Chapter 5, Problem 5.27P

(a)

Interpretation Introduction

Interpretation :

Calculate entropy change of the gas in a steady state flow process at approximately atmospheric pressure.

Concept Introduction :

The general equation for the steady state flow entropy change as the function of temperature and pressure is defined as:

  $\begin{array}{l}dS=C_P\frac{dT}{T}-{\left(\frac{\partial V}{\partial T}\right)}_{P}dP\\ \text{at constant pressure}\end{array}$

  $\Delta S=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}\frac{dT}{T}}$......(1)

Answer to Problem 5.27P

  $\Delta S\text{=548}\text{.12}\frac{\text{J}}{\text{K}}$

Explanation of Solution

Given information:

It is given that

  $n=10\text{mol of }S{O}_2$

  $T_0={200}^{\circ }\text{C}+273.15=473.15\text{K}$

  $T={1100}^{\circ }\text{C}+273.15=1373.15\text{K}$

From equation (1)

  $\Delta S=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}\frac{dT}{T}}$

Where

  ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=\left[A+\left[B{T}_0+\left(C{T}_0{}^2+\frac{D}{{\tau }^2{T}_0{}^2}\right)\left(\frac{\tau +1}{2}\right)\right]\left(\frac{\tau -1}{\ln \tau }\right)\right]\ln \tau$......(2)

And

  $\tau =\frac{T}{T_0}$

The Values of above constants for $S{O}_2$ in equation (2) are given in appendix C table C.1 and noted down below:

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ {\text{SO}}_{2}5.6990.8010-1.015\end{array}$

Hence

  $\begin{array}{l}\tau =\frac{T}{T_0}\\ \tau =\frac{1373.15\text{K}}{473.15\text{K}}=2.902\end{array}$

and

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=\left[A+\left[B{T}_0+\left(C{T}_0{}^2+\frac{D}{{\tau }^2{T}_0{}^2}\right)\left(\frac{\tau +1}{2}\right)\right]\left(\frac{\tau -1}{\ln \tau }\right)\right]\ln \tau \\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=\left[5.699+\left[0.801\times {10}^{-3}\times 473.15+\left(0\times {473.15}^2+\frac{-1.015\times {10}^5}{{2.902}^2\times {473.15}^2}\right)\left(\frac{2.902+1}{2}\right)\right]\left(\frac{2.902-1}{\ln 2.902}\right)\right]\ln 2.902\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=6.59279\end{array}$

Therefore, entropy change is

  $\begin{array}{l}\Delta S=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}\frac{dT}{T}}\\ \Delta S=8.314\frac{\text{J}}{\text{mol K}}\times 6.59279\\ \\ \Delta S=54.8124\frac{\text{J}}{\text{mol K}}\end{array}$

For $n=10\text{mol of }S{O}_2$

  $\begin{array}{l}\Delta S=54.8124\frac{\text{J}}{\text{mol K}}\times 10\text{mol}\\ \Delta S\text{=548}\text{.12}\frac{\text{J}}{\text{K}}\end{array}$

(b)

Interpretation Introduction

Interpretation :

Calculate entropy change of the gas in a steady state flow process at approximately atmospheric pressure.

Concept Introduction :

The general equation for the steady state flow entropy change as the function of temperature and pressure is defined as:

  $\begin{array}{l}dS=C_P\frac{dT}{T}-{\left(\frac{\partial V}{\partial T}\right)}_{P}dP\\ \text{at constant pressure}\end{array}$

  $\Delta S=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}\frac{dT}{T}}$......(1)

Answer to Problem 5.27P

  $\Delta S\text{=2018}\text{.79}\frac{\text{J}}{\text{K}}$

Explanation of Solution

Given information:

It is given that

  $n=12\text{mol of Propane}$

  $T_0={250}^{\circ }\text{C}+273.15=523.15\text{K}$

  $T={1200}^{\circ }\text{C}+273.15=1473.15\text{K}$

From equation (1)

