Question

What is the density of carbon dioxide at STP?

Answer 1

`d=1.79g*L^(-1)`

Explanation:

Assuming that carbon dioxide behaves ideally, then we can use the ideal gas law:

`PV=nRT`.

Since we are looking for the density of `CO_2`, we can modify the law as follows:

First we replace `n` by `n=m/(MM)` where, `m` is the mass and `MM=40g/(mol)` is the molar mass of `CO_2`.

`=>PV=nRT=>PV=(m)/(MM)RT`

Then rearrange the expression to become:

`P=m/V(RT)/(MM)` where `m/V=d` (`d` is the density).

`=>P=(dRT)/(MM)=>d=(PxxMM)/(RT)`

Therefore, `d=(1cancel(atm)xx40g/(cancel(mol)))/(0.08201(L*cancel(atm))/(cancel(K)*cancel(mol))xx273cancel(K))=1.79g*L^(-1)`

Answer 2

REFERENCE DENSITY

Wikipedia gives the density as `"0.001977 g/mL"` at `"1 atm"`, or if we convert it for `"1 bar"`, `color(blue)("0.001951 g/mL")`.

Or, one can calculate this from this website.

This also gives a real mass density of `color(blue)("0.001951 g/mL")` at `"1 bar"` and `0^@ "C"`.

DENSITY ASSUMING IDEALITY

To get an idea of how the density is like when assuming ideality, we can use the ideal gas law to compare.

`\mathbf(PV = nRT)`

where:

• `P` is the pressure in `"bar"`. STP currently involves `"1 bar"` pressure.
• `V` is the volume in `"L"`.
• `n` is the `\mathbf("mol")`s of gas .
• `R` is the universal gas constant, `"0.083145 L"cdot"bar/mol"cdot"K"`.
• `T` is the temperature in `K"`.

`P/(RT) = n/V`

Notice how `(nM_m)/V = rho`, where `M_m` is the molar mass of `"CO"_2` (`"44.009 g/mol"`, not `"40 g/mol"`...), and `rho` is the mass density in `"g/L"`. Thus:

`color(blue)(rho) = (PM_m)/(RT)`

`= (("1 bar")("44.009 g/mol"))/(("0.083145 L"cdot"bar/mol"cdot"K")("273.15 K"))`

`=` `"1.94 g/L"`

`=` `color(blue)("0.001937 g/mL")`

That is about `0.72%` error from the true density, which is quite good. Thus, `"CO"_2` is fairly ideal.

DENSITY WITHOUT ASSUMING IDEALITY

Let's calculate the density another way.

We can also use the compressibility factor `Z = (PV)/(nRT)`, which is an empirical constant related to how easily `"CO"_2` responds to compression. If `Z = 1`, then `"CO"_2` is perfectly ideal.

From this website again, I get `Z = 0.9934`.

Since `Z < 1`, `"CO"_2` is easier to compress than a comparable ideal gas (thus its molar volume is less than `22.711` at `"1 bar"` and `"273.15 K"`).

Let's see what its density is this time.

`color(green)(Z) = P/(RT)V/n`

`Z/(M_m) = P/(RT)V/(nM_m)`

`= color(green)(P/(RTrho))`

Thus...

`color(blue)(rho) = (PM_m)/(RTZ)`

`= (("1 bar")("44.009 g/mol"))/(("0.083145 L"cdot"bar/mol"cdot"K")("273.15 K")(0.9934))`

`=` `"1.9507 g/L"`

`~~` `color(blue)("0.001951 g/mL")`

Oh look at that... it's dead-on, and all I did was use `Z` as a correctional factor in the ideal gas law. :)