# What is the pressure of 1.00 mole of Cl_2 gas that is non-ideal, at a temperature of 0.0 deg C occupying a volume 34.6 L?

Jan 14, 2017

The pressure is 0.652 bar.

#### Explanation:

I assume that you are treating the chlorine as a van der Waals gas.

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,

$n = \text{1.00 mol}$
$R = \text{0.083 14"color(white)(l)"bar·L·K"^"-1""mol"^"-1}$
$T = \text{273.15 K}$
$V = \text{34.6 L}$
$a = \text{6.579 bar·L"^2"mol"^"-2}$
$b = \text{0.0562 L·mol"^"-1}$

P = (nRT)/(V-nb) – (n^2a)/V^2

$= {\left(1.00 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{mol"))) × "0.083 14 bar"color(red)(cancel(color(black)("L·""K"^(-1)"mol"^(-1))))× 273.15 color(red)(cancel(color(black)("K"))))/(34.6 color(red)(cancel(color(black)("L"))) – 1.00 color(red)(cancel(color(black)("mol")))× 0.0562 color(red)(cancel(color(black)("L·mol"^(-1))))) - ((1.00 color(red)(cancel(color(black)("mol"))))^2 × "6.579 bar" color(red)(cancel(color(black)("L"^2"mol"^"-2"))))/(34.6 color(red)(cancel(color(black)("L}}}}\right)}^{2}$

$= \text{22.71 bar"/(34.6 - 0.0562) - "0.005 50 bar "= "0.6574 bar" - "0.005 50 bar"= "0.652 bar}$

The pressure predicted by the van der Waals equation is 0.652 bar.

Note: The Ideal Gas Law predicts a pressure of 0.656 bar. Under these conditions, chlorine is behaving almost like an ideal gas.