Chapter 3, Problem 3A.3P

(a)

Interpretation Introduction

Interpretation: The volume of $0.1\text{mol}$ of the perfect gas which has initial volume of $1.00{\text{dm}}^{\text{3}}$ which doubles over the course of the first isothermal stage where temperature of hot source is $373\text{K}$ and that of the cold sink is $273\text{K}$ has to be calculated after Stage 1 and Stage 2.

Concept introduction: The Carnot cycle is an ideal process which includes a hot source $\left(T_{\text{h}}\right)$ and a cold sink $\left(T_{\text{c}}\right)$ and consists of the following four processes: a reversible adiabatic expansion, a reversible isothermal expansion, a reversible isothermal compression and a reversible adiabatic compression.

Answer to Problem 3A.3P

The volume of $0.1\text{mol}$ of perfect gas after Stage 1 is $\underset{\_}{2.00d{m}^3}$ and the volume after Stage 2 is $\underset{\_}{3.16d{m}^3}$.

Explanation of Solution

The pressure-volume diagram for the carnot cycle is represented as,

Where

• Stage $1$ represents isothermal expansion.
• Stage $2$ represents adiabatic expansion.
• Stage $3$ represents isothermal compression.
• Stage $4$ represents adiabatic compression.

Stage 1 represents reversible isothermal expansion from $p_{\text{A}}{V}_{\text{A}}$ to $p_{\text{B}}{V}_{\text{B}}$ at temperature $T_{\text{h}}$ (temperature of the hot source).

The initial volume of the perfect gas $\left(V_{\text{A}}\right)$ is $1.00{\text{dm}}^{\text{3}}$.

The volume of the perfect gas doubles over the course of the first isothermal stage

Hence, the volume of the perfect gas after Stage 1 $\left(V_{\text{B}}\right)$ is $\underset{\_}{2.00d{m}^3}$.

Stage $2$ represents adiabatic expansion from $p_{\text{B}}{V}_{\text{B}}$ to $p_{\text{C}}{V}_{\text{C}}$ and the temperature changes from $T_{\text{h}}$ (temperature of the hot source) to $T_{\text{c}}$ (temperature of the cold sink).

According to Poisson’s equation for adiabatic process, the relation between temperature and volume is given as,

    $T{V}^{\text{γ}-1}=\text{K}$                                                                                                      (1)

Where,

• $\text{K}$ is a constant.

For reversible adiabatic stages,

    $V{T}^{\frac{3}{2}}=\text{K}$                                                                                                        (2)

Multiplying the powers of the equation (2) by $\frac{2}{3}$.

    $V^{\frac{2}{3}}T=\text{K}$

Equate the equation (1) and (2).

    $\begin{array}{c}\text{γ}-1=\frac{2}{3}\\ \text{γ}\text{=}\frac{\text{5}}{3}\end{array}$

Substitute $T_{\text{h}}$, $T_{\text{c}}$, $V_{\text{B}}$, and $V_{\text{C}}$ for Stage 2 in equation (1).

    $T_{\text{h}}{V}_{\text{B}}^{\text{γ-1}}=T_{\text{c}}{V}_{\text{C}}^{\text{γ-1}}$

Or,

  $\frac{T_{\text{h}}}{T_{\text{c}}}={\left(\frac{V_{\text{C}}}{V_{\text{B}}}\right)}^{\text{γ-1}}$                                                                                                  (3)

Where

• $T_{\text{h}}$ is the temperature of the hot source.
• $T_{\text{c}}$ is the temperature of the cold sink.
• $V_{\text{B}}$ is the volume of the perfect gas after Stage 1 of isothermal expansion.
• $V_{\text{C}}$ is the volume of the perfect gas after Stage 2 of adiabatic expansion.

The temperature of the hot source $\left(T_{\text{h}}\right)$ is $373\text{K}$.

The temperature of the cold sink $\left(T_{\text{c}}\right)$ is $273\text{K}$.

The volume after Stage 1 $\left(V_{\text{B}}\right)$ is $2.00{\text{dm}}^{\text{3}}$.

Substitute the value of $T_{\text{h}}$, $T_{\text{c}}$, $V_{\text{B}}$, $V_{\text{C}}$ in equation (3).

    $\frac{\text{373}\text{K}}{\text{273}\text{K}}={\left(\frac{V_{\text{C}}}{2{\text{dm}}^{\text{3}}}\right)}^{\frac{5}{3}-\text{1}}$

Hence,

    $V_{\text{C}}=\underset{\_}{3.16d{m}^3}$

Thus the volume after Stage 2 is $\underset{\_}{3.16d{m}^3}$.

