Chapter 4, Problem 4.1P

(a)

Interpretation Introduction

Interpretation:

Heat transferred ( $Q$ ) by ${\text{SO}}_{\text{2}}$ on rise of temperature from ${\text{200}}^{\text{0}}\text{C}$ to ${\text{1100}}^{\text{0}}\text{C}$

Concept Introduction :

The transfer of heat in steady-flow exchangers for a gas is given as:

  $Q=n\Delta H=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}$......(1)

Where $\Delta H$ is heat of reaction at any temperature $T$ .

and

  ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$......(2)

Where $\tau =\frac{T}{T_0}$

Answer to Problem 4.1P

  $Q=470034.765\text{J}$

Explanation of Solution

Given information:

  $T_1{\text{=200}}^{\text{0}}\text{C}\Rightarrow T_1{\text{=200}}^{\text{0}}\text{C+273}\text{.15=473}\text{.15}\text{K}$

  $T_2{\text{= 1100}}^{\text{0}}\text{C}\Rightarrow T_2{\text{= 1100}}^{\text{0}}\text{C}+273.15=1373.15\text{K}$

  $n\text{=10mol}$

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

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

  $\text{gas constant R in SI unit is 8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}$

  $\tau =\frac{T}{T_0}=\frac{1373.15}{473.15}=2.902$

Put values in equation (2)

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{\text{473}\text{.15}}{\overset{1373.15}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=5.699\times \text{473}\text{.15}\times (2.902-1)+\frac{0.801\times {10}^{-3}}{2}\text{473}{\text{.15}}^2({2.902}^2-1)\\ +\frac{0}{3}\text{473}{\text{.15}}^3({2.902}^3-1)+\frac{-1.015\times {10}^5}{\text{473}\text{.15}}(\frac{2.902-1}{2.902})\\ {\displaystyle \underset{\text{473}\text{.15}}{\overset{1373.15}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=5653.533\text{K}\end{array}$

Now from equation (1),

  $\begin{array}{l}Q=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ Q=\text{8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}\times 5653.533\text{K}\\ Q=47003.4765\frac{\text{J}}{\text{mol}}\end{array}$

Now,

  $\begin{array}{l}n=10\text{mol}\\ \\ Q=47003.4765\frac{\text{J}}{\text{mol}}\times 10\text{mol }\\ Q=470034.765\text{J}\end{array}$

(b)

Interpretation Introduction

Interpretation:

Heat transferred ( $Q$ ) by propane on rise of temperature from ${\text{250 to 1200}}^{\text{0}}\text{C}$

Concept Introduction :

The transfer of heat in steady-flow exchangers for a gas is given as:

  $Q=n\Delta H=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}$......(1)

Where $\Delta H$ is heat of reaction at any temperature $T$ .

and

  ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$......(2)

Where $\tau =\frac{T}{T_0}$

Answer to Problem 4.1P

  $1942106.848\text{J}$

Explanation of Solution

Given information:

  $T_1{\text{=200}}^{\text{0}}\text{C}\Rightarrow T_1{\text{=250}}^{\text{0}}\text{C+273}\text{.15=523}\text{.15}\text{K}$

  $T_2{\text{= 1100}}^{\text{0}}\text{C}\Rightarrow T_2{\text{= 1200}}^{\text{0}}\text{C}+273.15=1473.15\text{K}$

  $n\text{=12mol}$

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

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

  $\text{gas constant R in SI unit is 8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}$

  $\tau =\frac{T}{T_0}=\frac{1473.15}{523.15}=2.816$

Put values in equation (2)

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=1.213\times \text{523}\text{.15}\times (2.816-1)+\frac{28.785\times {10}^{-3}}{2}\text{523}{\text{.15}}^2({2.816}^2-1)\\ +\frac{-8.824\times {10}^{-6}}{3}\times \text{523}{\text{.15}}^3({2.816}^3-1)+\frac{0}{\text{523}\text{.15}}(\frac{2.816-1}{2.816})\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=19466.23013\text{K}\end{array}$

