Chapter 22, Problem 22.32P

At point A in a Carnot cycle, 2.34 mol of a monatomic ideal gas has a pressure of 1 4000 kPa, a volume of 10.0 L, and a temperature of 720 K. The gas expands isothermally to point B and then expands adiabatically to point C, where its volume is 24.0 L. An isothermal compression brings it to point D, where its volume is 15.0 L. An adiabatic process returns the gas to point A. (a) Determine all the unknown pressures, volumes, and temperatures as you f ill in the following table:

(b) Find the energy added by heat, the work done by the engine, and the change in internal energy for each of the steps A → B, B → C, C → D, and D → A (c) Calculate the efficiency W_net/|Q_k|. (d) Show that the efficiency is equal to 1 - T_C/T_A, the Carnot efficiency.

To determine

The unknown pressures, volumes and the temperature in the table.

Answer to Problem 22.32P

The values of unknown pressures, volumes and the temperature in the table are,

	$P$	$V$	$T$
$A$	$1400\text{kPa}$	$10.0\text{L}$	$720\text{K}$
$B$	$875\text{kPa}$	$16.0\text{L}$	$720\text{K}$
$C$	$445\text{kPa}$	$24.0\text{L}$	$549\text{K}$
$D$	$712\text{kPa}$	$15.0\text{L}$	$549\text{K}$

Explanation of Solution

Given: The number of moles of a mono atomic ideal gas is $2.34\text{mol}$, the pressure of an ideal gas is $1400\text{kPa}$, the volume of an ideal gas is $10.0\text{L}$, the temperature of an ideal gas is $720\text{K}$, the volume of gas when it expands adiabatically to point $C$ is $24.0\text{L}$ and the volume of gas when it is isothermally compressed is $15.0\text{L}$.

Write the equation of adiabatic process $D\to A$.

$\begin{array}{c}{P}_{\text{D}}{V}_{\text{D}}^{\gamma }=P_{\text{A}}{V}_{\text{A}}^{\gamma }\\ P_{\text{D}}=P_{\text{A}}{\left(\frac{V_{\text{A}}}{V_{\text{D}}}\right)}^{\gamma }\end{array}$

Here,

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

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

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

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

$\gamma$ is the ratio of specific heats of the gas.

The value of $\gamma$ is $5/3$ for mono atomic gas.

Substitute $1400\text{kPa}$ for $P_{\text{A}}$, $10.0\text{L}$ for $V_{\text{A}}$, $15.0\text{L}$ for $V_{\text{D}}$ and $5/3$ for $\gamma$ in above equation to find $P_{\text{D}}$.

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

Thus, the pressure of the gas at point $D$ is $712\text{kPa}$.

Write the ideal gas equation.

$\begin{array}{c}{P}_{\text{D}}{V}_{\text{D}}=nR{T}_{\text{D}}\\ T_{\text{D}}=\frac{P_{\text{D}}{V}_{\text{D}}}{nR}\end{array}$

Here,

$n$ is the number of moles of the gas.

$R$ is the gas constant.

$T_{\text{D}}$ is the temperature of the gas at point $D$.

The value of gas constant is $8.314\text{J}/\text{mol}\cdot \text{K}$.

Substitute $712\text{kPa}$ for $P_{\text{D}}$, $15.0\text{L}$ for $V_{\text{D}}$, $2.34\text{mol}$ for $n$ and $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$ in above equation to find $T_{\text{D}}$.

$\begin{array}{c}{T}_{\text{D}}=\frac{\left\{712\text{kPa}\left(\frac{1000\text{Pa}}{1\text{kPa}}\right)\right\}\left\{15.0\text{L}\left(\frac{1{\text{m}}^3}{1000\text{L}}\right)\right\}}{\left(2.34\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)}\\ =548.96\text{K}\\ \approx \text{549}\text{K}\end{array}$

Thus, the temperature of the gas at point $D$ is $549\text{K}$.

In isothermal process, the temperature is constant.

For isothermal process $C\to D$, the temperature at point $C$ is equal to the temperature at point $D$.

The temperature of the gas at point $C$ is,

$T_{\text{C}}=549\text{K}$

Thus, the temperature of the gas at point $C$ is $549\text{K}$.

Write the ideal gas equation.