(b)

Interpretation Introduction

Interpretation: The volume of $0.1\text{mol}$ of the perfect gas which has initial volume of $1.00{\text{dm}}^{\text{3}}$ which doubles over the course of the first isothermal stage where temperature of hot source is $373\text{K}$ and that of the cold sink is $273\text{K}$ has to be calculated after Stage 3 considering reversible adiabatic expansion from starting point.

Concept introduction: The Carnot cycle is an ideal process which includes a hot source $\left(T_{\text{h}}\right)$ and a cold sink $\left(T_{\text{c}}\right)$ and consists of the following four processes.

1.         i.            A reversible adiabatic expansion.
2.       ii.            A reversible isothermal expansion.
3.     iii.            A reversible isothermal compression.
4.     iv.            A reversible adiabatic compression.

Answer to Problem 3A.3P

The volume of $0.1\text{mol}$ of the perfect gas after Stage 3 is $\underset{\_}{1.58d{m}^3}$.

Explanation of Solution

The pressure-volume diagram for the carnot cycle is represented as,

Where

• Stage $1$ represents isothermal expansion.
• Stage $2$ represents adiabatic expansion.
• Stage $3$ represents isothermal compression.
• Stage $4$ represents adiabatic compression.

Stage 2 (process BC) is reversible adiabatic expansion.

According to poisson’s equation for adiabatic process the relation between temperature and volume is given as,

  $\frac{T_{\text{h}}}{T_{\text{c}}}={\left(\frac{V_{\text{C}}}{V_{\text{B}}}\right)}^{\text{γ-1}}$                                                                                                  (3)

Stage 4 (DA) is reversible adiabatic compression.

According to poisson’s equation for adiabatic process the relation between temperature and volume is given as,

    $\frac{T_{\text{h}}}{T_{\text{c}}}={\left(\frac{V_{\text{D}}}{V_{\text{A}}}\right)}^{\text{γ-1}}$                                                                                                  (4)

Equate equation (3) and (4).

    ${\left(\frac{V_{\text{C}}}{V_{\text{B}}}\right)}^{\text{γ-1}}={\left(\frac{V_{\text{D}}}{V_{\text{A}}}\right)}^{\text{γ-1}}$

Or,

  $\left(\frac{V_{\text{C}}}{V_{\text{B}}}\right)=\left(\frac{V_{\text{D}}}{V_{\text{A}}}\right)$                                                                                                 (5)

Where

• $T_{\text{h}}$ is the temperature of the hot source.
• $T_{\text{c}}$ is the temperature of the cold sink.
• $V_{\text{B}}$ is the volume of the perfect gas after Stage 1 of isothermal expansion.
• $V_{\text{C}}$ is the volume of the perfect gas after Stage 2 of adiabatic expansion.
• $V_{\text{D}}$ is the volume of the perfect gas after Stage 3 of isothermal compression.
• $V_{\text{A}}$ is the initial volume of the perfect gas.

Substitute the value of $V_{\text{B}}$, $V_{\text{C}}$, $V_{\text{D}}$ and $V_{\text{A}}$ in the equation (5).

  $\begin{array}{c}\left(\frac{\text{3}\text{.16}\text{dm}{}^{\text{3}}}{2.00{\text{dm}}^{\text{3}}}\right)=\left(\frac{V_{\text{D}}}{1.00{\text{dm}}^{\text{3}}}\right)\\ V_{\text{D}}=\underset{\_}{1.58d{m}^3}\end{array}$

Thus the volume after Stage 3 is $\underset{\_}{1.58d{m}^3}$.

(c)

Interpretation Introduction

Interpretation: The heat transferred to or from $0.1\text{mol}$ of the perfect gas which has initial volume of $1.00{\text{dm}}^{\text{3}}$ which doubles over the course of the first isothermal stage where temperature of hot source is $373\text{K}$ and that of the cold sink is $273\text{K}$ has to be calculated for each stage.

Concept introduction: The Carnot cycle is an ideal process which includes a hot source $\left(T_{\text{h}}\right)$ and a cold sink $\left(T_{\text{c}}\right)$ and consists of the following four processes.

1.         i.            A reversible adiabatic expansion.
2.       ii.            A reversible isothermal expansion.
3.     iii.            A reversible isothermal compression.
4.     iv.            A reversible adiabatic compression.

