How do I calculate the molar volume and pressure correction terms in the van der Waals equation of state for ${CO}_{2}$ $"CO"_2$ if the density of ${CO}_{2}$ $"CO"_2$ at a certain temperature is $4.4 g/L$ $"4.4 g/L"$ , while $a = {3.6 L}^{2} \cdot {atm/mol}^{2}$ $a = "3.6 L"^2cdot"atm/mol"^2$ and $b = 0.04 L/mol$ $b = "0.04 L/mol"$ ?

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How do I calculate the molar volume and pressure correction terms in the van der Waals equation of state for ${CO}_{2}$ $"CO"_2$ if the density of ${CO}_{2}$ $"CO"_2$ at a certain temperature is $4.4 g/L$ $"4.4 g/L"$ , while $a = {3.6 L}^{2} \cdot {atm/mol}^{2}$ $a = "3.6 L"^2cdot"atm/mol"^2$ and $b = 0.04 L/mol$ $b = "0.04 L/mol"$ ?

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Answer 1 · 2017-12-07T17:51:25

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

$P = \frac{R T}{¯ ¯ ¯ V - b} - \frac{a}{{¯ ¯ ¯ V}^{2}}$ $P = (RT)/(barV - b) - a/(barV^2)$

are:

${¯ ¯ ¯ V}_{corr} = ¯ ¯ ¯ V - b$ $barV_"corr" = barV - b$

$P_{corr} = \frac{a}{{¯ ¯ ¯ V}^{2}}$ $P_"corr" = a/(barV^2)$

where $¯ ¯ ¯ V = \frac{V}{n}$ $barV = V/n$ is the molar volume, $a$ $a$ is the vdW term for intermolecular forces, and $b$ $b$ 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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