  $\Delta S=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}\frac{dT}{T}}$

Where

  ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=\left[A+\left[B{T}_0+\left(C{T}_0{}^2+\frac{D}{{\tau }^2{T}_0{}^2}\right)\left(\frac{\tau +1}{2}\right)\right]\left(\frac{\tau -1}{\ln \tau }\right)\right]\ln \tau$......(2)

And

  $\tau =\frac{T}{T_0}$

The Values of above constants for Propane in equation (2) are given in appendix C table C.1 and noted down below:

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ \text{Propane}1.21328.785-8.8240\end{array}$

Hence

  $\begin{array}{l}\tau =\frac{T}{T_0}\\ \tau =\frac{1473.15\text{K}}{523.15\text{K}}=2.816\end{array}$

and

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=\left[A+\left[B{T}_0+\left(C{T}_0{}^2+\frac{D}{{\tau }^2{T}_0{}^2}\right)\left(\frac{\tau +1}{2}\right)\right]\left(\frac{\tau -1}{\ln \tau }\right)\right]\ln \tau \\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=\left[1.213+\left[28.785\times {10}^{-3}\times 523.15+\left(-8.824\times {10}^{-6}\times {523.15}^2\right)\left(\frac{2.816+1}{2}\right)\right]\left(\frac{2.816-1}{\ln 2.816}\right)\right]\ln 2.816\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=20.2349\end{array}$

Therefore, entropy change is

  $\begin{array}{l}\Delta S=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}\frac{dT}{T}}\\ \Delta S=8.314\frac{\text{J}}{\text{mol K}}\times 20.2349\\ \\ \Delta S=168.23\frac{\text{J}}{\text{mol K}}\end{array}$

For $n=12\text{mol of Propane}$

  $\begin{array}{l}\Delta S=168.23\frac{\text{J}}{\text{mol K}}\times 12\text{mol}\\ \Delta S\text{=2018}\text{.79}\frac{\text{J}}{\text{K}}\end{array}$

(c)

Interpretation Introduction

Interpretation :

Calculate entropy change of the gas in a steady state flow process at approximately atmospheric pressure.

Concept Introduction :

The general equation for the steady state flow entropy change as the function of temperature and pressure is defined as:

  $\begin{array}{l}dS=C_P\frac{dT}{T}-{\left(\frac{\partial V}{\partial T}\right)}_{P}dP\\ \text{at constant pressure}\end{array}$

  $\Delta S=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}\frac{dT}{T}}$......(1)

Answer to Problem 5.27P

  $\Delta S\text{=73168}\text{.8}\frac{\text{J}}{\text{K}}$

Explanation of Solution

Given information:

It is given that

  $n=20\text{kg of methane}$

  $T_0={100}^{\circ }\text{C}+273.15=373.15\text{K}$

  $T={800}^{\circ }\text{C}+273.15=1073.15\text{K}$

  $M=16.043\frac{\text{gm}}{\text{mol}}\text{ of methane}$

From equation (1)

  $\Delta S=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}\frac{dT}{T}}$

Where

  ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=\left[A+\left[B{T}_0+\left(C{T}_0{}^2+\frac{D}{{\tau }^2{T}_0{}^2}\right)\left(\frac{\tau +1}{2}\right)\right]\left(\frac{\tau -1}{\ln \tau }\right)\right]\ln \tau$......(2)

And

  $\tau =\frac{T}{T_0}$

The Values of above constants for methane in equation (2) are given in appendix C table C.1 and noted down below:

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ \text{Mathane}1.7029.081-2.1640\end{array}$