Now from equation (1),

  $\begin{array}{l}Q=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ Q=\text{8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}\times 19466.23013\text{K}\\ Q=161842.2373\frac{\text{J}}{\text{mol}}\end{array}$

Now,

  $\begin{array}{l}n=12\text{mol}\\ \\ Q=161842.2373\frac{\text{J}}{\text{mol}}\times 12\text{mol }\\ Q=1942106.848\text{J}\end{array}$

(c)

Interpretation Introduction

Interpretation:

Heat transferred ( $Q$ ) by methane on rise of temperature from ${\text{100 to 800}}^{\text{0}}\text{C}$

Concept Introduction :

The transfer of heat in steady-flow exchangers for a gas is given as:

  $Q=n\Delta H=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}$......(1)

Where $\Delta H$ is heat of reaction at any temperature $T$ .

and

  ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$......(2)

Where $\tau =\frac{T}{T_0}$

Answer to Problem 4.1P

  $51153886.48\text{J}$

Explanation of Solution

Given information:

  $T_1{\text{=100}}^{\text{0}}\text{C}\Rightarrow T_1{\text{=100}}^{\text{0}}\text{C+273}\text{.15=373}\text{.15}\text{K}$

  $T_2{\text{= 800}}^{\text{0}}\text{C}\Rightarrow T_2{\text{= 800}}^{\text{0}}\text{C}+273.15=1073.15\text{K}$

  $m\text{=20}\text{kg}$

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

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

  $\text{gas constant R in SI unit is 8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}$

  $\tau =\frac{T}{T_0}=\frac{1073.15}{373.15}=2.876$

Put values in equation (2)

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=1.702\times 37\text{3}\text{.15}\times (2.876-1)+\frac{9.081\times {10}^{-3}}{2}37\text{3}{\text{.15}}^2({2.876}^2-1)\\ +\frac{-2.164\times {10}^{-6}}{3}\times 37\text{3}{\text{.15}}^3({2.876}^3-1)+\frac{0}{37\text{3}\text{.15}}(\frac{2.876-1}{2.876})\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=4934.49807\text{K}\end{array}$

Now from equation (1),

  $\begin{array}{l}Q=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ Q=\text{8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}\times 4934.49807\text{K}\\ Q=41025.41696\frac{\text{J}}{\text{mol}}\end{array}$

Now,

  $\begin{array}{l}m=20\text{kg}\Rightarrow n=\frac{20000\text{g}}{16.04\text{g}/\text{mol}}=1246.883\text{mol}\\ \\ Q=41025.41696\frac{\text{J}}{\text{mol}}\times 1246.883\text{mol }\\ Q=51153886.48\text{J}\end{array}$

(d)

Interpretation Introduction

Interpretation:

Heat transferred ( $Q$ ) by $n$ butane on rise of temperature from ${\text{150 to 1150}}^{\text{0}}\text{C}$

Concept Introduction :

The transfer of heat in steady-flow exchangers for a gas is given as:

  $Q=n\Delta H=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}$......(1)

Where $\Delta H$ is heat of reaction at any temperature $T$ .

and

  ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$......(2)

Where $\tau =\frac{T}{T_0}$

Answer to Problem 4.1P

  $2107019.862\text{J}$

Explanation of Solution

Given information:

  $T_1{\text{=150}}^{\text{0}}\text{C}\Rightarrow T_1{\text{=150}}^{\text{0}}\text{C+273}\text{.15=423}\text{.15}\text{K}$

  $T_2{\text{= 1150}}^{\text{0}}\text{C}\Rightarrow T_2{\text{= 1150}}^{\text{0}}\text{C}+273.15=1423.15\text{K}$

  $n\text{=10}\text{mol}$

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

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

  $\text{gas constant R in SI unit is 8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}$

  $\tau =\frac{T}{T_0}=\frac{1423.15}{423.15}=3.363$

Put values in equation (2)

