# 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?

Oct 19, 2016

The pressure is 21.4 bar.

#### Explanation:

The van der Waals equation is

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \left(P + \frac{{n}^{2} a}{V} ^ 2\right) \left(V - n b\right) = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

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

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

For this problem,

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

P = (nRT)/(V-nb) – (n^2a)/V^2

$= {\left(3.60 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{mol"))) × "0.083 14 bar"color(red)(cancel(color(black)("L·""K"^(-1)"mol"^(-1))))× 453 color(red)(cancel(color(black)("K"))))/(5.90 color(red)(cancel(color(black)("L"))) – 3.60 color(red)(cancel(color(black)("mol")))× "0.030 49" color(red)(cancel(color(black)("L·mol"^(-1))))) - ((3.60 color(red)(cancel(color(black)("mol"))))^2 × "5.536 bar" color(red)(cancel(color(black)("L"^2"mol"^"-2"))))/(5.90 color(red)(cancel(color(black)("L}}}}\right)}^{2}$
.
$= \text{135.6 bar"/(5.90 - 0.1098) - "2.061 bar "= "23.42 bar" - "2.061 bar"= "21.4 bar}$

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