Question

Figure 20-27 shows a reversible cycle through which 1.00 molof a monatomic ideal gas is taken. Assume that$\mathbf{p}\mathbf{=}\mathbf{2}{\mathbf{p}}_{\mathbf{0}}\mathbf{,}\mathbf{}\mathbf{V}\mathbf{=}\mathbf{2}{\mathbf{V}}_{\mathbf{0}}\mathbf{,}{\mathbf{p}}_{\mathbf{0}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{Pa}$, and${\mathbf{V}}_{\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0225}\mathbf{}{\mathbf{m}}^{\mathbf{3}}$. (a) Calculate the work done during the cycle, (b) Calculate the energy added as heat during stroke abc, and (c) Calculate the efficiency of the cycle. (d) What is the efficiency of a Carnot engine operating between the highest and lowest temperatures that occur in the cycle? (e) Is this greater than or less than the efficiency calculated in (c)?

Short answer

a) Work done during the cycle is$2.27\times {10}^3\mathrm{J}$

b) The energy added as heat during stroke abc is$14.8\times {10}^3\mathrm{J}$

c) The efficiency of the cycle is 15.4 %.

d) The efficiency of the Carnot engine operating between the highest and lowest temperatures that occur in the cycle is 75.0 %.

e) The efficiency calculated in part d) is greater than that in part c).

Step-by-step solution

The given data

No. of moles of a monatomic gas is, n = 1.00 mol

$$\begin{gathered} p_0=1.01\times {10}^5\mathrm{Pa} \\ {\mathrm{V}}_0=0.0225{\mathrm{m}}^3 \\ \mathrm{p}=2{\mathrm{p}}_0,=2{\mathrm{V}}_0 \end{gathered}$$

Understanding the concept of work done per cycle

We can find work done during the cycle by calculating the area under the curve. The energy added as heat during stroke abc can be calculated by adding heat during path ab and bc got from specific heats. The efficiency of the cycle can be calculated from the above two, using the corresponding relation. Then we can find the efficiency of the Carnot engine using the corresponding formula from temperatures. We can get the answer for part e) by comparing the answers of parts c) and d).

Formulae:

The efficiency of the Carnot engine,

$\epsilon =\frac{W}{Q_H}or\epsilon =\frac{T_H-T_L}{T_H}$ (1)

The work done per cycle of the gas,

$W=Q_H-Q_L$ (2)

The heat and temperature relation of a Carnot engine,

$\frac{Q_H}{Q_L}=\frac{T_H}{T_L}$ (3)

The energy absorbed by the gas,

$Q=nC∆T$ (4)

a) Calculation of the work done during the full cycle

Total work done during the cycle is calculated from the area under the curve.

$\mathrm{Area}\mathrm{of}\mathrm{rectangle}\mathrm{abcd}=\mathrm{ab}\times \mathrm{bc}$

Thus, the work done during the full cycle using the given values can be given as:

$$\begin{gathered} \mathrm{W}=\left(\mathrm{V}-{\mathrm{V}}_0\right)\left(\mathrm{p}-{\mathrm{p}}_0\right) \\ =\left(2\mathrm{V}-{\mathrm{V}}_0\right)\left(2\mathrm{p}-{\mathrm{p}}_0\right) \\ ={\mathrm{V}}_0{\mathrm{p}}_0 \\ =0.0225{\mathrm{m}}^3\times 1.01\times {10}^5\mathrm{Pa} \\ =2.27\times {10}^3\mathrm{J} \end{gathered}$$

Therefore, work done during the cycle is$2.27\times {10}^3\mathrm{J}$

(b) Calculation of the energy added as heat during path abc

The energy added as heat during stroke abc using equation (4) can be given as considering,$\left(\mathrm{ab}=\mathrm{constant}\mathrm{pressure}\mathrm{and}\mathrm{bc}=\mathrm{constant}\mathrm{volume}\right)$ such that

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{abc}}={\mathrm{Q}}_{\mathrm{ab}}+{\mathrm{Q}}_{\mathrm{bc}} \\ ={\mathrm{nC}}_{\mathrm{P}}∆{\mathrm{T}}_{\mathrm{ab}}+{\mathrm{nC}}_{\mathrm{v}}∆{\mathrm{T}}_{\mathrm{bc}} \end{gathered}$$

For a monoatomic gas,

$$\begin{gathered} {\mathrm{C}}_{\mathrm{v}}=\frac{3}{2}\mathrm{R}\mathrm{and}{\mathrm{C}}_{\mathrm{P}}=\frac{5}{2}\mathrm{R} \\ {\mathrm{Q}}_{\mathrm{abc}}=\left(1\right)\frac{3}{2}\mathrm{R}\left({\mathrm{T}}_{\mathrm{b}}-{\mathrm{T}}_{\mathrm{a}}\right)+\left(1\right)\frac{5}{2}\mathrm{R}\left({\mathrm{T}}_{\mathrm{c}}-{\mathrm{T}}_{\mathrm{b}}\right) \\ =\left(1\right)\frac{3}{2}{\mathrm{RT}}_{\mathrm{a}}\left(\frac{{\mathrm{T}}_{\mathrm{b}}}{{\mathrm{T}}_{\mathrm{a}}}-1\right)+\left(1\right)\left(1\right)\frac{5}{2}{\mathrm{RT}}_{\mathrm{a}}\left(\frac{{\mathrm{T}}_{\mathrm{c}}}{{\mathrm{T}}_{\mathrm{a}}}-\frac{{\mathrm{T}}_{\mathrm{b}}}{{\mathrm{T}}_{\mathrm{a}}}\right) \\ \frac{3}{2}{\mathrm{RT}}_{\mathrm{a}}\left(\frac{{\mathrm{T}}_{\mathrm{b}}}{{\mathrm{T}}_{\mathrm{a}}}-1\right)+\frac{5}{2}{\mathrm{RT}}_{\mathrm{a}}\left(\frac{{\mathrm{T}}_{\mathrm{c}}}{{\mathrm{T}}_{\mathrm{a}}}-\frac{{\mathrm{T}}_{\mathrm{b}}}{{\mathrm{T}}_{\mathrm{a}}}\right) \end{gathered}$$

