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"#

2 Answers
Oct 27, 2016

Answer:

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

Oct 27, 2016

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"#.