Chapter 22, Problem 32P

(a)

To determine

All unknown pressures, volumes, and temperatures in the table given in the question.

Answer to Problem 32P

The table completed with unknown pressures, volumes, and temperatures given beow.

state	P(kPa)	V (L)	T (K)
A	1400	10.0	720
B	875	16.0	720
C	445	24.0	549
D	712	15.0	549

Explanation of Solution

Consider the adiabatic process $D\to A$.

Write the adiabatic condition for the monoatomic ideal gas.

  $P_D{V}_D^{\gamma }=P_A{V}_A^{\gamma }$                                                                                                             (I)

Here, $P_A$ is the pressure at point $A$ in the Carnot cycle, $P_D$ is the pressure at point $D$ in the Carnot cycle, $V_A$  is the volume of the ideal gas at point $A$, $V_D$ is the volume of the ideal gas at point $D$ and $\gamma$ is the adiabatic constant.

Rearrange above equation to get $P_D$.

  $P_D=P_A{\left(\frac{V_A}{V_D}\right)}^{\gamma }$                                                                                                         (II)

Write ideal gas equation for ideal gas at point $A$ in the Cycle.

  $P_A{V}_A=nR{T}_A$

Here, $n$ is the number of moles of the gas and $R$ is the universal gas constant.

Rearrange above equation to get $P_A$.

  $P_A=\frac{nR{T}_A}{V_A}$                                                                                                              (III)

Write ideal gas equation for ideal gas at point $D$ in the Cycle.

  $P_D{V}_D=nR{T}_D$

Rearrange above equation to get $P_D$.

  $P_D=\frac{nR{T}_D}{V_D}$                                                                                                              (IV)

Use equation (III) in (IV) in equation (I) to get $T_D$.

  $\begin{array}{c}\left(\frac{nR{T}_D}{V_D}\right)V_D^{\gamma }=\left(\frac{nR{T}_A}{V_A}\right)V_A^{\gamma }\\ T_D=T_A{\left(\frac{V_A}{V_D}\right)}^{\gamma -1}\end{array}$                                                                                        (V)

Consider the isothermal process $C\to D$.

Write the condition for the isothermal process.

  $T_C=T_D$                                                                                                                   (VI)

Write the equation for an isothermal process.

  $PV=\text{constant}$

Here, $P$ is the pressure and $V$ is the volume.

Apply isothermal equation to process $C\to D$.

  $P_C{V}_C=P_D{V}_D$                                                                                                          (VII)

Rearrange above equation to get $P_C$.

  $P_C=P_D\left(\frac{V_D}{V_C}\right)$                                                                                                       (VIII)

Substitute $P_A{\left(\frac{V_A}{V_D}\right)}^{\gamma }$ for $P_D$ in above equation to get $P_C$.

  $\begin{array}{c}{P}_C=P_A{\left(\frac{V_A}{V_D}\right)}^{\gamma }\left(\frac{V_D}{V_C}\right)\\ =\frac{P_A{V}_A^{\gamma }}{V_C{V}_D^{\gamma -1}}\end{array}$                                                                                              (IX)

Consider the adiabatic process $B\to C$.

Write the adiabatic condition for the monoatomic ideal gas.

  $P_B{V}_B^{\gamma }=P_C{V}_C^{\gamma }$                                                                                                             (X)

Here, $P_B$ is the pressure at point $B$ in the Carnot cycle, $P_C$ is the pressure at point $C$ in the Carnot cycle, $V_B$ is the volume of the ideal gas at point $B$ and $V_C$ is the volume of the ideal gas at point $C$.

Apply isothermal equation to process $A\to B$.

  $P_A{V}_A=P_B{V}_B$                                                                                                           (XI)

Rearrange to get $P_B$.

  $P_B=P_A\left(\frac{V_A}{V_B}\right)$                                                                                                        (XII)

Use equation (XI) and (IX) in equation (X) to get $V_B$.

  $\begin{array}{c}{P}_A\left(\frac{V_A}{V_B}\right)V_B^{\gamma }=\left(\frac{P_A{V}_A^{\gamma }}{V_C{V}_D^{\gamma -1}}\right)V_C^{\gamma }\\ {\left(V_B\right)}^{{}^{\gamma -1}}={\left(\frac{V_A{V}_C}{V_D}\right)}^{{}^{\gamma -1}}\\ V_B=\frac{V_A{V}_C}{V_D}\end{array}$                                                                                  (XIII)

Conclusion:

Substitute $1400\text{kPa}$ for $P_A$, $10.0\text{L}$ for $V_A$ , $15.0\text{L}$ for $V_D$ and $\frac{5}{3}$ for $\gamma$ in equation (II) to get $P_D$.

