Chapter 21, Problem 37AP

A 1.00-mol sample of an ideal monatomic gas is taken through the cycle shown in Figure P21.37. The process A → B is a reversible isothermal expansion. Calculate (a) the net work done by the gas, (b) the energy added to the gas by heat, (c) the energy exhausted from the gas by heat, and (d) the efficiency of the cycle. (e) Explain how the efficiency compares with that of a Carnot engine operating between the same temperature extremes.

Figure P21.37

To determine

The net work done by the gas.

Answer to Problem 37AP

The net work done by the gas is $4.10\times {10}^3\text{J}$.

Explanation of Solution

Given Info: The pressure and volume at A are $5\text{atm}$ and $10\text{L}$ respectively, the pressure and volume at B are $1\text{atm}$ and $50\text{L}$ and the pressure and volume at C are $1\text{atm}$ and $10\text{L}$.

The formula for the net work done by the gas at $AB$ is,

$W_{\text{AB}}=-P_{\text{A}}{V}_{\text{A}}\ln \left(\frac{V_{\text{B}}}{V_{\text{A}}}\right)$

Here,

$P_{\text{A}}$ is the pressure at $A$.

$V_{\text{A}}$ is the volume at $A$.

$V_{\text{B}}$ is the volume at $B$.

Substitute $5\text{atm}$ for $P_{\text{A}}$, $10\text{L}$ for $V_{\text{A}}$ and $50\text{L}$ for $V_{\text{B}}$ in above equation to find the $W_{\text{AB}}$.

$\begin{array}{c}{W}_{\text{AB}}=-5\text{atm}\cdot \left(\frac{1.013\times {10}^5\text{Pa}}{1\text{atm}}\right)\cdot 10\text{L}\cdot \left(\frac{{10}^{-3}{\text{m}}^3}{1\text{L}}\right)\ln \left(\frac{50}{10}\right)\\ =-8.15\times {10}^3\text{J}\end{array}$

Thus, the net work done by the gas at $AB$ is $-8.15\times {10}^3\text{J}$.

The formula for the net work done by the gas at $BC$ is,

$\begin{array}{c}{W}_{\text{BC}}=-P_{\text{B}}\Delta V\\ =-P_{\text{B}}\left(V_C-V_{\text{B}}\right)\end{array}$

Here,

$P_{\text{A}}$ is the pressure at A.

$V_{\text{C}}$ is the volume at A.

$V_{\text{B}}$ is the volume at B.

Substitute $1\text{atm}$ for $P_{\text{B}}$, $10\text{L}$ for $V_{\text{C}}$ and $50\text{L}$ for $V_{\text{B}}$ in above equation to find the $W_{\text{BC}}$.

$\begin{array}{c}{W}_{\text{BC}}=-1\text{atm}\cdot \left(\frac{1.013\times {10}^5\text{Pa}}{1\text{atm}}\right)\cdot \left(10-50\right)\text{L}\cdot \left(\frac{{10}^{-3}{\text{m}}^3}{1\text{L}}\right)\\ =4.05\times {10}^3\text{J}\end{array}$

Thus, the net work done by the gas at BC is $4.05\times {10}^3\text{J}$.

The formula for the net work done by the gas at $CA$ is,

$W_{\text{CA}}=0\text{J}$

As the change in the volume between processes $CA$ is zero, the work done is zero.

Thus, the net work done by the gas at $CA$ is $0\text{J}$.

The formula for the total work done by the gas is,

$W_{\text{T}}=W_{\text{AB}}+W_{\text{BC}}+W_{\text{CA}}$

Substitute $-8.15\times {10}^3\text{J}$ for $W_{\text{AB}}$, $4.05\times {10}^3\text{J}$ for $W_{\text{BC}}$ and $0\text{J}$ for $W_{\text{CA}}$ in above equation to find the $W_{\text{T}}$.

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

Conclusion:

Therefore, the total work done is $4.10\times {10}^3\text{J}$.

