Chapter 18, Problem 77P

(a)

To determine

The P-V diagram.

Answer to Problem 77P

The P-V diagram is shown in figure 1.

Explanation of Solution

Calculation:

Consider the given data, the pressure versus volume graph for the process is shown below.

  

Figure 1

Conclusion:

Therefore, the P-V diagram is shown in figure 1.

(b)

To determine

The proof that $Q_{\text{h}}=nR{T}_{\text{h}}\ln \left(\frac{V_2}{V_1}\right)$ .

Answer to Problem 77P

It is proved that $Q_{\text{h}}=nR{T}_{\text{h}}\ln \left(\frac{V_2}{V_1}\right)$ .

Explanation of Solution

Formula used:

The expression for heat absorbed is given by,

  $Q_{\text{h}}=\Delta E_{\mathrm{int},1\to 2}+W_{1\to 2}$

The expression for work done is given by,

  $W_{1\to 2}=nR{T}_{\text{h}}\ln \left(\frac{V_2}{V_1}\right)$

Calculation:

The heat absorbed is calculated as,

  $\begin{array}{c}{Q}_{\text{h}}=\Delta E_{\mathrm{int},1\to 2}+W_{1\to 2}\\ =0+nR{T}_{\text{h}}\ln \left(\frac{V_2}{V_1}\right)\\ =nR{T}_{\text{h}}\ln \left(\frac{V_2}{V_1}\right)\end{array}$

Conclusion:

Therefore, it is proved that $Q_{\text{h}}=nR{T}_{\text{h}}\ln \left(\frac{V_2}{V_1}\right)$ .

(c)

To determine

The proof that $Q_{\text{c}}=nR{T}_{\text{c}}\ln \left(\frac{V_3}{V_4}\right)$ .

Answer to Problem 77P

It is proved that $Q_{\text{c}}=nR{T}_{\text{c}}\ln \left(\frac{V_3}{V_4}\right)$ .

Explanation of Solution

Formula used:

The expression for heat absorbed is given by,

  $Q_{\text{c}}=\Delta E_{\mathrm{int},3\to 4}-W_{3\to 4}$

The expression for work done is given by,

  $W_{3\to 4}=nR{T}_{\text{c}}\ln \left(\frac{V_3}{V_4}\right)$

Calculation:

The heat absorbed is calculated as,

  $\begin{array}{c}{Q}_{\text{c}}=\Delta E_{\mathrm{int},3\to 4}-W_{3\to 4}\\ =0-nR{T}_{\text{c}}\ln \left(\frac{V_3}{V_4}\right)\\ =-nR{T}_{\text{c}}\ln \left(\frac{V_3}{V_4}\right)\end{array}$

Conclusion:

Therefore, it is proved that $Q_{\text{c}}=nR{T}_{\text{c}}\ln \left(\frac{V_3}{V_4}\right)$ .

(d)

To determine

The proof that $\frac{V_2}{V_1}=\frac{V_3}{V_4}$ .

Answer to Problem 77P

It is proved that $\frac{V_2}{V_1}=\frac{V_3}{V_4}$ .

Explanation of Solution

Formula used:

The expression for quasistatic adiabatic process at point 4 is given by,

  $T_{\text{c}}{V}_4^{\gamma -1}=T_{\text{h}}{V}_1^{\gamma -1}$

The expression for quasistatic adiabatic process at point 2 and 3 is given by,

  $T_{\text{h}}{V}_2^{\gamma -1}=T_{\text{c}}{V}_3^{\gamma -1}$

Calculation:

The expression for quasistatic adiabatic process at point 4 is given by,

  $\begin{array}{c}{T}_{\text{c}}{V}_4^{\gamma -1}=T_{\text{h}}{V}_1^{\gamma -1}\\ \frac{V_1}{V_4}={\left(\frac{T_{\text{c}}}{T_{\text{h}}}\right)}^{\frac{1}{\gamma -1}}\end{array}$

The expression for quasistatic adiabatic process at point 2 and 3 is given by,

  $\begin{array}{c}{T}_{\text{h}}{V}_2^{\gamma -1}=T_{\text{c}}{V}_3^{\gamma -1}\\ \frac{V_2}{V_3}={\left(\frac{T_{\text{c}}}{T_{\text{h}}}\right)}^{\frac{1}{\gamma -1}}\\ =\frac{V_1}{V_4}\end{array}$

Conclusion:

Therefore, it is proved that $\frac{V_2}{V_1}=\frac{V_3}{V_4}$ .

(e)

To determine

The proof that efficiency is $1-\frac{Q_{\text{c}}}{Q_{\text{h}}}$ .

Answer to Problem 77P

It is proved that efficiency is $1-\frac{Q_{\text{c}}}{Q_{\text{h}}}$ .

Explanation of Solution

Formula used:

The expression for efficiency is given by,

  $\epsilon =\frac{-W_{\text{on}}}{Q_{\text{h}}}$

The expression for work done is given by,

  $W_{\text{on}}=-\left(Q_{\text{h}}-Q_{\text{c}}\right)$

Calculation:

The efficiency is calculated as,

  $\begin{array}{c}\epsilon =\frac{-\left(-\left(Q_{\text{h}}-Q_{\text{c}}\right)\right)}{Q_{\text{h}}}\\ =1-\frac{Q_{\text{c}}}{Q_{\text{h}}}\end{array}$

Conclusion:

Therefore, it is proved that efficiency is $1-\frac{Q_{\text{c}}}{Q_{\text{h}}}$ .

(f)

To determine

The proof that $\frac{Q_{\text{c}}}{Q_{\text{h}}}=\frac{T_{\text{c}}}{T_{\text{h}}}$ .

Answer to Problem 77P

It is proved that $\frac{Q_{\text{c}}}{Q_{\text{h}}}=\frac{T_{\text{c}}}{T_{\text{h}}}$ .

Explanation of Solution

Formula used:

The expression for heat is given by,

  $Q_{\text{h}}=nR{T}_{\text{h}}\ln \frac{V_2}{V_1}$

Calculation:

The ratio of $Q_{\text{c}}$ and $Q_{\text{h}}$ is calculated as,

  $\begin{array}{c}\frac{Q_{\text{c}}}{Q_{\text{h}}}=\frac{nR{T}_{\text{c}}\ln \frac{V_3}{V_4}}{nR{T}_{\text{h}}\ln \frac{V_2}{V_1}}\\ =\frac{T_{\text{c}}}{T_{\text{h}}}\left(\because \frac{V_3}{V_4}=\frac{V_2}{V_1}\right)\end{array}$

Conclusion:

Therefore, it is proved that $\frac{Q_{\text{c}}}{Q_{\text{h}}}=\frac{T_{\text{c}}}{T_{\text{h}}}$ .