Chapter 8, Problem ISP

In a typical automobile engine, a gasoline vapor-air mixture is compressed and ignited in the cylinders of the engine. This results in a combustion reaction that produces mainly carbon dioxide and water vapor. For simplicity, assume that the fuel is C₈H₁₈ and has a density of 0.760 g/mL.

1. (a) Calculate the partial pressures of N₂ and 0₂ in the air before it goes into the cylinder; assume the atmospheric pressure is 734 mmHg.
2. (b) Consider the case where the air, without any fuel added, is compressed in the cylinder to seven times atmospheric pressure, the compression ratio of many modem automobile engines. Calculate the partial pressures of N₂ and O₂ at this pressure.
3. (c) Now consider the case where 0.050 mL gasoline is added to the air in the cylinder just before compression and completely vaporized. Assume that the volume of the cylinder is 485 mL and the temperature is 150°C. Calculate the partial pressure of the gasoline vapor.
4. (d) Calculate the amount (mol) of oxygen required to bum the gasoline in part (c) completely to CO₂ and H₂O.
5. (e) The combustion reaction in the cylinder creates temperatures in excess of 1200K. Due to the high temperature, some of the nitrogen and oxygen in the air reacts to form nitrogen monoxide. If 10% of the nitrogen is converted to NO, calculate the mass (g) of NO produced by this combustion.
6. (f) Hot-rod cars use another oxide of nitrogen, dinitrogen monoxide, to create an extra burst in power. When such a power boost is needed, dinitrogen monoxide gas is injected into the cylinders where it reacts with oxygen to form NO. Calculate the mass of dinitrogen monoxide that would have to be injected to form the same quantity of NO as produced in part (e). Assume that sufficient oxygen is present to do so.

Interpretation Introduction

Interpretation:

Partial pressure of $N_2$ and $O_2$ has to be calculated when the atmospheric pressure is $\text{734 mmHg}\text{.}$

Concept Introduction:

Mole fraction: Quantity which defines the number of moles of a substance in a mixture divided by the total number of moles of all substances present.

${\text{X}}_{\text{a}}\text{=}\frac{{\text{n}}_{\text{a}}}{{\text{n}}_{\text{total}}}$

Partial pressure of a gas in the mixture of gases is the product of mole fraction of the gas and the total pressure.

  ${\text{P}}_{\text{a}}\text{=}{\text{X}}_{\text{a}}\text{×}{\text{P}}_{\text{total}}$

Answer to Problem ISP

Partial pressure of $N_2$ is $573\text{mmHg}$.

Partial pressure of $O_2$ is $154\text{mmHg}$.

Explanation of Solution

Percentage by volume of $N_2$ is 78.084%.

Percentage by volume of $O_2$ is 20.948%.

Mole fraction of $N_2$ and $O_2$ is calculated as follows,

  $\begin{array}{c}{\text{X}}_{{\text{N}}_2}=\frac{\%{\text{N}}_2}{100\%}\\ \\ =\frac{78.084\%}{100\%}=0.78084\\ \\ {\text{X}}_{O_2}=\frac{\%{\text{O}}_2}{100\%}\\ \\ =\frac{20.948\%}{100\%}=0.20948\end{array}$

Partial pressure of $N_2$ is calculated as follows,

  $\begin{array}{c}{\text{P}}_{{\text{N}}_2}={\text{X}}_{{\text{N}}_2}{\text{P}}_{\text{total }}\\ \\ =\left(0.78084\right)\left(734\text{mmHg}\right)=573\text{mmHg}\end{array}$

Partial pressure of $O_2$ is calculated as follows,

  $\begin{array}{c}{\text{P}}_{{\text{O}}_2}={\text{X}}_{{\text{O}}_2}{\text{P}}_{\text{total }}\\ \\ =\left(0.20948\right)\left(734\text{mmHg}\right)=154\text{mmHg}\end{array}$

Interpretation Introduction

Interpretation:

Partial pressure of $N_2$ and $O_2$ when the volume of gas is compressed by seven times has to be calculated.

Answer to Problem ISP

Partial pressure of $N_2$ is $4012\text{mmHg}$.

Partial pressure of $O_2$ is $1076\text{mmHg}$.

Explanation of Solution

At $734\text{mmHg}$, partial pressure of $N_2$ and $O_2$ is $573\text{mmHg}$ and $154\text{mmHg}$ respectively.

According to Boyle’s law, pressure and volume is inversely proportional to each other at constant temperature and number of molecules.

  $\text{P}\propto \frac{\text{1}}{\text{V}}$

Hence, when the volume of gas is compressed by seven times, then pressure of the sample is increased by seven times.

