Question b1cd1

Aug 27, 2017

Because gases really aren't always negligible in size compared to their container, particularly at high pressures and low temperatures.

Using ${\text{CO}}_{2}$ at $\text{230 K}$ and $\text{8 bar}$ as an example, I got:

${\overline{V}}_{\text{real" = "2.1508 L/mol}}$ (Actual value)
${\overline{V}}_{\text{RK" = "2.1827 L/mol}}$ (Redlich-Kwong, 1.48% error)
${\overline{V}}_{\text{vdW" = "2.2324 L/mol}}$ (van der Waals, 3.79% error)
${\overline{V}}_{\text{id" = "2.3904 L/mol}}$ (Ideal Gas Law, 11.1% error!!)

And it is clear that the ideal gas law overestimates the size of ${\text{CO}}_{2}$ in these conditions, whereas the van der Waals equation of state is significantly closer (and as an added bonus, the Redlich-Kwong is even closer!).

Under these conditions, according to its phase diagram: ...${\text{CO}}_{2}$ should still be a gas, but it would be a bit to the right of the triple point. If I had chosen $\text{220 K}$, that may have been too close to the triple point, and the density may be wildly different there.

VAN DER WAALS EQUATION OF STATE

The van der Waals (vdW) equation of state is given by:

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

where:

• $P$, $V$, $n$, $R$, and $T$ are known from the ideal gas law.
• $\overline{V} = \frac{V}{n}$ is the molar volume.
• $a$ is a vdW constant that accounts for the attractive and repulsive forces in the gas sample. Its units can be ${\text{L"^2cdot"bar/mol}}^{2}$ if pressure is in $\text{bar}$.
• $b$ is a vdW constant that accounts for the volume excluded by the gas particles ($b > 0$). Its units are $\text{L/mol}$.

This equation gives better agreement than the ideal gas law would, particularly at higher pressure and lower temperatures where the ideal gas law fails.

EXAMPLE: CO2

Consider ${\text{CO}}_{2} \left(g\right)$. Its density at $\text{8 bar}$ and $\text{230 K}$ is $\text{20.4617 g/L}$. That means its molar volume at that temperature and pressure is:

$\textcolor{g r e e n}{{\overline{V}}_{\text{real") = "1 L"/(20.4617 cancel"g") xx (44.009 cancel"g")/"mol" = ulcolor(green)("2.1508 L/mol}}}$

The ideal gas law would give:

$\textcolor{g r e e n}{{\overline{V}}_{\text{id}}} \equiv \frac{V}{n} = \frac{R T}{P}$

= (("0.083145 L"cdotcancel"bar""/mol"cdotcancel"K")(230 cancel"K"))/cancel"8 bar"#

$=$ $\underline{\textcolor{g r e e n}{\text{2.3904 L/mol}}}$

which is fairly off, though it isn't bad. It means we used the equation in a physically valid situation.

With the vdW equation, however, we may get a better result. For ${\text{CO}}_{2}$, $a = {\text{3.658 bar"cdot"L"^2"/mol}}^{2}$ and $b = \text{0.04286 L/mol}$.

One would solve for $\overline{V}$ to get:

$P {\overline{V}}^{3} - \left(b P + R T\right) {\overline{V}}^{2} + a \overline{V} - a b = 0$

In an iterative calculation (Newton-Raphson method), one could solve this equation to get three roots. Choosing the correct root would yield:

$\textcolor{b l u e}{{\overline{V}}_{\text{vdW" = ul"2.2324 L/mol}}}$

It's not perfect, but it is definitely closer than the ideal gas law was.

Alternatively, the more complicated Redlich-Kwong equation of state, with $A = {\text{64.597 bar"cdot"L"^2"/mol"^2cdot"K}}^{1 / 2}$ and $B = \text{0.029677 L/mol}$ for ${\text{CO}}_{2}$, solved in a similar manner, would give an even more accurate result, denoted ${\overline{V}}_{\text{RK}}$.

${\overline{V}}_{\text{real" = "2.1508 L/mol}}$
${\overline{V}}_{\text{RK" = "2.1827 L/mol}}$
${\overline{V}}_{\text{vdW" = "2.2324 L/mol}}$
${\overline{V}}_{\text{id" = "2.3904 L/mol}}$

This illustrates the lack of ideality that a gas could exhibit at high pressure and low temperature.