  $\begin{array}{c}{P}_D=\left(1400\text{kPa}\right){\left(\frac{10.0\text{L}}{15.0\text{L}}\right)}^{\frac{5}{3}}\\ =712\text{kPa}\end{array}$

Substitute $720\text{K}$ for $T_A$, $10.0\text{L}$ for $V_A$ , $15.0\text{L}$ for $V_D$ and $\frac{5}{3}$ for $\gamma$ in equation (V) to get $T_D$.

  $\begin{array}{c}{T}_D=\left(720\text{K}\right){\left(\frac{10.0\text{L}}{15.0\text{L}}\right)}^{\frac{5}{3}-1}\\ =\left(720\text{K}\right){\left(\frac{10.0\text{L}}{15.0\text{L}}\right)}^{\frac{2}{3}}\\ =549\text{K}\end{array}$

Substitute $1400\text{kPa}$ for $P_A$, $10.0\text{L}$ for $V_A$ , $15.0\text{L}$ for $V_D$, $24.0\text{L}$ for $V_C$ and $\frac{5}{3}$ for $\gamma$ in equation (IX) to get $P_C$.

  $\begin{array}{c}{P}_C=\frac{\left(1400\text{kPa}\right){\left(10.0\text{L}\right)}^{\frac{5}{3}}}{\left(24.0\text{L}\right){\left(15.0\text{L}\right)}^{\frac{5}{3}-1}}\\ =445\text{kPa}\end{array}$

Substitute $10.0\text{L}$ for $V_A$, $24.0\text{L}$ for $V_C$ and $15.0\text{L}$ for $V_D$ in equation (XIII) to get $V_B$.

  $\begin{array}{c}{V}_B=\frac{\left(10.0\text{L}\right)\left(24.0\text{L}\right)}{\left(15.0\text{L}\right)}\\ =16.0\text{L}\end{array}$

Substitute $1400\text{kPa}$ for $P_A$, $10.0\text{L}$ for $V_A$ and $16.0\text{L}$ for $V_B$ in equation (XII) to get $P_B$.

  $\begin{array}{c}{P}_B=\left(1400\text{kPa}\right)\left(\frac{10.0\text{L}}{16.0\text{L}}\right)\\ =875\text{kPa}\end{array}$

The table completed with unknown pressures, volumes, and temperatures given below.

state	P(kPa)	V (L)	T (K)
A	1400	10.0	720
B	875	16.0	720
C	445	24.0	549
D	712	15.0	549

(b)

To determine

The energy added by heat, work done by the engine, and the change in internal energy for each of the steps $A\to B$, $B\to C$, $C\to D$ , and $D\to A$.

Answer to Problem 32P

The energy added by heat, work done by the engine, and the change in internal energy for each of the steps $A\to B$, $B\to C$, $C\to D$ , and $D\to A$ are given below.

Process	Q(kJ)	W(kJ)	$\Delta E_{int}\left(kJ\right)$
$A\to B$	+6.58	$-6.58$	0
$B\to C$	0	$-4.98$	$-4.98$
$C\to D$	$-5.02$	$+5.02$	0
$D\to A$	0	$+4.98$	$+4.98$

Explanation of Solution

Consider the isothermal process $A\to B$.

Write the expression for the change in internal energy.

  $\Delta E_{\text{int}}=n{C}_V\Delta T$                                                                                                    (XIV)

Here, $\Delta E_{\text{int}}$ is the change in internal energy, $C_V$ is the specific heat capacity at constant volume, $\Delta T$ is the change in temperature.

Write the expression for the first law of thermodynamics.

  $Q=-W+\Delta E_{\text{int}}$                                                                                                     (XV)

Here, $Q$ is the heat added to the system, $W$ is the work done by the engine.

Write the expression for the work done in the isothermal process.

  $W=-nRT\ln \left(\frac{V_B}{V_A}\right)$                                                                                              (XVI)

Consider the adiabatic process $B\to C$ for which total heat change in zero.

Therefore, $Q=0$.

