Question

According to the ideal gas law, a 0.9766 mol sample of carbon dioxide gas in a 1.295 L container at 270.0 K should exert a pressure of 16.71 atm. By what percent does the pressure calculated using the van der Waals' equation differ from the ideal ?

Hint: % difference = 100×(P ideal - Pvan der Waals) / P ideal

Answer 1

The difference in pressure is 8.98 %.

(a) Ideal gas

The pressure predicted by the Ideal Gas Law is 16.71 atm.

(b) van der Waals Equation

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 = "0.9766 mol"`
`R = "0.083 14"color(white)(l)"bar·L·K"^"-1""mol"^"-1"`
`T = "270.0 K"`
`V = "1.295 L"`
`a = "3.658 bar·L"^2"mol"^"-2"`
`b = "0.042 86 L·mol"^"-1"`

`P = (nRT)/(V-nb) – (n^2a)/V^2`

`= (0.9766 color(red)(cancel(color(black)("mol"))) × "0.083 14 bar"color(red)(cancel(color(black)("L·""K"^"-1""mol"^"-1")))× 270.0 color(red)(cancel(color(black)("K"))))/(1.295 color(red)(cancel(color(black)("L"))) – 0.9766 color(red)(cancel(color(black)("mol")))× "0.042 860" color(red)(cancel(color(black)("L·mol"^(-1))))) - ((0.9766 color(red)(cancel(color(black)("mol"))))^2 × "3.658 bar" color(red)(cancel(color(black)("L"^2"mol"^"-2"))))/(1.295 color(red)(cancel(color(black)("L"))))^2`
.
`= "21.92 bar"/1.253 - "14.38 atm "= "17.49 bar" - "2.080 bar"= "15.41 bar"`

`p_text(vdW) = 15.41 color(red)(cancel(color(black)("bar"))) × "1 atm"/(1.01325 color(red)(cancel(color(black)("atm")))) = bb"15.21 atm"`

`"% difference" = (|p_text(ideal) - p_text(vdW)|)/p_text(ideal) × 100 % = (|"16.71 atm - 15.21 atm"|)/("16.71 atm") × 100 % = 1.50/16.71 ×100 % = 8.98 %`