# 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?

##### 1 Answer

Each gas has a **compressibility factor**

#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_("methane") = 0.99825#

#Z_("ethane") = 0.99240#

#Z_("propane") = 0.99381#

#Z_("butane") = 0.96996#

We therefore assume that **varies little compared to** at

#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 *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

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

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)#