# What is the density of a mixture of methane, ethane, propane, and n-butane at 130^@ "C" and "1 atm"? What is its specific gravity?

May 24, 2017

$D = \text{0.7867 g/L}$
$\text{SG} = 0.898$

Each gas has a compressibility factor $Z$:

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

• When $Z > 1$, the average distance between gases is farther than if they were ideal gases. i.e. it is more difficult to compress than if it were ideal.
• When $Z < 1$, the average distance between gases is closer than if they were ideal gases. i.e. it is easier to compress than if it were ideal.
• When $Z = 1$, the gas is ideal.

Regardless of whether it is ideal or not, $Z$ can be used in the above formula (which resembles the ideal gas law), since $\frac{V}{n}$ is the molar volume, which is the only variable in the ideal gas law dependent on the identity of the gas.

$Z$ can be looked up or calculated, and from there, $\overline{V} = \frac{V}{n}$ can be calculated to determine the density contribution. At ${25}^{\circ} \text{C}$ and $\text{1 atm}$, we reference $Z$ to be:

${Z}_{\text{methane}} = 0.99825$
${Z}_{\text{ethane}} = 0.99240$
${Z}_{\text{propane}} = 0.99381$
${Z}_{\text{butane}} = 0.96996$

We therefore assume that $Z$ at ${25}^{\circ} \text{C}$ varies little compared to at ${130}^{\circ} \text{C}$ (which does not give much error compared to assuming ideality). So:

${\overline{V}}_{\text{methane") = (Z_"methane"RT)/P = ((0.99825)("0.082057 L"cdot"atm/mol"cdot"K")("403.15 K"))/("1 atm}}$

$=$ $\text{33.02 L/mol}$

${\overline{V}}_{\text{ethane") = (Z_"ethane"RT)/P = ((0.99240)("0.082057 L"cdot"atm/mol"cdot"K")("403.15 K"))/("1 atm}}$

$=$ $\text{32.83 L/mol}$

${\overline{V}}_{\text{propane") = (Z_"propane"RT)/P = ((0.99381)("0.082057 L"cdot"atm/mol"cdot"K")("403.15 K"))/("1 atm}}$

$=$ $\text{32.88 L/mol}$

${\overline{V}}_{\text{butane") = (Z_"butane"RT)/P = ((0.96996)("0.082057 L"cdot"atm/mol"cdot"K")("403.15 K"))/("1 atm}}$

$=$ $\text{32.09 L/mol}$

So, in general, the density of the mixture would then be, not assuming the gas molar volumes are identical (which would have been true for ideal gases):

$\boldsymbol{{D}_{\text{mixture}}} = \frac{{\sum}_{i} {m}_{i}}{{\sum}_{i} {V}_{i}}$

$= \boldsymbol{\frac{{\sum}_{i} {n}_{i} {M}_{i}}{{\sum}_{i} {n}_{i} {\overline{V}}_{i}}}$

where $n$, $m$, and $M$ are the mols, mass in $\text{g}$, and molar mass in $\text{g/mol}$, respectively.

So, we then get, for $\text{1 mol}$ of sample gas, and writing methane, ethane, propane, and n-butane in the sum in that order:

color(blue)(D_"mixture") = (0.6cdot16.0426 + 0.2cdot30.069 + 0.1cdot44.0962 + 0.1cdot58.123)/(0.6cdot33.02 + 0.2cdot32.83 + 0.1cdot32.88 + 0.1cdot32.09) "g"/"L"

$=$ $\textcolor{b l u e}{\text{0.7867 g/L}}$

(As a note, if we had assumed the molar volumes were identical, then the density would have turned out to be a simple average weighted by the mol fractions, and would be $\text{0.7899 g/L}$, which is only a little different.)

As for specific gravity, there is more than one definition, so I will assume you mean the true specific gravity for gases, which is the ratio of the gas density to the density of air at this $T$ and $P$ ($\text{0.876 g/L}$, ${130}^{\circ} \text{C}$, $\text{1 atm}$).

Now that we have its density though, the calculation is quite easy:

$\textcolor{b l u e}{\text{SG") = (D_"mixture")/(D_"air}}$

$= \text{0.7867 g/L"/"0.876 g/L}$

$= \textcolor{b l u e}{0.898}$