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?

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.