Question

Consider the combustion of octane (C8H18), a primary component of gasoline.

2C8H18 (l) + 25O2(g) ---> 18H2O(g) + 16 CO2 (g)

a. Calculate the change in volume(in L) due to the total number of moles of gases produced when 1.000 gal (~3.784 L) of octane undergoes combustion at 1.00 atm and 25.00 C The density of octane at 25.00 C is 0.703g/mL

b. Use your answer in part A to calculate the volume of gases produced at 475K (the temperature of a car engine)

c. Use the equation , w= -P delta V, to calculate the work (in kJ) done by the system (the reaction) during the combustion of 1.000 gal of octane (Use 1L*atm = 101.325 J)

d. If 1 kJ = 1 kWs (kilo watt*second), then calculate the energy (in kJ) in a kilowatt hour.

e. Consider a new kind of vehicle that could be powered by an electrical current similar to that used in our homes. Calculate the cost of electricity needed to produce the same amount of energy (determined in part C) as the combustion of 1.00gal of gasoline if Seattle City Light currently charges on average about 11.35 cents/kWh

f. The combustion of 1.00 gal of octane produces about 1.2x105 kJ of heat. Compare the change in internal energy due to work calculated in part C with the heat of the reaction. How much of the internal energy change is due to the work done by the system due to gas expansion? (Hint: Compare the absolute value for heat versus work)

Answer 1

a. 2562 L

b. 4082 L

c. 413,6 kJ

d. 0,115 kWh

e. 1,3 cents

f. 0,34 %

Step-by-step explanation:

a. To calculate the volume we should obtain the moles of Octane to know the reactant moles and produced moles. Then, with ideal gas law obtain the change in volume:

3,784 L ≡ 3784 mL octane × (0,703 g / 1 mL) = 2660 g Octane

density

2660 g octane × ( 1 mol / 114,23 g octane) = 23,29 mol octane

molar mass of C₈H₁₈

With the combustion reaction of octane we can know how many moles are produced from 23,29 mol of octane, thus, in (1) :

2 C₈H₁₈ (l) + 25 O₂(g) ---> 18 H₂O(g) + 16 CO₂ (g) (1)

23,29 mol octane × ( 25 mol O₂ / 2 mol octane) = 291,1 mol O₂ -reactant moles-

23,29 mol octane × ( 18 + 16 produced mol / 2 mol octane) = 395,9 moles produced

Ideal gas formula says:

V = nRT/ P

Where:

n = Δmoles number (produced-reactant) → 104,8 moles

R = Ideal gas constant → 0,082 atm·L/mol·K

T = Temperature → 25°C, 298,15 K

P = Pressure → 1 atm

Thus, replacing in the equation:

ΔV = 2562 L

b. To calculate the gas volume we should use the same values of ideal gas formula just changing the temperature value for 475 K -Because the produced moles of gas and presure are the same and R is constant.

Thus, the volume of produced gases is:

ΔV = 4082 L

c. The work, w, is equal to -pressure times Δ Volume:

w = - P×ΔV

The pressure is 1 atm and ΔV in the system is 4082 L

So, w = 4082 atm·L (101,325 J / 1 atm·L) = 413,6 kJ

d. As kJ is equal to kWs, 413,6 kJ ≡ 413,6 kWs × ( 1 hour / 3600 s) =

0,115 kWh

e. In Seattle 1kWh cost 11,35 cents. So, 0,115 kWh cost:

0,115 kWh × (11,35 cents/ 1kWh) = 1,3 cents

f. The energy calculated in part C, 413,6 kJ is due to the work done by the system in gas expansion but total of heat produced in (1) are 1,2 ×10⁵ kJ. Thus, the proportion of work in gas expansion in total energy in combustion of octane is:

413,6 kJ / 1,2×10 ⁵ kJ × 100 = 0,34 %

I hope it helps!