Chapter 22, Problem 22.76AP

A 1.00-mol sample of a monatomic ideal gas is taken through the cycle shown in Figure P22.76. At point A, the pressure, volume, and temperature are P_i, V_i, and T_i, respectively. In terms of R and T_i, find (a) the total energy entering the system by heat per cycle, (b) the total energy leaving the system by heat per cycle, and (c) the efficiency of an engine operating in this cycle. (d) Explain how the efficiency compares with that of an engine operating in a Carnot cycle between the same temperature extremes.

To determine

The total energy entering the system by heat per cycle.

Answer to Problem 22.76AP

The total energy entering the system by heat per cycle is $10.5P_{\text{i}}{V}_{\text{i}}$.

Explanation of Solution

For work done from point $\text{A}$ to point $\text{B}$:

The work done is,

$w_1=0$

The ideal gas equation for temperature at point $\text{A}$ is,

$\begin{array}{c}{P}_{\text{A}}{V}_{\text{A}}=nR{T}_{\text{A}}\\ T_{\text{A}}=\frac{P_{\text{A}}{V}_{\text{A}}}{nR}\end{array}$ . (1)

Here,

$P_{\text{A}}$ is the pressure at point $\text{A}$.

$V_{\text{A}}$ is the volume at point $\text{A}$.

$T_{\text{A}}$ is the temperature at point $\text{A}$.

$n$ is the number of mole of gas.

$R$ is the ideal gas constant.

Consider the point $\text{A}$ is the initial point and the number of mole is 1.

Substitute $P_{\text{i}}$ for $P_{\text{A}}$, $V_{\text{i}}$ for $V_{\text{A}}$, $T_{\text{i}}$ for $T_{\text{A}}$ and $1$ for $n$ in equation (1).

$T_{\text{i}}=\frac{P_{\text{i}}{V}_{\text{i}}}{R}$

The ideal gas equation for temperature at point $\text{B}$ is,

$T_{\text{B}}=\frac{P_{\text{B}}{V}_{\text{B}}}{nR}$

Here,

$P_{\text{B}}$ is the pressure at point $\text{B}$.

$V_{\text{B}}$ is the volume at point $\text{B}$.

$T_{\text{B}}$ is the temperature at point $\text{B}$.

Substitute $3P_{\text{i}}$ for $P_{\text{B}}$, $V_{\text{i}}$ for $V_{\text{B}}$ and $1$ for $n$ in above equation.

$\begin{array}{c}{T}_{\text{B}}=\frac{3P_{\text{i}}{V}_{\text{i}}}{R}\\ =3T_{\text{i}}\end{array}$

The heat equation between point $\text{A}$ to point $\text{B}$ is,

$Q_1=n{C}_{\text{v}}\left(T_{\text{B}}-T_{\text{A}}\right)$

Here,

$C_{\text{v}}$ is the specific heat at constant volume.

Substitute $3T_{\text{i}}$ for $T_{\text{B}}$, $T_{\text{i}}$ for $T_{\text{A}}$, and $\frac{3R}{2}$ for $C_{\text{v}}$ and $1$ for $n$ in above equation.

$\begin{array}{c}{Q}_1=1\times \frac{3R}{2}\times \left(3T_{\text{i}}-T_{\text{i}}\right)\\ =3R{T}_{\text{i}}\\ =3R\left(\frac{P_{\text{i}}{V}_{\text{i}}}{R}\right)\\ =3P_{\text{i}}{V}_{\text{i}}\end{array}$

For work done from point $\text{B}$ to point $\text{C}$:

The work done is,

$w_2=P_{\text{B}}\left(V_{\text{B}}-V_{\text{C}}\right)$

Here,

$V_{\text{C}}$ is the volume at point $\text{C}$.

Substitute $3P_{\text{i}}$ for $P_{\text{B}}$, $V_{\text{i}}$ for $V_{\text{B}}$ and $2V_{\text{i}}$ for $V_{\text{C}}$ in above equation.

$\begin{array}{c}{w}_2=3P_{\text{i}}\left(V_{\text{i}}-2V_{\text{i}}\right)\\ =-3P_{\text{i}}{V}_{\text{i}}\end{array}$

The formula for the internal energy is,

$\Delta E_{\mathrm{int}}=n{C}_{\text{v}}\left(T_C-T_{\text{B}}\right)$ (2)

Here,

$T_{\text{C}}$ is the temperature at point $\text{C}$.