Hence

  $\begin{array}{l}\tau =\frac{T}{T_0}\\ \tau =\frac{1073.15\text{K}}{373.15\text{K}}=2.876\end{array}$

and

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=\left[A+\left[B{T}_0+\left(C{T}_0{}^2+\frac{D}{{\tau }^2{T}_0{}^2}\right)\left(\frac{\tau +1}{2}\right)\right]\left(\frac{\tau -1}{\ln \tau }\right)\right]\ln \tau \\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=\left[1.702+\left[9.081\times {10}^{-3}\times 373.15+\left(-2.164\times {10}^{-6}\times {373.15}^2\right)\left(\frac{2.876+1}{2}\right)\right]\left(\frac{2.876-1}{\ln 2.876}\right)\right]\ln 2.876\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=7.05946\end{array}$

Therefore, entropy change is

  $\begin{array}{l}\Delta S=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}\frac{dT}{T}}\\ \Delta S=8.314\frac{\text{J}}{\text{mol K}}\times 7.05946\\ \\ \Delta S=58.69\frac{\text{J}}{\text{mol K}}\end{array}$

For $n=20\text{kg of methane}$ and $M=16.043\frac{\text{gm}}{\text{mol}}\text{ of methane}$

  $\begin{array}{l}\Delta S=58.69\frac{\text{J}}{\text{mol K}}\times 20\text{kg of methane}\times \frac{1000\text{gm}}{1\text{}\text{kg}}\times \frac{\text{mol}}{16.043\text{gm}}\\ \Delta S\text{=73168}\text{.8}\frac{\text{J}}{\text{K}}\end{array}$

(d)

Interpretation Introduction

Interpretation :

Calculate entropy change of the gas in a steady state flow process at approximately atmospheric pressure.

Concept Introduction :

The general equation for the steady state flow entropy change as the function of temperature and pressure is defined as:

  $\begin{array}{l}dS=C_P\frac{dT}{T}-{\left(\frac{\partial V}{\partial T}\right)}_{P}dP\\ \text{at constant pressure}\end{array}$

  $\Delta S=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}\frac{dT}{T}}$......(1)

Answer to Problem 5.27P

  $\Delta S\text{=2388}\text{.95}\frac{\text{J}}{\text{K}}$

Explanation of Solution

Given information:

It is given that

  $n=\text{10 mol of }n\text{-butane}$

  $T_0={150}^{\circ }\text{C}+273.15=423.15\text{K}$

  $T={1150}^{\circ }\text{C}+273.15=1423.15\text{K}$

From equation (1)

  $\Delta S=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}\frac{dT}{T}}$

Where

  ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=\left[A+\left[B{T}_0+\left(C{T}_0{}^2+\frac{D}{{\tau }^2{T}_0{}^2}\right)\left(\frac{\tau +1}{2}\right)\right]\left(\frac{\tau -1}{\ln \tau }\right)\right]\ln \tau$......(2)

And

  $\tau =\frac{T}{T_0}$

The Values of above constants for $n\text{-butane}$ in equation (2) are given in appendix C table C.1 and noted down below:

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ n-\text{Butane}1.93536.915-11.4020\end{array}$

Hence

  $\begin{array}{l}\tau =\frac{T}{T_0}\\ \tau =\frac{1423.15\text{K}}{423.15\text{K}}=3.363\end{array}$

and

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=\left[A+\left[B{T}_0+\left(C{T}_0{}^2+\frac{D}{{\tau }^2{T}_0{}^2}\right)\left(\frac{\tau +1}{2}\right)\right]\left(\frac{\tau -1}{\ln \tau }\right)\right]\ln \tau \\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=\left[1.935+\left[36.915\times {10}^{-3}\times 423.15+\left(-11.402\times {10}^{-6}\times {423.15}^2\right)\left(\frac{3.363+1}{2}\right)\right]\left(\frac{3.363-1}{\ln 3.363}\right)\right]\ln 3.363\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=28.7341\end{array}$

Therefore, entropy change is

  $\begin{array}{l}\Delta S=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}\frac{dT}{T}}\\ \Delta S=8.314\frac{\text{J}}{\text{mol K}}\times 28.7341\\ \\ \Delta S=238.895\frac{\text{J}}{\text{mol K}}\end{array}$

For $n=\text{10 mol of }n\text{-butane}$

  $\begin{array}{l}\Delta S=238.895\frac{\text{J}}{\text{mol K}}\times \text{10 mol of }n\text{-butane}\\ \Delta S\text{=2388}\text{.95}\frac{\text{J}}{\text{K}}\end{array}$

(e)

Interpretation Introduction

Interpretation :

Calculate entropy change of the gas in a steady state flow process at approximately atmospheric pressure.

