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
#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
#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)#