Chapter 4, Problem 4.3P

(a)

Interpretation Introduction

Interpretation:

Final temperature ${\text{(T}}_2\text{)}$ of the methane is to be calculated when some amount of heat is given to the to the methane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

  $Q=n\Delta H=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}$.     ...(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 4.3P

  $T=\text{626}\text{.817}\text{K}$

Explanation of Solution

Given information:

  $\text{Q= 12 kJ/mol}$

  ${\text{initial temperature(T}}_{\text{1}}{\text{) =100}}^{\text{0}}\text{C}$. So,

  $T_0{\text{=100}}^{\text{0}}\text{C}\Rightarrow T_0{\text{=100}}^{\text{0}}\text{C+273}\text{.15=373}\text{.15}\text{K}$

  $T={?}^{\text{o}}\text{C=}T\text{+273}\text{.15}\text{K}$

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

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ methane1.7029.081-2.1640\end{array}$

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

Put values in equation (2)

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=1.702\times 373.15\times (\tau -1)+\frac{9.081\times {10}^{-3}}{2}{373.15}^2({\tau }^2-1)\\ +\frac{-2.164\times {10}^{-6}}{3}\times {373.15}^3({\tau }^3-1)+\frac{0}{373.15}\times (\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=635.1013(\tau -1)+632.223({\tau }^2-1)-37.479({\tau }^3-1)\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}635.1013\tau -635.1013+632.223{\tau }^2-632.223-37.479{\tau }^3+37.479\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}-37.479{\tau }^3+632.223{\tau }^2+635.1013\tau -1229.8453\end{array}$

Given that

  $\begin{array}{l}\text{Q=12}\frac{\text{kJ}}{\text{mol}}\\ \Rightarrow Q=12\frac{\text{kJ}}{\text{mol}}\times \frac{1000\text{ J}}{\text{1}\text{kJ}}=12000\frac{\text{J}}{\text{mol}}\end{array}$

Now from equation (1),

  $\begin{array}{l}12000\frac{\text{J}}{\text{mol}}=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 12000\frac{\text{J}}{\text{mol}}=\text{8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}\text{×}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 1443.349\text{K}={\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\end{array}$

Put in abovE

  $\begin{array}{l}1443.349\text{K}\text{=}-37.479{\tau }^3+632.223{\tau }^2+635.1013\tau -1229.8453\\ \\ 37.479{\tau }^3-632.223{\tau }^2-635.1013\tau +2673.194=0\end{array}$

On solving,

  $\tau =1.6798,-2.412,17.6$

Considering positive value of temperature $\tau =1.6798$

Hence,

  $\begin{array}{l}\tau =\frac{T}{T_0}=\frac{T}{100+273.15}=1.6798\\ \\ T=\text{626}\text{.817}\text{K}\end{array}$

(b)

Interpretation Introduction

Interpretation:

Final temperature ${\text{(T}}_2\text{)}$ of the ethane is to be calculated when some amount of heat is given to the to the ethane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

  $Q=n\Delta H=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}$.     ...(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 4.3P

  $T=\text{538}\text{.455}\text{K}$

Explanation of Solution

Given information:

  $\text{Q= 12 kJ/mol}$

  ${\text{initial temperature(T}}_{\text{1}}{\text{) =100}}^{\text{0}}\text{C}$. So,

  $T_0{\text{=100}}^{\text{0}}\text{C}\Rightarrow T_0{\text{=100}}^{\text{0}}\text{C+273}\text{.15=373}\text{.15}\text{K}$

  $T={?}^{\text{o}}\text{C=}T\text{+273}\text{.15}\text{K}$

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

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ ethane1.13119.225-5.5610\end{array}$

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

Put values in equation (2)

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=1.131\times 373.15\times (\tau -1)+\frac{19.225\times {10}^{-3}}{2}{373.15}^2({\tau }^2-1)\\ +\frac{-5.561\times {10}^{-6}}{3}\times {373.15}^3({\tau }^3-1)+\frac{0}{373.15}\times (\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=422.033(\tau -1)+1338.453({\tau }^2-1)-96.31({\tau }^3-1)\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}422.033\tau -422.033+1338.453{\tau }^2-1338.453-96.31{\tau }^3+96.31\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}-96.31{\tau }^3+1338.453{\tau }^2+422.033\tau -1664.176\end{array}$

Given that

  $\begin{array}{l}\text{Q=12}\frac{\text{kJ}}{\text{mol}}\\ \Rightarrow Q=12\frac{\text{kJ}}{\text{mol}}\times \frac{1000\text{ J}}{\text{1}\text{kJ}}=12000\frac{\text{J}}{\text{mol}}\end{array}$

Now from equation (1),

  $\begin{array}{l}12000\frac{\text{J}}{\text{mol}}=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 12000\frac{\text{J}}{\text{mol}}=\text{8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}\text{×}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 1443.349\text{K}={\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\end{array}$

Put in abovE

  $\begin{array}{l}1443.349\text{K=}-96.31{\tau }^3+1338.453{\tau }^2+422.033\tau -1664.176\\ \\ 96.31{\tau }^3-1338.453{\tau }^2-422.033\tau +3107.525=0\end{array}$

On solving,

  $\tau =1.443,-1.592,14.0458$

Considering positive value of temperature $\tau =1.443$

Hence,

  $\begin{array}{l}\tau =\frac{T}{T_0}=\frac{T}{100+273.15}=1.443\\ \\ T=\text{538}\text{.455}\text{K}\end{array}$

(c)

Interpretation Introduction

Interpretation:

Final temperature ${\text{(T}}_2\text{)}$ of the propane is to be calculated when some amount of heat is given to the to the propane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

  $Q=n\Delta H=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}$.     ...(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 4.3P

  $T=\text{493}\text{.3043}\text{K}$

Explanation of Solution

Given information:

  $\text{Q= 12 kJ/mol}$

  ${\text{initial temperature(T}}_{\text{1}}{\text{) =100}}^{\text{0}}\text{C}$. So,

