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?

Question

Caleb Adams · Accepted Answer

#D = "0.7867 g/L"#
#"SG" = 0.898# Each gas has a compressibility factor #Z#: #Z = (PV)/(nRT)# 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)#

Wyatt Adkins · Answer

undefined To calculate the density of the mixture, you can use the ideal gas law. The specific gravity of the mixture can be calculated by comparing its density to that of air. Given the composition of the mixture, you'll need to know the molar masses of each component to perform the calculations.