To determine

The energy added to the gas by heat.

Answer to Problem 37AP

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

Explanation of Solution

Given Info: The pressure and volume at $A$ are $5\text{atm}$ and $10\text{L}$ respectively, the pressure and volume at $B$ are $1\text{atm}$ and $50\text{L}$ and the pressure and volume at $C$ are $1\text{atm}$ and $10\text{L}$.

The formula for the energy added by the gas in process $AB$ is,

$Q_{\text{AB}}=-W_{\text{AB}}$

Substitute $-8.15\times {10}^3\text{J}$ for $W_{\text{AB}}$ in above equation to find the $Q_{\text{AB}}$.

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

Thus, the energy added by the gas in process $AB$ is $8.15\times {10}^3\text{J}$.

The formula for temperature at $A$ and $B$ are,

$\begin{array}{l}{T}_{\text{A}}=T_{\text{B}}=\frac{P_{\text{B}}{V}_{\text{B}}}{nR}\\ \end{array}$

Substitute $1\text{atm}$ for $P_{\text{B}}$, $1\text{mole}$ for $n$ and $50\text{L}$ for $V_{\text{B}}$ in above equation to find the $T_{\text{A}}$.

$\begin{array}{c}{T}_{\text{A}}=T_{\text{B}}=\frac{1\text{atm}\left(\frac{1.013\times {10}^5\text{Pa}}{1\text{atm}}\right)\times 50\text{L}\times \left(\frac{1{\text{m}}^3}{1\text{L}}\right)}{1\times R}\\ =\frac{5.05\times {10}^3}{R}\text{K}\end{array}$

Thus, the temperature at $A$ and $B$ is $\frac{5.05\times {10}^3}{R}\text{K}$.

The formula for temperature at $C$ is,

$\begin{array}{l}{T}_{\text{C}}=\frac{P_{\text{C}}{V}_{\text{C}}}{nR}\\ \end{array}$

Substitute $1\text{atm}$ for $P_{\text{C}}$, $1\text{mole}$ for $n$ and $10\text{L}$ for $V_{\text{C}}$ in above equation to find the $T_{\text{C}}$.

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

Thus, the temperature at $C$ is $\frac{1.01\times {10}^3}{R}\text{K}$.

The formula for the energy added by the gas in process $CA$ is,

$Q_{\text{CA}}=n{C}_{\text{V}}\left(T_{\text{A}}-T_{\text{C}}\right)$

Here,

$C_{\text{V}}$ is the specific heat at constant volume.

Substitute $\frac{1.01\times {10}^3}{R}\text{K}$ for $T_{\text{C}}$, $\frac{5.05\times {10}^3}{R}\text{K}$ for $T_{\text{A}}$ and $\frac{3}{2}R$ for $C_{\text{V}}$ in above equation to find the $Q_{\text{CA}}$.

$\begin{array}{c}{Q}_{\text{CA}}=\left(1\text{mol}\right)\left(\frac{3}{2}R\right)\left(\frac{5.05\times {10}^3}{R}\text{K}-\frac{1.01\times {10}^3}{R}\text{K}\right)\\ =6.08\times {10}^3\text{J}\end{array}$

Thus, the energy added in process $CA$ is $6.08\times {10}^3\text{J}$.

The formula for the total energy added is,

$Q_{\text{T}}=Q_{\text{CA}}+Q_{\text{AB}}$

Substitute $8.15\times {10}^3\text{J}$ for $Q_{\text{AB}}$ and $6.08\times {10}^3\text{J}$ for $Q_{\text{CA}}$ in above equation to find $Q_{\text{T}}$.

$\begin{array}{c}{Q}_{\text{T}}=6.08\times {10}^3\text{J}+8.15\times {10}^3\text{J}\\ =1.42\times {10}^4\text{J}\end{array}$

Thus, the total energy added is $1.42\times {10}^4\text{J}$.

Conclusion:

Therefore, the total energy added is $1.42\times {10}^4\text{J}$.