  $T_0{\text{=100}}^{\text{0}}\text{C}\Rightarrow T_0{\text{=100}}^{\text{0}}\text{C+273}\text{.15=373}\text{.15}\text{K}$

  $T={?}^{\text{o}}\text{C=}T\text{+273}\text{.15}\text{K}$

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

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ propane1.21328.785-8.8240\end{array}$

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

Put values in equation (2)

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=1.213\times 373.15\times (\tau -1)+\frac{28.785\times {10}^{-3}}{2}{373.15}^2({\tau }^2-1)\\ +\frac{-8.824\times {10}^{-6}}{3}\times {373.15}^3({\tau }^3-1)+\frac{0}{373.15}\times (\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=452.631(\tau -1)+2004.025({\tau }^2-1)-152.825({\tau }^3-1)\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}452.631\tau -452.631+2004.025{\tau }^2-2004.025-152.825{\tau }^3+152.825\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}-152.825{\tau }^3+2004.025{\tau }^2+452.631\tau -2303.831\end{array}$

Given that

  $\begin{array}{l}\text{Q=12}\frac{\text{kJ}}{\text{mol}}\\ \Rightarrow Q=12\frac{\text{kJ}}{\text{mol}}\times \frac{1000\text{ J}}{\text{1}\text{kJ}}=12000\frac{\text{J}}{\text{mol}}\end{array}$

Now from equation (1),

  $\begin{array}{l}12000\frac{\text{J}}{\text{mol}}=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 12000\frac{\text{J}}{\text{mol}}=\text{8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}\text{×}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 1443.349\text{K}={\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\end{array}$

Put in abovE

  $\begin{array}{l}1443.349\text{K=}-152.825{\tau }^3+2004.025{\tau }^2+452.631\tau -2303.831\\ \\ 152.825{\tau }^3-2004.025{\tau }^2-452.631\tau +3747.18=0\end{array}$

On solving,

  $\tau =1.322,-1.405,13.197$

Considering positive value of temperature $\tau =1.322$

Hence,

  $\begin{array}{l}\tau =\frac{T}{T_0}=\frac{T}{100+273.15}=1.322\\ \\ T=\text{493}\text{.3043}\text{K}\end{array}$

(d)

Interpretation Introduction

Interpretation:

Final temperature ${\text{(T}}_2\text{)}$ of the $n$ -butane is to be calculated when some amount of heat is given to the to the $n$ -butane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

  $Q=n\Delta H=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}$.     ...(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 4.3P

  $T=\text{466}\text{.81}\text{K}$

Explanation of Solution

Given information:

  $\text{Q= 12 kJ/mol}$

  ${\text{initial temperature(T}}_{\text{1}}{\text{) =100}}^{\text{0}}\text{C}$. So,

  $T_0{\text{=100}}^{\text{0}}\text{C}\Rightarrow T_0{\text{=100}}^{\text{0}}\text{C+273}\text{.15=373}\text{.15}\text{K}$

  $T={?}^{\text{o}}\text{C=}T\text{+273}\text{.15}\text{K}$

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

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ n-butane1.93536.915-11.4020\end{array}$

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

Put values in equation (2)

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=1.935\times 373.15\times (\tau -1)+\frac{36.915\times {10}^{-3}}{2}{373.15}^2({\tau }^2-1)\\ +\frac{-11.402\times {10}^{-6}}{3}\times {373.15}^3({\tau }^3-1)+\frac{0}{373.15}\times (\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=722.045(\tau -1)+2570.039({\tau }^2-1)-197.474({\tau }^3-1)\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}722.045\tau -722.045+2570.039{\tau }^2-2570.039-197.474{\tau }^3+197.474\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}-197.474{\tau }^3+2570.039{\tau }^2+722.045\tau -3094.61\end{array}$

Given that

  $\begin{array}{l}\text{Q=12}\frac{\text{kJ}}{\text{mol}}\\ \Rightarrow Q=12\frac{\text{kJ}}{\text{mol}}\times \frac{1000\text{ J}}{\text{1}\text{kJ}}=12000\frac{\text{J}}{\text{mol}}\end{array}$

Now from equation (1),

  $\begin{array}{l}12000\frac{\text{J}}{\text{mol}}=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 12000\frac{\text{J}}{\text{mol}}=\text{8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}\text{×}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 1443.349\text{K}={\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\end{array}$

Put in abovE

  $\begin{array}{l}1443.349\text{K=}-197.474{\tau }^3+2570.039{\tau }^2+722.045\tau -3094.61\\ \\ 197.474{\tau }^3-2570.039{\tau }^2-722.045\tau +4537.96=0\end{array}$

On solving,

  $\tau =1.251,-1.396,13.16$

Considering positive value of temperature $\tau =1.251$

Hence,

  $\begin{array}{l}\tau =\frac{T}{T_0}=\frac{T}{100+273.15}=1.251\\ \\ T=\text{466}\text{.81}\text{K}\end{array}$

(e)

Interpretation Introduction

Interpretation:

Final temperature ${\text{(T}}_2\text{)}$ of the $n$ -hexane is to be calculated when some amount of heat is given to the $n$ -hexane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

  $Q=n\Delta H=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}$.     ...(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 4.3P

  $T=\text{438}\text{.45}\text{K}$

Explanation of Solution

Given information:

  $\text{Q= 12 kJ/mol}$

  ${\text{initial temperature(T}}_{\text{1}}{\text{) =100}}^{\text{0}}\text{C}$. So,

  $T_0{\text{=100}}^{\text{0}}\text{C}\Rightarrow T_0{\text{=100}}^{\text{0}}\text{C+273}\text{.15=373}\text{.15}\text{K}$

  $T={?}^{\text{o}}\text{C=}T\text{+273}\text{.15}\text{K}$

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

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ n-hexane3.02553.722-16.7910\end{array}$