Write the expression for the change in internal energy.

  $\Delta E_{\text{int}}=n{C}_V\left(T_C-T_B\right)$                                                                                         (XVII)

Here, $T_C$ is the temperature of the ideal gas at point $C$ in the Cycle and $T_B$ is the temperature of the ideal gas at point $B$ of the Cycle.

Consider the isothermal process $C\to D$.

Write the expression for the work done in the isothermal process.

  $W=-nRT\ln \left(\frac{V_D}{V_C}\right)$                                                                                           (XVIII)

Consider the adiabatic process $D\to A$ for which total heat change in zero.

Therefore, $Q=0$.

Write the expression for the change in internal energy.

  $\Delta E_{\text{int}}=n{C}_V\left(T_A-T_D\right)$                                                                                          (XIX)

Here, $T_A$ is the temperature of the ideal gas at point $A$ in the Cycle and $T_D$ is the temperature of the ideal gas at point $D$ of the Cycle.

Conclusion:

For the isothermal process $A\to B$.

For isothermal process $\Delta T=0$.

Substitute $0\text{K}$ for $\Delta T$ in equation (XIV) to get $\Delta E_{\text{int}}$.

  $\begin{array}{c}\Delta E_{\text{int}}=n{C}_V\left(0\text{K}\right)\\ =\text{0}\text{J}\end{array}$

Substitute $2.34\text{mol}$ for $n$, $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$, $720\text{K}$ for $T$, $16.0\text{L}$ for $V_B$ and $10.0\text{L}$ for $V_A$ in equation (XVI) to get $W$.

  $\begin{array}{c}W=-\left(2.34\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(720\text{K}\right)\ln \left(\frac{16.0\text{L}}{10.0\text{L}}\right)\\ =-6.58\text{kJ}\end{array}$

Substitute $-6.58\text{kJ}$ for $W$ and $0\text{J}$ for $\Delta E_{\text{int}}$ in (XV) to get $Q$.

  $\begin{array}{c}Q=-\left(-6.58\text{kJ}\right)+\left(0\text{J}\right)\\ =+6.58\text{kJ}\end{array}$

For the adiabatic process $B\to C$.

Substitute $2.34\text{mol}$ for $n$, $\frac{3}{2}\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)$ for $C_V$, $549\text{K}$ for $T_C$ and $720\text{K}$ for $T_B$ in equation (XVII) to get $\Delta E_{\text{int}}$.

  $\begin{array}{c}\Delta E_{\text{int}}=\left(2.34\text{mol}\right)\left(\frac{3}{2}\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\right)\left(549\text{K}-720\text{K}\right)\\ =-4.98\text{kJ}\end{array}$

Substitute $0\text{J}$ for $Q$ and $-4.98\text{kJ}$ for $\Delta E_{\text{int}}$ in (XVIII) to get $W$.

  $\begin{array}{c}0\text{J}=-W+-4.98\text{kJ}\\ \text{W}=-4.98\text{kJ}\end{array}$

Consider the isothermal process $C\to D$.

Substitute $0\text{K}$ for $\Delta T$ in equation (XIV) to get $\Delta E_{\text{int}}$.

  $\begin{array}{c}\Delta E_{\text{int}}=n{C}_V\left(0\text{K}\right)\\ =\text{0}\text{J}\end{array}$

Substitute $2.34\text{mol}$ for $n$, $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$, $549\text{K}$ for $T$, $15.0\text{L}$ for $V_D$ and $24.0\text{L}$ for $V_C$ in equation (XVI) to get $W$.

  $\begin{array}{c}W=-\left(2.34\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(549\text{K}\right)\ln \left(\frac{15.0\text{L}}{24.0\text{L}}\right)\\ =5.02\text{kJ}\end{array}$

Substitute $5.02\text{kJ}$ for $W$ and $0\text{J}$ for $\Delta E_{\text{int}}$ in (XV) to get $Q$.

  $\begin{array}{c}Q=-\left(5.02\text{kJ}\right)+\left(0\text{J}\right)\\ =-5.02\text{kJ}\end{array}$

For the adiabatic process $D\to A$.

Substitute $2.34\text{mol}$ for $n$, $\frac{3}{2}\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)$ for $C_V$, $720\text{K}$ for $T_A$ and $549\text{K}$ for $T_D$ in equation (XVII) to get $\Delta E_{\text{int}}$.

