Chapter 4, Problem 123QRT

Interpretation Introduction

Interpretation:

The miles of travelled by a taxi so that decreased fuel cost would balance the added cost of modified taxi’s motor has to be calculated.

Answer to Problem 123QRT

The miles travelled by taxi are approximately $400,000$ miles.

Explanation of Solution

First, the octane in milliliters is determined.

    $\begin{array}{c}\text{Volume }\Rightarrow \text{1 gal×}\frac{\text{4 quart}}{\text{1 gal}}\text{×}\frac{\text{1}{\text{.00×10}}^{\text{-3}}{\text{ m}}^{\text{3}}}{\text{1}\text{.056710 quart}}\text{×}{\left(\frac{\text{100 cm}}{\text{1 m}}\right)}^{\text{3}}\times \frac{1\text{ mL}}{1{\text{ cm}}^{\text{3}}}\\ \\ =\text{ 3}\text{.79 }\times {\text{10}}^{\text{3}}\text{ mL}\end{array}$

The energy released from combustion of one gram of octane is calculated as follows,

    $\begin{array}{l}{\text{2C}}_{\text{8}}{\text{H}}_{\text{18}}\left(l\right){\text{ + 25O}}_{\text{2}}\left(g\right)\to 16{\text{CO}}_{\text{2}}\left(g\right){\text{ + 18H}}_{\text{2}}\text{O}\left(g\right){\text{Δ}}_{\text{r}}{\text{H}}^{\text{o}}=-10992\text{ kJ}\\ \\ {\text{C}}_{\text{8}}{\text{H}}_{\text{18}}\left(l\right)\text{ + }\frac{\text{25}}{2}{\text{O}}_{\text{2}}\left(g\right)\to 8{\text{CO}}_{\text{2}}\left(g\right){\text{ + 9H}}_{\text{2}}\text{O}\left(g\right){\text{Δ}}_{\text{r}}{\text{H}}^{\text{o}}=-5496\raisebox{1ex}{$\text{kJ}$}\!\left/ \!\raisebox{-1ex}{${\text{mol C}}_{\text{8}}{\text{H}}_{\text{18}}$}\right.\\ \\ \Rightarrow \frac{5496\text{ kJ produced }}{1{\text{ molC}}_{\text{8}}{\text{H}}_{\text{18}}}\times \frac{1{\text{ molC}}_{\text{8}}{\text{H}}_{\text{18}}}{114.2278{\text{ g C}}_{\text{8}}{\text{H}}_{\text{18}}}\\ \\ =\text{ 48}\text{.11 }\frac{\text{kJ}}{{\text{g C}}_{\text{8}}{\text{H}}_{\text{18}}}\\ \\ \left[\text{Molar mass of propane = }114.2278\text{ g/mol}\right]\end{array}$

    $\begin{array}{l}\Rightarrow \frac{\text{48}\text{.11 kJ produced}}{{\text{1 g C}}_{\text{8}}{\text{H}}_{\text{18}}}\text{×}\frac{\text{0}{\text{.703 g C}}_{\text{8}}{\text{H}}_{\text{18}}}{{\text{1 mLC}}_{\text{8}}{\text{H}}_{\text{18}}}\text{×}\frac{3.79\times {\text{10}}^{\text{3}}\text{ mL}}{1\text{ gal}}\\ \\ =\text{ 1}\text{.28}\times {\text{10}}^{\text{5}}\frac{\text{kJ}}{\text{gal}}\\ \\ \Rightarrow \text{ 1}\text{.28}\times {\text{10}}^{\text{5}}\frac{\text{kJ}}{\text{gal}}\times \frac{1{\text{ gal C}}_{\text{8}}{\text{H}}_{\text{18}}}{\$\text{ 3}\text{.50}}\\ \\ =\text{ 3}\text{.66 }\times {\text{10}}^{\text{4}}\frac{\text{kJ}}{\text{\$}}\end{array}$