Answer to Problem 3A.3P

The heat absorbed by the perfect gas over isothermal expansion is $\underset{\_}{214.99J}$, over the adiabatic expansion is $\underset{\_}{0}J$, over the isothermal compression is $\underset{\_}{-157.353J}$ and over adiabatic compression is $\underset{\_}{0J}$.

Explanation of Solution

The pressure-volume diagram for the carnot cycle is represented as,

Where

• Stage $1$ represents isothermal expansion.
• Stage $2$ represents adiabatic expansion.
• Stage $3$ represents isothermal compression.
• Stage $4$ represents adiabatic compression.

Stage 1 represents isothermal expansion from $p_{\text{A}}{V}_{\text{A}}$ to $p_{\text{B}}{V}_{\text{B}}$ at temperature $T_{\text{h}}$.

According to the first law of thermodynamics,

    $\Delta U=q+w$

For a reversible isothermal process, the temperature remains constant.

Hence, the value of $\Delta U=0$.  Thus,

  $q=-w$                                                                                                           (6)

The expression for the work done in the process AB (reversible isothermal expansion) is given as,

    $w=-2.303nR{T}_{\text{h}}\log \frac{V_{\text{B}}}{V_{\text{A}}}$                                                                                (7)

Where

• $T_{\text{h}}$ is the temperature of the hot source.
• $V_{\text{B}}$ is the volume of the perfect gas after Stage 1 of isothermal expansion.
• $V_{\text{A}}$ is the initial volume of the perfect gas.
• $n$ is the number of moles of the perfect gas.
• $R$ is the gas constant $\left(\text{8}\text{.314}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)$.

The number of moles of the perfect gas is $\text{0}\text{.1}\text{mol}$

The temperature of the hot source $\left(T_{\text{h}}\right)$ is $373\text{K}$.

The volume after Stage 1 $\left(V_{\text{B}}\right)$ is $2.00{\text{dm}}^{\text{3}}$.

The initial volume $\left(V_{\text{A}}\right)$ is $1.00{\text{dm}}^{\text{3}}$

Substitute the value of $n$, $R$, $V_{\text{A}}$, $V_{\text{B}}$ and $T_{\text{h}}$ in equation (7).

    $\begin{array}{c}{w}_1=-\text{2}\text{.303×0}\text{.10}\text{mol×8}\text{.314}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{×373}\text{K×log}\frac{\text{2}{\text{dm}}^{\text{3}}}{\text{1}{\text{dm}}^{\text{3}}}\\ \text{=}-\text{214}\text{.99}\text{J}\end{array}$

Substitute the value of $w_1$ in equation (6).

  $\begin{array}{c}{q}_1=-w_1\\ =-\left(-\text{214}\text{.99}\text{J}\right)\\ \underset{\_}{=214.99J}\end{array}$

Stage $2$ represents adiabatic expansion from $p_{\text{B}}{V}_{\text{B}}$ to $p_{\text{C}}{V}_{\text{C}}$ and the temperature changes from $T_{\text{h}}$ to $T_{\text{c}}$.

For adiabatic expansion process in the above diagram, $q_2$ is $\underset{\_}{0J}$

Stage 3 represents isothermal compression from $p_{\text{C}}{V}_{\text{C}}$ to $p_{\text{D}}{V}_{\text{D}}$ at temperature $T_{\text{c}}$.

For a reversible isothermal process the temperature remains constant.

Hence the value of $\Delta U=0$.  Thus,

  $q=-w$                                                                                                           (6)

The expression for the work done in the process CD (reversible isothermal compression) is given as,

    $w=-2.303nR{T}_{\text{c}}\log \frac{V_{\text{D}}}{V_{\text{C}}}$                                                                                 (8)

Where

• $T_{\text{c}}$ is the temperature of the hot source.
• $V_{\text{c}}$ is the volume of the perfect gas after Stage 2 of adiabatic expansion.
• $V_{\text{D}}$ is the volume of the perfect gas after Stage 3 of isothermal compression.
• $n$ is the number of moles of the perfect gas,
• $R$ is the gas constant $\left(\text{8}\text{.314}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)$.

The number of moles of the perfect gas is $\text{0}\text{.1}\text{mol}$

The temperature of the hot source $\left(T_{\text{c}}\right)$ is $273\text{K}$.

The volume after Stage 2 $\left(V_{\text{C}}\right)$ is $3.16{\text{dm}}^{\text{3}}$.

The volume after Stage 3 $\left(V_{\text{D}}\right)$ is $1.58{\text{dm}}^{\text{3}}$

Substitute the value of $n$, $R$, $V_{\text{C}}$, $V_{\text{D}}$ and $T_{\text{h}}$ in equation (8).

