# What is the density of carbon dioxide at STP?

May 17, 2016

$d = 1.79 g \cdot {L}^{- 1}$

#### Explanation:

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

$P V = n R T$.

Since we are looking for the density of $C {O}_{2}$, we can modify the law as follows:

First we replace $n$ by $n = \frac{m}{M M}$ where, $m$ is the mass and $M M = 40 \frac{g}{m o l}$ is the molar mass of $C {O}_{2}$.

$\implies P V = n R T \implies P V = \frac{m}{M M} R T$

Then rearrange the expression to become:

$P = \frac{m}{V} \frac{R T}{M M}$ where $\frac{m}{V} = d$ ($d$ is the density).

$\implies P = \frac{\mathrm{dR} T}{M M} \implies d = \frac{P \times M M}{R T}$

Therefore, $d = \frac{1 \cancel{a t m} \times 40 \frac{g}{\cancel{m o l}}}{0.08201 \frac{L \cdot \cancel{a t m}}{\cancel{K} \cdot \cancel{m o l}} \times 273 \cancel{K}} = 1.79 g \cdot {L}^{- 1}$

May 18, 2016

REFERENCE DENSITY

Wikipedia gives the density as $\text{0.001977 g/mL}$ at $\text{1 atm}$, or if we convert it for $\text{1 bar}$, $\textcolor{b l u e}{\text{0.001951 g/mL}}$.

Or, one can calculate this from this website.

This also gives a real mass density of $\textcolor{b l u e}{\text{0.001951 g/mL}}$ at $\text{1 bar}$ and ${0}^{\circ} \text{C}$.

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.

$\setminus m a t h b f \left(P V = n R T\right)$

where:

• $P$ is the pressure in $\text{bar}$. STP currently involves $\text{1 bar}$ pressure.
• $V$ is the volume in $\text{L}$.
• $n$ is the $\setminus m a t h b f \left(\text{mol}\right)$s of gas .
• $R$ is the universal gas constant, $\text{0.083145 L"cdot"bar/mol"cdot"K}$.
• $T$ is the temperature in K".

$\frac{P}{R T} = \frac{n}{V}$

Notice how $\frac{n {M}_{m}}{V} = \rho$, where ${M}_{m}$ is the molar mass of ${\text{CO}}_{2}$ ($\text{44.009 g/mol}$, not $\text{40 g/mol}$...), and $\rho$ is the mass density in $\text{g/L}$. Thus:

$\textcolor{b l u e}{\rho} = \frac{P {M}_{m}}{R T}$

$= \left(\left(\text{1 bar")("44.009 g/mol"))/(("0.083145 L"cdot"bar/mol"cdot"K")("273.15 K}\right)\right)$

$=$ $\text{1.94 g/L}$

$=$ $\textcolor{b l u e}{\text{0.001937 g/mL}}$

That is about 0.72% error from the true density, which is quite good. Thus, ${\text{CO}}_{2}$ is fairly ideal.

DENSITY WITHOUT ASSUMING IDEALITY

Let's calculate the density another way.

We can also use the compressibility factor $Z = \frac{P V}{n R T}$, which is an empirical constant related to how easily ${\text{CO}}_{2}$ responds to compression. If $Z = 1$, then ${\text{CO}}_{2}$ is perfectly ideal.

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

Since $Z < 1$, ${\text{CO}}_{2}$ is easier to compress than a comparable ideal gas (thus its molar volume is less than $22.711$ at $\text{1 bar}$ and $\text{273.15 K}$).

Let's see what its density is this time.

$\textcolor{g r e e n}{Z} = \frac{P}{R T} \frac{V}{n}$

$\frac{Z}{{M}_{m}} = \frac{P}{R T} \frac{V}{n {M}_{m}}$

$= \textcolor{g r e e n}{\frac{P}{R T \rho}}$

Thus...

$\textcolor{b l u e}{\rho} = \frac{P {M}_{m}}{R T Z}$

$= \left(\left(\text{1 bar")("44.009 g/mol"))/(("0.083145 L"cdot"bar/mol"cdot"K")("273.15 K}\right) \left(0.9934\right)\right)$

$=$ $\text{1.9507 g/L}$

$\approx$ $\textcolor{b l u e}{\text{0.001951 g/mL}}$

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