How do I calculate the molar volume and pressure correction terms in the van der Waals equation of state for #"CO"_2# if the density of #"CO"_2# at a certain temperature is #"4.4 g/L"#, while #a = "3.6 L"^2cdot"atm/mol"^2# and #b = "0.04 L/mol"#?
1 Answer
Dec 7, 2017
The van der Waals (vdW) volume correction and pressure correction terms in
#P = (RT)/(barV - b) - a/(barV^2)#
are:
#barV_"corr" = barV - b#
#P_"corr" = a/(barV^2)# where
#barV = V/n# is the molar volume,#a# is the vdW term for intermolecular forces, and#b# is the vdW term for excluded volume of a non-point-mass particle.
Therefore:
#barV_"corr" = "1 L"/(4.4 cancel"g" cdot "1 mol"/(44.01 cancel("g CO"_2))) - "0.04 L"/"mol"#
#= ul"9.96 L/mol"#
#P_"corr" = (3.6 cancel("L"^2)cdot"atm"/cancel("mol"^2))/([1 cancel"L"//4.4 cancel"g" cdot cancel"1 mol"/(44.01 cancel("g CO"_2))]^2)#
#=# #ul"0.360 atm"#