Chapter 22, Problem 73AP

(a)

To determine

The net work done by the gas.

Answer to Problem 73AP

The net work done by the gas is $4.10\text{kJ}$.

Explanation of Solution

In the figure P22.73, the curve $AB$ represents isothermal expansion, The curve $BC$ represents isobaric compression since volume of the gas decreases while pressure remains the same and the curve $CA$ represents an isochoric process.

Write the expression for the work done on the gas during path $AB$.

  $W_{AB}=-P_A{V}_A\ln \left(\frac{V_B}{V_A}\right)$                                                                                                 (I)

Here, $W_{AB}$ is the work done on the gas during the path $AB$ , $P_A$ is the pressure of the gas at point $A$, $V_A$ is the volume of the gas at point $A$ and $V_{B;}$ is the volume of the gas at point $B$.

Write the expression for the work done on the gas during path $BC$.

  $W_{BC}=-P_B\left(V_C-V_B\right)$                                                                                                (II)

Here, $W_{BC}$ is the work done by the gas during path BC, $P_B$ is the pressure at point $B$ and $V_C$ is the volume of the gas at point $C$.

Since the process along $CA$ is isochoric, work done along that path is zero.

  $W_{CA}=0$                                                                                                                    (III)

Here, $W_{CA}$ is the work done on the gas during isochoric process.

Write the expression for the total work done on the gas.

  $W=-\left(W_{AB}+W_{BC}+W_{CA}\right)$                                                                                        (IV)

Conclusion:

From figure, $P_A$ is $5\text{atm}$, $P_B$ is $1\text{atm}$ , $V_A$ is $10\text{liters}$, $V_B$ is $50\text{liters}$ and $V_C$ is $10\text{liters}$.

Substitute $5\text{atm}$ for $P_A$, $10.0\text{L}$ for $V_A$ and $50.0\text{L}$ for $V_B$ in equation (I) to get $W_{AB}$.

  $\begin{array}{c}{W}_{AB}=-\left(5\text{atm}\right)\times \frac{1.013\times {10}^5\text{Pa}}{1\text{atm}}\left(10.0\text{L}\frac{1{\text{ m}}^3}{{10}^3\text{ L}}\right)\ln \left(\frac{50.0\text{L}}{10.0\text{L}}\right)\\ =-8.15\times {10}^3\text{J}\end{array}$

Substitute $1\text{atm}$ for $P_B$, $10.0\text{L}$ for $V_C$ and $50.0\text{L}$ for $V_B$ in equation (II) to get $W_{BC}$.

  $\begin{array}{c}{W}_{BC}=-\left(1\text{atm}\frac{1.013\times {10}^5\text{Pa}}{1\text{atm}}\right)\left[\left(10.0\text{L}-50.0\text{L}\right)\times \frac{1{\text{ m}}^3}{{10}^3\text{ L}}\right]\\ =+4.05\times {10}^3\text{J}\end{array}$

Substitute $-8.15\times {10}^3\text{J}$ for $W_{AB}$, $+4.05\times {10}^3\text{J}$ for $W_{BC}$ and $0\text{J}$ for $W_{CA}$ in equation (IV) to get total work done on the gas.

  $\begin{array}{c}W=-\left(-8.15\times {10}^3\text{J}+4.05\times {10}^3\text{J}+0\text{J}\right)\\ =4.10\times {\text{10}}^3\text{J}\times \frac{1\text{kJ}}{1000\text{J}}\\ =4.10\text{kJ}\end{array}$

Therefore, the net work done by the gas is $4.10\text{kJ}$.

(b)

To determine

The energy added to the gas by heat.

Answer to Problem 73AP

The energy added to the gas by heat is $1.42\times {10}^4\text{J}$.

Explanation of Solution

In figure $AB$ is a isothermal process so that change in internal energy during the process will be zero, $\Delta E_{\text{int,}AB}=0$.

Write the expression for the heat absorbed by the gas in isothermal process.

  $Q_{AB}=\Delta E_{\text{int,}AB}-W_{AB}$                                                                                                  (V)

Here, $Q_{AB}$ is the heat absorbed by the gas during process $AB$, $\Delta E_{\text{int,}AB}$ is the change in internal energy during $AB$ and $W_{AB}$ is the work done the gas.

Substitute $0$ for $\Delta E_{\text{int,}AB}$ in above equation to get $Q_{AB}$.

  $\begin{array}{c}{Q}_{AB}=0-W_{AB}\\ =-W_{AB}\end{array}$                                                                                                          (VI)

Write the expression for the specific heat capacity at constant volume of monoatomic gas.

  $C_V=\frac{3R}{2}$                                                                                                               (VII)

Here, $C_V$ is the specific heat capacity at constant volume and $R$ is the universal gas constant.