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

Put values in equation (2)

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=3.025\times 373.15\times (\tau -1)+\frac{53.722\times {10}^{-3}}{2}{373.15}^2({\tau }^2-1)\\ +\frac{-16.791\times {10}^{-6}}{3}\times {373.15}^3({\tau }^3-1)+\frac{0}{373.15}\times (\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=1128.78(\tau -1)+3740.15({\tau }^2-1)-290.807({\tau }^3-1)\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}1128.78\tau -1128.78+3740.15{\tau }^2-3740.15-290.807{\tau }^3+290.807\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}-290.807{\tau }^3+3740.15{\tau }^2+1128.78\tau -4578.12\end{array}$

Given that

  $\begin{array}{l}\text{Q=12}\frac{\text{kJ}}{\text{mol}}\\ \Rightarrow Q=12\frac{\text{kJ}}{\text{mol}}\times \frac{1000\text{ J}}{\text{1}\text{kJ}}=12000\frac{\text{J}}{\text{mol}}\end{array}$

Now from equation (1),

  $\begin{array}{l}12000\frac{\text{J}}{\text{mol}}=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 12000\frac{\text{J}}{\text{mol}}=\text{8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}\text{×}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 1443.349\text{K}={\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\end{array}$

Put in abovE

  $\begin{array}{l}1443.349\text{K=}-290.807{\tau }^3+3740.15{\tau }^2+1128.78\tau -4578.12\\ \\ 290.807{\tau }^3-3740.15{\tau }^2-1128.78\tau +6021.471=0\end{array}$

On solving,

  $\tau =1.175,-1.351,13.04$

Considering positive value of temperature $\tau =1.175$

Hence,

  $\begin{array}{l}\tau =\frac{T}{T_0}=\frac{T}{100+273.15}=1.175\\ \\ T=\text{438}\text{.45}\text{K}\end{array}$

(f)

Interpretation Introduction

Interpretation:

Final temperature ${\text{(T}}_2\text{)}$ of the $n$ -octane is to be calculated when some amount of heat is given to the to the $n$ -octane.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

  $Q=n\Delta H=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}$.     ...(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 4.3P

  $T=\text{371}\text{.695}\text{K}$

Explanation of Solution

Given information:

  $\text{Q= 12 kJ/mol}$

  ${\text{initial temperature(T}}_{\text{1}}{\text{) =100}}^{\text{0}}\text{C}$. So,

  $T_0{\text{=100}}^{\text{0}}\text{C}\Rightarrow T_0{\text{=100}}^{\text{0}}\text{C+273}\text{.15=373}\text{.15}\text{K}$

  $T={?}^{\text{o}}\text{C=}T\text{+273}\text{.15}\text{K}$

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

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ n-hexane4.10870.567-22.2080\end{array}$

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

Put values in equation (2)

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=4.108\times 373.15\times (\tau -1)+\frac{70.567\times {10}^{-3}}{2}{373.15}^2({\tau }^2-1)\\ +\frac{-22.208\times {10}^{-6}}{3}\times {373.15}^3({\tau }^3-1)+\frac{0}{373.15}\times (\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=1532.9002(\tau -1)+4912.91({\tau }^2-1)-384.63({\tau }^3-1)\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}1532.9002\tau -1532.9002+4912.91{\tau }^2-4912.91-384.63{\tau }^3+384.63\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}-384.63{\tau }^3+4912.91{\tau }^2+1532.9002\tau -6061.184\end{array}$

Given that

  $\begin{array}{l}\text{Q=12}\frac{\text{kJ}}{\text{mol}}\\ \Rightarrow Q=12\frac{\text{kJ}}{\text{mol}}\times \frac{1000\text{ J}}{\text{1}\text{kJ}}=12000\frac{\text{J}}{\text{mol}}\end{array}$

Now from equation (1),

  $\begin{array}{l}12000\frac{\text{J}}{\text{mol}}=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 12000\frac{\text{J}}{\text{mol}}=\text{8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}\text{×}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 1443.349\text{K}={\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\end{array}$

Put in abovE

  $\begin{array}{l}1443.349\text{K=}-384.63{\tau }^3+4912.91{\tau }^2+1532.9002\tau -6061.184\\ \\ 384.63{\tau }^3-4912.91{\tau }^2-1532.9002\tau +6021.471=0\end{array}$

On solving,

  $\tau =0.9961,-1.21,12.987$

Considering positive value of temperature $\tau =0.9961$

Hence,

  $\begin{array}{l}\tau =\frac{T}{T_0}=\frac{T}{100+273.15}=0.9961\\ \\ T=\text{371}\text{.695}\text{K}\end{array}$

(g)

Interpretation Introduction

Interpretation:

Final temperature ${\text{(T}}_2\text{)}$ of the propylene is to be calculated when some amount of heat is given to the to the propylene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

  $Q=n\Delta H=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}$.     ...(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 4.3P

  $T=\text{513}\text{.08}\text{K}$

Explanation of Solution

Given information:

  $\text{Q= 12 kJ/mol}$

  ${\text{initial temperature(T}}_{\text{1}}{\text{) =100}}^{\text{0}}\text{C}$. So,

  $T_0{\text{=100}}^{\text{0}}\text{C}\Rightarrow T_0{\text{=100}}^{\text{0}}\text{C+273}\text{.15=373}\text{.15}\text{K}$

  $T={?}^{\text{o}}\text{C=}T\text{+273}\text{.15}\text{K}$

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

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ propylene1.63722.706-6.9150\end{array}$

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

Put values in equation (2)

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=1.637\times 373.15\times (\tau -1)+\frac{22.706\times {10}^{-3}}{2}{373.15}^2({\tau }^2-1)\\ +\frac{-6.915\times {10}^{-6}}{3}\times {373.15}^3({\tau }^3-1)+\frac{0}{373.15}\times (\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=610.847(\tau -1)+1580.8({\tau }^2-1)-119.763({\tau }^3-1)\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}610.847\tau -610.847+1580.8{\tau }^2-1580.8-119.763{\tau }^3+119.763\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}-119.763{\tau }^3+1580.8{\tau }^2+610.847\tau -2071.88\end{array}$