    $\begin{array}{c}{w}_3=-\text{2}\text{.303×0}\text{.10}\text{mol×8}\text{.314}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{×273}\text{K×log}\frac{1.58{\text{dm}}^{\text{3}}}{3.16{\text{dm}}^{\text{3}}}\\ \text{=157}\text{.353}\text{J}\end{array}$

Substitute the value of $w_1$ in equation (6).

  $\begin{array}{c}{q}_3=-w_1\\ =-\left(157.353\text{J}\right)\\ =\underset{\_}{-157.353}\text{J}\end{array}$

Path $4$ represents adiabatic expansion from $p_{\text{D}}{V}_{\text{D}}$ to $p_{\text{A}}{V}_{\text{A}}$ and the temperature changes from $T_{\text{c}}$ to $T_{\text{h}}$.

For adiabatic compression process in the above diagram, $q_4$ is $0J$.

The total heat absorbed or given out is the sum of the heat absorbed or given out in individual paths.

  $q=q_1+q_2+q_3+q_4$

    $\begin{array}{c}q=214.99\text{J}+0+\left(-157.353\text{J}\right)+0\\ =\underset{\_}{57.637J}\end{array}$

The heat absorbed by the system is $\underset{\_}{57.637J}$.

(d)

Interpretation Introduction

Interpretation: The statement: the work done is equal to the difference between the heat extracted from the hot source and deposited at the cold source has to be explained.

Concept introduction: The Carnot cycle is an ideal process which includes a hot source $\left(T_{\text{h}}\right)$ and a cold sink $\left(T_{\text{c}}\right)$ and consists of the following four processes.

1.                     i.            A reversible adiabatic expansion.
2.                   ii.            A reversible isothermal expansion.
3.                 iii.            A reversible isothermal compression.
4.                 iv.            A reversible adiabatic compression.

Answer to Problem 3A.3P

The statement: the work done is equal to the difference between the heat extracted from the hot source and deposited at the cold source has to be explained.

Explanation of Solution

The Carnot cycle is represented by the diagram below.

Where

• $Q_1$ is the heat absorbed from the hot source.
• $Q_2$ is the heat deposited at the cold sink.
• $w$ is the total work done.

After the substance returns to initial conditions, the net change in internal energy is zero.. Thus,

    $\Delta U=0$

According to first law of thermodynamics,

  $dQ=dU+dw$

Hence,

    $\begin{array}{c}dw=dQ\\ =Q_2-Q_1\end{array}$

Hence the work done is equal to the difference to the difference between the heat extracted from the hot source and deposited at the cold source.

(e)

Interpretation Introduction

Interpretation: The work done over the cycle and the efficiency of the cycle for $0.1\text{mol}$ of the perfect gas which has initial volume of $1.00{\text{dm}}^{\text{3}}$ which doubles over the course of the first isothermal stage where temperature of hot source is $373\text{K}$ and that of the cold sink is $273\text{K}$ has to e calculated.

Concept introduction: The Carnot cycle is an ideal process which includes a hot source $\left(T_{\text{h}}\right)$ and a cold sink $\left(T_{\text{c}}\right)$ and consists of the following four processes.

1.                   v.            A reversible adiabatic expansion.
2.                 vi.            A reversible isothermal expansion.
3.               vii.            A reversible isothermal compression.
4.             viii.            A reversible adiabatic compression.

Answer to Problem 3A.3P

The work done over the cycle is $\underset{\_}{-57.632J}$ and the efficiency is $\underset{\_}{0.26}$.

Explanation of Solution

The pressure-volume diagram for the carnot cycle is represented as,

Where

• Stage $1$ represents isothermal expansion.
• Stage $2$ represents adiabatic expansion.
• Stage $3$ represents isothermal compression.
• Stage $4$ represents adiabatic compression.

Stage 1 represents isothermal expansion from $p_{\text{A}}{V}_{\text{A}}$ to $p_{\text{B}}{V}_{\text{B}}$ at temperature $T_{\text{h}}$.

The expression for the work done in the process AB (reversible isothermal expansion) is given as,

    $w_1=-2.303nR{T}_{\text{h}}\log \frac{V_{\text{B}}}{V_{\text{A}}}$                                                                               (7)

Where

• $T_{\text{h}}$ is the temperature of the hot source.
• $V_{\text{B}}$ is the volume of the perfect gas after Stage 1 of isothermal expansion.
• $V_{\text{A}}$ is  the initial volume of the perfect gas.
• $n$ is the number of moles of the perfect gas,
• $R$ is the gas constant $\left(\text{8}\text{.314}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)$.

