How much work is done at 1.2 atm and 316 K when 1 mole of propane undergoes combustion?
C3H8 (l) + 5O2 (g) → 3CO2 (g) + 4H2O (l)
mols gas on left = 6 (for 1 mol propane) and 7 on the right. Vi then is 6 and Vf is 7
W = -p*delta V or -p(Vf-Vi)
C3H8 (l) + 5O2 (g) → 3CO2 (g) + 4H2O (l)
W = -p*delta V or -p(Vf-Vi)