The ideal gas equation for temperature at point $\text{C}$ is,

$T_{\text{C}}=\frac{P_{\text{C}}{V}_{\text{C}}}{nR}$

Here,

$P_{\text{C}}$ is the pressure at point $\text{C}$.

$V_{\text{C}}$ is the volume at point $\text{C}$.

Substitute $3P_{\text{i}}$ for $P_{\text{C}}$, $2V_{\text{i}}$ for $V_{\text{C}}$ and $1$ for $n$ in above equation.

$\begin{array}{c}{T}_{\text{C}}=\frac{3P_{\text{i}}\times 2V_{\text{i}}}{R}\\ =6\times \frac{P_{\text{i}}{V}_{\text{i}}}{R}\\ =6T_{\text{i}}\end{array}$

Substitute $3T_{\text{i}}$ for $T_{\text{B}}$, $6T_{\text{i}}$ for $T_{\text{C}}$, and $\frac{3R}{2}$ for $C_{\text{v}}$ and $1$ for $n$ in equation (2).

$\begin{array}{c}\Delta E_{\mathrm{int}}=\frac{3R}{2}\left(6T_{\text{i}}-3T_{\text{i}}\right)\\ =4.5R{T}_{\text{i}}\end{array}$

Substitute $\frac{P_{\text{i}}{V}_{\text{i}}}{R}$ for $T_{\text{i}}$ in above equation.

$\begin{array}{c}\Delta E_{\mathrm{int}}=4.5R\times \frac{P_{\text{i}}{V}_{\text{i}}}{R}\\ =4.5P_{\text{i}}{V}_{\text{i}}\end{array}$

The energy transferred is,

$Q_2=\Delta E_{\mathrm{int}}-w_2$

Substitute $4.5P_{\text{i}}{V}_{\text{i}}$ for $\Delta E_{\mathrm{int}}$ and $-3P_{\text{i}}{V}_{\text{i}}$ for $w_2$ in above equation.

$\begin{array}{c}{Q}_2=4.5P_{\text{i}}{V}_{\text{i}}-\left(-3P_{\text{i}}{V}_{\text{i}}\right)\\ =7.5P_{\text{i}}{V}_{\text{i}}\end{array}$

For work done from point $\text{C}$ to point $\text{D}$:

The work done is,

$w_3=0$

The ideal gas equation for temperature at point $\text{D}$ is,

$T_{\text{D}}=\frac{P_{\text{D}}{V}_{\text{D}}}{nR}$

Here,

$P_{\text{D}}$ is the pressure at point $\text{D}$.

$V_{\text{D}}$ is the volume at point $\text{D}$.

Substitute $P_{\text{i}}$ for $P_{\text{D}}$, $2V_{\text{i}}$ for $V_{\text{D}}$ and $1$ for $n$ in above equation.

$\begin{array}{c}{T}_{\text{D}}=\frac{P_{\text{i}}\times 2V_{\text{i}}}{R}\\ =\frac{2P_{\text{i}}{V}_{\text{i}}}{R}\\ =2T_{\text{i}}\end{array}$

The energy transferred is,

$Q_3=n{C}_{\text{v}}\left(T_{\text{D}}-T_{\text{C}}\right)$

Substitute $2T_{\text{i}}$ for $T_{\text{D}}$, $6T_{\text{i}}$ for $T_{\text{C}}$, and $\frac{3R}{2}$ for $C_{\text{v}}$ and $1$ for $n$ in above equation.

$\begin{array}{c}{Q}_3=1\times \frac{3R}{2}\times \left(2T_{\text{i}}-6T_{\text{i}}\right)\\ =-6R{T}_{\text{i}}\end{array}$

Substitute $\frac{P_{\text{i}}{V}_{\text{i}}}{R}$ for $T_{\text{i}}$ in above equation.

$\begin{array}{c}{Q}_3=-6R\times \frac{P_{\text{i}}{V}_{\text{i}}}{R}\\ =-6P_{\text{i}}{V}_{\text{i}}\end{array}$

For work done from point $\text{D}$ to point $\text{A}$:

The work done is,

$w_4=P_{\text{D}}\left(V_{\text{D}}-V_{\text{A}}\right)$

Substitute $P_{\text{i}}$ for $P_{\text{D}}$, $2V_{\text{i}}$ for $V_{\text{D}}$ and $V_{\text{i}}$ for $V_{\text{A}}$ in above equation.

