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 CO2 + 18 H2O

C8H18 + 25/2 O2 -------------------> 8CO2 + 9 H2O

1 mol octane needs -----------------> 25/2 moles O2

1 mol any gas can occupy --------------> 22.4 L

so 25/2 mol O2 volume = 22.4 x 25/2 = 280 L

V1 = 280 L

next product side

1 mol octane gives ---------------- > 8 mol CO2

                          --------------------> 8 x 22.4 = 179.2 L

1 mol octane gives -----------------> 9 mol H2O

                                --------------> 9 x 22.4 = 201.6 L

total volume on product side = 179.2 + 201.6

                                      V2     = 380.8 L

now work = p (V2-V1)

reaction is carried out at atmspheric pressure = 1 atm

work = P (V2-V1) =1 (380.8- 280) = 100.8 L-atm

work = 100.8 L-atm per mol of octane