Chapter 22, Problem 81CP

(a)

To determine

The pressures and volumes at points $A,B,C,\text{and }D$.

Answer to Problem 81CP

Pressure at $A$ is $25.0\text{atm}$, pressure at $B$ is $4.14\text{atm}$, pressure at $C$ is $1.00\text{atm}$, pressure at $D$ is $6.03\text{atm}$, volume at $A$ is $1.97\times {10}^{-3}{\text{m}}^3$, volume at $B$ is $11.9\times {10}^{-3}{\text{ m}}^3$ , volume at $C$ is $32.8\times {10}^{-3}{\text{m}}^3$ and volume at $D$ is $5.44\times {10}^{-3}{\text{m}}^3$.

Explanation of Solution

It is given that pressure at $A$ is $25.0\text{atm}$, pressure at $C$ is $1.00\text{atm}$, temperature at $A$ is $600\text{K}$ and temperature at $C$ is $400\text{K}$.

Write ideal gas equation.

  $PV=nRT$                                                                                                                 (I)

Here, $P$ is the pressure of gas, $V$ is the volume of the gas, $n$ is the number of moles, $R$ is the universal gas constant and $T$ is the temperature of the gas.

Rearrange above equation to get $V$.

  $V=\frac{nRT}{P}$                                                                                                                  (II)

Use equation (II) to get pressure at point $A$ of the Carnot cycle.

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

Here, $P_A$ is the pressure of gas at point $A$ , $V_A$ is the volume of the gas at point $A$  and $T_A$ is the temperature of the gas at point $A$.

Use equation (II) to get pressure at point $C$ of the Carnot cycle.

  $V_C=\frac{nR{T}_C}{P_C}$                                                                                                              (IV)

Here, $P_C$ is the pressure of gas at point $C$ , $V_C$ is the volume of the gas at point $C$ and $T_C$ is the temperature of the gas at point $C$.

Write the condition of isothermal process.

  $PV=\text{constant}$                                                                                                         (V)

The curve $AB$ represents isothermal expansion.

Use equation (V) for the points $A$ and $B$.

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

Here, $P_B$ is the pressure of gas at point $B$ and $V_B$ is the volume of the gas at point $B$.

Write the condition of adiabatic process.

  $P{V}^{\gamma }=\text{constant}$                                                                                                     (VII)

Here, $\gamma$ is the adiabatic constant.

The curve $BC$ represents adiabatic expansion.

Use equation (VII) for the points $B$ and $C$.

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

Rearrange above equation to get $P_B$.

  $P_B=\frac{P_C{V}_C^{\gamma }}{V_B^{\gamma }}$                                                                                                              (IX)

Substitute $\frac{P_C{V}_C^{\gamma }}{V_B^{\gamma }}$ for $P_B$ in equation (VI) to get $V_B$.

  $\begin{array}{c}{P}_A{V}_A=\left(\frac{P_C{V}_C^{\gamma }}{V_B^{\gamma }}\right)V_B\\ V_B^{1-\gamma }=\left(\frac{P_A}{P_C}\right)\frac{V_A}{V_C^{\gamma }}\\ V_B^{\gamma -1}=\left(\frac{P_C}{P_A}\right)\frac{V_C^{\gamma }}{V_A}\\ V_B={\left[\left(\frac{P_C}{P_A}\right)\frac{V_C^{\gamma }}{V_A}\right]}^{1/\left(\gamma -1\right)}\end{array}$                                                                                            (X)

The curve $CD$ represents isothermal compression.

Use equation (V) for the points $C$ and $D$.

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

Here, $P_D$ is the pressure of gas at point $D$ and $V_D$ is the volume of the gas at point.

The curve $DA$  represents adiabatic compression.

Use equation (VII) for the points $D$ and $A$.

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

Rearrange above equation to get $P_D$.

  $P_D=\frac{P_A{V}_A^{\gamma }}{V_D^{\gamma }}$                                                                                                            (XIII)

Substitute $\frac{P_A{V}_A^{\gamma }}{V_D^{\gamma }}$ for $P_D$ in equation (XI) to get $V_D$.

