Find Van der Waals constants and what is the pressure for C2H6?

Find van der waals constants for C2H6 if critical temperature=32,1 deegres, critical pressure is 494, 76kPa. Find pressure for 5g C2H6 in the container of volume=1L and on 15 degrees. NOTE: it seems that ${P}_{c}$ should be closer to $4947.6$ $k P a$. - Truong-Son

Jan 6, 2017

$a = {\text{5.493 L"^2cdot"bar/mol}}^{2}$
$b = \text{0.06412 L/mol}$
$P = \text{3.875 bar}$

Using the result derived in full here that:

$a = \frac{27 {R}^{2} {T}_{c}^{2}}{64 {P}_{c}}$
$b = \frac{R {T}_{c}}{8 {P}_{c}}$

and the van der Waals equation of state:

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

First, I'd convert ${P}_{c}$ to $\text{bar}$ and ${T}_{c}$ to $\text{K}$ since I am familiar with the universal gas constant with pressure units of $\text{bar}$ (though it seems like ${P}_{c}$ is off by a factor of $10$, so I corrected it):

${P}_{c} = \text{4947.6 kPa" xx "0.01 bar"/"1 kPa" = "49.476 bar}$
(compared to $\text{49 bar}$ from the NIST database)

${T}_{c} = {32.1}^{\circ} \text{C" = "305.25 K}$

Then:

$\textcolor{b l u e}{a} = \left(\left(27\right) \left(\text{0.083145 L"cdot"bar/mol"cdot"K")^2("305.25 K")^2)/(64("49.476 bar}\right)\right)$

$=$ $\textcolor{b l u e}{{\text{5.493 L"^2cdot"bar/mol}}^{2}}$
(compared to ${\text{5.570 L"^2cdot"bar/mol}}^{2}$ from here.)

The $b$ constant is:

$\textcolor{b l u e}{b} = \left(\left(\text{0.083145 L"cdot"bar/mol"cdot"K")("305.25 K"))/(8("49.476 bar}\right)\right)$

$=$ $\textcolor{b l u e}{\text{0.06412 L/mol}}$
(compared to $\text{0.06499 L/mol}$ here.)

Using this information, we can find the pressure of $\text{5 g}$ of ethane in a $\text{1-L}$ container at ${15}^{\circ} \text{C}$, or $\text{288.15 K}$:

$\text{5 g ethane" xx "1 mol ethane"/"30.0694 g" = "0.1663 mols}$

giving us a molar volume of (exactly):

$\overline{V} = \text{1 L"/"0.1663 mols" = "6.01388 L/mol}$

Finally, we can solve for the pressure to get:

color(blue)(P) = (("0.083145 L"cdot"bar/mol"cdot"K")("288.15 K"))/("6.01388 L/mol" - "0.06412 L/mol") - ("5.493 L"^2cdot"bar/mol"^2)/("6.01388 L/mol")^2

$=$ $\textcolor{b l u e}{\text{3.875 bar}}$

Using the ideal gas law, we can check our answer:

P = (nRT)/V = (RT)/(barV) = (("0.083145 L"cdot"bar/mol"cdot"K")("288.15 K"))/("6.01388 L/mol")

$=$ $\text{3.984 bar}$

which is indeed close! The real gas actually takes up less volume (its attractive forces dominate), meaning that there is more empty space in the container than we expected, and thus, a slightly smaller pressure is exerted.

Note: some textbooks ask you to go through and derive $a$ and $b$ in terms of ${P}_{c}$ and ${T}_{c}$. Due to the length of the derivation, I've instead moved it to a separate scratchpad.