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?

1 Answer
Jan 14, 2017

Answer:

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.