Some automobiles and buses have been equipped to burn propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right) .\) Compare the amounts of energy that can be obtained per gram of \(\mathrm{C}_{3} \mathrm{H}_{8}(g)\) and per gram of gasoline, assuming that gasoline is pure octane, \(\mathrm{C}_{8} \mathrm{H}_{18}(l) .\) (See Example \(6.11 .\) ) Look up the boiling point of propane. What disadvantages are there to using propane instead of gasoline as a fuel?

Short Answer

Expert verified
The energy per gram obtained from propane (C3H8) is -50.33 kJ/g, while the energy per gram obtained from gasoline (assumed as pure octane) is -44.4 kJ/g. Thus, propane provides slightly more energy than gasoline. The boiling point of propane is -42.1 °C or -43.8 °F. However, there are several disadvantages of using propane as fuel: it has to be stored under pressure, availability of filling stations is limited, energy conversion efficiency for propane engines is lower, and it's less suitable for cold climates.

Step by step solution

01

1. Calculate energy obtained from propane (C3H8)

To find the energy obtained from one gram of propane, we need to find the heat of combustion of propane. The balanced equation for the combustion of propane is: C3H8 (g) + 5 O2 (g) -> 3 CO2 (g) + 4 H2O (g) The heat of combustion ∆H_comb, for propane can be found from standard heats of formation (∆H_f°) values. Here are the ∆H_f° values we need: ∆H_f° (C3H8) = -104 kJ/mol ∆H_f° (CO2) = -393.5 kJ/mol ∆H_f° (H2O) = -241.8 kJ/mol Now, we can calculate the ∆H_comb for propane: ∆H_comb = [ 3 * ∆H_f° (CO2) + 4 * ∆H_f° (H2O) ] - ∆H_f° (C3H8) = [ 3 * (-393.5) + 4 * (-241.8) ] - (-104) = -2219.4 kJ/mol To find energy per gram, we divide the energy obtained per mole by the molar mass of propane. \( Energy\ per\ gram\ of\ propane\ (C3H8) = \frac{-2219.4\ kJ/mol}{(3\times12.01 + 8\times1.008)\ g/mol} = -50.33 \frac{kJ}{g} \)
02

2. Calculate energy obtained from gasoline (octane, C8H18)

Similarly, for gasoline assumed as pure octane, we calculate the heat of combustion of octane. The balanced equation for the combustion of octane is: C8H18 (l) + 25/2 O2 (g) -> 8 CO2 (g) + 9 H2O (g) Here are the ∆H_f° values we need: ∆H_f° (C8H18) = -249.9 kJ/mol Now, we can calculate the ∆H_comb for octane: ∆H_comb = [ 8 * ∆H_f° (CO2) + 9 * ∆H_f° (H2O) ] - ∆H_f° (C8H18) = [ 8 * (-393.5) + 9 * (-241.8) ] - (-249.9) = -5471 kJ/mol To find energy per gram, we divide the energy obtained per mole by the molar mass of octane. \( Energy\ per\ gram\ of\ gasoline\ (octane) = \frac{-5471\ kJ/mol}{(8\times12.01 + 18\times1.008)\ g/mol} = -44.4 \frac{kJ}{g} \)
03

3. Boiling point of propane

The boiling point of propane can be found in a reference table. The boiling point of propane is approximately -42.1 °C or -43.8 °F.
04

4. Disadvantages of using propane as a fuel

Some disadvantages of using propane instead of gasoline as a fuel are: 1. Propane has a much lower boiling point, which means it needs to be stored under pressure, adding complexity and weight to the fuel system. 2. Propane has a lower energy density (50.33 kJ/g) than gasoline (44.4 kJ/g), meaning it will provide slightly less power than gasoline. 3. The availability of propane filling stations is limited compared to gasoline, making it less convenient for refueling. 4. The energy conversion efficiency for propane engines is lower than that of gasoline engines. 5. Propane is less suitable for cold climates, as its low boiling point may lead to problems with fuel delivery and cause engine cold-start issues.

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