# Problem in ideal gases

1. Sep 20, 2009

### Gil-H

1. The problem statement, all variables and given/known data
What volume of air (T=25C, P=1atm) is required
for complete combustion of one litter of gasoline?
The partial pressure of oxygen in the air is 0.205 atm.
One litter of gasoline contains 6.15 moles octane C8H18.

2. Relevant equations
The unbalanced reaction is:
C8H18 + O2 --> CO2 + H2O
The ideal gas law:
PV = nRT

3. The attempt at a solution
The balanced reaction is:
C8H18 + (25/2)O2 --> 8CO2 + 9H2O
So 6.15 moles octane requires 6.15*12.5 = 76.875 moles O2.
With the ideal gas law I get:
PV = nRT
(0.205)V=(76.875)(0.082)(298)
V = 9,163.5 L

Is this the correct answer?
My friend beleives it is,
but I think that this value is just the volume of oxygen needed,
and the volume of air needed is V = (9,163.5/0.2) = 45,817.5 L

2. Sep 20, 2009

### Staff: Mentor

Your friend is right.

Your approach would be correct if you would use not partial pressure of the oxygen, but 1 atm.

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