`=??L`

Do the units cancel out to give an answer in `"litres"`? They should!

The key to doing problems such as these is to recognize that `1*atm-=760*mm*Hg`. A column of mercury is traditionally used in a laboratory to measure pressure (mind you, these days, they are disappearing due to safety concerns).

`"1 atmosphere"` pressure will support a column of mercury `760*mm` high. And thus in the problem we use the quotient, `(231*mm*Hg)/(760*mm*Hg*atm^-1)` to give an answer in `1/(atm^-1)=1/(1/(atm))=atm` as required.