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

Conclusion:

For ideal gas the value of $\gamma$ is $1.40$.

Substitute $1.00\text{mol}$ for $n$, $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$, $600\text{K}$ for $T_A$, $25.0\text{atm}$ for $P_A$ in in equation (III) to get $V_A$.

  $\begin{array}{c}{V}_A=\frac{\left(1.00\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(600\text{K}\right)}{\left(25.0\text{atm}\times \frac{1.013\times {10}^5\text{Pa}}{1\text{atm}}\right)}\\ =1.97\times {10}^{-3}{\text{m}}^3\end{array}$

Substitute $1.00\text{mol}$ for $n$, $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$, $400\text{K}$ for $T_C$, $1.00\text{atm}$ for $P_C$ in in equation (IV) to get $V_C$.

  $\begin{array}{c}{V}_C=\frac{\left(1.00\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(400\text{K}\right)}{\left(1.00\text{atm}\times \frac{1.013\times {10}^5\text{Pa}}{1\text{atm}}\right)}\\ =32.8\times {10}^{-3}{\text{m}}^3\end{array}$

Substitute $25.0\text{atm}$ for $P_A$, $1.00\text{atm}$ for $P_C$, $1.97\times {10}^{-3}{\text{m}}^3$ for $V_A$ , $32.8\times {10}^{-3}{\text{m}}^3$ for $V_C$ and $1.40$ for $\gamma$ in equation (X) to get $V_B$.

  $\begin{array}{c}{V}_B={\left[\left(\frac{1.00\text{atm}}{25.0\text{atm}}\right)\frac{{\left(32.8\times {10}^{-3}{\text{m}}^3\right)}^{1.40}}{\left(1.97\times {10}^{-3}{\text{m}}^3\right)}\right]}^{1/\left(1.40-1\right)}\\ =11.9\times {10}^{-3}{\text{m}}^3\end{array}$

Substitute $25.0\text{atm}$ for $P_A$, $1.00\text{atm}$ for $P_C$, $1.97\times {10}^{-3}{\text{m}}^3$ for $V_A$ , $32.8\times {10}^{-3}{\text{m}}^3$ for $V_C$ and $1.40$ for $\gamma$ in equation (XIV) to get $V_D$.

  $\begin{array}{c}{V}_D={\left[\left(\frac{25.0\text{atm}}{1.00\text{atm}}\right)\frac{{\left(1.97\times {10}^{-3}{\text{m}}^3\right)}^{1.40}}{\left(32.8\times {10}^{-3}{\text{m}}^3\right)}\right]}^{1/\left(1.40-1\right)}\\ =5.44\times {10}^{-3}{\text{m}}^3\end{array}$

Substitute $25.0\text{atm}$ for $P_A$, $11.9\times {10}^{-3}{\text{m}}^3$ for $V_B$, $1.97\times {10}^{-3}{\text{m}}^3$ for $V_A$ in equation (VI) to get $P_B$.

  $\begin{array}{c}\left(25.0\text{atm}\right)\left(1.97\times {10}^{-3}{\text{m}}^3\right)=P_B\left(11.9\times {10}^{-3}{\text{m}}^3\right)\\ P_B=\frac{\left(25.0\text{atm}\right)\left(1.97\times {10}^{-3}{\text{m}}^3\right)}{\left(11.9\times {10}^{-3}{\text{m}}^3\right)}\\ =4.14\text{atm}\end{array}$

Substitute $1.00\text{atm}$ for $P_C$, $32.8\times {10}^{-3}{\text{m}}^3$ for $V_C$, $5.44\times {10}^{-3}{\text{m}}^3$ for $V_D$ in equation (XI) to get $P_D$.