Given that

  $\begin{array}{l}\text{Q=12}\frac{\text{kJ}}{\text{mol}}\\ \Rightarrow Q=12\frac{\text{kJ}}{\text{mol}}\times \frac{1000\text{ J}}{\text{1}\text{kJ}}=12000\frac{\text{J}}{\text{mol}}\end{array}$

Now from equation (1),

  $\begin{array}{l}12000\frac{\text{J}}{\text{mol}}=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 12000\frac{\text{J}}{\text{mol}}=\text{8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}\text{×}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 1443.349\text{K}={\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\end{array}$

Put in abovE

  $\begin{array}{l}1443.349\text{K=}-119.763{\tau }^3+1580.8{\tau }^2+610.847\tau -2071.88\\ \\ 119.763{\tau }^3-1580.8{\tau }^2-610.847\tau +3515.233=0\end{array}$

On solving,

  $\tau =1.375,-1.592,13.416$

Considering positive value of temperature $\tau =1.375$

Hence,

  $\begin{array}{l}\tau =\frac{T}{T_0}=\frac{T}{100+273.15}=1.375\\ \\ T=\text{513}\text{.08}\text{K}\end{array}$

(h)

Interpretation Introduction

Interpretation:

Final temperature ${\text{(T}}_2\text{)}$ of the 1-pentene is to be calculated when some amount of heat is given to the to the 1-pentene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

  $Q=n\Delta H=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}$.     ...(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 4.3P

  $T=\text{457}\text{.855}\text{K}$

Explanation of Solution

Given information:

  $\text{Q= 12 kJ/mol}$

  ${\text{initial temperature(T}}_{\text{1}}{\text{) =100}}^{\text{0}}\text{C}$. So,

  $T_0{\text{=100}}^{\text{0}}\text{C}\Rightarrow T_0{\text{=100}}^{\text{0}}\text{C+273}\text{.15=373}\text{.15}\text{K}$

  $T={?}^{\text{o}}\text{C=}T\text{+273}\text{.15}\text{K}$

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

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ 1-pentene2.69139.753-12.4470\end{array}$

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

Put values in equation (2)

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=2.691\times 373.15\times (\tau -1)+\frac{39.753\times {10}^{-3}}{2}{373.15}^2({\tau }^2-1)\\ +\frac{-12.447\times {10}^{-6}}{3}\times {373.15}^3({\tau }^3-1)+\frac{0}{373.15}\times (\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=1004.147(\tau -1)+2767.62({\tau }^2-1)-215.57({\tau }^3-1)\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}1004.147\tau -1004.147+2767.62{\tau }^2-2767.62-215.57{\tau }^3+215.57\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}-215.57{\tau }^3+2767.62{\tau }^2+1004.147\tau -3556.197\end{array}$

Given that

  $\begin{array}{l}\text{Q=12}\frac{\text{kJ}}{\text{mol}}\\ \Rightarrow Q=12\frac{\text{kJ}}{\text{mol}}\times \frac{1000\text{ J}}{\text{1}\text{kJ}}=12000\frac{\text{J}}{\text{mol}}\end{array}$

Now from equation (1),

  $\begin{array}{l}12000\frac{\text{J}}{\text{mol}}=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 12000\frac{\text{J}}{\text{mol}}=\text{8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}\text{×}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 1443.349\text{K}={\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\end{array}$

Put in abovE

  $\begin{array}{l}1443.349\text{K=}-215.57{\tau }^3+2767.62{\tau }^2+1004.147\tau -3556.197\\ \\ 215.57{\tau }^3-2767.62{\tau }^2-1004.147\tau +4999.546=0\end{array}$

On solving,

  $\tau =1.227,-1.447,13.059$

Considering positive value of temperature $\tau =1.227$

Hence,

  $\begin{array}{l}\tau =\frac{T}{T_0}=\frac{T}{100+273.15}=1.227\\ \\ T=\text{457}\text{.855}\text{K}\end{array}$

(i)

Interpretation Introduction

Interpretation:

Final temperature ${\text{(T}}_2\text{)}$ of the 1-heptene is to be calculated when some amount of heat is given to the to the 1-heptene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

  $Q=n\Delta H=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}$.     ...(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 4.3P

  $T=\text{433}\text{.97}\text{K}$

Explanation of Solution

Given information:

  $\text{Q= 12 kJ/mol}$

  ${\text{initial temperature(T}}_{\text{1}}{\text{) =100}}^{\text{0}}\text{C}$. So,

  $T_0{\text{=100}}^{\text{0}}\text{C}\Rightarrow T_0{\text{=100}}^{\text{0}}\text{C+273}\text{.15=373}\text{.15}\text{K}$

  $T={?}^{\text{o}}\text{C=}T\text{+273}\text{.15}\text{K}$

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

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ 1-heptene3.76856.588-17.8470\end{array}$

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

Put values in equation (2)

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=3.768\times 373.15\times (\tau -1)+\frac{56.588\times {10}^{-3}}{2}{373.15}^2({\tau }^2-1)\\ +\frac{-17.847\times {10}^{-6}}{3}\times {373.15}^3({\tau }^3-1)+\frac{0}{373.15}\times (\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=1406.03(\tau -1)+3939.68({\tau }^2-1)-309.097({\tau }^3-1)\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}1406.03\tau -1406.03+3939.68{\tau }^2-3939.68-309.097{\tau }^3+309.097\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}-309.097{\tau }^3+3939.68{\tau }^2+1406.03\tau -5036.61\end{array}$

Given that

  $\begin{array}{l}\text{Q=12}\frac{\text{kJ}}{\text{mol}}\\ \Rightarrow Q=12\frac{\text{kJ}}{\text{mol}}\times \frac{1000\text{ J}}{\text{1}\text{kJ}}=12000\frac{\text{J}}{\text{mol}}\end{array}$