To determine

The energy exhausted from the gas by heat.

Answer to Problem 37AP

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

Explanation of Solution

Given Info: The pressure and volume at A are $5\text{atm}$ and $10\text{L}$ respectively, the pressure and volume at B are $1\text{atm}$ and $50\text{L}$ and the pressure and volume at C are $1\text{atm}$ and $10\text{L}$.

The formula for the energy added by the gas in process $BC$ is,

$Q_{\text{BC}}=\frac{5}{2}{P}_{\text{B}}\left(V_{\text{C}}-V_{\text{B}}\right)$

Substitute $1\text{atm}$ for $P_{\text{B}}$, $50\text{L}$ for $V_{\text{B}}$ and $50\text{L}$ for $V_{\text{C}}$ in above equation to find $Q_{\text{BC}}$.

$\begin{array}{c}{Q}_{\text{BC}}=\frac{5}{2}\left(1\text{atm}\right)\left(\frac{1.013\times {10}^5\text{Pa}}{1\text{atm}}\right)\left(10-50\right)\text{L}\left(\frac{{10}^{-3}{\text{m}}^3}{1\text{L}}\right)\\ \text{=-1}\text{.01}\times {\text{10}}^4\text{J}\end{array}$

Thus, the energy exhausted in process $BC$ is $1.01\times {10}^4\text{J}$.

Conclusion:

Therefore, the energy exhausted in process $BC$ is $1.01\times {10}^4\text{J}$.

To determine

The efficiency of the cycle.

Answer to Problem 37AP

The efficiency of the cycle is $28.8\%$.

Explanation of Solution

Given Info: The pressure and volume at A are $5\text{atm}$ and $10\text{L}$ respectively, the pressure and volume at B are $1\text{atm}$ and $50\text{L}$ and the pressure and volume at C are $1\text{atm}$ and $10\text{L}$.

Formula to calculate the efficiency of the engine is,

$e=\frac{W}{Q_{\text{T}}}$

Here,

$e$ is the efficiency of engine .

$W$ is the work done by the engine.

Substitute $4.10\times {10}^3\text{J}$ for $W$ and $1.42\times {10}^4\text{J}$ for $Q_{\text{T}}$ in above equation to find the $e$.

$\begin{array}{c}e=\frac{4.10\times {10}^3\text{J}}{1.42\times {10}^4\text{J}}\\ =0.288\left(\frac{100\%}{1}\right)\\ =28.8\%\end{array}$

Thus, the efficiency of the engine is $28.8\%$.

Conclusion:

Therefore, the efficiency of the engine is $28.8\%$.

To determine

The comparison of efficiency to that of the Carnot engine.

Answer to Problem 37AP

The efficiency of Carnot engine is $80.0\%$, the efficiency of this cycle is much less than that of the Carnot cycle operating within same temperature range.

Explanation of Solution

Given Info: The pressure and volume at A are $5\text{atm}$ and $10\text{L}$ respectively, the pressure and volume at B are $1\text{atm}$ and $50\text{L}$ and the pressure and volume at C are $1\text{atm}$ and $10\text{L}$.

Formula to calculate the efficiency of the Carnot engine is,

$e=1-\frac{T_{\text{C}}}{T_{\text{A}}}$

Substitute $\frac{1.01\times {10}^3}{R}\text{K}$ for $T_{\text{C}}$, $\frac{5.05\times {10}^3}{R}\text{K}$ for $T_{\text{A}}$ in above equation to find the $e$

$\begin{array}{c}e=1-\frac{\frac{1.01\times {10}^3}{R}\text{K}}{\frac{5.05\times {10}^3}{R}\text{K}}\\ \text{=1-0}\text{.2}\\ \text{=0}\text{.8}\left(\frac{100\%}{1}\right)\\ =80\%\end{array}$

Conclusion:

Therefore, the efficiency of Carnot engine is $80.0\%$, the efficiency of this cycle is much less than that of the Carnot cycle operating within same temperature range