# 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.)