  $\begin{array}{c}\Delta E_{\text{int}}=\left(2.34\text{mol}\right)\left(\frac{3}{2}\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\right)\left(549\text{K}-720\text{K}\right)\\ =+4.98\text{kJ}\end{array}$

Substitute $0\text{J}$ for $Q$ and $+4.98\text{kJ}$ for $\Delta E_{\text{int}}$ in (XVIII) to get $W$.

  $\begin{array}{c}0\text{J}=-W++4.98\text{kJ}\\ \text{W}=+4.98\text{kJ}\end{array}$

Therefore,

Process	Q(kJ)	W(kJ)	$\Delta E_{int}\left(kJ\right)$
$A\to B$	+6.58	$-6.58$	0
$B\to C$	0	$-4.98$	$-4.98$
$C\to D$	$-5.02$	$+5.02$	0
$D\to A$	0	$+4.98$	$+4.98$

(c)

To determine

The efficiency $W_{\text{net}}/\left|Q_h\right|$.

Answer to Problem 32P

The efficiency $W_{\text{net}}/\left|Q_h\right|$ is $23.7\%$.

Explanation of Solution

Write the expression for the efficiency.

  $e=\frac{W_{\text{eng}}}{\left|Q_h\right|}$                                                                                                                (XX)

Here, $W_{\text{eng}}$ is the efficiency and $\left|Q_h\right|$ is the heat extracted from the hot reservoir.

The total work done by the engine is the negative of the work input.

Write the expression for the net work done by the engine.

  $W_{ABCD}=W_{AB}+W_{BC}+W_{CD}+W_{DA}$                                                                          (XXI)

Here, $W_{AB}$ is the work done during the process $A\to B$, $W_{BC}$ is the work done during the process $B\to C$, $W_{CD}$ is the work done during the process $C\to D$, $W_{DA}$ is the work done during the process $D\to A$.

Write the expression for the $W_{\text{eng}}$.

  $W_{\text{eng}}=-W_{ABCD}$                                                                                                     (XXII)

Here, $W_{ABCD}$ is the total work input.

Write the expression for $\left|Q_h\right|$.

  $\left|Q_h\right|=Q_{A\to B}$                                                                                                       (XXIII)

Use equation (XXII) and (XXIII) in equation (XX) to get $e$.

  $e=\frac{-W_{ABCD}}{Q_{A\to B}}$

Conclusion:

Substitute $-6.58\text{kJ}$ for $W_{AB}$, $-4.98\text{kJ}$ for $W_{BC}$, $+5.02\text{kJ}$ for $W_{CD}$ and $+4.98\text{kJ}$ for $W_{DA}$ in (XXI) to get $W_{ABCD}$.

  $\begin{array}{c}{W}_{ABCD}=-6.58\text{kJ}+-4.98\text{kJ}+5.02\text{kJ}+4.98\text{kJ}\\ \text{=}-1.56\text{kJ}\end{array}$

Substitute $-1.56\text{kJ}$ for $W_{ABCD}$ and $6.58\text{kJ}$ for $Q_{A\to B}$ in equation (XXIV) to get $e$.

  $\begin{array}{c}e=\frac{-\left(-1.56\text{kJ}\right)}{6.58\text{kJ}}\\ =0.237\times 100\\ =23.7\%\end{array}$

Therefore, the efficiency $W_{\text{net}}/\left|Q_h\right|$ is $23.7\%$.

(d)

To determine

To show that the efficiency is equal to Carnot efficiency.

Answer to Problem 32P

It is showed that the efficiency of the engine calculated in part(c) is equal to Carnot efficiency.

Explanation of Solution

Write the expression for the Carnot efficiency.

  $e_c=1-\frac{T_c}{T_h}$

Here, $e_c$ is the Carnot efficiency.

Conclusion:

Substitute $549\text{K}$ for $T_c$ and $720\text{K}$ for $T_h$ in above equation to get $e_c$.

  $\begin{array}{c}{e}_c=1-\frac{549\text{K}}{720\text{K}}\\ =0.237\times 100\%\\ =23.7\%\end{array}$

The above calculated efficiency is equal to that obtained in part(c). Thus, efficiency of the engine is equal to Carnot efficiency.

Therefore, it is showed that the efficiency of the engine calculated in part(c) is equal to Carnot efficiency.