Question

If 4.00 moles of gasoline are burned, what volume of oxygen is needed if the pressure is 0.953 atm, and the temperature is 35.0°C?

Answer 1

Well we will treat gasoline as octanes......`C_8H_18`, and we get a volume of

Explanation:

We need (i) a stoichiometric equation........

`C_8H_18(l) +25/2O_2(g) rarr 8CO_2(g) + 9H_2O(l)`

And so we need `25/2` equiv dioxygen gas.......

And thus if `4*mol` of octane are combusted we need `25/2xx4=50*mol` `"dioxygen gas."`

And (ii) given `V=(nRT)/P=(50*molxx0.0821*(L*atm)/(K*mol)xx308*K)/(0.953*atm)`

Approx....`1300*L` or `1.3*m^3` `"dioxygen gas"`.