Chapter 3, Problem 3B.10P

(a)

Interpretation Introduction

Interpretation: The work for the process has to be evaluated.  The heat involved in the each of the four steps has to be calculated in terms of $C_{\text{V,m}}$ and temperatures, $T_{\text{A}}-T_{\text{D}}$.

Concept introduction: The gas power cycles that are used in the spark-ignition internal combustion engines is known as Otto cycle.  It consists of two reversible adiabatic processes and two isochoric processes.

Answer to Problem 3B.10P

The total work done is $C_{\text{V,m}}\left[\left(T_{\text{B}}-T_{\text{A}}\right)+\left(T_{\text{D}}-T_{\text{C}}\right)\right]$.  The heat $q_{\text{I}}=0$, $q_{\text{III}}=0$, $q_{\text{II}}=C_{\text{V,m}}\left(T_{\text{C}}-T_{\text{B}}\right)$ and $q_{\text{IV}}=C_{\text{V,m}}\left(T_{\text{D}}-T_{\text{A}}\right)$.

Explanation of Solution

The work done for isochoric process is zero.  Hence, the work done for (II) and (IV) process is zero.  The general formula to calculate work, $W$ is,

    $W=U+q$

Where,

• • $U$ is the internal energy.
• • $q$ is the heat.

For adiabatic process, heat is zero.  Therefore,

    $W=U$

Hence, the total work done is calculated as,

    $W_{\text{tot}}=W_{\text{I}}+W_{\text{III}}$

    $W_{\text{tot}}=U_{\text{I}}+U_{\text{III}}$                                                                                               (1)

Where,

• • $W_{\text{tot}}$ is the total work done.
• • $W_{\text{I}}$ is the work done in process (I).
• • $W_{\text{II}}$ is the work done in process (III).
• • $U_{\text{I}}$ is the internal energy for process (I).
• • $U_{\text{III}}$ is the internal energy for process (III).

The general formula for internal energy is,

    $U=C_{\text{V,m}}dT$

Where,

• • $U$ is the internal energy.
• • $C_{\text{V,m}}$ is the specific heat at constant volume.
• • $dT$ is the change in temperature.

Substitute the value of internal energy in equation (1).

    $\begin{array}{c}{W}_{\text{tot}}=C_{\text{V,m}}\left(T_{\text{B}}-T_{\text{A}}\right)+C_{\text{V,m}}\left(T_{\text{D}}-T_{\text{C}}\right)\\ =C_{\text{V,m}}\left[\left(T_{\text{B}}-T_{\text{A}}\right)+\left(T_{\text{D}}-T_{\text{C}}\right)\right]\end{array}$

Where,

• • $T_{\text{B}}$ is the temperature at point B.
• • $T_{\text{A}}$ is the temperature at point A.
• • $T_{\text{D}}$ is the temperature at point D.
• • $T_{\text{C}}$ is the temperature at point C.

Hence, the total work done is $C_{\text{V,m}}\left[\left(T_{\text{B}}-T_{\text{A}}\right)+\left(T_{\text{D}}-T_{\text{C}}\right)\right]$.

The heat for the adiabatic process, (I) and (III), $q_{\text{I}}$ and $q_{\text{III}}$ will be zero.  Hence, the heat of the process (II), $q_2$ is given by,

    $\begin{array}{c}{q}_{\text{II}}=U_{\text{II}}=C_{\text{V,m}}dT\\ =C_{\text{V,m}}\left(T_{\text{C}}-T_{\text{B}}\right)\end{array}$

Similarly, heat for the process (IV) is,

    $\begin{array}{c}{q}_{\text{IV}}=U_{\text{IV}}=C_{\text{V,m}}dT\\ =C_{\text{V,m}}\left(T_{\text{D}}-T_{\text{A}}\right)\end{array}$

Hence, the heat, $q_{\text{I}}=0$, $q_{\text{III}}=0$, $q_{\text{II}}=C_{\text{V,m}}\left(T_{\text{C}}-T_{\text{B}}\right)$ and $q_{\text{IV}}=C_{\text{V,m}}\left(T_{\text{D}}-T_{\text{A}}\right)$.

(b)

Interpretation Introduction

Interpretation: The expression for efficiency, $\eta$ has to be derived in terms of $T_{\text{A}}-T_{\text{D}}$.

Concept introduction: The gas power cycles that are used in the spark-ignition internal combustion engines is known as Otto cycle.  It consists of two reversible adiabatic processes and two isochoric processes.

Answer to Problem 3B.10P

The efficiency of the cycle is $1-\frac{\left(T_{\text{D}}-T_{\text{A}}\right)}{\left(T_{\text{C}}-T_{\text{B}}\right)}$.

Explanation of Solution

The total work done, $W_{\text{tot}}$ is $C_{\text{V,m}}\left[\left(T_{\text{B}}-T_{\text{A}}\right)+\left(T_{\text{D}}-T_{\text{C}}\right)\right]$.  The heat, $q_{\text{II}}=C_{\text{V,m}}\left(T_{\text{C}}-T_{\text{B}}\right)$ .

