Chapter 17, Problem 48P

An ideal gas with specific heat ratio γ confined to a cylinder is put through a closed cycle. Initially, the gas is at P_i, V_i, and T_i. First, its pressure is tripled under constant volume. It then expands adiabatically to its original pressure and finally is compressed isobarically to its original volume. (a) Draw a PV diagram of this cycle. (b) Determine the volume at the end of the adiabatic expansion. Find (c) the temperature of the gas at the start of the adiabatic expansion and (d) the temperature at the end of the cycle. (e) What was the net work done on the gas for this cycle?

To determine

Draw the $PV$ diagram of the cycle.

Answer to Problem 48P

The $PV$ diagram of the cycle is drawn.

Explanation of Solution

In this cycle, from $C$ to $A$ the pressure remains same, only the volume is changed.

From $A$ to $B$ the volume remains same, only the pressure changed, but from $B$ to $C$ the pressure and volume both are changed.

Figure 1 is the $PV$ diagram of the cycle.

Conclusion:

Therefore, the $PV$ diagram of the cycle is drawn.

To determine

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

Answer to Problem 48P

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

Explanation of Solution

Write the expression for the adiabatic process,

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

Here, $P_B$ is the pressure of the gas at point $B$, $P_C$ is the pressure of the gas at point $C$, $V_B$ is the volume of the gas at point $B$, $V_C$ is the volume of the gas at point $C$ and $\gamma$ is the adiabatic constant.

Conclusion:

Substitute $3P_i$ for $P_B$, $P_i$ for $P_C$ and $V_i$ for $V_B$ in (I),

    $3P_i{V}_i^{\gamma }=P_i{V}_C^{\gamma }$

Rewrite the above equation for $V_C$ ,

    $V_C=\left(3^{1/\gamma }\right)V_i$

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

To determine

The temperature of the gas at the start of the expansion.

Answer to Problem 48P

The temperature of the gas at the start of the expansion is $\underset{\_}{3T_i}$.

Explanation of Solution

Write the expression for the ideal gas law,

    $P_B{V}_B=nR{T}_B$        (II)

Conclusion:

Substitute $3P_i$ for $P_B$ and $V_i$ for $V_B$ in (II),

    $\begin{array}{c}3P_i{V}_i=nR{T}_B\\ 3nR{T}_i=nR{T}_B\end{array}$

    $T_B=3T_i$

Therefore, the temperature of the gas at the start of the expansion is $\underset{\_}{3T_i}$.

To determine

The temperature at the end of the cycle.

Answer to Problem 48P

The temperature at the end of the cycle is $\underset{\_}{T_i}$.

Explanation of Solution

In this case, starting point is $A$, after one whole cycle $T_A$ is equal to $T_i$.

Write the expression for the temperature at the end of the cycle,

    $T_A=T_i$        (III)

Conclusion:

Therefore, the temperature at the end of the cycle is $\underset{\_}{T_i}$.

To determine

The net work done on the gas during the cycle.

Answer to Problem 48P

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

Explanation of Solution

Write the expression for the heat transferred during the cycle $AB$,

    $Q_{AB}=n{C}_V\Delta T$        (IV)

Here, $Q_{AB}$ is the heat transferred during the cycle $AB$, $n$ is the number of molecule, $C_V$ is the specific heat at constant volume and $\Delta T$ is the temperature change.

Substitute $\frac{R}{1-\gamma }$ for $C_V$ and $3T_i-T_i$ for $\Delta T$ in the above equation,

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

In an adiabatic process,

    $Q_{BC}=0$        (VI)

Here, $Q_{BC}$  is the heat transferred during the cycle $BC$.

Write the expression for the ideal gas law,

    $P_C{V}_C=nR{T}_C$        (VII)

Substitute $P_i$ for $P_C$  and $3^{1/\gamma }{V}_i$ for $V_C$ in the above equation,

    $\begin{array}{c}{P}_i\left(3^{1/\gamma }\right)V_i=nR{T}_C\\ \left(3^{1/\gamma }\right)nR{T}_i=nR{T}_C\end{array}$

    $T_C=\left(3^{1/\gamma }\right)T_i$        (VIII)

Write the expression for the heat transferred during the cycle $CA$ ,

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

Here, $Q_{CA}$ is the heat transferred during the cycle $CA$, $n$ is the number of molecule, $C_P$ is the specific heat at constant pressure and $\Delta T$ is the temperature change.

Substitute $\frac{R}{1-\gamma }$ for $C_P$ and $T_i-\left(3^{1/\gamma }\right)T_i$ for $\Delta T$ in the above equation,

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

Write the expression for the heat transferred for whole cycle,

    $Q_{ABCA}=Q_{AB}+Q_{BC}+Q_{CA}$        (XIII)

Here, $Q_{ABCA}$ is the heat transferred for whole cycle.

Substitute (V), (VI) and (XII) in (XIII),

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

Write the expression for the internal energy change in the whole cycle,

    $\begin{array}{c}\Delta E_{\mathrm{int},ABCA}=Q_{ABCA}+W_{ABCA}\\ 0=Q_{ABCA}+W_{ABCA}\end{array}$

Write the expression for the net work done on the gas during the cycle,

    $W_{ABCA}=-Q_{ABCA}$

Conclusion:

Substitute $P_i{V}_i\left[\left(1-\left(3^{1/\gamma }\right)+\left(\frac{3-3^{1/\gamma }}{\gamma -1}\right)\right)\right]$ for $Q_{ABCA}$ in the above equation to find $W_{ABCA}$,

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

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