  $\begin{array}{c}\left(1.00\text{atm}\right)\left(32.8\times {10}^{-3}{\text{m}}^3\right)=P_D\left(5.44\times {10}^{-3}{\text{m}}^3\right)\\ P_D=\frac{\left(1.00\text{atm}\right)\left(32.8\times {10}^{-3}{\text{m}}^3\right)}{\left(5.44\times {10}^{-3}{\text{m}}^3\right)}\\ =6.03\text{atm}\end{array}$

Therefore, the pressure at $A$ is $25.0\text{atm}$, pressure at $B$ is $4.14\text{atm}$, pressure at $C$ is $1.00\text{atm}$, pressure at $D$ is $6.03\text{atm}$, volume at $A$ is $1.97\times {10}^{-3}{\text{m}}^3$, volume at $B$ is $11.9\times {10}^{-3}{\text{ m}}^3$ , volume at $C$ is $32.8\times {10}^{-3}{\text{m}}^3$ and volume at $D$ is $5.44\times {10}^{-3}{\text{m}}^3$.

(b)

To determine

The net work done per cycle.

Answer to Problem 81CP

The net work done per cycle is $2.99\text{kJ}$.

Explanation of Solution

In the Carnot cycle energy is added by heat to the gas during isothermal expansion, in the figure it is the process $AB$. For isothermal process, change in internal energy is zero. Thus, according to first law of thermodynamics energy added by heat to the gas is equal to work done by the gas.

Write the expression for work done by the gas.

  $W_{AB}=nR{T}_h\ln \left(\frac{V_B}{V_A}\right)$                                                                                              (XV)

Here, $W_{AB}$ is the work done by the gas during isothermal expansion and $T_h$ is the temperature of the hot reservoir which is equal to $T_A$.

Write the relation between heat added and work done according to first law of thermodynamics.

  $Q_{AB}=-W_{AB}$                                                                                                          (XVI)

Here, $Q_{AB}$ is the energy added by heat during process $AB$.

The energy added by heat during the process $AB$ is equal to heat absorbed by the gas from hot reservoir.

Write the expression for the Carnot efficiency.

  $e_C=1-\frac{T_c}{T_h}$                                                                                                           (XVII)

Here, $e_C$ is the Carnot efficiency, $T_c$ is the temperature of cold reservoir and $T_h$ is the temperature of hot reservoir.

Write the expression for the efficiency of the in terms of work done.

  $e_C=\frac{W_{\text{eng}}}{\left|Q_h\right|}$                                                                                                       (XVIII)

Here, $W_{\text{eng}}$ is the work done per cycle.

Conclusion:

Substitute $1.00\text{mol}$ for $n$, $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$, $600\text{K}$ for $T_h$, $11.9\times {\text{10}}^{-3}{\text{m}}^3$ for $V_B$ and $1.97\times {\text{10}}^{-3}{\text{m}}^3$ for $V_A$ in equation (XV) to get $W_{AB}$.

  $\begin{array}{c}{W}_{AB}=\left(1.00\text{mol}\right)\left(8.314\text{J}/\text{mol}\cdot \text{K}\right)\left(600\text{K}\right)\ln \left(\frac{11.9\times {\text{10}}^{-3}{\text{m}}^3}{1.97\times {\text{10}}^{-3}{\text{m}}^3}\right)\\ =8.97\times {10}^3\text{J}\times \frac{1\text{kJ}}{1000\text{J}}\\ =8.97\text{kJ}\end{array}$

Substitute $8.97\text{kJ}$ in equation (XVI) to get $Q_{AB}$.

  $Q_{AB}=-8.97\text{kJ}$

Since heat added to gas is equal to $Q_{AB}$, $\left|Q_h\right|=8.97\text{kJ}$.

Substitute $600\text{K}$ for $T_h$ and $400\text{K}$ for $T_c$ in equation (XVII) to get $e_C$.

  $\begin{array}{c}{e}_C=1-\frac{400\text{K}}{600\text{K}}\\ =0.333\end{array}$

Substitute $8.97\text{kJ}$ for $\left|Q_h\right|$ and $0.333$ for $e_C$ in equation (XVIII) to get $W_{\text{eng}}$.

  $\begin{array}{c}0.333=\frac{W_{\text{eng}}}{8.97\text{kJ}}\\ W_{\text{eng}}=\left(0.333\right)\left(8.97\text{kJ}\right)\\ =2.99\text{kJ}\end{array}$

Therefore, the net work done per cycle is $2.99\text{kJ}$.