$\begin{array}{c}{P}_{\text{C}}{V}_{\text{C}}=nR{T}_{\text{C}}\\ P_{\text{C}}=\frac{nR{T}_{\text{C}}}{V_{\text{C}}}\end{array}$

Here,

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

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

Substitute $2.34\text{mol}$ for $n$, $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$, $549\text{K}$ for $T_{\text{C}}$ and $24.0\text{L}$ for $V_{\text{C}}$ in above equation to find $P_{\text{C}}$.

$\begin{array}{c}{P}_{\text{C}}=\frac{\left(2.34\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(549\text{K}\right)}{\left\{24.0\text{L}\left(\frac{1{\text{m}}^3}{1000\text{L}}\right)\right\}}\\ =445\times {10}^3\text{Pa}\left(\frac{1\text{kPa}}{1000\text{Pa}}\right)\\ =445\text{kPa}\end{array}$

Thus, the pressure of the gas at point $C$ is $445\text{kPa}$.

Write the equation of adiabatic process $B\to C$.

$\begin{array}{c}{T}_{\text{B}}{V}_{\text{B}}^{\gamma -1}=T_{\text{C}}{V}_{\text{C}}^{\gamma -1}\\ V_{\text{B}}=V_{\text{C}}{\left(\frac{T_{\text{C}}}{T_{\text{B}}}\right)}^{1/\left(\gamma -1\right)}\end{array}$

Here,

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

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

Substitute $24.0\text{L}$ for $V_{\text{C}}$, $549\text{K}$ for $T_{\text{C}}$, $720\text{K}$ for $T_{\text{B}}$ and $5/3$ for $\gamma$ in above equation to find $V_{\text{B}}$.

$\begin{array}{c}{V}_{\text{B}}=\left(24.0\text{L}\right){\left(\frac{\left(549\text{K}\right)}{\left(720\text{K}\right)}\right)}^{1/\left(5/3-1\right)}\\ =15.979\text{L}\\ \approx \text{16}\text{.0}\text{L}\end{array}$

Thus, the volume of the gas at point $B$ is $16.0\text{L}$.

In isothermal process, the temperature is constant.

For isothermal process $A\to B$, the temperature at point $A$ is equal to the temperature at point $B$.

The temperature of the gas at point $B$ is,

$T_{\text{B}}=720\text{K}$

Thus, the temperature of the gas at point $B$ is $720\text{K}$.

Write the ideal gas equation.

$\begin{array}{c}{P}_{\text{B}}{V}_{\text{B}}=nR{T}_{\text{B}}\\ P_{\text{B}}=\frac{nR{T}_{\text{B}}}{V_{\text{B}}}\end{array}$

Here,

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

Substitute $2.34\text{mol}$ for $n$, $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$, $720\text{K}$ for $T_{\text{B}}$ and $16.0\text{L}$ for $V_{\text{B}}$ in above equation to find $P_{\text{B}}$.

$\begin{array}{c}{P}_{\text{B}}=\frac{\left(2.34\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(720\text{K}\right)}{\left\{16.0\text{L}\left(\frac{1{\text{m}}^3}{1000\text{L}}\right)\right\}}\\ =875464.2\text{Pa}\left(\frac{1\text{kPa}}{1000\text{Pa}}\right)\\ \approx 875\text{kPa}\end{array}$

Thus, the pressure of the gas at point $B$ is $875\text{kPa}$.

Form a table and show the unknown value of pressures, volumes and temperatures.

	$P$	$V$	$T$
$A$	$1400\text{kPa}$	$10.0\text{L}$	$720\text{K}$
$B$	$875\text{kPa}$	$16.0\text{L}$	$720\text{K}$
$C$	$445\text{kPa}$	$24.0\text{L}$	$549\text{K}$
$D$	$712\text{kPa}$	$15.0\text{L}$	$549\text{K}$

Conclusion:

Therefore, the values of unknown pressures, volumes and the temperature in the table are,

	$P$	$V$	$T$
$A$	$1400\text{kPa}$	$10.0\text{L}$	$720\text{K}$
$B$	$875\text{kPa}$	$16.0\text{L}$	$720\text{K}$
$C$	$445\text{kPa}$	$24.0\text{L}$	$549\text{K}$
$D$	$712\text{kPa}$	$15.0\text{L}$	$549\text{K}$

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 22.32P

The values of energy added by heat, work done by the engine and the change in internal energy for each of the steps in the table are,