Stage $2$ represents adiabatic expansion from $p_{\text{B}}{V}_{\text{B}}$ to $p_{\text{C}}{V}_{\text{C}}$ and the temperature changes from $T_{\text{h}}$ to $T_{\text{c}}$.

According to the first law of thermodynamics,

    $\Delta U=q+w$

For adiabatic expansion process in the above diagram $q_2$ is 0.  Thus,

  $w=\Delta U$                                                                                                       (9)

The expression for $\Delta U$ is given as,

    $\begin{array}{c}\Delta U_2=n{C}_{\text{v}}dT\\ =n{C}_{\text{v}}(T_{\text{c}}-T_{\text{h}})\end{array}$

Substitute the value of $\Delta U$ from equation (3)

    $w_2=n{C}_{\text{v}}(T_{\text{c}}-T_{\text{h}})$

Stage 3 represents isothermal compression from $p_{\text{C}}{V}_{\text{C}}$ to $p_{\text{D}}{V}_{\text{D}}$ at temperature $T_{\text{c}}$.

The expression for the work done in the process CD (reversible isothermal compression) is given as,

    $w_3=-2.303nR{T}_{\text{c}}\log \frac{V_{\text{D}}}{V_{\text{C}}}$                                                                             (10)

Where

• $T_{\text{c}}$ is the temperature of the hot source.
• $V_{\text{c}}$ is the volume of the perfect gas after Stage 2 of adiabatic expansion.
• $V_{\text{D}}$ is the volume of the perfect gas after Stage 3 of isothermal compression.
• $n$ is the number of moles of the perfect gas,
• $R$ is the gas constant $\left(\text{8}\text{.314}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)$

Path $4$ represents adiabatic expansion from $p_{\text{D}}{V}_{\text{D}}$ to $p_{\text{A}}{V}_{\text{A}}$ and the temperature changes from $T_{\text{c}}$ to $T_{\text{h}}$.

According to the first law of thermodynamics,

    $\Delta U=q+w$

For adiabatic expansion process in the above diagram, $q_4$ is 0. Thus,

  $w=\Delta U$                                                                                                       (9)

The expression for $\Delta U$ is given as,

    $\begin{array}{c}\Delta U_4=n{C}_{\text{v}}dT\\ =n{C}_{\text{v}}(T_{\text{h}}-T_{\text{c}})\end{array}$

Substitute the value of $\Delta U$ from equation (3)

    $w_4=n{C}_{\text{v}}(T_{\text{h}}-T_{\text{c}})$

The total work done is the sum of the work done in individual paths.

  $w=w_1+w_2+w_3+w_4$

Hence,

  $w=-2.303nR{T}_{\text{h}}\log \frac{V_{\text{B}}}{V_{\text{A}}}+n{C}_{\text{v}}(T_{\text{c}}-T_{\text{h}})-2.303nR{T}_{\text{c}}\log \frac{V_{\text{D}}}{V_{\text{C}}}+n{C}_{\text{v}}(T_{\text{h}}-T_{\text{c}})$

Or,

  $w=-2.303nR{T}_{\text{h}}\log \frac{V_{\text{B}}}{V_{\text{A}}}-2.303nR{T}_{\text{c}}\log \frac{V_{\text{D}}}{V_{\text{C}}}$                                               (11)

Where

• $T_{\text{c}}$ is the temperature of the hot source.
• $T_{\text{h}}$ is the temperature of the hot source.
• $n$ is the number of moles of the perfect gas,
• $R$ is the gas constant $\left(\text{8}\text{.314}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)$
• $V_{\text{c}}$ is the volume of the perfect gas after Stage 2 of adiabatic expansion.
• $V_{\text{D}}$ is the volume of the perfect gas after Stage 3 of isothermal compression.
• $V_{\text{B}}$ is the volume of the perfect gas after Stage 1 of isothermal expansion.
• $V_{\text{A}}$ is  the initial volume of the perfect gas.

The temperature of the hot source $\left(T_{\text{h}}\right)$ is $373\text{K}$.

The temperature of the cold sink $\left(T_{\text{c}}\right)$ is $273\text{K}$.

The volume after Stage 1 $\left(V_{\text{B}}\right)$ is $2.00{\text{dm}}^{\text{3}}$.

The initial volume $\left(V_{\text{A}}\right)$ is $1.00{\text{dm}}^{\text{3}}$.

