Question

Question: A young chemist buys a "one-lung" motorcycle but, before learning how to drive it, wants to understand the processes that occur in its engine. The manual says the cylinder has a radius of $\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathit{c}\mathit{m}$, a piston stroke of 12.00 cm , and a (volume) compression ratio of 8:1 . If a mixture of gasoline vapour (taken to be${\mathbf{C}}_{\mathbf{8}}{\mathbf{H}}_{\mathbf{18}}$ ) and air in mole ratio 1:62.5 is drawn into the cylinder at 80°Cand 1.00atm , calculate:

(a) The temperature of the compressed gases just before the spark plug ignites them. (Assume that the gases are ideal, that the compression is adiabatic, and that the average heat capacity of the mixture of gasoline vapour and air is ${\mathit{c}}_{\mathbf{P}}\mathbf{=}\mathbf{35}\mathbf{}\mathbf{J}\mathbf{}{\mathbf{K}}^{\mathbf{-}\mathbf{1}}\mathbf{}{\mathbf{mol}}^{\mathbf{-}\mathbf{1}}$.)

(b) The volume of the compressed gases just before ignition.

(c) The pressure of the compressed gases just before ignition.

(d) The maximum temperature of the combustion products, assuming combustions completed before the piston begins its down stroke. Takerole="math" localid="1663436252694" $\mathit{\Delta }{\mathit{H}}_{\mathbf{\text{f}}}^{\mathbf{°}}\left({\text{C}}_8{\text{H}}_{18}\right)\mathbf{=}\mathbf{-}\mathbf{57}\mathbf{.4}\mathbf{}\mathbf{\text{kJ}}\mathbf{}{\mathbf{\text{mol}}}^{\mathbf{-}\mathbf{1}}$

(e) The temperature of the exhaust gases, assuming the expansion stroke to be adiabatic.

Short answer

(a) The temperature of the compressed gases just before ignition is,$T_2=672.56\mathrm{K}$.

(b) The volume of the compressed gases just before ignition is $V_2=0.118\mathrm{L}$ .

(c) The pressure of the compressed gases just before ignition is,$p_2=15.24\mathrm{atm}$.

(d) The maximum temperature of the combustion productsis$T_3=3898.81\mathrm{K}$.

(e) The temperature of the exhaust gases is ,$T_4=2046.32\mathrm{K}$ .

Step-by-step solution

Given data

Volume compression ratio $=1:8$. The ratio of mixture of ${\mathrm{C}}_8{\mathrm{H}}_{18}$ and air is s drawn into the cylinder is given as, ${\mathrm{C}}_8{\mathrm{H}}_{18}:air=1:62.5$.

The radius and piston stroke are given as:

$\begin{array}{l}r=5.00\mathrm{cm}\\ l=12.00\mathrm{cm}\end{array}$

Given temperature,$T={80}^{o}C$.

Given pressure, $p=1.00\mathrm{atm}$.

Concept of principle of the combustion engine

The principle of the combustion engine

The processes that occur:

1 Adiabatic compression

2 Combustion

3 Adiabatic expansions

4 Heat being extracted

The engine is made up from single cylinder and a piston.

The piston moves, so the gas inside the cylinder can occupy two possible volumes - the expanded volume and the compressed volume.

First, a mixture of air and gas is introduced in the cylinder. The piston then moves compressing this mixture adiabatically and increasing its temperature. An electric spark causes the gasoline to combust with oxygen. This process raises the temperature which causes the piston to push back in a fast adiabatic expansion. The remaining gases are then "extracted" through the motorcycle’s exhaust

Calculate temperature before ignition

(a)

The volume compression ratio is = 1:8.

That means that the compressed volume is:

$V_c=\frac{1}{8}{V}_i$

This would mean that the volume ratio is:

$\frac{V_1}{V_2}=\frac{V_1}{\frac{1}{8}{V}_1}=8$

For determination of γ first the value of $c_v$is required to calculate.

The relation between the capacities:

$\begin{array}{l}{c}_p=c_v+R\\ c_v=c_p-R\\ c_v=35-8.314\\ c_v=26.69\mathrm{J}{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}\end{array}$

The γ is obtained by, $\gamma =\frac{c_p}{c_v}$.

$\begin{array}{l}\gamma =\frac{35}{26.69}\\ \gamma =1.31\end{array}$

The relation between the temperature and volume for adiabatic processes:

$\frac{T_2}{T_1}={\left(\frac{V_1}{V_2}\right)}^{\gamma -1}$ ……… (1)

Where γ is the relation of the heat capacities.

Put the value of capacities in equation (1).

$\begin{array}{c}\frac{T_2}{80+273}=8^{1.31-1}\\ T_2=353\times 8^{0.31}\\ T_2=672.56\mathrm{K}\end{array}$

Calculate volume of the compressed gas before ignition

(b)

The compression takes place in the cylinder, therefore use the formula for the volume of a cylinder.

$V=r^2\pi l$

Where ris the radius, the l is the piston stroke.