The efficiency is calculated by the expression,

    $\begin{array}{c}\eta =\frac{W_{\text{tot}}}{q_{\text{II}}}\\ =\frac{C_{\text{V,m}}\left[\left(T_{\text{B}}-T_{\text{A}}\right)+\left(T_{\text{D}}-T_{\text{C}}\right)\right]}{C_{\text{V,m}}\left(T_{\text{C}}-T_{\text{B}}\right)}\\ =1-\frac{\left(T_{\text{D}}-T_{\text{A}}\right)}{\left(T_{\text{C}}-T_{\text{B}}\right)}\end{array}$

Hence, the efficiency of the cycle is $1-\frac{\left(T_{\text{D}}-T_{\text{A}}\right)}{\left(T_{\text{C}}-T_{\text{B}}\right)}$.

(c)

Interpretation Introduction

Interpretation: The expression for efficiency has to be derived as $1-{\left(\frac{V_{\text{B}}}{V_{\text{A}}}\right)}^{1/\text{c}}$.

Concept introduction: The gas power cycles that are used in the spark-ignition internal combustion engines is known as Otto cycle.  It consists of two reversible adiabatic processes and two isochoric processes.

Answer to Problem 3B.10P

The expression for efficiency can be written as $1-{\left(\frac{V_{\text{B}}}{V_{\text{A}}}\right)}^{1/\text{c}}$.

Explanation of Solution

The efficiency of the cycle is $1-\frac{\left(T_{\text{D}}-T_{\text{A}}\right)}{\left(T_{\text{C}}-T_{\text{B}}\right)}$.

Since,

    $P{V}^{\text{c}}=\text{Constant}$ and $T{V}^{1/\text{c}}=\text{Constant}$

Therefore,

    $\begin{array}{l}{T}_{\text{A}}{V}_{\text{A}}^{1/\text{c}}=T_{\text{B}}{V}_{\text{B}}^{1/\text{c}}\\ T_{\text{C}}{V}_{\text{C}}^{1/\text{c}}=T_{\text{D}}{V}_{\text{D}}^{1/\text{c}}\end{array}$

Hence, dividing the above two equations.

    $\frac{T_{\text{A}}{V}_{\text{A}}^{1/\text{c}}}{T_{\text{D}}{V}_{\text{D}}^{1/\text{c}}}=\frac{T_{\text{B}}{V}_{\text{B}}^{1/\text{c}}}{T_{\text{C}}{V}_{\text{C}}^{1/\text{c}}}$                                                                                              (2)

Since, $V_{\text{C}}=V_{\text{B}}$ and $V_{\text{D}}=V_{\text{A}}$.

Therefore,

    $\frac{T_{\text{A}}}{T_{\text{D}}}=\frac{T_{\text{B}}}{T_{\text{C}}}$

  $T_{\text{D}}=\frac{T_{\text{C}}{T}_{\text{A}}}{T_{\text{B}}}$                                                                                                      (3)

Hence, the efficiency becomes,

    $\begin{array}{c}\eta =1-\frac{\left(\frac{T_{\text{C}}{T}_{\text{A}}}{T_{\text{B}}}-T_{\text{A}}\right)}{\left(T_{\text{C}}-T_{\text{B}}\right)}\\ =1-\frac{T_{\text{A}}}{T_{\text{B}}}\end{array}$

From equation (2),

    $\frac{T_{\text{A}}}{T_{\text{B}}}={\left(\frac{V_{\text{B}}}{V_{\text{A}}}\right)}^{1/\text{c}}$

Therefore, the efficiency is,

    $\eta =1-{\left(\frac{V_{\text{B}}}{V_{\text{A}}}\right)}^{1/\text{c}}$

Hence, the expression for efficiency can be written as $1-{\left(\frac{V_{\text{B}}}{V_{\text{A}}}\right)}^{1/\text{c}}$.

(d)

Interpretation Introduction

Interpretation: The expression for the change in entropy for each step in terms of $C_{\text{V,m}}$ and temperatures has to be calculated.

Concept introduction: The gas power cycles that are used in the spark-ignition internal combustion engines is known as Otto cycle.  It consists of two reversible adiabatic processes and two isochoric processes.

Answer to Problem 3B.10P

The entropy for each process is, $\Delta S_{\text{I}}=0,\Delta S_{\text{III}}=0$, $\Delta S_{\text{II}}=C_{\text{V,m}}\ln \frac{T_{\text{C}}}{T_{\text{B}}}$ and $\Delta S_{\text{IV}}=C_{\text{V,m}}\ln \frac{T_{\text{A}}}{T_{\text{D}}}$.

