What volume of air at 1.0 atm and 25 degree Celsius must be taken to an engine to burn 1 gallon of gasoline?
Assuming that:
2 C8H18 + 25 O2 --> 16 CO2 + 18 H2O
provides reasonable model of the actual combustion process. ( The partial pressure of oxygen in air is 0.21 atm and the density of liquid octane is 0.70g/mL)
To determine the volume of air required to burn 1 gallon of gasoline, we need to calculate the moles of gasoline burned and then use stoichiometry to find the moles of air required. Finally, we can convert the moles of air to volume using the ideal gas law.
1. Convert the volume of gasoline from gallons to liters:
Since 1 gallon = 3.78541 liters, 1 gallon of gasoline is equal to 3.78541 liters of gasoline.
2. Convert the volume of gasoline from liters to milliliters:
Since 1 liter = 1000 milliliters, 1 gallon of gasoline is equal to 3785.41 milliliters.
3. Calculate the moles of octane (C8H18):
Using the density given, we can calculate the mass of octane in grams. The density of liquid octane is 0.70 g/mL, so the mass of 3785.41 mL of octane is (0.70 g/mL) × (3785.41 mL) = 2649.79 grams.
The molar mass of octane (C8H18) is [(12.01 g/mol × 8) + (1.01 g/mol × 18)] = 114.23 g/mol.
The moles of octane (C8H18) can be calculated as: (2649.79 g) / (114.23 g/mol) = 23.16 moles.
4. Use stoichiometry to find the moles of air (O2) required:
According to the balanced equation, 2 moles of octane react with 25 moles of oxygen gas (O2) to produce 16 moles of carbon dioxide (CO2) and 18 moles of water (H2O).
Therefore, the moles of oxygen gas required to burn 23.16 moles of octane are: (23.16 moles C8H18) × (25 moles O2 / 2 moles C8H18) = 289.50 moles O2.
5. Determine the moles of air required:
The mole ratio between oxygen gas (O2) and air is 1:1 (since oxygen makes up about 21% of air).
Therefore, the moles of air required to burn 23.16 moles of octane are also 289.50 moles.
6. Convert moles of air to volume using the ideal gas law:
The ideal gas law equation is PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature in Kelvin.
The pressure (P) is given as 1.0 atm.
The temperature (T) is given as 25 degrees Celsius, which needs to be converted to Kelvin: T(K) = T(°C) + 273.15 = 25 + 273.15 = 298.15 K.
The ideal gas constant (R) is 0.0821 L × atm / (mol × K).
So, using the ideal gas law equation, the volume of air required can be calculated as follows:
V = (n × R × T) / P
= (289.50 moles × 0.0821 L × atm / (mol × K) × 298.15 K) / 1.0 atm
= 68.24 liters
Therefore, approximately 68.24 liters of air at 1.0 atm and 25 degrees Celsius are required to burn 1 gallon of gasoline.