Problem in ideal gases

  • Thread starter Gil-H
  • Start date
  • #1
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Homework Statement


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.


Homework Equations


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


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
 

Answers and Replies

  • #2
Borek
Mentor
28,886
3,431
Your friend is right.

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

--
methods
 

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