Question

Assuming all volume change is due to gases, calculate the work for combustion of 1 mol of octane producing gaseous water. Determine the work per liter of octane (density = 0.6986 g/cm³).
____L*atm

Answer 1

2 C8H18 + 25 O2 ---------------------> 16 O2 + 18 H2O

we know that 1 mol of gas can occupy 22.4 L volume

2 mol C8H18 require = 25 mol O2

                                  = 25 x 22.4 L

                                 = 560 L

so initial volume (V1) = 560 L

increases on product side . this can be considered as V2

2 mol C8H18 = 16 moles of CO2

                       = 16 x 22.4

                      = 358.4 L

2 mol C8H18 = 18 mol H2O

                       = 18 x 22.4 L

                        = 403.2 L

total volume on products side = 358.4 + 403.2 = 716.6 L

V2 = 716.6 L

work = P (V2-V1)

          = 1 x (716.6-560 )

           = 201.6 L-atm

for 2 mol C8H18 work = 201.6

for 1 mol C8H18 work = 100.8 L-atm

work = 100.8 L-atm ------------------------> answer