Explanation of Solution

It is given that,

    $C_{\text{V,m}}=\frac{5}{2}R$

The entropy for process (I) and (III) is zero.  The entropy change for the process (II) and (IV) can be calculated as,

    $\begin{array}{c}\Delta S_{\text{II}}={\displaystyle \underset{T_{\text{B}}}{\overset{T_{\text{C}}}{\int }}{C}_{\text{V,m}}\frac{dT}{T}}\\ =C_{\text{V,m}}\ln \frac{T_{\text{C}}}{T_{\text{B}}}\end{array}$

Similarly, the entropy change for process (IV) is,

    $\begin{array}{c}\Delta S_{\text{IV}}={\displaystyle \underset{T_{\text{D}}}{\overset{T_{\text{A}}}{\int }}{C}_{\text{V,m}}\frac{dT}{T}}\\ =C_{\text{V,m}}\ln \frac{T_{\text{A}}}{T_{\text{D}}}\end{array}$

Hence, the entropy for each process is, $\Delta S_{\text{I}}=0,\Delta S_{\text{III}}=0$, $\Delta S_{\text{II}}=C_{\text{V,m}}\ln \frac{T_{\text{C}}}{T_{\text{B}}}$ and $\Delta S_{\text{IV}}=C_{\text{V,m}}\ln \frac{T_{\text{A}}}{T_{\text{D}}}$.

(e)

Interpretation Introduction

Interpretation: The entropy changes and efficiency has to be calculated from the given data.

Concept introduction: The gas power cycles that are used in the spark-ignition internal combustion engines is known as Otto cycle.  It consists of two reversible adiabatic processes and two isochoric processes.

Answer to Problem 3B.10P

The efficiency of the cycle is $\underset{\_}{0.60}$.  The entropy of the each process is $\Delta S_{\text{I}}=0,\Delta S_{\text{III}}=0$, $\Delta S_{\text{II}}=\underset{\_}{33.4J{K}^{-1}}$ and $\Delta S_{\text{IV}}=\underset{\_}{-33.4J{K}^{-1}}$.

Explanation of Solution

It is given that,

• • $V_{\text{A}}=4{\text{dm}}^3$.
• • $P_{\text{A}}=1\text{atm}$.
• • $T_{\text{A}}=300\text{K}$.
• • $V_{\text{A}}=10V_{\text{B}}$.
• • $C_{\text{V,m}}=\frac{5}{2}R$.
• • $\frac{P_{\text{C}}}{P_{\text{B}}}=5$.

The formula for efficiency is $1-{\left(\frac{V_{\text{B}}}{V_{\text{A}}}\right)}^{1/\text{c}}$.  Substitute the values of $V_{\text{A}}$ in the formula of efficiency to calculate its value.

    $\eta =1-{\left(\frac{V_{\text{B}}}{V_{\text{A}}}\right)}^{1/\text{c}}$

  $\eta =1-{\left(\frac{V_{\text{B}}}{10V_{\text{B}}}\right)}^{1/\text{c}}$                                                                                            (4)

The value of $c$ is given by the expression,

    $c=\frac{C_{\text{V,m}}}{R}$

Calculate the value $c$ by substituting the values of $C_{\text{V,m}}$ in the above equation.

    $\begin{array}{c}c=\frac{\frac{5}{2}R}{R}\\ =\frac{5}{2}\end{array}$

Substitute the value of $c$ to calculate efficiency.

    $\begin{array}{c}\eta =1-{\left(\frac{V_{\text{B}}}{10V_{\text{B}}}\right)}^{2/5}\\ =\underset{\_}{0.60}\end{array}$

Hence, the efficiency of the cycle is $\underset{\_}{0.60}$.

The entropy of the process (II) is calculated by,

  $\Delta S_{\text{II}}=C_{\text{V,m}}\ln \frac{T_{\text{C}}}{T_{\text{B}}}$                                                                                            (5)

At constant volume,

    $\frac{T_{\text{C}}}{T_{\text{B}}}=\frac{P_{\text{C}}}{P_{\text{B}}}=5$

Substitute the value of $C_{\text{V,m}}$ and $\frac{T_{\text{C}}}{T_{\text{B}}}$ in equation (5) to calculate the value of entropy.

    $\begin{array}{c}\Delta S_{\text{II}}=\frac{5}{2}\times 8.314\times \ln 5\\ =\underset{\_}{33.4J{K}^{-1}}\end{array}$

  $\Delta S_{\text{surr}}=\Delta S_{\text{II}}+\Delta S_{\text{IV}}$

Where,

• • $\Delta S_{\text{surr}}$ is the entropy of the surroundings.

The entropy, $\Delta S_{\text{surr}}$ is zero.  Hence, the entropy of $\Delta S_{\text{IV}}$ is calculated as,

    $\begin{array}{c}0=33.4\text{J}{\text{K}}^{-1}+\Delta S_{\text{IV}}\\ \Delta S_{\text{IV}}=\underset{\_}{-33.4J{K}^{-1}}\end{array}$

Hence, the entropy of the each process is $\Delta S_{\text{I}}=0,\Delta S_{\text{III}}=0$, $\Delta S_{\text{II}}=\underset{\_}{33.4J{K}^{-1}}$ and $\Delta S_{\text{IV}}=\underset{\_}{-33.4J{K}^{-1}}$.