Write the expression for the specific heat capacity at constant pressure of monoatomic gas.

  $C_P=\frac{5R}{2}$                                                                                                              (VIII)

Here, $C_P$ is the specific heat capacity at constant pressure.

Apply ideal gas equation at point $B$ of the cycle.

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

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

Rearrange above equation to get $T_B$.

  $T_B=\frac{P_B{V}_B}{nR}$                                                                                                                (X)

Since $AB$ is isothermal process, temperatures at points A and B will be equal.

  $T_A=T_B$

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

Substitute $T_A$ for $T_B$ in above equation to get $T_A$.

  $T_A=\frac{P_B{V}_B}{nR}$                                                                                                               (XI)

Apply ideal gas equation at point $B$ of the cycle to get $T_C$.

  $\begin{array}{c}{P}_C{V}_C=nR{T}_C\\ T_C=\frac{P_C{V}_C}{nR}\end{array}$                                                                                                          (XII)

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

$BC$ in the cycle is isobaric compression and in this process, heat is liberated from the gas.

Write the expression for the energy absorbed by heat during $CA$.

  $Q_{CA}=n{C}_V\left(T_A-T_C\right)$ (XIII)

Here, $Q_{CA}$ is the energy absorbed by heat during $CA$.

Write the expression for the total energy absorbed by heat.

  $Q_{\text{tot\_abs}}=Q_{AB}+Q_{CA}$                                                                                              (XIV)

Here, $Q_{\text{tot\_abs}}$ is the total energy absorbed by heat.

Conclusion:

Substitute $-8.15\times {10}^3\text{J}$ for $W_{AB}$ in equation (VI) to get $Q_{AB}$.

  $\begin{array}{c}{Q}_{AB}=-\left(-8.15\times {10}^3\text{J}\right)\\ =8.15\times {10}^3\text{J}\times \frac{1\text{kJ}}{1000\text{J}}\\ =8.15\text{kJ}\end{array}$

Substitute $1\text{atm}$ for $P_B$ , $50.0\text{L}$ for $V_B$ and $1.00\text{ mol}$ for $n$ in equation (X) to get $T_B$.

  $\begin{array}{c}{T}_B=\frac{\left(1\text{atm}\right)\times \frac{1.013\times {10}^5\text{Pa}}{1\text{atm}}\times \left(50.0\text{L}\right)\times \frac{1{\text{m}}^3}{{10}^3\text{L}}}{\left(1.00\text{ mol}\right)R}\\ =\frac{5.06\times {10}^3}{R}\end{array}$

Since $T_A=T_B$, $T_A=\frac{5.06\times {10}^3}{R}$.

Substitute $1\text{atm}$ for $P_C$ , $10.0\text{L}$ for $V_C$ and $1.00\text{ mol}$ for $n$ in equation (XII) to get $T_C$.

  $\begin{array}{c}{T}_C=\frac{\left(1\text{atm}\right)\times \frac{1.013\times {10}^5\text{Pa}}{1\text{atm}}\times \left(10.0\text{L}\right)\times \frac{1{\text{m}}^3}{{10}^3\text{L}}}{\left(1.00\text{ mol}\right)R}\\ =\frac{1.01\times {10}^3}{R}\end{array}$

Substitute $\frac{5.06\times {10}^3}{R}$ for $T_A$ , $\frac{1.01\times {10}^3}{R}$ for $T_C$ , and $\frac{3}{2}R$ for $C_V$ in equation (XIII) to get $Q_{CA}$.

  $\begin{array}{c}{Q}_{CA}=\left(1.00\text{ mol}\right)\left(\frac{3}{2}R\right)\left(\frac{5.06\times {10}^3}{R}-\frac{1.01\times {10}^3}{R}\right)\\ =6.08\times {10}^3\text{J}\times \frac{1\text{kJ}}{1000\text{J}}\\ =6.08\text{kJ}\end{array}$

Substitute $8.15\text{kJ}$ for $Q_{AB}$ and $6.08\text{kJ}$ for $Q_{CA}$ in equation (XIV) to get $Q_{\text{tot\_abs}}$.

  $\begin{array}{c}{Q}_{\text{tot\_abs}}=8.15\text{kJ}+6.08\text{kJ}\\ \text{=14}\text{.2}\text{kJ}\times \frac{1000\text{J}}{1\text{kJ}}\\ =1.42\times {10}^4\text{J}\end{array}$

Therefore, the energy added to the gas by heat is $1.42\times {10}^4\text{J}$.

(c)

To determine

The energy exhausted from the gas by heat.

Answer to Problem 73AP

The energy exhausted from the gas by heat is $1.01\times {10}^4\text{J}$.

Explanation of Solution

Write the expression for the energy exhausted from the gas by heat.