Now from equation (1),

  $\begin{array}{l}12000\frac{\text{J}}{\text{mol}}=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 12000\frac{\text{J}}{\text{mol}}=\text{8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}\text{×}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 1443.349\text{K}={\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\end{array}$

Put in abovE

  $\begin{array}{l}1443.349\text{K=}-309.097{\tau }^3+3939.68{\tau }^2+1406.03\tau -5036.61\\ \\ 309.097{\tau }^3-3939.68{\tau }^2-1406.03\tau +6479.96=0\end{array}$

On solving,

  $\tau =1.163,-1.389,12.97$

Considering positive value of temperature $\tau =1.163$

Hence,

  $\begin{array}{l}\tau =\frac{T}{T_0}=\frac{T}{100+273.15}=1.163\\ \\ T=\text{433}\text{.97}\text{K}\end{array}$

(j)

Interpretation Introduction

Interpretation:

Final temperature ${\text{(T}}_2\text{)}$ of the 1-Octene is to be calculated when some amount of heat is given to the to the 1-Octene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

  $Q=n\Delta H=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}$.     ...(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 4.3P

  $T=\text{426}\text{.51}\text{K}$

Explanation of Solution

Given information:

  $\text{Q= 12 kJ/mol}$

  ${\text{initial temperature(T}}_{\text{1}}{\text{) =100}}^{\text{0}}\text{C}$. So,

  $T_0{\text{=100}}^{\text{0}}\text{C}\Rightarrow T_0{\text{=100}}^{\text{0}}\text{C+273}\text{.15=373}\text{.15}\text{K}$

  $T={?}^{\text{o}}\text{C=}T\text{+273}\text{.15}\text{K}$

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

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ 1-Octene4.32464.96-20.5210\end{array}$

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

Put values in equation (2)

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=4.324\times 373.15\times (\tau -1)+\frac{64.96\times {10}^{-3}}{2}{373.15}^2({\tau }^2-1)\\ +\frac{-20.521\times {10}^{-6}}{3}\times {373.15}^3({\tau }^3-1)+\frac{0}{373.15}\times (\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=1613.5(\tau -1)+4522.55({\tau }^2-1)-355.41({\tau }^3-1)\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}1613.5\tau -1613.5+4522.55{\tau }^2-4522.55-355.41{\tau }^3+355.41\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}-355.41{\tau }^3+4522.55{\tau }^2+1613.5\tau -5780.64\end{array}$

Given that

  $\begin{array}{l}\text{Q=12}\frac{\text{kJ}}{\text{mol}}\\ \Rightarrow Q=12\frac{\text{kJ}}{\text{mol}}\times \frac{1000\text{ J}}{\text{1}\text{kJ}}=12000\frac{\text{J}}{\text{mol}}\end{array}$

Now from equation (1),

  $\begin{array}{l}12000\frac{\text{J}}{\text{mol}}=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 12000\frac{\text{J}}{\text{mol}}=\text{8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}\text{×}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 1443.349\text{K}={\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\end{array}$

Put in abovE

  $\begin{array}{l}1443.349\text{K=}-355.41{\tau }^3+4522.55{\tau }^2+1613.5\tau -5780.64\\ \\ 355.41{\tau }^3-4522.55{\tau }^2-1613.5\tau +7223.99=0\end{array}$

On solving,

  $\tau =1.143,-1.37,12.95$

Considering positive value of temperature $\tau =1.143$

Hence,

  $\begin{array}{l}\tau =\frac{T}{T_0}=\frac{T}{100+273.15}=1.143\\ \\ T=\text{426}\text{.51}\text{K}\end{array}$

(k)

Interpretation Introduction

Interpretation:

Final temperature ${\text{(T}}_2\text{)}$ of the acetyleneis to be calculated when some amount of heat is given to the to the acetylene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

  $Q=n\Delta H=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}$.     ...(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 4.3P

  $T=\text{595}\text{.55}\text{K}$

Explanation of Solution

Given information:

  $\text{Q= 12 kJ/mol}$

  ${\text{initial temperature(T}}_{\text{1}}{\text{) =100}}^{\text{0}}\text{C}$. So,

  $T_0{\text{=100}}^{\text{0}}\text{C}\Rightarrow T_0{\text{=100}}^{\text{0}}\text{C+273}\text{.15=373}\text{.15}\text{K}$

  $T={?}^{\text{o}}\text{C=}T\text{+273}\text{.15}\text{K}$

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

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ acetylene6.1321.9520-1.299\end{array}$

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

Put values in equation (2)

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=6.132\times 373.15\times (\tau -1)+\frac{1.952\times {10}^{-3}}{2}{373.15}^2({\tau }^2-1)\\ +\frac{0}{3}\times {373.15}^3({\tau }^3-1)+\frac{-1.299\times {10}^5}{373.15}\times (\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=2288.15(\tau -1)+135.9({\tau }^2-1)-348.12(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}2288.15\tau -2288.15+135.9{\tau }^2-135.9-348.12+\frac{348.12}{\tau }\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}135.9{\tau }^2+2288.15\tau -2772.17+\frac{348.12}{\tau }\end{array}$

Given that

  $\begin{array}{l}\text{Q=12}\frac{\text{kJ}}{\text{mol}}\\ \Rightarrow Q=12\frac{\text{kJ}}{\text{mol}}\times \frac{1000\text{ J}}{\text{1}\text{kJ}}=12000\frac{\text{J}}{\text{mol}}\end{array}$

Now from equation (1),

  $\begin{array}{l}12000\frac{\text{J}}{\text{mol}}=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 12000\frac{\text{J}}{\text{mol}}=\text{8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}\text{×}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 1443.349\text{K}={\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\end{array}$