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=1.935\times 4\text{23}\text{.15}\times (3.363-1)+\frac{36.915\times {10}^{-3}}{2}\text{423}{\text{.15}}^2({3.363}^2-1)\\ +\frac{-11.402\times {10}^{-6}}{3}\times 4\text{23}{\text{.15}}^3({3.363}^3-1)+\frac{0}{\text{423}\text{.15}}(\frac{3.363-1}{3.363})\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=25343.03418\text{K}\end{array}$

Now from equation (1),

  $\begin{array}{l}Q=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ Q=\text{8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}\times 25343.03418\text{K}\\ Q=210701.9862\frac{\text{J}}{\text{mol}}\end{array}$

Now,

  $\begin{array}{l}n=10\text{mol}\\ \\ Q=210701.9862\frac{\text{J}}{\text{mol}}\times 10\text{mol }\\ Q=2107019.862\text{J}\end{array}$

(e)

Interpretation Introduction

Interpretation:

Heat transferred ( $Q$ ) by air on rise of temperature from ${\text{25 to 1000}}^{\text{0}}\text{C}$

Concept Introduction :

The transfer of heat in steady-flow exchangers for a gas is given as:

  $Q=n\Delta H=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}$......(1)

Where $\Delta H$ is heat of reaction at any temperature $T$ .

and

  ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$......(2)

Where $\tau =\frac{T}{T_0}$

Answer to Problem 4.1P

  $1064048522\text{J}$

Explanation of Solution

Given information:

  $T_1{\text{=25}}^{\text{0}}\text{C}\Rightarrow T_1{\text{=25}}^{\text{0}}\text{C+273}\text{.15=298}\text{.15}\text{K}$

  $T_2{\text{= 1000}}^{\text{0}}\text{C}\Rightarrow T_2{\text{= 1000}}^{\text{0}}\text{C}+273.15=1273.15\text{K}$

  $m\text{=1000}\text{kg}$

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

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

  $\text{gas constant R in SI unit is 8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}$

  $\tau =\frac{T}{T_0}=\frac{1073.15}{373.15}=4.27017$

Put values in equation (2)

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=3.355\times 298.15\times (4.2701-1)+\frac{0.575\times {10}^{-3}}{2}{298.15}^2({4.2701}^2-1)\\ +\frac{0}{3}\times {298.15}^3({4.2701}^3-1)+\frac{-0.016}{298.15}\times (\frac{4.2701-1}{4.2701})\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=3711.4995\text{K}\end{array}$

Now from equation (1),

  $\begin{array}{l}Q=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ Q=\text{8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}\times 3711.4995\text{K}\\ Q=30857.40715\frac{\text{J}}{\text{mol}}\end{array}$

Now,

  $\begin{array}{l}m=1000\text{kg}\Rightarrow n=\frac{1000\times 1000\text{g}}{29\text{g}/\text{mol}}=34482.75862\text{mol}\\ \\ Q=30857.40715\frac{\text{J}}{\text{mol}}\times 34482.75862\text{mol}\\ Q=1064048522\text{J}\end{array}$

(f)

Interpretation Introduction

Interpretation:

Heat transferred ( $Q$ ) by ammonia on rise of temperature from ${\text{100 to 800}}^{\text{0}}\text{C}$

Concept Introduction :

The transfer of heat in steady-flow exchangers for a gas is given as:

  $Q=n\Delta H=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}$......(1)

Where $\Delta H$ is heat of reaction at any temperature $T$ .

and

  ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$......(2)

Where $\tau =\frac{T}{T_0}$

Answer to Problem 4.1P

  $665290.495\text{J}$

Explanation of Solution

Given information:

  $T_1{\text{=100}}^{\text{0}}\text{C}\Rightarrow T_1{\text{=100}}^{\text{0}}\text{C+273}\text{.15=373}\text{.15}\text{K}$

  $T_2{\text{= 800}}^{\text{0}}\text{C}\Rightarrow T_2{\text{= 800}}^{\text{0}}\text{C}+273.15=1073.15\text{K}$

