Question

How would you use the van der Waals equation of state to calculate the pressure of 3.60 mol of `H2O` at 453 K in a 5.90-L vessel?

Answer 1

The pressure is 21.4 bar.

Explanation:

The van der Waals equation is

`color(blue)(bar(ul(|color(white)(a/a) (P + (n^2a)/V^2)(V - nb) = nRTcolor(white)(a/a)|)))" "`

`P + (n^2a)/V^2 = (nRT)/(V - nb)`

`P = (nRT)/(V - nb)- (n^2a)/V^2`

For this problem,

`n = "3.60 mol"`
`R = "0.083 14"color(white)(l)"bar·L·K"^"-1""mol"^"-1"`
`T = "453 K"`
`V = "5.90 L"`
`a = "5.536 bar·L"^2"mol"^"-2"`
`b = "0.030 49 L·mol"^"-1"`

`P = (nRT)/(V-nb) – (n^2a)/V^2`

`= (3.60 color(red)(cancel(color(black)("mol"))) × "0.083 14 bar"color(red)(cancel(color(black)("L·""K"^(-1)"mol"^(-1))))× 453 color(red)(cancel(color(black)("K"))))/(5.90 color(red)(cancel(color(black)("L"))) – 3.60 color(red)(cancel(color(black)("mol")))× "0.030 49" color(red)(cancel(color(black)("L·mol"^(-1))))) - ((3.60 color(red)(cancel(color(black)("mol"))))^2 × "5.536 bar" color(red)(cancel(color(black)("L"^2"mol"^"-2"))))/(5.90 color(red)(cancel(color(black)("L"))))^2`
.
`= "135.6 bar"/(5.90 - 0.1098) - "2.061 bar "= "23.42 bar" - "2.061 bar"= "21.4 bar"`

The pressure predicted by the van der Waals equation is 21.4 bar.