The reaction enthalpy for combustion of propane is determined as follows,

    $\begin{array}{l}{\text{C}}_{\text{3}}{\text{H}}_{\text{18}}\left(g\right)+5{\text{O}}_{\text{2}}\left(g\right)\to {\text{3CO}}_{\text{2}}\left(g\right)+{\text{4H}}_{\text{2}}\text{O}\left(g\right)\\ \\ {\text{Δ}}_{\text{r}}{\text{H}}^{\text{o}}{\text{ = 3Δ}}_{\text{f}}{\text{H}}^{\text{o}}\left[{\text{CO}}_{\text{2}}\left(\text{g}\right)\right]{\text{+4Δ}}_{\text{f}}{\text{H}}^{\text{o}}\left[{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)\right]\text{-}\left\{{\text{Δ}}_{\text{f}}{\text{H}}^{\text{o}}\left[{\text{C}}_{\text{3}}{\text{H}}_{\text{18}}\left(\text{g}\right)\right]{\text{+5Δ}}_{\text{f}}{\text{H}}^{\text{o}}\left[{\text{O}}_{\text{2}}\left(\text{g}\right)\right]\right\}\\ \\ =\text{ 3}\left[\text{-393}\text{.509 kJ/mol}\right]\text{+4}\left[\text{-241}\text{.818kJ/mol}\right]\text{-}\left[\text{-103}\text{.8kJ/mol}\right]\text{-5}\left[\text{0kJ/mol}\right]\\ \\ \text{ = -2044}\text{.0kJ}\end{array}$

    $\begin{array}{l}\Rightarrow \frac{\text{2044 kJ produced}}{{\text{1 mol C}}_{\text{3}}{\text{H}}_{\text{8}}}\text{×}\frac{{\text{1 mol C}}_{\text{3}}{\text{H}}_{\text{8}}}{\text{44}{\text{.0953 g C}}_{\text{3}}{\text{H}}_{\text{8}}}\\ \\ =46.354\frac{\text{kJ}}{{\text{g C}}_{\text{3}}{\text{H}}_{\text{8}}}\\ \\ \Rightarrow \frac{46.354\text{ kJ produced}}{{\text{1 g C}}_{\text{3}}{\text{H}}_{\text{8}}}\times \frac{0.531{\text{ g C}}_{\text{3}}{\text{H}}_{\text{8}}}{1{\text{ mLC}}_{\text{3}}{\text{H}}_{\text{8}}}\times \frac{3.79\times {10}^3\text{ mL}}{1\text{ gal}}\times \frac{1{\text{ galC}}_{\text{3}}{\text{H}}_{\text{8}}}{\$2.50}\\ \\ =\text{ 3}{\text{.73×10}}^{\text{4}}\frac{\text{kJ}}{\text{\$}}\end{array}$

The energy required for taxi to travel per mile is determined as follows,

Consider the taxi gets average of $15.0$ miles per gallon using gasoline.

  $\begin{array}{l}\Rightarrow \frac{1.28\times {10}^5\text{ kJ}}{1\text{ gal}}\times \frac{1\text{ gal}}{15.0\text{ miles}}\\ \\ =\text{ 8}\text{.53}\times {\text{10}}^{\text{3}}\frac{\text{kJ}}{\text{miles}}\end{array}$

Then the cost of fuel used per mile is determined.

    $\begin{array}{l}\Rightarrow \frac{1\text{ gal gasoline}}{15.0\text{ miles}}\times \frac{\$3.50}{1\text{ gal gasoline}}\\ \\ =\frac{\text{\$ 0}\text{.233}}{\text{miles}}\end{array}$

Assume the modified taxi also used same energy per mile the fuel cost for travelling one mile using propane is shown as follows,

    $\begin{array}{l}\Rightarrow \frac{8.53\times {10}^3\text{ kJ}}{1\text{ miles}}\times \frac{\$1}{3.73\times {10}^4\text{ kJ}}\\ \\ =\frac{\text{\$ 0}\text{.229}}{\text{miles}}\end{array}$

The miles required to recouped with the conversion cost is shown as follows,

    $\begin{array}{l}\$2000\times \frac{1\text{ mile}}{\$0.005}\\ \\ =\text{ 400,000 miles}\end{array}$