Put in abovE

  $\begin{array}{l}1443.349\text{K=}135.9{\tau }^2+2288.15\tau -2772.17+\frac{348.12}{\tau }\\ \\ 135.9{\tau }^2+2288.15\tau -4215.519+\frac{348.12}{\tau }=0\\ \\ \frac{135.9{\tau }^3+2288.15{\tau }^2-4215.519\tau +348.12}{\tau }=0\end{array}$

On solving,

  $\tau =1.596,0.099,-18.52$

Considering positive value of temperature $\tau =1.596$

Hence,

  $\begin{array}{l}\tau =\frac{T}{T_0}=\frac{T}{100+273.15}=1.596\\ \\ T=\text{595}\text{.55}\text{K}\end{array}$

(l)

Interpretation Introduction

Interpretation:

Final temperature ${\text{(T}}_2\text{)}$ of the benzene is to be calculated when some amount of heat is given to the to the benzene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

  $Q=n\Delta H=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}$.     ...(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 4.3P

  $T=\text{476}\text{.51}\text{K}$

Explanation of Solution

Given information:

  $\text{Q= 12 kJ/mol}$

  ${\text{initial temperature(T}}_{\text{1}}{\text{) =100}}^{\text{0}}\text{C}$. So,

  $T_0{\text{=100}}^{\text{0}}\text{C}\Rightarrow T_0{\text{=100}}^{\text{0}}\text{C+273}\text{.15=373}\text{.15}\text{K}$

  $T={?}^{\text{o}}\text{C=}T\text{+273}\text{.15}\text{K}$

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

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ benzene-0.20639.064-13.3010\end{array}$

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

Put values in equation (2)

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=-0.206\times 373.15\times (\tau -1)+\frac{39.064\times {10}^{-3}}{2}{373.15}^2({\tau }^2-1)\\ +\frac{-13.301\times {10}^{-6}}{3}\times {373.15}^3({\tau }^3-1)+\frac{0}{373.15}\times (\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=-76.87(\tau -1)+2719.65({\tau }^2-1)-230.36({\tau }^3-1)\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}-76.87\tau +76.87+2719.65{\tau }^2-2719.65-230.36{\tau }^3+230.36\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}-230.36{\tau }^3+2719.65{\tau }^2-76.87\tau -2412.42\end{array}$

Given that

  $\begin{array}{l}\text{Q=12}\frac{\text{kJ}}{\text{mol}}\\ \Rightarrow Q=12\frac{\text{kJ}}{\text{mol}}\times \frac{1000\text{ J}}{\text{1}\text{kJ}}=12000\frac{\text{J}}{\text{mol}}\end{array}$

Now from equation (1),

  $\begin{array}{l}12000\frac{\text{J}}{\text{mol}}=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 12000\frac{\text{J}}{\text{mol}}=\text{8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}\text{×}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 1443.349\text{K}={\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\end{array}$

Put in abovE

  $\begin{array}{l}1443.349\text{K=}-230.36{\tau }^3+2719.65{\tau }^2-76.87\tau -2412.42\\ \\ 230.36{\tau }^3-2719.65{\tau }^2+76.87\tau +3855.769=0\end{array}$

On solving,

  $\tau =1.277,-1.12,11.65$

Considering positive value of temperature $\tau =1.277$

Hence,

  $\begin{array}{l}\tau =\frac{T}{T_0}=\frac{T}{100+273.15}=1.277\\ \\ T=\text{476}\text{.51}\text{K}\end{array}$

(m)

Interpretation Introduction

Interpretation:

Final temperature ${\text{(T}}_2\text{)}$ of the ethanolis to be calculated when some amount of heat is given to the to the ethanol.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

  $Q=n\Delta H=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}$.     ...(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 4.3P

  $T=\text{380}\text{.613}\text{K}$

Explanation of Solution

Given information:

  $\text{Q= 12 kJ/mol}$

  ${\text{initial temperature(T}}_{\text{1}}{\text{) =100}}^{\text{0}}\text{C}$. So,

  $T_0{\text{=100}}^{\text{0}}\text{C}\Rightarrow T_0{\text{=100}}^{\text{0}}\text{C+273}\text{.15=373}\text{.15}\text{K}$

  $T={?}^{\text{o}}\text{C=}T\text{+273}\text{.15}\text{K}$

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

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ ethanol3.51820.001-6.0020\end{array}$

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

Put values in equation (2)

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=3.518\times 373.15\times (\tau -1)+\frac{20.001\times {10}^{-3}}{2}{373.15}^2({\tau }^2-1)\\ +\frac{-6.002\times {10}^{-6}}{3}\times {373.15}^3({\tau }^3-1)+\frac{0}{373.15}\times (\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=1312.74(\tau -1)+1392.48({\tau }^2-1)-103.95({\tau }^3-1)\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}1312.74\tau -1312.74+1392.48{\tau }^2-1392.48-103.95{\tau }^3+103.95\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}-103.95{\tau }^3+2719.65{\tau }^2+1312.74\tau -2601.27\end{array}$

Given that

  $\begin{array}{l}\text{Q=12}\frac{\text{kJ}}{\text{mol}}\\ \Rightarrow Q=12\frac{\text{kJ}}{\text{mol}}\times \frac{1000\text{ J}}{\text{1}\text{kJ}}=12000\frac{\text{J}}{\text{mol}}\end{array}$

Now from equation (1),

  $\begin{array}{l}12000\frac{\text{J}}{\text{mol}}=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 12000\frac{\text{J}}{\text{mol}}=\text{8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}\text{×}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 1443.349\text{K}={\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\end{array}$

Put in abovE

  $\begin{array}{l}1443.349\text{K=}-103.95{\tau }^3+2719.65{\tau }^2+1312.74\tau -2601.27\\ \\ 103.95{\tau }^3-2719.65{\tau }^2-1312.74\tau +4044.62=0\end{array}$

