Question

A closed container has a mixture of methane, CH₄, ethane, C₂H₆, and propane, C₃H₈. If 26.5 g of methane, 29.5 g of ethane, and 31.0 g of propane are placed in the 8.60 L container at 58.5°C, what is the total pressure in the container? Assume the gases do not react with each other.
atm

Answer 1

moles of ethane= mass/molar mass= 29.5 g/30.07 g/mol= 0.981 moles

moles of methane= mass/molar mass= 26.5/16.04 = 1.652 moles

moles of propane= mass/molar mass= 31/44.1= 0.7029 moles

total moles= moles of (methane +ethane+propane)= 0.981+1.652+0.7029= 3.3359 moles of gas

now using ideal gas equation PV=nRT

P= pressure= ? ; V volume= 8.60 L ; n= moles= 3.3359 moles

R= gas constant= 0.0821 Latm/molK ; T= temperature= 58.5+273.15=331.65 K

rearranging the equation for P and using the values

P= nRT/V= (3.3359*0.0821*331.65)/8.60= 10.6 atm