Therefore,

Partial pressure of $N_2$ and $O_2$ when the pressure is raised by a factor of 7 is calculated as shown below,

  $\begin{array}{c}{\text{P}}_{{\text{N}}_2}=7\times 573\text{mmHg}\\ \\ =4012\text{mmHg}\\ \\ {\text{P}}_{{\text{O}}_2}=7\times 154\text{mmHg}\\ \\ =1076\text{mmHg}\end{array}$

Interpretation Introduction

Interpretation:

Partial pressure of the gasoline vapor has to be calculated.

Concept Introduction:

Ideal gas Equation:

Any gas is described by using four terms namely pressure, volume, temperature and the amount of gas. Thus combining three laws namely Boyle’s, Charles’s Law and Avogadro’s Hypothesis the following equation could be obtained. It is referred as ideal gas equation.

  $\text{PV = nRT}$

Here,

  n is the moles of gas

    P is the Pressure

  V is the Volume

    T is the Temperature

    R is the gas constant

Answer to Problem ISP

Partial pressure of the gasoline vapor is $\text{18 mmHg}\text{.}$

Explanation of Solution

Given information is shown below,

  $\begin{array}{c}{\text{V}}_{{\text{C}}_{\text{8}}{\text{H}}_{\text{18}}}\text{ = 0}\text{.050 mL}\\ \\ \text{Density= 0}\text{.760 g/mL}\\ \\ {\text{V}}_{\text{cylinder}}\text{ = 485 mL = 0}\text{.485 L}\\ \\ \text{T = 150}°\text{C = }\left(150+273\right)\text{ = 423 K}\end{array}$

Number of moles of gasoline $\left(C_8{H}_{18}\right)$ can be determined form the density and molar mass $C_8{H}_{18}$ as shown below,

  $\begin{array}{c}n\text{ = }0.050{\text{ mL C}}_8{\text{H}}_{18}\times \frac{0.760{\text{ g C}}_8{\text{H}}_{18}}{1{\text{ mL C}}_8{\text{H}}_{18}}\times \frac{1{\text{ mol C}}_8{\text{H}}_{18}}{114.2{\text{ g C}}_8{\text{H}}_{18}}\\ \\ =3.3\times {10}^{-4}{\text{molC}}_8{\text{H}}_{18}\end{array}$

Partial pressure of the gasoline vapor can be calculated using Ideal Gas equation as follows,

  $\begin{array}{c}\text{PV}=\text{nRT}\\ \\ P\left(0.485\text{L}\right)=\left(3.3\times {10}^{-4}{\text{molC}}_8{\text{H}}_{18}\right)\left(0.0821\text{ L}\text{.atm}.{\text{mol}}^{-1}{\text{K}}^{-1}\right)\left(423\text{}\text{}\text{}\text{ K}\right)\\ \\ =\frac{\left(3.3\times {10}^{-4}\text{mol}{\text{C}}_8{\text{H}}_{18}\right)\left(0.0821\text{ L}\text{.atm}.{\text{mol}}^{-1}{\text{K}}^{-1}\right)\left(423\text{}\text{}\text{}\text{ K}\right)}{\left(0.485\text{}\text{}\text{ L}\right)}\\ \\ =0.024\text{ atm}{\text{C}}_8{\text{H}}_{18}\\ \\ =0.024\text{ atm}\times \frac{760\text{ mmHg}}{1\text{ atm}}\\ \\ =\text{ 18 mmHg}{\text{C}}_8{\text{H}}_{18}\end{array}$

Interpretation Introduction

Interpretation:

Number of moles of ${\text{O}}_2$ required to burn $C_8{H}_{18}$ has to be calculated.

Answer to Problem ISP

Number of moles of ${\text{O}}_2$ reacted is $0.0041\text{ mol}\text{.}$

Explanation of Solution

Balanced equation for the combustion of gasoline is given below,

  $2{\text{C}}_8{\text{H}}_{18}\left(l\right)+25{\text{O}}_2\left(\text{g}\right)\underset{}{\to }16{\text{CO}}_2\left(\text{g}\right)+18{\text{H}}_2\text{O}\left(\text{g}\right)$

Number of moles of gasoline $\left(C_8{H}_{18}\right)$ can be determined form the density and molar mass $C_8{H}_{18}$ as shown below,

  $\begin{array}{c}n\text{ = }0.050{\text{ mL C}}_8{\text{H}}_{18}\times \frac{0.760{\text{ g C}}_8{\text{H}}_{18}}{1{\text{ mL C}}_8{\text{H}}_{18}}\times \frac{1{\text{ mol C}}_8{\text{H}}_{18}}{114.2{\text{ g C}}_8{\text{H}}_{18}}\\ \\ =3.3\times {10}^{-4}{\text{mol C}}_8{\text{H}}_{18}\end{array}$

From the balanced equation, it is clear that ${\text{2 mol C}}_8{H}_{18}$ reacts with $25{\text{ mol O}}_2$. Hence, number of moles of ${\text{O}}_2$ reacted is determined as follows,

  $\begin{array}{c}{n}_{O_2}\text{ = }3.3\times {10}^{-4}{\text{mol C}}_8{\text{H}}_{18}\times \frac{25{\text{ mol O}}_2}{2{\text{ mol C}}_8{\text{H}}_{18}}\\ \\ =0.0041{\text{ mol O}}_2\end{array}$

Interpretation Introduction

Interpretation:

Mass of $NO$ formed from $N_2$ has to be calculated.

