Chapter 17, Problem 47P

(a)

To determine

Draw the $PV$ diagram of the cycle.

Answer to Problem 47P

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.

The Figure 1 shown the $PV$ diagram of the cycle.

Conclusion:

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

(b)

To determine

Thevolume of the gas at the end of the adiabatic expansion.

Answer to Problem 47P

The volume of the gas at the end of the adiabatic expansion is $\underset{\_}{8.77\text{L}}$.

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.

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

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

Rewrite the above equation for $V_C$ ,

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

Conclusion:

Substitute $4.00\text{L}$ for $V_i$ and $7/5$ for $\gamma$ in (III) to find $V_C$,

    $\begin{array}{c}{V}_C=\left(3^{5/7}\right)\left(4.00\text{L}\right)\\ =2.19\left(\text{4}\text{.00}\text{L}\right)\\ =8.77\text{L}\end{array}$

Therefore, the volume of the gas at the end of the adiabatic expansion is $\underset{\_}{8.77\text{L}}$.

(c)

To determine

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

Answer to Problem 47P

Thetemperature of the gas at the start of the expansion is $\underset{\_}{900\text{K}}$.

Explanation of Solution

Write the expression for the ideal gas law,

    $P_B{V}_B=nR{T}_B$        (IV)

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

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

    $T_B=3T_i$        (V)

Conclusion:

Substitute $300\text{K}$ for $T_i$ in the above equation to find $T_B$ ,

    $\begin{array}{c}{T}_B=3\left(300\text{K}\right)\\ =900\text{K}\end{array}$

Therefore, the temperature of the gas at the start of the expansion is $\underset{\_}{900\text{K}}$.

(d)

To determine

The temperature at the end of the cycle.

Answer to Problem 47P

Thetemperature at the end of the cycle is $\underset{\_}{300\text{K}}$.

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,

    $T_A=T_i$        (VI)

Conclusion:

Substitute $300\text{K}$ for $T_i$ in (VI) to find $T_A$,

    $T_A=300\text{K}$

Therefore, the temperature at the end of the cycle is $\underset{\_}{300\text{K}}$.

(e)

To determine

The net work done on the gas during the cycle.

Answer to Problem 47P

Thenet work done on the gas during the cycle is $\underset{\_}{-336\text{J}}$ .

Explanation of Solution

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

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

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 $\frac{5}{2}R$ for $C_V$ and $3T_i-T_i$ for $\Delta T$ in the above equation,

    $\begin{array}{c}{Q}_{AB}=n\left(\frac{5}{2}R\right)\left(3T_i-T_i\right)\\ =5nR{T}_i\end{array}$        (VIII)

In an adiabatic process, $Q_{BC}=0$.

Write the expression for the ideal gas law,

    $P_C{V}_C=nR{T}_C$        (IX)

Substitute $P_i$ for $P_C$  and $2.19V_i$ for $V_C$ in the above equation,

    $\begin{array}{c}{P}_i\left(2.19V_i\right)=nR{T}_C\\ 2.19nR{T}_i=nR{T}_C\\ T_C=2.19T_i\end{array}$        (X)

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

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

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 $\frac{7}{2}R$ for $C_P$ and $T_i-2.19T_i$ for $\Delta T$ in the above equation,

    $\begin{array}{c}{Q}_{CA}=n\left(\frac{7}{2}R\right)\left(T_i-2.19T_i\right)\\ =-4.17nR{T}_i\end{array}$        (XII)

Write the expression for the heat transferredfor whole cycle,

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

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

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,

    $W_{ABCA}=-Q_{ABCA}$

    $\begin{array}{c}{W}_{ABCA}=-\left(0.829\right)nR{T}_i\\ =-\left(0.829\right)P_i{V}_i\end{array}$

Conclusion:

Substitute $4.00\times {10}^{-3}{\text{m}}^3$ for $V_i$ and $1.013\times {10}^5\text{Pa}$ for $P_i$ in the above equation to find $W_{ABCA}$,

    $\begin{array}{c}{W}_{ABCA}=-\left(0.829\right)\left(1.013\times {10}^5\text{Pa}\right)\left(4.00\times {10}^{-3}{\text{m}}^3\right)\\ =-336\text{J}\end{array}$

Therefore, the net work done on the gas during the cycle is $\underset{\_}{-336\text{J}}$.