  $n\text{=20}\text{mol}$

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

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

  $\text{gas constant R in SI unit is 8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}$

  $\tau =\frac{T}{T_0}=\frac{1073.15}{373.15}=2.876$

Put values in equation (2)

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=3.578\times 37\text{3}\text{.15}\times (2.876-1)+\frac{3.020\times {10}^{-3}}{2}37\text{3}{\text{.15}}^2({2.876}^2-1)\\ +\frac{0}{3}\times 37\text{3}{\text{.15}}^3({2.876}^3-1)+\frac{-0.186\times {10}^5}{37\text{3}\text{.15}}(\frac{2.876-1}{2.876})\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=4001.025\text{K}\end{array}$

Now from equation (1),

  $\begin{array}{l}Q=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ Q=\text{8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}\times 4001.025\text{K}\\ Q=33264.525\frac{\text{J}}{\text{mol}}\end{array}$

Now,

  $\begin{array}{l}n=20\text{mol}\\ \\ Q=33264.525\frac{\text{J}}{\text{mol}}\times 20\text{mol }\\ Q=665290.495\text{J}\end{array}$

(g)

Interpretation Introduction

Interpretation:

Heat transferred ( $Q$ ) by water on rise of temperature from ${\text{150 to 300}}^{\text{0}}\text{C}$

Concept Introduction :

The transfer of heat in steady-flow exchangers for a gas is given as:

  $Q=n\Delta H=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}$......(1)

Where $\Delta H$ is heat of reaction at any temperature $T$ .

and

  ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$......(2)

Where $\tau =\frac{T}{T_0}$

Answer to Problem 4.1P

  $52983.298\text{J}$

Explanation of Solution

Given information:

  $T_1{\text{=150}}^{\text{0}}\text{C}\Rightarrow T_1{\text{=150}}^{\text{0}}\text{C+273}\text{.15=423}\text{.15}\text{K}$

  $T_2{\text{= 300}}^{\text{0}}\text{C}\Rightarrow T_2{\text{= 300}}^{\text{0}}\text{C}+273.15=573.15\text{K}$

  $n\text{=10}\text{mol}$

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

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

  $\text{gas constant R in SI unit is 8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}$

  $\tau =\frac{T}{T_0}=\frac{573.15}{423.15}=1.355$

Put values in equation (2)

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=3.470\times 4\text{23}\text{.15}\times (1.355-1)+\frac{1.45\times {10}^{-3}}{2}\text{423}{\text{.15}}^2({1.355}^2-1)\\ +\frac{0}{3}\times 4\text{23}{\text{.15}}^3({1.355}^3-1)+\frac{0.121\times {10}^5}{\text{423}\text{.15}}(\frac{1.355-1}{1.355})\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=637.278\text{K}\end{array}$

Now from equation (1),

  $\begin{array}{l}Q=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ Q=\text{8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}\times 637.278\text{K}\\ Q=5298.3298\frac{\text{J}}{\text{mol}}\end{array}$

Now,

  $\begin{array}{l}n=10\text{mol}\\ \\ Q=5298.3298\frac{\text{J}}{\text{mol}}\times 10\text{mol }\\ Q=52983.298\text{J}\end{array}$

(h)

Interpretation Introduction

Interpretation:

Heat transferred ( $Q$ ) by chlorine on rise of temperature from ${\text{200}}^{\text{0}}\text{C}$ to ${\text{500}}^{\text{0}}\text{C}$

Concept Introduction :

The transfer of heat in steady-flow exchangers for a gas is given as:

  $Q=n\Delta H=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}$......(1)

Where $\Delta H$ is heat of reaction at any temperature $T$ .

and

  ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$......(2)

Where $\tau =\frac{T}{T_0}$

Answer to Problem 4.1P

  $54910.861\text{J}$

Explanation of Solution

Given information:

