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

1 Answer
Feb 2, 2017

Answer:

#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.