Question

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?

Answer 1

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

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

`P = (nRT)/(V-nb) – (n^2a)/V^2`

`= (1.00 color(red)(cancel(color(black)("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"))))^2`

`= "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.