# 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"

Oct 27, 2016

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:

$\boldsymbol{P = \frac{R T}{\overline{V} - b} - \frac{a}{{\overline{V}}^{2}}}$

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

You might have also seen it as:

$\left[P + a {\left(\frac{n}{V}\right)}^{2}\right] \left(V - n b\right) = n R T$

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

${\textcolor{b l u e}{{P}_{\text{vdW") = (("0.082057 L"cdot"atm/mol"cdot"K")("30.50 + 273.15 K"))/(0.75/0.400 "L/mol" - "0.0391 L/mol") - ("1.39 atm"cdot"L"^2"/mol"^2)/(0.75/0.400 "L/mol}}}}^{2}$

$= \left({\text{24.917 L"cdot"atm/mol")/(1.8359 "L/mol") - ("1.39 atm"cdot"L"^2"/mol"^2)/(3.515625 "L"^2"/mol}}^{2}\right)$

$=$ $\textcolor{b l u e}{\text{13.18 atm}}$

For the ideal gas law you would get $\text{13.29 atm}$.