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

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

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Using the result derived in full that:

$a = (27R^2T_c^2)/(64P_c)$
$b = (RT_c)/(8P_c)$

and the van der Waals equation of state:

$P = (RT)/(barV - b) - a/(barV^2)$,

First, I'd convert $P_c$ to $"bar"$ and $T_c$ to $"K"$ since I am familiar with the universal gas constant with pressure units of $"bar"$ (though it seems like $P_c$ is off by a factor of $10$, so I corrected it):

$P_c = "4947.6 kPa" xx "0.01 bar"/"1 kPa" = "49.476 bar"$
(compared to $"49 bar"$ from the )

$T_c = 32.1^@ "C" = "305.25 K"$

Then:

$color(blue)(a) = ((27)("0.083145 L"cdot"bar/mol"cdot"K")^2("305.25 K")^2)/(64("49.476 bar"))$

$=$ $color(blue)("5.493 L"^2cdot"bar/mol"^2)$
(compared to $"5.570 L"^2cdot"bar/mol"^2$ from .)

The $b$ constant is:

$color(blue)(b) = (("0.083145 L"cdot"bar/mol"cdot"K")("305.25 K"))/(8("49.476 bar"))$

$=$ $color(blue)("0.06412 L/mol")$
(compared to $"0.06499 L/mol"$ .)

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

$"5 g ethane" xx "1 mol ethane"/"30.0694 g" = "0.1663 mols"$

giving us a molar volume of (exactly):

$barV = "1 L"/"0.1663 mols" = "6.01388 L/mol"$

Finally, we can solve for the pressure to get:

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

$=$ $color(blue)("3.875 bar")$

Using the , we can check our answer:

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

$=$ $"3.984 bar"$

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.

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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 .