$V_2=\frac{r^2\pi l}{8}$ ……. (2)

Put the value of r and l

$\begin{array}{l}{V}_2=\frac{{5.00}^2\times \pi \times 12.00}{8}\\ V_2=\frac{{5.00}^2\times \pi \times 12.00}{8}\\ V_2=117.81{\mathrm{cm}}^3\\ V_2=0.118\mathrm{L}\end{array}$

Calculate pressure of compressed gas

(c)

The calculated data is, γ = 1.31 .

$\begin{array}{c}{V}_1=8V_2\\ =0.944\\ V_2=0.118\\ P_1=1.00\mathrm{atm}\end{array}$

Put the value of all above data in below equation to calculate pressure before ignition.

$\begin{array}{l}{p}_2=p_1{\left(\frac{V_1}{V_2}\right)}^{\gamma }\\ p_2=1.00\times 8^{1.31}\\ p_2=15.24\mathrm{atm}\end{array}$

Calculate total number of moles

(d)

The total number of moles can be calculated through the ideal gas equation.

$\begin{array}{c}pV=nRT\\ n=\frac{p_2{V}_2}{R{T}_2}\\ n=\frac{15.21\times 0.118}{0.08206\times 672.56}\\ n=0.0325\mathrm{mol}\end{array}$

Note: this value is for the mixture of air and gas.

To calculate the amount of moles of ${\mathrm{C}}_8{\mathrm{H}}_{18}$.

$\begin{array}{c}{\mathrm{C}}_8{\mathrm{H}}_{18}:air=1:62.5\\ n\left({\mathrm{C}}_8{\mathrm{H}}_{18}\right)=\frac{0.0325}{62.5}\\ n\left({\mathrm{C}}_8{\mathrm{H}}_{18}\right)=5.2\times {10}^{-4}\mathrm{mol}\end{array}$

Calculate change in enthalpy

The reaction, ${\mathrm{C}}_8{\mathrm{H}}_{18}+25/2{\mathrm{O}}_2\to 8{\mathrm{CO}}_2+9{\mathrm{H}}_2\mathrm{O}$.

The standard enthalpies of formation:

$\begin{array}{c}\Delta H_f\left({\mathrm{H}}_2\mathrm{O}\right)=-241.82\mathrm{kJ}/\mathrm{mol}\\ \Delta H_f\left({\mathrm{CO}}_2\right)=-393.51\mathrm{kJ}/\mathrm{mol}\\ \Delta H_f\left({\mathrm{C}}_8{\mathrm{H}}_{18}\right)=-57.4\mathrm{kJ}{\mathrm{mol}}^{-1}\end{array}$

The enthalpy change for the combustion of octane is calculated as:

$\begin{array}{c}\Delta H=8\Delta H_f\left({\mathrm{CO}}_2\right)+9\Delta H_f\left({\mathrm{H}}_2\mathrm{O}\right)-\Delta H_f\left({\mathrm{C}}_8{\mathrm{H}}_{18}\right)\\ =8\times (-393.51)+9\times (-241.82)+(-57.4)\\ =-5381.78\mathrm{kJ}/\mathrm{mol}\end{array}$

$\begin{array}{c}\Delta H_f\left({\mathrm{H}}_2\mathrm{O}\right)=-241.82\mathrm{kJ}/\mathrm{mol}\\ \Delta H_f\left({\mathrm{CO}}_2\right)=-393.51\mathrm{kJ}/\mathrm{mol}\\ \Delta H_f\left({\mathrm{C}}_8{\mathrm{H}}_{18}\right)=-57.4\mathrm{kJ}{\mathrm{mol}}^{-1}\end{array}$

Calculate heat generate

The heat is calculated as follows:

$\begin{array}{l}q=n\left({\mathrm{C}}_8{\mathrm{H}}_{18}\right)\times \Delta H\left({\mathrm{C}}_8{\mathrm{H}}_{18}\right)\\ q=5.2\times {10}^{-4}\times 5381.78\times {10}^3\\ q=2798.53\mathrm{J}.\end{array}$

Calculate maximum temperature of combustion product

The temperature change is calculated as:

$\begin{array}{l}\Delta T=\frac{q}{n\left(\mathrm{mixture}\right)\times c_v}\\ \Delta T=\frac{2798.53}{0.0325\times 26.69}\\ \Delta T=3226.25\mathrm{K}\end{array}$

The temperature after ignition is calculated below.

$\begin{array}{l}{T}_3=T_2+\Delta T\\ T_3=T_2+\Delta T\\ T_3=672.56+3226.25\\ T_3=3898.81\mathrm{K}\end{array}$

Calculate temperature of the exhaust gases

(e)

The temperature volume relation for adiabatic processes is shown below.

$\frac{T_4}{T_3}={\left(\frac{V_3}{V_4}\right)}^{\gamma -1}$

Put all of the necessary information which is calculated in the previous steps.

The temperature of the exhaust gases is calculated as:

$\begin{array}{l}{T}_4=3898.81\times {\left(\frac{1}{8}\right)}^{0.31}\\ T_4=2046.32\mathrm{K}\end{array}$