Concept Introduction :

The general equation for the steady state flow entropy change as the function of temperature and pressure is defined as:

  $\begin{array}{l}dS=C_P\frac{dT}{T}-{\left(\frac{\partial V}{\partial T}\right)}_{P}dP\\ \text{at constant pressure}\end{array}$

  $\Delta S=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}\frac{dT}{T}}$......(1)

Answer to Problem 5.27P

  $\Delta S\text{=1562580}\text{.153}\frac{\text{J}}{\text{K}}$

Explanation of Solution

Given information:

It is given that

  $n=1000\text{kg of air}$

  $T_0={25}^{\circ }\text{C}+273.15=298.15\text{K}$

  $T={1000}^{\circ }\text{C}+273.15=1273.15\text{K}$

  $M=28.851\frac{\text{gm}}{\text{mol}}\text{ of air}$

From equation (1)

  $\Delta S=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}\frac{dT}{T}}$

Where

  ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=\left[A+\left[B{T}_0+\left(C{T}_0{}^2+\frac{D}{{\tau }^2{T}_0{}^2}\right)\left(\frac{\tau +1}{2}\right)\right]\left(\frac{\tau -1}{\ln \tau }\right)\right]\ln \tau$......(2)

And

  $\tau =\frac{T}{T_0}$

The Values of above constants for air in equation (2) are given in appendix C table C.1 and noted down below:

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ \text{air}3.3550.5750-0.016\end{array}$

Hence

  $\begin{array}{l}\tau =\frac{T}{T_0}\\ \tau =\frac{1273.15\text{K}}{298.15\text{K}}=4.2702\end{array}$

and

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=\left[A+\left[B{T}_0+\left(C{T}_0{}^2+\frac{D}{{\tau }^2{T}_0{}^2}\right)\left(\frac{\tau +1}{2}\right)\right]\left(\frac{\tau -1}{\ln \tau }\right)\right]\ln \tau \\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=\left[3.355+\left[0.575\times {10}^{-3}\times 298.15+\left(0+\frac{-0.016\times {10}^5}{{4.2702}^2\times {298.15}^2}\right)\left(\frac{4.2702+1}{2}\right)\right]\left(\frac{4.2702-1}{\ln 4.2702}\right)\right]\ln 4.2702\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=5.42245\end{array}$

Therefore, entropy change is

  $\begin{array}{l}\Delta S=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}\frac{dT}{T}}\\ \Delta S=8.314\frac{\text{J}}{\text{mol K}}\times 5.42245\\ \\ \Delta S=45.082\frac{\text{J}}{\text{mol K}}\end{array}$

For $n=1000\text{kg of air}$ and $M=28.851\frac{\text{gm}}{\text{mol}}\text{ of air}$

  $\begin{array}{l}\Delta S=45.082\frac{\text{J}}{\text{mol K}}\times 1000\text{kg of air}\times \frac{1000\text{gm}}{1\text{}\text{kg}}\times \frac{\text{mol}}{28.851\text{gm}}\\ \Delta S\text{=1562580}\text{.153}\frac{\text{J}}{\text{K}}\end{array}$

(f)

Interpretation Introduction

Interpretation :

Calculate entropy change of the gas in a steady state flow process at approximately atmospheric pressure.

