What is the volume expressed by a #74.3*g# mass of carbon dioxide at #294.9*K# enclosed in a piston at a pressure of #439*"mm Hg"#?

1 Answer
Mar 23, 2017

Answer:

#V=(nRT)/P~=70*L#

Explanation:

The key to this question is to express the pressure, which is given in #"mm Hg"#, to atmospheres. We know (or should know) that #1*atm# will support a column of mercury that is #760*mm# high. Mercury columns (which are fast disappearing in laboratories these days) provides a highly visual and accessible measurement of pressures BELOW atmospheric.

And thus #P=("439 mm Hg")/("760 mm Hg"*atm^-1)=0.578*atm#

And thus #V=((74.3*g)/(44.0*g*mol^-1)xx0.0821*L*atm*K^-1*mol^-1xx294.9K)/(0.578*atm)#

#~=70*L#

Do the units in the calculation cancel out to give an answer in #L#? If not they should!