	$Q$	$W$	$\Delta E_{\mathrm{int}}$
$A$	$+6.58\text{kJ}$	$-6.58\text{kJ}$	$0$
$B$	$0$	$-4.99\text{kJ}$	$-4.99\text{kJ}$
$C$	$-5.02\text{kJ}$	$+5.02\text{kJ}$	$0$
$D$	$0$	$+4.99\text{kJ}$	$+4.99\text{kJ}$

Explanation of Solution

Given: The number of moles of a mono atomic ideal gas is $2.34\text{mol}$, the pressure of an ideal gas is $1400\text{kPa}$, the volume of an ideal gas is $10.0\text{L}$, the temperature of an ideal gas is $720\text{K}$, the volume of gas when it expands adiabatically to point $C$ is $24.0\text{L}$ and the volume of gas when it is isothermally compressed is $15.0\text{L}$.

The process $A\to B$ is an isothermal process in which the temperature is constant due to which the change in temperature $\Delta T$ is $0$.

Write the equation of change in temperature for process $A\to B$.

$\Delta E_{\mathrm{int}}=n{C}_{\text{v}}\Delta T$

Here,

$C_{\text{v}}$ is the molar heat capacity of the gas at constant volume.

Substitute $0$ for $\Delta T$ in above equation to find $\Delta E_{\mathrm{int}}$.

$\begin{array}{c}\Delta E_{\mathrm{int}}=n{C}_{\text{v}}\left(0\right)\\ =0\end{array}$

Thus, the change in internal energy for process $A\to B$ is $0$.

Write the equation of work done by the engine for process $A\to B$.

$W=-nR{T}_{\text{A}}\ln \left(\frac{V_{\text{B}}}{V_{\text{A}}}\right)$

Substitute $2.34\text{mol}$ for $n$, $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$, $720\text{K}$ for $T_{\text{A}}$, $16.0\text{L}$ for $V_{\text{B}}$ and $10.0\text{L}$ for $V_{\text{A}}$ in above equation to find $W$.

$\begin{array}{c}W=-\left(2.34\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(720\text{K}\right)\ln \left(\frac{\left(16.0\text{L}\right)}{\left(10.0\text{L}\right)}\right)\\ =-6583.54\text{J}\left(\frac{1\text{kJ}}{1000\text{J}}\right)\\ \approx -6.58\text{kJ}\end{array}$

Thus, the work done by the engine for process $A\to B$ is $-6.58\text{kJ}$.

Write the equation of isothermal process $A\to B$.

$Q=-W$

Substitute $-6.58\text{kJ}$ for $W$ in above equation to find $Q$.

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

Thus, the energy added by heat for process $A\to B$ is $+6.58\text{kJ}$.

Write the equation of change in temperature for process $B\to C$.

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

Here,

$C_{\text{v}}$ is the molar heat capacity of the gas at constant volume.

The value of $C_{\text{v}}$ for mono atomic gas is $3R/2$.

Substitute $3R/2$ for $C_{\text{v}}$ in above equation.

$\Delta E_{\mathrm{int}}=n\left(\frac{3}{2}R\right)\left(T_{\text{C}}-T_{\text{B}}\right)$

Substitute $2.34\text{mol}$ for $n$, $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$, $549\text{K}$ for $T_{\text{C}}$ and $720\text{K}$ for $T_{\text{B}}$ in above equation to find $\Delta E_{\mathrm{int}}$.

$\begin{array}{c}\Delta E_{\mathrm{int}}=\left(2.34\text{mol}\right)\left\{\frac{3}{2}\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\right\}\left\{\left(549\text{K}\right)-\left(720\text{K}\right)\right\}\\ =-4990.14\text{J}\left(\frac{1\text{kJ}}{1000\text{J}}\right)\\ \approx -4.99\text{kJ}\end{array}$

Thus, the change in internal energy for process $B\to C$ is $-4.99\text{kJ}$.

The process $B\to C$ is an adiabatic process. In adiabatic process, the transfer of heat is zero due to which $Q$ is zero.

Thus, the energy added by heat for process $B\to C$ is $0$.

Write the equation of change in internal energy for process $B\to C$.

$\begin{array}{c}\Delta E_{\mathrm{int}}=Q+W\\ W=\Delta E_{\mathrm{int}}-Q\end{array}$

Substitute $-4.99\text{kJ}$ for $\Delta E_{\mathrm{int}}$ and $0$ for $Q$ in above equation to find $W$.

