Question

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?

Answer 1

`D = "0.7867 g/L"`
`"SG" = 0.898`

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Each gas has a compressibility factor `Z`:

`Z = (PV)/(nRT)`

• 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 `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, `barV = V/n` can be calculated to determine the density contribution. At `25^@ "C"` and `"1 atm"`, we reference `Z` to be:

`Z_("methane") = 0.99825`
`Z_("ethane") = 0.99240`
`Z_("propane") = 0.99381`
`Z_("butane") = 0.96996`

We therefore assume that `Z` at `25^@ "C"` varies little compared to at `130^@ "C"` (which does not give much error compared to assuming ideality). So:

`barV_("methane") = (Z_"methane"RT)/P = ((0.99825)("0.082057 L"cdot"atm/mol"cdot"K")("403.15 K"))/("1 atm")`

`=` `"33.02 L/mol"`

`barV_("ethane") = (Z_"ethane"RT)/P = ((0.99240)("0.082057 L"cdot"atm/mol"cdot"K")("403.15 K"))/("1 atm")`

`=` `"32.83 L/mol"`

`barV_("propane") = (Z_"propane"RT)/P = ((0.99381)("0.082057 L"cdot"atm/mol"cdot"K")("403.15 K"))/("1 atm")`

`=` `"32.88 L/mol"`

`barV_("butane") = (Z_"butane"RT)/P = ((0.96996)("0.082057 L"cdot"atm/mol"cdot"K")("403.15 K"))/("1 atm")`

`=` `"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):

`bb(D_"mixture") = (sum_i m_i)/(sum_i V_i)`

`= bb((sum_i n_iM_i)/(sum_i n_i barV_i))`

where `n`, `m`, and `M` are the mols, mass in `"g"`, and molar mass in `"g/mol"`, respectively.

So, we then get, for `"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"`

`=` `color(blue)("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 `"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` (`"0.876 g/L"`, `130^@ "C"`, `"1 atm"`).

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

`color(blue)("SG") = (D_"mixture")/(D_"air")`

`= "0.7867 g/L"/"0.876 g/L"`

`= color(blue)(0.898)`