Question

Question #b1cd1

Answer 1

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

Using `"CO"_2` at `"230 K"` and `"8 bar"` as an example, I got:

`barV_"real" = "2.1508 L/mol"` (Actual value)
`barV_"RK" = "2.1827 L/mol"` (Redlich-Kwong, `1.48%` error)
`barV_"vdW" = "2.2324 L/mol"` (van der Waals, `3.79%` error)
`barV_"id" = "2.3904 L/mol"` (Ideal Gas Law, `11.1%` error!!)

And it is clear that the ideal gas law overestimates the size of `"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:

...`"CO"_2` should still be a gas, but it would be a bit to the right of the triple point. If I had chosen `"220 K"`, that may have been too close to the triple point, and the density may be wildly different there.

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VAN DER WAALS EQUATION OF STATE

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

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

where:

• `P`, `V`, `n`, `R`, and `T` are known from the ideal gas law.
• `barV = 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 `"L"^2cdot"bar/mol"^2` if pressure is in `"bar"`.
• `b` is a vdW constant that accounts for the volume excluded by the gas particles (`b > 0`). Its units are `"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 `"CO"_2(g)`. Its density at `"8 bar"` and `"230 K"` is `"20.4617 g/L"`. That means its molar volume at that temperature and pressure is:

`color(green)(barV_"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:

`color(green)(barV_"id") -= V/n = (RT)/P`

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

`=` `ulcolor(green)("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 `"CO"_2`, `a = "3.658 bar"cdot"L"^2"/mol"^2` and `b = "0.04286 L/mol"`.

One would solve for `barV` to get:

`PbarV^3 - (bP + RT)barV^2 + abarV - ab = 0`

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

`color(blue)(barV_"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 = "64.597 bar"cdot"L"^2"/mol"^2cdot"K"^(1//2)` and `B = "0.029677 L/mol"` for `"CO"_2`, solved in a similar manner, would give an even more accurate result, denoted `barV_"RK"`.

`barV_"real" = "2.1508 L/mol"`
`barV_"RK" = "2.1827 L/mol"`
`barV_"vdW" = "2.2324 L/mol"`
`barV_"id" = "2.3904 L/mol"`

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