$\begin{array}{c}{w}_4=P_{\text{i}}\left(2V_{\text{i}}-V_{\text{i}}\right)\\ =P_{\text{i}}{V}_{\text{i}}\end{array}$

The energy transferred as heat is,

$Q_4=n{C}_{\text{P}}\left(T_{\text{A}}-T_{\text{D}}\right)$

Here,

$C_{\text{P}}$ is the specific heat at constant pressure.

Substitute $T_{\text{i}}$ for $T_{\text{A}}$, $2T_{\text{i}}$ for $T_{\text{D}}$, and $\frac{5R}{2}$ for $C_{\text{p}}$ and $1$ for $n$ in above equation.

$\begin{array}{c}{Q}_4=1\times \frac{5R}{2}\times \left(1T_{\text{i}}-2T_{\text{i}}\right)\\ =-2.5R{T}_{\text{i}}\end{array}$

Substitute $\frac{P_{\text{i}}{V}_{\text{i}}}{R}$ for $T_{\text{i}}$ in above equation.

$\begin{array}{c}{Q}_4=-2.5R\times \frac{P_{\text{i}}{V}_{\text{i}}}{R}\\ =-2.5P_{\text{i}}{V}_{\text{i}}\end{array}$

The net energy entering the system is,

$Q_{\text{net}}=Q_1+Q_2$

Substitute $3P_{\text{i}}{V}_{\text{i}}$ for $Q_1$ and $7.5P_{\text{i}}{V}_{\text{i}}$ for $Q_2$ in above equation.

$\begin{array}{c}{Q}_{\text{net}}=3P_{\text{i}}{V}_{\text{i}}+7.5P_{\text{i}}{V}_{\text{i}}\\ =10.5P_{\text{i}}{V}_{\text{i}}\end{array}$

Conclusion:

Therefore, the total energy entering the system by heat per cycle is $10.5P_{\text{i}}{V}_{\text{i}}$.

To determine

The total energy leaving the system by heat per cycle.

Answer to Problem 22.76AP

The total energy leaving the system by heat per cycle is $8.5P_{\text{i}}{V}_{\text{i}}$.

Explanation of Solution

The total energy leaving the system is,

$Q_{\text{leave}}=\left|Q_3\right|+\left|Q_4\right|$

Substitute $-6P_{\text{i}}{V}_{\text{i}}$ for $Q_3$ and $-2.5P_{\text{i}}{V}_{\text{i}}$ for $Q_4$ in above equation.

$\begin{array}{c}{Q}_{\text{leave}}=\left|-6P_{\text{i}}{V}_{\text{i}}\right|+\left|-2.5P_{\text{i}}{V}_{\text{i}}\right|\\ =8.5P_{\text{i}}{V}_{\text{i}}\end{array}$

Conclusion:

Therefore, the total energy leaving the system by heat per cycle is $8.5P_{\text{i}}{V}_{\text{i}}$.

To determine

The efficiency of an engine operating in this cycle.

Answer to Problem 22.76AP

The efficiency of an engine operating in this cycle is $0.19$.

Explanation of Solution

The formula of efficiency of the engine is,

$\begin{array}{c}e=\frac{Q_{\text{net}}-Q_{\text{leave}}}{Q_{\text{net}}}\\ =\frac{10.5P_{\text{i}}{V}_{\text{i}}-8.5P_{\text{i}}{V}_{\text{i}}}{10.5P_{\text{i}}{V}_{\text{i}}}\\ =0.19\end{array}$

Conclusion:

Therefore, the efficiency of an engine operating in this cycle is $0.19$.

To determine

The efficiency comparison with that of an engine operating in a Carnot cycle between the same temperature extremes.

Answer to Problem 22.76AP

The efficiency of an engine is $0.833$.

Explanation of Solution

The maximum temperature in cycle is,

$T_{\text{C}}=6T_{\text{i}}$

The minimum temperature in cycle is,

$T_{\text{A}}=T_{\text{i}}$

The formula of the Carnot efficiency is,

$e_{\text{C}}=1-\frac{T_{\text{A}}}{T_{\text{C}}}$

Substitute $T_{\text{i}}$ for $T_{\text{A}}$ and $6T_{\text{i}}$ for $T_C$ in above equation.

$\begin{array}{c}{e}_{\text{C}}=1-\frac{T_{\text{i}}}{6T_{\text{i}}}\\ =\frac{5}{6}\\ =0.833\end{array}$

Conclusion:

Therefore, the efficiency of an engine is $0.833$.