  $Q_{BC}=n{C}_P\left(T_C-T_B\right)$                                                                                             (XV)

Here, $Q_{BC}$ is the energy exhausted from the gas by heat during the process BC.

The specific heat capacity of the gas at constant pressure is $C_P=\frac{5}{2}R$.

Substitute $\frac{5}{2}R$ for $C_P$ in equation (XV) to get $Q_{BC}$.

  $Q_{BC}=\frac{5}{2}nR\left(T_C-T_B\right)$                                                                                          (XVI)

Apply ideal gas equation during the isobaric process $BC$.

  $P_B\left(V_C-V_B\right)=nR\left(T_C-T_B\right)$                                                                                (XVII)

Substitute (XVII) in equation (XVI) to get $Q_{BC}$.

  $Q_{BC}=\frac{5}{2}{P}_B\left(V_C-V_B\right)$                                                                                        (XVIII)

Conclusion:

Substitute $1.00\text{atm}$ for $P_B$, $50.0\text{L}$ for $V_B$ and $10.0\text{L}$ for $V_C$ in equation (XVIII) to get $Q_{BC}$.

  $\begin{array}{c}{Q}_{BC}=\frac{5}{2}\left(1.00\text{atm}\times \frac{1.013\times {10}^5\text{Pa}}{1\text{atm}}\right)\left(10.0\text{L}-50.0\text{L}\right)\left(\times \frac{{10}^{-3}}{1\text{L}}\right)\\ =-1.01\times {10}^4\text{J}\end{array}$

Then energy exhausted from the gas by heat is,

  $\begin{array}{c}\left|Q_{BC}\right|=\left|-1.01\times {10}^4\text{J}\right|\\ =1.01\times {10}^4\text{J}\end{array}$

Therefore, the energy exhausted from the gas by heat is $1.01\times {10}^4\text{J}$.

(d)

To determine

The efficiency of the cycle.

Answer to Problem 73AP

The efficiency of the cycle is $28.9\%$.

Explanation of Solution

Write the expression for the efficiency of the cycle.

  $e=\frac{W_{\text{eng}}}{\left|Q_h\right|}$                                                                                                              (XIX)

Here, $e$ is the efficiency of the cycle, $W_{\text{eng}}$ is the work done by the engine and $Q_h$ is the heat exhausted at the hot reservoir.

The total heat exhausted is equal to the sum of the heat liberated through the process described by the curves $Q_{AB}$ and $Q_{CA}$.

Write the expression for the total heat exhausted at hot reservoir.

  $\left|Q_h\right|=Q_{AB}+Q_{CA}$                                                                                                    (XX)

Substitute (XX) in (XIX) to get $e$.

  $e=\frac{W_{\text{eng}}}{\left|Q_{AB}+Q_{CA}\right|}$

Conclusion:

From part(b), $Q_{AB}+Q_{CA}=1.42\times {10}^4\text{J}$.

Substitute $1.42\times {10}^4\text{J}$ for $Q_{AB}+Q_{CA}$ and $4.10\times {10}^3\text{J}$ for $W$ in equation (XX) to get $e$.

  $\begin{array}{c}e=\frac{4.10\times {10}^3\text{J}}{1.42\times {10}^4\text{J}}\\ =0.289\end{array}$

Convert $0.289$ into percentage.

  $\begin{array}{c}e=0.289\times 100\\ 28.9\%\end{array}$

Therefore the efficiency of the cycle is $28.9\%$.

(e)

To determine

The comparison for the efficiency of the engine with efficiency of Carnot engine operating between same temperature extremes.

Answer to Problem 73AP

The efficiency of the cycle is much lower than that of a Carnot engine operating between the same temperature extremes.

Explanation of Solution

The temperature of the cold reservoir is equal to temperature at point $C$. The temperature of hot reservoir is the temperature at point $A$.

From part(a). $T_A=\frac{5.06\times {10}^3}{R}$ and $T_C=\frac{1.01\times {10}^3}{R}$.

Write the expression for the efficiency Carnot engine.

  $e_C=1-\frac{T_c}{T_h}$

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

Conclusion:

Substitute $\frac{1.01\times {10}^3}{R}$ for $T_c$ and $\frac{5.06\times {10}^3}{R}$ for $T_h$ in above equation to get $e_C$.

  $\begin{array}{c}{e}_C=1-\frac{\frac{1.01\times {10}^3}{R}}{\frac{5.06\times {10}^3}{R}}\\ =0.80\%\end{array}$

Efficiency of the cycle is only $28.9\%$.

Compared to efficiency of Carnot engine , efficiency of the cycle is much lower.

Therefore, the efficiency of the cycle is much lower than that of a Carnot engine operating between the same temperature extremes.