Concept Introduction :

The general equation for the steady state flow entropy change as the function of temperature and pressure is defined as:

  $\begin{array}{l}dS=C_P\frac{dT}{T}-{\left(\frac{\partial V}{\partial T}\right)}_{P}dP\\ \text{at constant pressure}\end{array}$

  $\Delta S=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}\frac{dT}{T}}$......(1)

Answer to Problem 5.27P

  $\Delta S\text{=970}\text{.272}\frac{\text{J}}{\text{K}}$

Explanation of Solution

Given information:

It is given that

  $n=2\text{0 mol of ammonia}$

  $T_0={100}^{\circ }\text{C}+273.15=373.15\text{K}$

  $T={800}^{\circ }\text{C}+273.15=1073.15\text{K}$

From equation (1)

  $\Delta S=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}\frac{dT}{T}}$

Where

  ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=\left[A+\left[B{T}_0+\left(C{T}_0{}^2+\frac{D}{{\tau }^2{T}_0{}^2}\right)\left(\frac{\tau +1}{2}\right)\right]\left(\frac{\tau -1}{\ln \tau }\right)\right]\ln \tau$......(2)

And

  $\tau =\frac{T}{T_0}$

The Values of above constants for ammonia in equation (2) are given in appendix C table C.1 and noted down below:

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ \text{Ammonia}3.5783.0200-0.186\end{array}$

Hence

  $\begin{array}{l}\tau =\frac{T}{T_0}\\ \tau =\frac{1073.15\text{K}}{373.15\text{K}}=2.876\end{array}$

and

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=\left[A+\left[B{T}_0+\left(C{T}_0{}^2+\frac{D}{{\tau }^2{T}_0{}^2}\right)\left(\frac{\tau +1}{2}\right)\right]\left(\frac{\tau -1}{\ln \tau }\right)\right]\ln \tau \\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=\left[3.578+\left[3.020\times {10}^{-3}\times 373.15+\left(0+\frac{-0.186\times {10}^5}{{2.876}^2\times {373.15}^2}\right)\left(\frac{2.876+1}{2}\right)\right]\left(\frac{2.876-1}{\ln 2.876}\right)\right]\ln 2.876\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=5.83517\end{array}$

Therefore, entropy change is

  $\begin{array}{l}\Delta S=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}\frac{dT}{T}}\\ \Delta S=8.314\frac{\text{J}}{\text{mol K}}\times 5.83517\\ \\ \Delta S=48.514\frac{\text{J}}{\text{mol K}}\end{array}$

For $n=2\text{0 mol of ammonia}$

  $\begin{array}{l}\Delta S=48.514\frac{\text{J}}{\text{mol K}}\times 2\text{0 mol of ammonia}\\ \Delta S\text{=970}\text{.272}\frac{\text{J}}{\text{K}}\end{array}$

(g)

Interpretation Introduction

Interpretation :

Calculate entropy change of the gas in a steady state flow process at approximately atmospheric pressure.

Concept Introduction :

The general equation for the steady state flow entropy change as the function of temperature and pressure is defined as:

  $\begin{array}{l}dS=C_P\frac{dT}{T}-{\left(\frac{\partial V}{\partial T}\right)}_{P}dP\\ \text{at constant pressure}\end{array}$

  $\Delta S=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}\frac{dT}{T}}$......(1)

Answer to Problem 5.27P

  $\Delta S\text{=2388}\text{.95}\frac{\text{J}}{\text{K}}$

Explanation of Solution

Given information:

It is given that

  $n=\text{10 mol of water}$

  $T_0={150}^{\circ }\text{C}+273.15=423.15\text{K}$

  $T={300}^{\circ }\text{C}+273.15=573.15\text{K}$

From equation (1)

  $\Delta S=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}\frac{dT}{T}}$

Where

  ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=\left[A+\left[B{T}_0+\left(C{T}_0{}^2+\frac{D}{{\tau }^2{T}_0{}^2}\right)\left(\frac{\tau +1}{2}\right)\right]\left(\frac{\tau -1}{\ln \tau }\right)\right]\ln \tau$......(2)

And

  $\tau =\frac{T}{T_0}$

The Values of above constants for water in equation (2) are given in appendix C table C.1 and noted down below:

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ \text{water}3.4701.45000.121\end{array}$

Hence

  $\begin{array}{l}\tau =\frac{T}{T_0}\\ \tau =\frac{573.15\text{K}}{423.15\text{K}}=1.354\end{array}$

and

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=\left[A+\left[B{T}_0+\left(C{T}_0{}^2+\frac{D}{{\tau }^2{T}_0{}^2}\right)\left(\frac{\tau +1}{2}\right)\right]\left(\frac{\tau -1}{\ln \tau }\right)\right]\ln \tau \\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=\left[3.47+\left[1.45\times {10}^{-3}\times 423.15+\left(0+\frac{0.121\times {10}^5}{{1.354}^2\times {423.15}^2}\right)\left(\frac{1.354+1}{2}\right)\right]\left(\frac{1.354-1}{\ln 1.354}\right)\right]\ln 1.354\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=1.28419\end{array}$

Therefore, entropy change is

  $\begin{array}{l}\Delta S=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}\frac{dT}{T}}\\ \Delta S=8.314\frac{\text{J}}{\text{mol K}}\times 1.28419\\ \\ \Delta S=10.677\frac{\text{J}}{\text{mol K}}\end{array}$

For $n=\text{10 mol of water}$

  $\begin{array}{l}\Delta S=10.677\frac{\text{J}}{\text{mol K}}\times \text{10 mol of water}\\ \Delta S\text{=106}\text{.77}\frac{\text{J}}{\text{K}}\end{array}$

(h)

Interpretation Introduction

Interpretation :

Calculate entropy change of the gas in a steady state flow process at approximately atmospheric pressure.

Concept Introduction :

The general equation for the steady state flow entropy change as the function of temperature and pressure is defined as:

  $\begin{array}{l}dS=C_P\frac{dT}{T}-{\left(\frac{\partial V}{\partial T}\right)}_{P}dP\\ \text{at constant pressure}\end{array}$

  $\Delta S=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}\frac{dT}{T}}$......(1)

Answer to Problem 5.27P

  $\Delta S\text{=89}\text{.783}\frac{\text{J}}{\text{K}}$

Explanation of Solution

Given information:

It is given that

  $n=5\text{mol of }C{l}_2$

  $T_0={200}^{\circ }\text{C}+273.15=473.15\text{K}$

  $T={500}^{\circ }\text{C}+273.15=773.15\text{K}$

From equation (1)

  $\Delta S=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}\frac{dT}{T}}$

Where

  ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=\left[A+\left[B{T}_0+\left(C{T}_0{}^2+\frac{D}{{\tau }^2{T}_0{}^2}\right)\left(\frac{\tau +1}{2}\right)\right]\left(\frac{\tau -1}{\ln \tau }\right)\right]\ln \tau$......(2)

And

  $\tau =\frac{T}{T_0}$

The Values of above constants for $C{l}_2$ in equation (2) are given in appendix C table C.1 and noted down below:

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ \text{Chlorine}4.4420.0890-0.344\end{array}$

Hence

  $\begin{array}{l}\tau =\frac{T}{T_0}\\ \tau =\frac{773.15\text{K}}{473.15\text{K}}=1.634\end{array}$

and

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=\left[A+\left[B{T}_0+\left(C{T}_0{}^2+\frac{D}{{\tau }^2{T}_0{}^2}\right)\left(\frac{\tau +1}{2}\right)\right]\left(\frac{\tau -1}{\ln \tau }\right)\right]\ln \tau \\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=\left[4.442+\left[0.089\times {10}^{-3}\times 473.15+\left(0+\frac{-0.344\times {10}^5}{{1.634}^2\times {473.15}^2}\right)\left(\frac{1.634+1}{2}\right)\right]\left(\frac{1.634-1}{\ln 1.634}\right)\right]\ln 1.634\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=2.15980\end{array}$