Concept Introduction:

Refer to (c)

Answer to Problem ISP

Mass of $NO$ formed is $\text{1}\text{.56}\text{}\text{ g}$

Explanation of Solution

Given information is shown below,

  $\begin{array}{c}\text{P = 4012 mm Hg}\\ \\ =\text{ 4012 mm Hg}\times \frac{1\text{ atm}}{\text{760 mm Hg}}\text{ = 5}\text{.279e atm}\\ \\ {\text{V}}_{\text{cylinder}}\text{ = 485 mL = 0}\text{.485 L}\\ \\ \text{T = 1}200\text{ K}\end{array}$

Number of moles of $N_2$ is calculated using Ideal gas law as follows,

  $\begin{array}{c}\text{PV}={\text{n}}_{{\text{N}}_2}\text{RT}\\ \\ \left(5.279\text{ atm}\right)\left(0.485\text{ L}\right)={\text{ n}}_{{\text{N}}_2}\left(0.0821\text{ L}\text{.atm}{\text{.mol}}^{-1}{\text{.K}}^{-1}\right)\left(1200\text{ K}\right)\\ \\ {\text{n}}_{{\text{N}}_2}=\frac{\left(5.279\text{ atm}\right)\left(0.485\text{ L}\right)}{\left(0.0821\text{ L}\text{.atm}{\text{.mol}}^{-1}{\text{.K}}^{-1}\right)\left(1200\text{ K}\right)}\\ \\ =2.599\times {10}^{-2}{\text{mol N}}_2\end{array}$

The reaction that shows the formation of $NO$ from $N_2$ is given below,

  ${\text{N}}_2\left(\text{g}\right)+{\text{O}}_2\left(\text{g}\right)\underset{}{\to }2\text{NO}\left(\text{g}\right)$

From the balanced equation, it is clear that $1\text{}\text{ mol}\text{}{\text{ N}}_2$ reacts to give $2\text{}\text{}\text{ mol}\text{}\text{ NO}$. Hence, number of moles of $NO$ formed is determined,

  $\begin{array}{c}n\text{ = }2.599\times {10}^{-2}\text{mol}\text{}{\text{ N}}_2\times \frac{2\text{}\text{}\text{ mol}\text{}\text{ NO}}{1\text{}\text{ mol}\text{}{\text{ N}}_2}\\ \\ =0.052\text{}\text{ mol}\text{}\text{ NO}\end{array}$

Mass of $NO$ formed is determined as follows,

  $\begin{array}{c}Mass=\text{ No}\text{. of moles}\times \text{Molar mass}\\ \\ =0.052\text{}\text{ mol}\text{}\text{ NO}\times \frac{30.01\text{}\text{ g}\text{}\text{ NO}}{1\text{}\text{ mol}\text{}\text{ NO}}\\ \\ =\text{ 1}\text{.56}\text{}\text{ g}\text{}\text{ NO}\end{array}$

Interpretation Introduction

Interpretation:

Mass of $N_{2}O$ required has to be calculated.

Answer to Problem ISP

Mass of $N_{2}O$ required is $\text{1}\text{.14 g}$.

Explanation of Solution

The reaction that shows the formation of $NO$ from $N_{2}O$ is given below,

  ${\text{N}}_{2}O\left(\text{g}\right)+\text{}\frac{1}{2}{\text{O}}_2\left(\text{g}\right)\underset{}{\to }2\text{NO}\left(\text{g}\right)$

Number of moles of $NO$ formed is $0.052\text{}\text{ mol}$.

From the balanced equation, it is clear that $1\text{}\text{ mol}\text{}{\text{ N}}_{2}O$ reacts to give $2\text{}\text{}\text{ mol}\text{}\text{ NO}$. Hence, mass of $N_{2}O$ required is determined as follows,

  $\begin{array}{c}Mass\text{ = }0.052\text{}\text{ mol NO}\times \left(\frac{{\text{1 mol N}}_{\text{2}}\text{O}}{\text{2 mol NO}}\right)\times \left(\frac{\text{44}{\text{.01 g N}}_{\text{2}}\text{O}}{{\text{1 mol N}}_{\text{2}}\text{O}}\right)\\ \\ =\text{ 1}{\text{.14 g N}}_{2}O\end{array}$