A rigid sealed #"3.70 L"# vessel contains #"0.12 mol"# of #He(g)# and #"0.24 mol"# #Ne(g)# at #25^@ "C"#. What is the total pressure in the vessel?
1 Answer
"Rigid" containers have constant volume, and we were given that
So, we can use the ideal gas law for total volume, total pressure, and total
#bb(PV = nRT)#
#=> color(blue)(P) = (nRT)/V = ((n_1 + n_2)RT)/V#
#= (("0.12 + 0.24 mols")("0.082057 L"cdot"atm/mol"cdot"K")("298.15 K"))/("3.70 L")#
#=# #color(blue)("2.38 atm")#
Another way to do it is via Dalton's Law of partial pressures for an ideal two-gas mixture:
#bb(P = P_1 + P_2 + . . .)#
To find
#P_1V = n_1RT#
#P_1 = (n_1RT)/V = (("0.12 mols")("0.082057 L"cdot"atm/mol"cdot"K")("298.15 K"))/("3.70 L")#
#=# #"0.79 atm"#
Similarly:
#P_2 = (n_2RT)/V = (("0.24 mols")("0.082057 L"cdot"atm/mol"cdot"K")("298.15 K"))/("3.70 L")#
#=# #"1.59 atm"#
So that the sum is, again:
#color(blue)(P) = P_1 + P_2#
#= (n_1RT)/V + (n_2RT)/V#
#= "0.79 + 1.59 atm" = color(blue)("2.38 atm")#
(In a way, we implied Dalton's Law of partial pressures in the first method.)