$\begin{array}{c}W=\left(-4.99\text{kJ}\right)-0\\ =-4.99\text{kJ}\end{array}$

Thus, the work done by the engine for process $B\to C$ is $-4.99\text{kJ}$.

The process $C\to D$ is an isothermal process in which the temperature is constant due to which the change in temperature $\Delta T$ is $0$.

Write the equation of change in temperature for process $C\to D$.

$\Delta E_{\mathrm{int}}=n{C}_{\text{v}}\Delta T$

Substitute $0$ for $\Delta T$ in above equation to find $\Delta E_{\mathrm{int}}$.

$\begin{array}{c}\Delta E_{\mathrm{int}}=n{C}_{\text{v}}\left(0\right)\\ =0\end{array}$

Thus, the change in internal energy for process $C\to D$ is $0$.

Write the equation of work done by the engine for process $C\to D$.

$W=-nR{T}_{\text{C}}\ln \left(\frac{V_{\text{D}}}{V_{\text{C}}}\right)$

Substitute $2.34\text{mol}$ for $n$, $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$, $549\text{K}$ for $T_{\text{C}}$, $15.0\text{L}$ for $V_{\text{D}}$ and $24.0\text{L}$ for $V_{\text{C}}$ in above equation to find $W$.

$\begin{array}{c}W=-\left(2.34\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(549\text{K}\right)\ln \left(\frac{\left(15.0\text{L}\right)}{\left(24.0\text{L}\right)}\right)\\ =-\left\{-5019.95\text{J}\left(\frac{1\text{kJ}}{1000\text{J}}\right)\right\}\\ \approx +5.02\text{kJ}\end{array}$

Thus, the work done by the engine for process $C\to D$ is $+5.02\text{kJ}$.

Write the equation of isothermal process $C\to D$.

$Q=-W$

Substitute $+5.02\text{kJ}$ for $W$ in above equation to find $Q$.

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

Thus, the energy added by heat for process $C\to D$ is $-5.02\text{kJ}$.

Write the equation of change in temperature for process $D\to A$.

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

Substitute $3R/2$ for $C_{\text{v}}$ in above equation.

$\Delta E_{\mathrm{int}}=n\left(\frac{3}{2}R\right)\left(T_{\text{A}}-T_{\text{D}}\right)$

Substitute $2.34\text{mol}$ for $n$, $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$, $720\text{K}$ for $T_{\text{A}}$ and $549\text{K}$ for $T_{\text{D}}$ in above equation to find $\Delta E_{\mathrm{int}}$.

$\begin{array}{c}\Delta E_{\mathrm{int}}=\left(2.34\text{mol}\right)\left\{\frac{3}{2}\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\right\}\left\{\left(720\text{K}\right)-\left(549\text{K}\right)\right\}\\ =+4990.14\text{J}\left(\frac{1\text{kJ}}{1000\text{J}}\right)\\ \approx +4.99\text{kJ}\end{array}$

Thus, the change in internal energy for process $D\to A$ is $+4.99\text{kJ}$.

The process $D\to A$ is an adiabatic process. In adiabatic process, the transfer of heat is zero due to which $Q$ is zero.

Thus, the energy added by heat for process $D\to A$ is $0$.

Write the equation of change in internal energy for process $D\to A$.

$\begin{array}{c}\Delta E_{\mathrm{int}}=Q+W\\ W=\Delta E_{\mathrm{int}}-Q\end{array}$

Substitute $+4.99\text{kJ}$ for $\Delta E_{\mathrm{int}}$ and $0$ for $Q$ in above equation to find $W$.

$\begin{array}{c}W=\left(+4.99\text{kJ}\right)-0\\ =+4.99\text{kJ}\end{array}$

Thus, the work done by the engine for process $D\to A$ is $+4.99\text{kJ}$.

Form a table and show the value of energy added by heat, work done by the engine and the change in internal energy.