During path ab, the volume is constant. Hence, Boyle’s law gives

$\mathrm{p}\propto \mathrm{T}$

Hence, the ratio of the temperatures can be given as:

role="math" $$\begin{gathered} \frac{{\mathrm{p}}_{\mathrm{a}}}{{\mathrm{p}}_{\mathrm{b}}}=\frac{{\mathrm{T}}_{\mathrm{a}}}{{\mathrm{T}}_{\mathrm{b}}} \\ \frac{{\mathrm{T}}_{\mathrm{a}}}{{\mathrm{T}}_{\mathrm{b}}}=\frac{{\mathrm{p}}_0}{\mathrm{p}} \\ \frac{{\mathrm{T}}_{\mathrm{a}}}{{\mathrm{T}}_{\mathrm{b}}}=\frac{1}{2} \\ \frac{{\mathrm{T}}_{\mathrm{b}}}{{\mathrm{T}}_{\mathrm{a}}}=2 \end{gathered}$$

During path bc, the pressure is constant. Hence, Charles’s law gives.

$\mathrm{V}\propto \mathrm{T}$

Hence, the ratio of the temperatures can be given as:

$$\begin{gathered} \frac{{\mathrm{V}}_{\mathrm{b}}}{{\mathrm{V}}_{\mathrm{c}}}=\frac{{\mathrm{T}}_{\mathrm{b}}}{{\mathrm{T}}_{\mathrm{c}}} \\ \frac{{\mathrm{T}}_{\mathrm{b}}}{{\mathrm{T}}_{\mathrm{c}}}=\frac{{\mathrm{V}}_0}{\mathrm{V}} \\ \frac{{\mathrm{T}}_{\mathrm{b}}}{{\mathrm{T}}_{\mathrm{c}}}=\frac{1}{2} \end{gathered}$$

Further,

$$\begin{gathered} \frac{{\mathrm{T}}_{\mathrm{a}}}{{\mathrm{T}}_{\mathrm{c}}}=\frac{{\mathrm{T}}_{\mathrm{a}}}{{\mathrm{T}}_{\mathrm{b}}}\times \frac{{\mathrm{T}}_{\mathrm{b}}}{{\mathrm{T}}_{\mathrm{c}}} \\ =\frac{1}{2}\times \frac{1}{2} \\ =\frac{1}{4} \end{gathered}$$

Hence, substituting the above values, we can get the energy transferred as heat as given:

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{abc}}=\frac{3}{2}{\mathrm{RT}}_{\mathrm{a}}\left(2-1\right)+\frac{5}{2}{\mathrm{RT}}_{\mathrm{a}}\left(4-2\right) \\ =\frac{3}{2}{\mathrm{p}}_0{\mathrm{V}}_0\left(2-1\right)+\frac{5}{2}{\mathrm{p}}_0{\mathrm{V}}_0\left(4-2\right)\left({\mathrm{RT}}_{\mathrm{a}}={\mathrm{p}}_0{\mathrm{V}}_0\right) \\ =\frac{13}{2}{\mathrm{p}}_0{\mathrm{V}}_0 \\ =\frac{13}{2}\left(0.0225{\mathrm{m}}^3\times 1.01\times {10}^5\mathrm{Pa}\right) \\ =14.8\times {10}^3\mathrm{J} \end{gathered}$$

Hence, the value of the energy added as heat is$14.8\times {10}^3\mathrm{J}$

c) Calculation of the efficiency of the engine

The efficiency of the cycle using equation (1) is given as:

$$\begin{gathered} \mathrm{\epsilon }=\left(\frac{22.27\times {10}^3\mathrm{J}}{14.76\times {10}^3\mathrm{J}}\right)\times 100\% \\ =\left(0.154\right)\times 100\% \\ =15.4\% \end{gathered}$$

Therefore, the efficiency of the cycle is 15.4%.

d) Calculation of the efficiency of the Carnot engine between temperatures

The efficiency of the Carnot engine using equation (1) is given as:

$$\begin{gathered} \mathrm{\epsilon }=\frac{{\mathrm{T}}_{\mathrm{c}}-{\mathrm{T}}_{\mathrm{a}}}{{\mathrm{T}}_{\mathrm{c}}} \\ =1-\frac{{\mathrm{T}}_{\mathrm{a}}}{{\mathrm{T}}_{\mathrm{c}}} \\ =1-\frac{1}{4} \\ =\frac{3}{4} \\ =0.75 \\ =75.0\% \end{gathered}$$

Therefore, the efficiency of the Carnot engine is 75.0 %

e) Comparing efficiencies of the above parts that is (c) and (d)

From calculations of parts (c) and (d), we can conclude that the efficiency calculated in part (d) is greater than that in part (c).