Question

A civil engineering student is considering buying an electric car. However, the conscious student wants to make sure her carbon footprint will decrease with her new investment. Her current car runs on gasoline (C7H16) and has quite good fuel efficiency at 36.5mi/gal. The electric vehicle she is looking to buy can perform at 5.2mi/kWh. The student will recharge her car battery at school and the energy there comes from an 85% efficient power plant that runs on natural gas (CH4). Help the student make her decision by comparing the volume of CO2 (in gal) produced in a month by both the gasoline and the electric cars. Her commute is 16 miles per day and she drives to school 5 days of the week.

Answer 1

Answer:

$\text{An gasoline-powered vehicle has }\\\boxed{\text{five times the carbon footprint}}\text{ of an electric vehicle.}$

Explanation:

1. Miles travelled in an average month  

$\text{Miles} = \text{1 mo} \times \dfrac{\text{52 wk}}{\text{12 mo}} \times \dfrac{\text{5 da}}{\text{1 wk}} \times \dfrac{\text{16 mi}}{\text{1 da}} = \text{347 mi}$

2. Using a gasoline powered vehicle  

(a) Moles of heptane used  

$n = \text{347 mi} \times \dfrac{\text{1 gal}}{\text{36.5 mi}}\times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{679.5 g}}{\text{1L}} \times \dfrac{ \text{1 mol}}{\text{100.20 g}}= \text{244 mol}$

(b) Equation for combustion  

C₇H₁₆ + O₂ ⟶ 7CO₂ + 8H₂O  

(c) Moles of CO₂ formed  

$n = \text{244 mol heptane} \times \dfrac{\text{7 mol CO _{2} }}{\text{1 mol heptane}} = \text{1710 mol CO _{2} }$

(d) Volume of CO₂ formed  

At 20 °C and 1 atm, the molar volume of a gas is 24.0 L.  

$V = \text{1710 mol } \times \dfrac{\text{24.0 L}}{\text{1 mol}} \times \dfrac{\text{1 gal}}{\text{3.785 L}} = \textbf{10 800 gal}$

3. Using an electric vehicle  

(a) Theoretical energy used  

$\text{Theor. Energy} = \text{347 mi} \times \dfrac{\text{1 kWh}}{\text{5.2 mi}} = \text{66.7 kWh theor.}$

(b) Actual energy used  

The power station is only 85 % efficient.  

$\text{Actual energy used} = \text{66.7 kWh theor.} \times \dfrac{\text{100 kWh actual}}{\text{ 85 kWh theor.}}\times \dfrac{\text{3600 kJ}}{\text{1 kWh}}\\\\ = 2.82\times 10^{5} \text{ kJ}$

(c) Combustion of CH₄

CH₄ + 2O₂ ⟶ CO₂ +2 H₂O

(d) Equivalent volume of CO₂

The heat of combustion of methane is -802.3 kJ·mol⁻¹  

$V= 2.82\times 10^{5}\text{ kJ} \times \dfrac{\text{1 mol methane}}{\text{802.3 kJ}} \times \dfrac{\text{1 mol CO _{2} } }{\text{1 mol methane}} \times \dfrac{ \text{24.0 L}}{ \text{1 mol CO _{2} }}\\\\ \times \dfrac{\text{1 gal}}{\text{3.875 L}} = \textbf{2180 gal}$

4. Comparison  

$\dfrac{V_{\text{gasoline}}}{V_{\text{electric}}} = \dfrac{10800}{2180} = 5.0\\\\ \text{An gasoline-powered vehicle has }\\ \boxed{\textbf{ five times the carbon footprint}}\text{ of an electric vehicle.}$