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
Truong-Son N.
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"#

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