On solving,

  $\tau =1.02,-1.44,26.58$

Considering positive value of temperature $\tau =1.02$

Hence,

  $\begin{array}{l}\tau =\frac{T}{T_0}=\frac{T}{100+273.15}=1.02\\ \\ T=\text{380}\text{.613}\text{K}\end{array}$

(n)

Interpretation Introduction

Interpretation:

Final temperature ${\text{(T}}_2\text{)}$ of the styreneis to be calculated when some amount of heat is given to the to the styrene.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

  $Q=n\Delta H=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}$.     ...(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 4.3P

  $T=\text{445}\text{.91}\text{K}$

Explanation of Solution

Given information:

  $\text{Q= 12 kJ/mol}$

  ${\text{initial temperature(T}}_{\text{1}}{\text{) =100}}^{\text{0}}\text{C}$. So,

  $T_0{\text{=100}}^{\text{0}}\text{C}\Rightarrow T_0{\text{=100}}^{\text{0}}\text{C+273}\text{.15=373}\text{.15}\text{K}$

  $T={?}^{\text{o}}\text{C=}T\text{+273}\text{.15}\text{K}$

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

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ styrene2.05050.192-16.6620\end{array}$

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

Put values in equation (2)

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=2.050\times 373.15\times (\tau -1)+\frac{50.192\times {10}^{-3}}{2}{373.15}^2({\tau }^2-1)\\ +\frac{-16.662\times {10}^{-6}}{3}\times {373.15}^3({\tau }^3-1)+\frac{0}{373.15}\times (\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=764.96(\tau -1)+3494.39({\tau }^2-1)-288.57({\tau }^3-1)\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}764.96\tau -764.96+3494.39{\tau }^2-3494.39-288.57{\tau }^3+288.57\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}-288.57{\tau }^3+3494.39{\tau }^2+764.96\tau -3970.78\end{array}$

Given that

  $\begin{array}{l}\text{Q=12}\frac{\text{kJ}}{\text{mol}}\\ \Rightarrow Q=12\frac{\text{kJ}}{\text{mol}}\times \frac{1000\text{ J}}{\text{1}\text{kJ}}=12000\frac{\text{J}}{\text{mol}}\end{array}$

Now from equation (1),

  $\begin{array}{l}12000\frac{\text{J}}{\text{mol}}=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 12000\frac{\text{J}}{\text{mol}}=\text{8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}\text{×}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 1443.349\text{K}={\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\end{array}$

Put in abovE

  $\begin{array}{l}1443.349\text{K=}-288.57{\tau }^3+3494.39{\tau }^2+764.96\tau -3970.78\\ \\ 288.57{\tau }^3-3494.39{\tau }^2-764.96\tau +5414.126=0\end{array}$

On solving,

  $\tau =1.195,-1.287,12.2$

Considering positive value of temperature $\tau =1.195$

Hence,

  $\begin{array}{l}\tau =\frac{T}{T_0}=\frac{T}{100+273.15}=1.195\\ \\ T=\text{445}\text{.91}\text{K}\end{array}$

(o)

Interpretation Introduction

Interpretation:

Final temperature ${\text{(T}}_2\text{)}$ of the formaldehydeis to be calculated when some amount of heat is given to the to the formaldehyde.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

  $Q=n\Delta H=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}$.     ...(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 4.3P

  $T=\text{643}\text{.68}\text{K}$

Explanation of Solution

Given information:

  $\text{Q= 12 kJ/mol}$

  ${\text{initial temperature(T}}_{\text{1}}{\text{) =100}}^{\text{0}}\text{C}$. So,

  $T_0{\text{=100}}^{\text{0}}\text{C}\Rightarrow T_0{\text{=100}}^{\text{0}}\text{C+273}\text{.15=373}\text{.15}\text{K}$

  $T={?}^{\text{o}}\text{C=}T\text{+273}\text{.15}\text{K}$

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

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ formaldehyde2.2647.022-1.8770\end{array}$

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

Put values in equation (2)

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=2.264\times 373.15\times (\tau -1)+\frac{7.022\times {10}^{-3}}{2}{373.15}^2({\tau }^2-1)\\ +\frac{-1.877\times {10}^{-6}}{3}\times {373.15}^3({\tau }^3-1)+\frac{0}{373.15}\times (\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=844.81(\tau -1)+488.88({\tau }^2-1)-32.51({\tau }^3-1)\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}844.81\tau -844.81+488.88{\tau }^2-488.88-32.51{\tau }^3+32.51\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=-}32.51{\tau }^3+488.88{\tau }^2+844.81\tau -1301.18\end{array}$

Given that

  $\begin{array}{l}\text{Q=12}\frac{\text{kJ}}{\text{mol}}\\ \Rightarrow Q=12\frac{\text{kJ}}{\text{mol}}\times \frac{1000\text{ J}}{\text{1}\text{kJ}}=12000\frac{\text{J}}{\text{mol}}\end{array}$

Now from equation (1),

  $\begin{array}{l}12000\frac{\text{J}}{\text{mol}}=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 12000\frac{\text{J}}{\text{mol}}=\text{8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}\text{×}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 1443.349\text{K}={\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\end{array}$

Put in abovE

  $\begin{array}{l}1443.349\text{K=-}32.51{\tau }^3+488.88{\tau }^2+844.81\tau -1301.18\\ \\ 32.51{\tau }^3-488.88{\tau }^2-844.81\tau +2744.53=0\end{array}$

On solving,

  $\tau =1.725,-3,16.31$

Considering positive value of temperature $\tau =1.725$

Hence,

  $\begin{array}{l}\tau =\frac{T}{T_0}=\frac{T}{100+273.15}=1.725\\ \\ T=\text{643}\text{.68}\text{K}\end{array}$

(p)

Interpretation Introduction

Interpretation:

Final temperature ${\text{(T}}_2\text{)}$ of the ammoniais to be calculated when some amount of heat is given to the to the ammonia.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

  $Q=n\Delta H=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}$.     ...(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 4.3P

