Question

What is the pressure for `"N"_2` gas if `"0.400 mols"` of it occupies `"0.75 L"` at `30.50^@ "C"`? `a = "1.39 atm"cdot"L"^2"/mol"^2`, `b = "0.0391 L/mol"`

Answer 1

11,09 atm

Explanation:

for N2 a=1,39Litri^2 atm / mole^2 b=0,0391 litro/ mole
P = nRT/(V-nb) -an^2/b^2 =
=0,4 mol x0,082 L atm/mol K x (303,66K)/(0,75L-0,4 mol x0,0391 L/mol)-1,39Litri^2 atm / mole^2 x (0,4 molx1,39 mol/L)^2 =11,09 atm

Answer 2

I write the van der Waals equation like this:

`bb(P = (RT)/(barV - b) - a/(barV^2))`

where `barV = V/n` is the molar volume in `"L/mol"`, `P` is pressure in, say, `"atm"`, `R` would thus be in `"L"cdot"atm/mol"cdot"K"`, `T` is temperature in `"K"`, and `a` and `b` are the intermolecular-interaction and excluded-volume constants, respectively.

You might have also seen it as:

`[P + a(n/V)^2](V - nb) = nRT`

Convince yourself that they are the same equation. That aside, the pressure is:

`color(blue)(P_"vdW") = (("0.082057 L"cdot"atm/mol"cdot"K")("30.50 + 273.15 K"))/(0.75/0.400 "L/mol" - "0.0391 L/mol") - ("1.39 atm"cdot"L"^2"/mol"^2)/(0.75/0.400 "L/mol")^2`

`= ("24.917 L"cdot"atm/mol")/(1.8359 "L/mol") - ("1.39 atm"cdot"L"^2"/mol"^2)/(3.515625 "L"^2"/mol"^2)`

`=` `color(blue)("13.18 atm")`

For the ideal gas law you would get `"13.29 atm"`.