  $T_1{\text{=200}}^{\text{0}}\text{C}\Rightarrow T_1{\text{=200}}^{\text{0}}\text{C+273}\text{.15=473}\text{.15}\text{K}$

  $T_2{\text{= 500}}^{\text{0}}\text{C}\Rightarrow T_2{\text{= 500}}^{\text{0}}\text{C}+273.15=773.15\text{K}$

  $n\text{=5}\text{mol}$

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

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ Chlorine4.4420.0890-0.344\end{array}$

  $\text{gas constant R in SI unit is 8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}$

  $\tau =\frac{T}{T_0}=\frac{773.15}{473.15}=1.634$

Put values in equation (2)

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{\text{473}\text{.15}}{\overset{773.15}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=4.442\times \text{473}\text{.15}\times (1.634-1)+\frac{0.089\times {10}^{-3}}{2}\text{473}{\text{.15}}^2({1.634}^2-1)\\ +\frac{0}{3}\text{473}{\text{.15}}^3({1.634}^3-1)+\frac{-0.344\times {10}^5}{\text{473}\text{.15}}(\frac{1.634-1}{\begin{array}{l}1.634\\ \end{array}})\\ {\displaystyle \underset{\text{473}\text{.15}}{\overset{773.15}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=1320.925\text{K}\end{array}$

Now from equation (1),

  $\begin{array}{l}Q=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ Q=\text{8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}\times 1320.925\text{K}\\ Q=10982.172\frac{\text{J}}{\text{mol}}\end{array}$

Now,

  $\begin{array}{l}n=5\text{mol}\\ \\ Q=10982.172\frac{\text{J}}{\text{mol}}\times 5\text{mol }\\ Q=54910.861\text{J}\end{array}$

(i)

Interpretation Introduction

Interpretation:

Heat transferred ( $Q$ ) by ethylbenzene on rise of temperature from ${\text{300 to 700}}^{\text{0}}\text{C}$

Concept Introduction :

The transfer of heat in steady-flow exchangers for a gas is given as:

  $Q=n\Delta H=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}$......(1)

Where $\Delta H$ is heat of reaction at any temperature $T$ .

and

  ${\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })$......(2)

Where $\tau =\frac{T}{T_0}$

Answer to Problem 4.1P

  $10228945.09\text{J}$

Explanation of Solution

Given information:

  $T_1{\text{=300}}^{\text{0}}\text{C}\Rightarrow T_1{\text{=300}}^{\text{0}}\text{C+273}\text{.15=573}\text{.15}\text{K}$

  $T_2{\text{= 700}}^{\text{0}}\text{C}\Rightarrow T_2{\text{= 700}}^{\text{0}}\text{C}+273.15=973.15\text{K}$

  $m\text{=10}\text{kg}$

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

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ \text{ethylbenzene}1.12455.380-18.4760\end{array}$

  $\text{gas constant R in SI unit is 8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}$

  $\tau =\frac{T}{T_0}=\frac{973.15}{573.15}=1.698$

Put values in equation (2)

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=1.124\times 57\text{3}\text{.15}\times (1.698-1)+\frac{55.38\times {10}^{-3}}{2}57\text{3}{\text{.15}}^2({1.698}^2-1)\\ +\frac{-18.476\times {10}^{-6}}{3}\times 57\text{3}{\text{.15}}^3({1.698}^3-1)+\frac{0}{57\text{3}\text{.15}}(\frac{1.698-1}{1.698})\\ {\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=13062.39\text{K}\end{array}$

Now from equation (1),

  $\begin{array}{l}Q=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ Q=\text{8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}\times 13062.39\text{K}\\ Q=108600.71\frac{\text{J}}{\text{mol}}\end{array}$

Now,

  $\begin{array}{l}m=10\text{kg}\Rightarrow n=\frac{10000\text{g}}{106.17\text{g}/\text{mol}}=94.189\text{mol}\\ \\ Q=108600.71\frac{\text{J}}{\text{mol}}\times 94.189\text{mol}\\ Q=10228945.09\text{J}\end{array}$