Question

15.0 moles of gas are in a 6.00 L tank at 24.3 ∘C . What is the difference in pressure between methane and an ideal gas under these conditions?

The van der Waals constants for methane are a=2.300L2⋅atm/mol2 and b=0.0430 L/mol.

Answer 1

The difference in pressure is 7.0 atm.

Explanation:

(a) Ideal gas

The equation for the Ideal Gas Law is

`color(blue)(bar(ul(|color(white)(a/a)pV = nRTcolor(white)(a/a)|)))" "`

Then,

`p = (nRT)/V`

`n ="15.0 mol"`
`R = "0.082 06 L·atm·K"^"-1""mol"^"-1"`
`T = "(24.3 + 273.15) K = 297.45"`
`V = "6.00 L"`

`p = (15.0 color(red)(cancel(color(black)("mol"))) × "0.082 06" color(red)(cancel(color(black)("L")))·"atm"·color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 297.45 color(red)(cancel(color(black)("K"))))/(6.00 color(red)(cancel(color(black)("L")))) = "61.0 atm"`

The pressure predicted by the Ideal Gas Law is 61.0 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 = "6.00 mol"`
`R = "0.082 06"color(white)(l)"L·atm·K"^"-1""mol"^"-1"`
`T = "297.45 K"`
`V = "6.00 L"`
`a = "2.300 atm·L"^2"mol"^"-2"`
`b = "0.0430 L·mol"^"-1"`

`P = (nRT)/(V-nb) – (n^2a)/V^2`

`= (15.0 color(red)(cancel(color(black)("mol"))) × "0.082 06 atm"color(red)(cancel(color(black)("L·""K"^"-1""mol"^"-1")))× 297.45 color(red)(cancel(color(black)("K"))))/(6.00 color(red)(cancel(color(black)("L"))) – 15.0 color(red)(cancel(color(black)("mol")))× 0.0430 color(red)(cancel(color(black)("L·mol"^(-1))))) - ((15.0 color(red)(cancel(color(black)("mol"))))^2 × "2.300 atm" color(red)(cancel(color(black)("L"^2"mol"^"-2"))))/(6.00 color(red)(cancel(color(black)("L"))))^2`
.
`= "366.15 atm"/5.355 - "14.38 atm "= "68.37 atm" - "14.38 atm"= "54.0 atm"`

The pressure predicted by the van der Waals equation is 54.0 atm.

`p_text(ideal) - p_text(vdW) = "61.0 atm - 54.0 atm = 7.0 atm"`