	$Q$	$W$	$\Delta E_{\mathrm{int}}$
$A$	$+6.58\text{kJ}$	$-6.58\text{kJ}$	$0$
$B$	$0$	$-4.99\text{kJ}$	$-4.99\text{kJ}$
$C$	$-5.02\text{kJ}$	$+5.02\text{kJ}$	$0$
$D$	$0$	$+4.99\text{kJ}$	$+4.99\text{kJ}$

Conclusion:

Therefore, the values of energy added by heat, work done by the engine and the change in internal energy for each of the steps in the table are,

	$Q$	$W$	$\Delta E_{\mathrm{int}}$
$A$	$+6.58\text{kJ}$	$-6.58\text{kJ}$	$0$
$B$	$0$	$-4.99\text{kJ}$	$-4.99\text{kJ}$
$C$	$-5.02\text{kJ}$	$+5.02\text{kJ}$	$0$
$D$	$0$	$+4.99\text{kJ}$	$+4.99\text{kJ}$

To determine

The value of efficiency $W_{\text{net}}/\left|Q_{\text{h}}\right|$.

Answer to Problem 22.32P

The value of efficiency $W_{\text{net}}/\left|Q_{\text{h}}\right|$ is $0.237$.

Explanation of Solution

Given: The number of moles of a mono atomic ideal gas is $2.34\text{mol}$, the pressure of an ideal gas is $1400\text{kPa}$, the volume of an ideal gas is $10.0\text{L}$, the temperature of an ideal gas is $720\text{K}$, the volume of gas when it expands adiabatically to point $C$ is $24.0\text{L}$ and the volume of gas when it is isothermally compressed is $15.0\text{L}$.

Calculate the net work done from the table is,

$\begin{array}{c}{W}_{\text{net}}=-\left\{\left(-6.58\text{kJ}\right)+\left(-4.99\text{kJ}\right)+\left(5.02\text{kJ}\right)+\left(+4.99\text{kJ}\right)\right\}\\ =1.56\text{kJ}\end{array}$

Write the equation for efficiency.

$e=\frac{W_{\text{net}}}{\left|Q_{\text{h}}\right|}$

Here,

$Q_{\text{h}}$ is the energy from the hot reservoir which the energy in process $A\to B$.

Substitute $1.56\text{kJ}$ for $W_{\text{net}}$ and $-6.58\text{kJ}$ for $Q_{\text{h}}$ in above equation.

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

The value of efficiency $W_{\text{net}}/\left|Q_{\text{h}}\right|$ is $0.237$.

Conclusion:

Therefore, the value of efficiency $W_{\text{net}}/\left|Q_{\text{h}}\right|$ is $0.237$.

To determine

To show: The efficiency is equal to the Carnot efficiency $1-T_{\text{C}}/T_{\text{A}}$.

Answer to Problem 22.32P

The efficiency is equal to the Carnot efficiency $1-T_{\text{C}}/T_{\text{A}}$.

Explanation of Solution

Given: The number of moles of a mono atomic ideal gas is $2.34\text{mol}$, the pressure of an ideal gas is $1400\text{kPa}$, the volume of an ideal gas is $10.0\text{L}$, the temperature of an ideal gas is $720\text{K}$, the volume of gas when it expands adiabatically to point $C$ is $24.0\text{L}$ and the volume of gas when it is isothermally compressed is $15.0\text{L}$.

Write the equation for Carnot efficiency.

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

Here,

$T_{\text{c}}$ is the temperature for cold reservoir.

$T_{\text{h}}$ is the temperature for hot reservoir.

The value of $T_{\text{c}}$ is $549\text{K}$ and $T_h$ is $720\text{K}$.

Substitute $549\text{K}$ for $T_{\text{c}}$ and $720\text{K}$ for $T_h$ in above equation to find $e_{\text{C}}$.

$\begin{array}{c}{e}_{\text{C}}=1-\frac{\left(549\text{K}\right)}{\left(720\text{K}\right)}\\ =0.2375\\ \approx 0.237\end{array}$

Thus, the Carnot efficiency is $0.237$.

Write the equation for efficiency.

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

Substitute $549\text{K}$ for $T_{\text{C}}$ and $720\text{K}$ for $T_{\text{A}}$ in above equation to find $e$.

$\begin{array}{c}e=1-\frac{\left(549\text{K}\right)}{\left(720\text{K}\right)}\\ =0.2375\\ \approx 0.237\end{array}$

The value of efficiency is $0.237$ which is equal to the Carnot efficiency.

Conclusion:

Therefore, the efficiency is equal to the Carnot efficiency $1-T_{\text{C}}/T_{\text{A}}$.