The volume after Stage 2 $\left(V_{\text{C}}\right)$ is $3.16{\text{dm}}^{\text{3}}$.

The volume after Stage 3 $\left(V_{\text{D}}\right)$ is $1.58{\text{dm}}^{\text{3}}$.

The number of moles $\left(n\right)$ is $\text{0}\text{.10}\text{mol}$.

Substitute the values of $T_{\text{c}}$, $T_{\text{h}}$, $n$, $R$, $V_{\text{c}}$, $V_{\text{D}}$, $V_{\text{D}}$, $V_{\text{B}}$, $V_{\text{A}}$ In equation (11).

    $\begin{array}{c}w=\left(\begin{array}{c}-\text{2}\text{.303×0}\text{.10}\text{mol×8}\text{.314}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{×373}\text{Klog}\frac{\text{2}{\text{dm}}^{\text{3}}}{\text{1}{\text{dm}}^{\text{3}}}-\text{2}\text{.303×}\\ \text{0}\text{.10}\text{mol×8}\text{.314}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{×273}\text{Klog}\frac{\text{1}\text{.58}{\text{dm}}^{\text{3}}}{3.16{\text{dm}}^{\text{3}}}\end{array}\right)\\ =\underset{\_}{-57.632J}\end{array}$

The efficiency $\left(\eta \right)$ of a cycle is given by the relation,

  $\eta =\frac{-w}{q_h}$                                                                                                        (12)

Where

  $q_{\text{h}}=2.303nR{T}_{\text{h}}\log \frac{V_{\text{B}}}{V_{\text{A}}}$                                                                                    (7)

Substitute the values of $T_{\text{h}}$, $n$, $R$, $V_{\text{B}}$, $V_{\text{A}}$ In equation (7)

    $\begin{array}{c}{q}_{\text{h}}=\text{2}\text{.303×0}\text{.10}\text{mol×8}\text{.314}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{×}\text{373}\text{K}\times \text{log}\frac{\text{2}{\text{dm}}^{\text{3}}}{\text{1}{\text{dm}}^{\text{3}}}\\ =\text{214}\text{.992}\text{J}\end{array}$

Substitute the value of $q_{\text{h}}$ and $w$ in equation (12).

    $\begin{array}{c}\eta =\frac{\text{57}\text{.632}\text{J}}{\text{214}\text{.992}\text{J}}\\ =\underset{\_}{0.26}\end{array}$

Thus the work done over the cycle is $\underset{\_}{-57.632J}$ and the efficiency is $\underset{\_}{0.26}$.

(f)

Interpretation Introduction

Interpretation: The value of efficiency in sub-part (e) and the value of heat involved .have to be confirmed with the given equations.

Concept introduction: The Carnot cycle is an ideal process which includes a hot source $\left(T_{\text{h}}\right)$ and a cold sink $\left(T_{\text{c}}\right)$ and consists of the following four processes.

1.                     i.            A reversible adiabatic expansion.
2.                   ii.            A reversible isothermal expansion.
3.                 iii.            A reversible isothermal compression.
4.                 iv.            A reversible adiabatic compression.

Answer to Problem 3A.3P

The value of efficiency and heat involved in isothermal equation agrees with given equation.

Explanation of Solution

The equation for Carnot efficiency $\left(\eta \right)$ is given as,

    $\eta =1-\frac{T_{\text{c}}}{T_{\text{h}}}$                                                                                                     (13)

Where

• $T_{\text{c}}$ is the temperature of the hot source.
• $T_{\text{h}}$ is the temperature of the hot source.

The temperature of the hot source $\left(T_{\text{h}}\right)$ is $373\text{K}$.

The temperature of the cold sink $\left(T_{\text{c}}\right)$ is $273\text{K}$.

Substitute the values of $T_{\text{c}}$, $T_{\text{h}}$, in equation (1).

    $\begin{array}{c}\eta =1-\frac{273\text{K}}{373\text{K}}\\ =\underset{\_}{0.26}\end{array}$

The equation for Carnot efficiency $\left(\eta \right)$ is given as,

    $\eta =1-\frac{q_{\text{c}}}{q_{\text{h}}}$                                                                                                    (14)

Substitute the value of $q_{\text{c}}$ and $q_{\text{h}}$ in the above equation

    $\begin{array}{c}\eta =1-\frac{157.353\text{J}}{214.99\text{J}}\\ =\underset{\_}{0.26}\end{array}$

Thus the answer agrees with the equations given.