What is the volume expressed by a 74.3*g74.3⋅g mass of carbon dioxide at 294.9*K294.9⋅K enclosed in a piston at a pressure of 439*"mm Hg"439⋅mm Hg?

1 Answer
anor277
Mar 23, 2017

V=(nRT)/P~=70*LV=nRTP≅70⋅L

Explanation:

The key to this question is to express the pressure, which is given in "mm Hg"mm Hg, to atmospheres. We know (or should know) that 1*atm1⋅atm will support a column of mercury that is 760*mm760⋅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*atmP=439 mm Hg760 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)V=74.3⋅g44.0⋅g⋅mol−1×0.0821⋅L⋅atm⋅K−1⋅mol−1×294.9K0.578⋅atm

~=70*L≅70⋅L

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

Related questions
See all questions in Ideal Gas Law
Impact of this question
1800 views around the world
You can reuse this answer
Creative Commons License