# A 9.380 mol sample of nitrogen gas is maintained in a 0.8198 L container at 301.8 K. What is the pressure in atm calculated using the van der Waals' equation for N2 gas under these conditions? For N2, a = 1.390 L2atm/mol2 and b = 3.910×10-2 L/mol. atm?

Jun 5, 2018

$p = \text{331.0 atm}$

#### 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,

$n = \text{9.380 mol}$
$R = \text{0.082 06"color(white)(l)"L·atm·K"^"-1""mol"^"-1}$
$T = \text{301.8 K}$
$V = \text{0.8198 L}$
$a = \text{1.390 atm·L"^2"mol"^"-2}$
b = 3.912 × 10^"-2"color(white)(l) "L·mol"^"-1"

P = (nRT)/(V-nb) – (n^2a)/V^2

$= {\left(9.380 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{mol"))) × "0.082 06 atm"color(red)(cancel(color(black)("L·""K"^"-1""mol"^"-1")))× 301.8 color(red)(cancel(color(black)("K"))))/(0.8198 color(red)(cancel(color(black)("L"))) – 9.380 color(red)(cancel(color(black)("mol")))× 3.912 × 10^"-2" color(red)(cancel(color(black)("L·mol"^(-1))))) - ((9.380 color(red)(cancel(color(black)("mol"))))^2 × "1.390 atm" color(red)(cancel(color(black)("L"^2"mol"^"-2"))))/(0.8198 color(red)(cancel(color(black)("L}}}}\right)}^{2}$
$= \text{232.3 atm"/0.4528 - "182.1 atm "= "513.0 atm" - "182.0 atm"= "331.0 atm}$

The pressure predicted by the van der Waals equation is 331.0 atm.

Note: The pressure of an ideal gas would be 283.4 atm.

The calculation shows that the volume occupied by the molecules increases the pressure more than the intermolecular attractive forces decreased it.