The motor gasoline, assumed consisting of pure heptane, has a density of $0.700 $. Calculated depending on the pressure (atm) and the ambient temperature (K) the volume of air that has passed through the engine when $ 1 \;L $ gasoline was consumed, knowing it spends 10 times more air that is strictly necessary for the complete combustion?
Application: $ t =27 ^o C $ and P = $ 1 \; atm $
$ \ rho_{heptane} = 700\frac{g}{L} $
$ 1 \;L $ heptane therefore has a mass of $ 700 \; g$ and thus contains $ \frac{700}{100} = 7.00 \;mol $
$ C_7H_{16}+ 11O_2 \; \rightarrow \; 7CO_2 + 8H_2O$
$ 7.00 \; mol $ require $ 77 \; mol \; O_2$
$ \frac{7RT}{P} \; L $ require $ \frac {77RT}{P} \; L \; O_2$, strictly $ 5 \cdot \frac {77RT}{P} \; L \; $ air, which amounts actually to $ 10\cdot 5 \cdot \frac{77RT}{P} = 316\frac{T}{P} \; L$ air.
For $ t = 27 ^ o C $ and P = $ 1 \;atm$ it does well $94800 \; L $ air!
