Chapter 21, Problem 34P

(a)

To determine

Draw the PV diagram of the closed cycle.

Answer to Problem 34P

The PV diagram for the closed cycle is drawn.

Explanation of Solution

An adiabatic process occurs without transfer of heat or mass of substance between a thermodynamic system and its surroundings. In an adiabatic process energy is transferred to the surroundings only as work.

An Isobaric process is a thermodynamic process in which the pressure stays constant. The heat transferred to the system does work, but also changes the internal energy of the system.

Conclusion:

The figure 1 shows the PV diagram for the ideal gas expands adiabatically to its original pressure and compressed isobarically to its original volume.

(b)

To determine

The volume of the gas at the end of the adiabatic expansion.

Answer to Problem 34P

The volume of the gas at the end of the adiabatic expansion is $V_C=\left(3^{1/\gamma }\right)V_i$.

Explanation of Solution

Write the expression from PV diagram at the point B and C.

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

Here, $P_B$ is the pressure at point B, $P_C$ is the pressure at point C, $V_B$ is the volume at point B, $V_C$ is the volume at point C.

Conclusion:

Rewrite the above expression from the PV diagram to find volume.

  $\begin{array}{c}3P_i{V}_i^{\gamma }=P_i{V}_C^{\gamma }\\ V_C=\left(3^{1/\gamma }\right)V_i\end{array}$                                                                                           (I)

Here, $P_i$ is the initial pressure and $V_i$ is the initial volume.

Therefore, the volume of the gas at the end of the adiabatic expansion is $V_C=\left(3^{1/\gamma }\right)V_i$.

(c)

To determine

The temperature of the gas in an adiabatic expansion.

Answer to Problem 34P

The temperature of the gas in an adiabatic expansion is $T_B=3T_i$.

Explanation of Solution

Write the expression from ideal gas law.

  $P_B{V}_B=nR{T}_B$                                                                                            (II)

Here, $n$ is the number of moles, $R$ is the gas constant, and $T_B$ is the temperature at point B.

Write the expression from the PV diagram

  $3P_i{V}_i=3nR{T}_i$                                                                                            (III)

Here, $T_i$ is the initial temperature.

Conclusion:

Compare equation (II) and (III) to find $T_B$.

  $T_B=3T_i$

Therefore, the temperature of the gas in an adiabatic expansion is $T_B=3T_i$.

(d)

To determine

The temperature at the end of the cycle.

Answer to Problem 34P

The temperature at the end of the cycle is $T_A=T_i$.

Explanation of Solution

Write the expression for temperature for whole cycle.

  $T_A=T_i$                                                                                       (IV)

Here, $T_A$ is the temperature at point A.

Conclusion:

Therefore, the temperature at the end of the cycle is $T_A=T_i$.

(e)

To determine

The net work done on the gas.

Answer to Problem 34P

The net work done on the gas is $W_{ABCA}=-P_i{V}_i\left[\left(\frac{1}{\gamma -1}\right)\left(1-3^{1/\gamma }\right)+\left(1-3^{1/\gamma }\right)\right]$.

Explanation of Solution

Write the expression for energy transfer between the points AB.

  $Q_{AB}=n{C}_V\Delta T$

Here, $Q_{AB}$ is the energy transfer between the points AB, $C_V$ is the specific heat capacity at constant volume, and $\Delta T$ is the change in temperature.

Replace $\raisebox{1ex}{$R$}\!\left/ \!\raisebox{-1ex}{$\gamma -1$}\right.$ for $C_V$ and $3T_i-T_i$ for $\Delta T$ in above equation.

  $\begin{array}{c}{Q}_{AB}=n\left(\frac{R}{\gamma -1}\right)\left(3T_i-T_i\right)\\ =\frac{2nR{T}_i}{\gamma -1}\\ =\frac{2P_i{V}_i}{\gamma -1}\end{array}$                                                                                        (V)

Write the expression for energy transfer for adiabatic process between the points BC.

  $Q_{BC}=0$                                                                                                                (VI)

Write the expression from ideal gas law.

  $P_C{V}_C=nR{T}_C$

Here, $T_C$ is the temperature at point C.

Replace $\left(3^{1/\gamma }\right)V_i$ for $V_C$, $P_i$ for $P_C$, and $\left(3^{1/\gamma }\right)T_i$ for $T_C$ in above equation.

  $P_i\left(3^{1/\gamma }\right)V_i=\left(3^{1/\gamma }\right)nR{T}_i$

Write the expression for energy transfer between the points CA.

$Q_{CA}=n{C}_P\Delta T$

Here, $Q_{CA}$ is the energy transfer between the points CA, $C_P$ is the specific heat capacity at constant pressure, and $\Delta T$ is the change in temperature.

Replace $\gamma C_V$ for $C_P$ and $T_i-\left(3^{1/\gamma }\right)T_i$ for $\Delta T$ in above equation.

  $\begin{array}{c}{Q}_{CA}=n\gamma C_V\left(T_i-\left(3^{1/\gamma }\right)T_i\right)\\ =\gamma \frac{R}{\gamma -1}n{T}_i\left[1-\left(3^{1/\gamma }\right)\right]\\ =P_i{V}_i\gamma \left(\frac{1}{\gamma -1}\right)\left[1-\left(3^{1/\gamma }\right)\right]\end{array}$ (VII)

Conclusion:

Find the energy transfer for whole cycle.

  $Q_{ABCA}=Q_{AB}+Q_{BC}+Q_{CA}$

Substitute the equation (V), (VI) and (VII) in above equation.

  $\begin{array}{c}{Q}_{ABCA}=\frac{2P_i{V}_i}{\gamma -1}+0+P_i{V}_i\gamma \left(\frac{1}{\gamma -1}\right)\left[1-\left(3^{1/\gamma }\right)\right]\\ =P_i{V}_i\left[\frac{2}{\gamma -1}+\gamma \left(\frac{1}{\gamma -1}\right)\left[1-\left(3^{1/\gamma }\right)\right]\right]\\ =P_{i}V{\left[\frac{2}{\gamma -1}+\left(\frac{\gamma -1+1}{\gamma -1}\right)\left[1-\left(3^{1/\gamma }\right)\right]\right]}_i\end{array}$

Rewrite the above equation.

  $\begin{array}{c}{Q}_{ABCA}=P_i{V}_i\left[\frac{2}{\gamma -1}+\left(1-3^{1/\gamma }\right)+\frac{1-\left(3^{1/\gamma }\right)}{\gamma -1}\right]\\ =P_i{V}_i\left[\left(1-3^{1/\gamma }\right)+\frac{3-\left(3^{1/\gamma }\right)}{\gamma -1}\right]\end{array}$                                                          (VIII)

Find the internal energy for whole cycle.

  $\begin{array}{c}{\left(\Delta E_{\mathrm{int}}\right)}_{ABCA}=0\\ Q_{ABCA}+W_{ABCA}=0\\ W_{ABCA}=-Q_{ABCA}\end{array}$

Here, $W_{ABCA}$ is the work done in whole cycle process.

Substitute equation (VIII) in the above equation.

  $W_{ABCA}=-P_i{V}_i\left[\left(\frac{1}{\gamma -1}\right)\left(1-3^{1/\gamma }\right)+\left(1-3^{1/\gamma }\right)\right]$

Therefore, the net work done on the gas is $W_{ABCA}=-P_i{V}_i\left[\left(\frac{1}{\gamma -1}\right)\left(1-3^{1/\gamma }\right)+\left(1-3^{1/\gamma }\right)\right]$.