# 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"?

Dec 7, 2017

The van der Waals (vdW) volume correction and pressure correction terms in

$P = \frac{R T}{\overline{V} - b} - \frac{a}{{\overline{V}}^{2}}$

are:

${\overline{V}}_{\text{corr}} = \overline{V} - b$

${P}_{\text{corr}} = \frac{a}{{\overline{V}}^{2}}$

where $\overline{V} = \frac{V}{n}$ 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:

${\overline{V}}_{\text{corr" = "1 L"/(4.4 cancel"g" cdot "1 mol"/(44.01 cancel("g CO"_2))) - "0.04 L"/"mol}}$

$= \underline{\text{9.96 L/mol}}$

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)

$=$ $\underline{\text{0.360 atm}}$