Therefore, entropy change is

  $\begin{array}{l}\Delta S=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}\frac{dT}{T}}\\ \Delta S=8.314\frac{\text{J}}{\text{mol K}}\times 2.15980\\ \\ \Delta S=17.956\frac{\text{J}}{\text{mol K}}\end{array}$

For $n=5\text{mol of }C{l}_2$

  $\begin{array}{l}\Delta S=17.956\frac{\text{J}}{\text{mol K}}\times 5\text{mol of }C{l}_2\\ \Delta S\text{=89}\text{.783}\frac{\text{J}}{\text{K}}\end{array}$

(i)

Interpretation Introduction

Interpretation :

Calculate entropy change of the gas in a steady state flow process at approximately atmospheric pressure.

Concept Introduction :

The general equation for the steady state flow entropy change as the function of temperature and pressure is defined as:

  $\begin{array}{l}dS=C_P\frac{dT}{T}-{\left(\frac{\partial V}{\partial T}\right)}_{P}dP\\ \text{at constant pressure}\end{array}$

  $\Delta S=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}\frac{dT}{T}}$......(1)

Answer to Problem 5.27P

  $\Delta S\text{=13340}\text{.52}\frac{\text{J}}{\text{K}}$

Explanation of Solution

Given information:

It is given that

  $n=10\text{kg of ethylbenzene}$

  $T_0={300}^{\circ }\text{C}+273.15=573.15\text{K}$

  $T={700}^{\circ }\text{C}+273.15=973.15\text{K}$

  $M=106.167\frac{\text{gm}}{\text{mol}}\text{ of ethylbenzene}$

From equation (1)

  $\Delta S=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}\frac{dT}{T}}$

Where

  ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=\left[A+\left[B{T}_0+\left(C{T}_0{}^2+\frac{D}{{\tau }^2{T}_0{}^2}\right)\left(\frac{\tau +1}{2}\right)\right]\left(\frac{\tau -1}{\ln \tau }\right)\right]\ln \tau$......(2)

And

  $\tau =\frac{T}{T_0}$

The Values of above constants for $\text{ethylbenzene}$ in equation (2) are given in appendix C table C.1 and noted down below:

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ \text{ethylbenzene}1.12455.38-18.4760\end{array}$

Hence

  $\begin{array}{l}\tau =\frac{T}{T_0}\\ \tau =\frac{973.15\text{K}}{573.15\text{K}}=1.698\end{array}$

and

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=\left[A+\left[B{T}_0+\left(C{T}_0{}^2+\frac{D}{{\tau }^2{T}_0{}^2}\right)\left(\frac{\tau +1}{2}\right)\right]\left(\frac{\tau -1}{\ln \tau }\right)\right]\ln \tau \\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=\left[1.124+\left[55.38\times {10}^{-3}\times 573.15+\left(-18.476\times {10}^{-6}\times {573.15}^2\right)\left(\frac{1.698+1}{2}\right)\right]\left(\frac{1.698-1}{\ln 1.698}\right)\right]\ln 1.698\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}\frac{dT}{T}}=17.0354\end{array}$

Therefore, entropy change is

  $\begin{array}{l}\Delta S=R{\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{C_P}{R}\frac{dT}{T}}\\ \Delta S=8.314\frac{\text{J}}{\text{mol K}}\times 17.0354\\ \\ \Delta S=141.632\frac{\text{J}}{\text{mol K}}\end{array}$

For $n=10\text{kg of ethylbenzene}$ and $M=106.167\frac{\text{gm}}{\text{mol}}\text{ of ethylbenzene}$

  $\begin{array}{l}\Delta S=141.632\frac{\text{J}}{\text{mol K}}\times 10\text{kg of ethylbenzene}\times \frac{1000\text{gm}}{1\text{}\text{kg}}\times \frac{\text{mol}}{106.167\text{gm}}\\ \Delta S\text{=13340}\text{.52}\frac{\text{J}}{\text{K}}\end{array}$