  $T=\text{416}\text{.06}\text{K}$

Explanation of Solution

Given information:

  $\text{Q= 12 kJ/mol}$

  ${\text{initial temperature(T}}_{\text{1}}{\text{) =100}}^{\text{0}}\text{C}$. So,

  $T_0{\text{=100}}^{\text{0}}\text{C}\Rightarrow T_0{\text{=100}}^{\text{0}}\text{C+273}\text{.15=373}\text{.15}\text{K}$

  $T={?}^{\text{o}}\text{C=}T\text{+273}\text{.15}\text{K}$

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

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ ammonia3.5783.0200-0.186\end{array}$

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

Put values in equation (2)

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=3.578\times 373.15\times (\tau -1)+\frac{3.020\times {10}^{-3}}{2}{373.15}^2({\tau }^2-1)\\ +\frac{0}{3}\times {373.15}^3({\tau }^3-1)+\frac{-0.186\times {10}^5}{373.15}\times (\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=1335.13(\tau -1)+210.25({\tau }^2-1)-49.85(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}1335.13\tau -1335.13+210.25{\tau }^2-210.25-49.85+\frac{49.85}{\tau }\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}210.25{\tau }^2+1335.13\tau -1595.23+\frac{49.85}{\tau }\end{array}$

Given that

  $\begin{array}{l}\text{Q=12}\frac{\text{kJ}}{\text{mol}}\\ \Rightarrow Q=12\frac{\text{kJ}}{\text{mol}}\times \frac{1000\text{ J}}{\text{1}\text{kJ}}=12000\frac{\text{J}}{\text{mol}}\end{array}$

Now from equation (1),

  $\begin{array}{l}12000\frac{\text{J}}{\text{mol}}=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 12000\frac{\text{J}}{\text{mol}}=\text{8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}\text{×}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 1443.349\text{K}={\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\end{array}$

Put in abovE

  $\begin{array}{l}1443.349\text{K=}210.25{\tau }^2+1335.13\tau -1595.23+\frac{49.85}{\tau }\\ \\ 1210.25{\tau }^2+1335.13\tau -3038.579+\frac{49.85}{\tau }=0\\ \\ \frac{1210.25{\tau }^3+1335.13{\tau }^2-3038.579\tau +49.85}{\tau }=0\end{array}$

On solving,

  $\tau =1.115,0.0165,-2.23$

Considering positive value of temperature $\tau =1.596$

Hence,

  $\begin{array}{l}\tau =\frac{T}{T_0}=\frac{T}{100+273.15}=1.115\\ \\ T=\text{416}\text{.06}\text{K}\end{array}$

(q)

Interpretation Introduction

Interpretation:

Final temperature ${\text{(T}}_2\text{)}$ of the carbon monoxideis to be calculated when some amount of heat is given to the to the carbon monoxide.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

  $Q=n\Delta H=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}$.     ...(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 4.3P

  $T=\text{764}\text{.96}\text{K}$

Explanation of Solution

Given information:

  $\text{Q= 12 kJ/mol}$

  ${\text{initial temperature(T}}_{\text{1}}{\text{) =100}}^{\text{0}}\text{C}$. So,

  $T_0{\text{=100}}^{\text{0}}\text{C}\Rightarrow T_0{\text{=100}}^{\text{0}}\text{C+273}\text{.15=373}\text{.15}\text{K}$

  $T={?}^{\text{o}}\text{C=}T\text{+273}\text{.15}\text{K}$

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

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ CO3.3760.5570-0.031\end{array}$

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

Put values in equation (2)

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=3.376\times 373.15\times (\tau -1)+\frac{0.557\times {10}^{-3}}{2}{373.15}^2({\tau }^2-1)\\ +\frac{0}{3}\times {373.15}^3({\tau }^3-1)+\frac{-0.031\times {10}^5}{373.15}\times (\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=1259.75(\tau -1)+38.78({\tau }^2-1)-8.31(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}1259.75\tau -1259.75+38.78{\tau }^2-38.78-8.31+\frac{8.31}{\tau }\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}38.78{\tau }^2+1259.75\tau -1306.84+\frac{8.31}{\tau }\end{array}$

Given that

  $\begin{array}{l}\text{Q=12}\frac{\text{kJ}}{\text{mol}}\\ \Rightarrow Q=12\frac{\text{kJ}}{\text{mol}}\times \frac{1000\text{ J}}{\text{1}\text{kJ}}=12000\frac{\text{J}}{\text{mol}}\end{array}$

Now from equation (1),

  $\begin{array}{l}12000\frac{\text{J}}{\text{mol}}=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 12000\frac{\text{J}}{\text{mol}}=\text{8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}\text{×}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 1443.349\text{K}={\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\end{array}$

Put in abovE

  $\begin{array}{l}1443.349\text{K=}38.78{\tau }^2+1259.75\tau -1306.84+\frac{8.31}{\tau }\\ \\ 38.78{\tau }^2+1259.75\tau -2750.19+\frac{8.31}{\tau }=0\\ \\ \frac{38.78{\tau }^3+1259.75{\tau }^2-2750.19\tau +8.31}{\tau }=0\end{array}$

On solving,

  $\tau =2.05,0.003,-34.54$

Considering positive value of temperature $\tau =2.05$

Hence,

  $\begin{array}{l}\tau =\frac{T}{T_0}=\frac{T}{100+273.15}=2.05\\ \\ T=\text{764}\text{.96}\text{K}\end{array}$

(r)

Interpretation Introduction

Interpretation:

Final temperature ${\text{(T}}_2\text{)}$ of the carbon dioxide is to be calculated when some amount of heat is given to the to the carbon dioxide.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

  $Q=n\Delta H=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}$.     ...(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 4.3P

  $T=\text{634}\text{.355}\text{K}$

Explanation of Solution

Given information:

  $\text{Q= 12 kJ/mol}$

  ${\text{initial temperature(T}}_{\text{1}}{\text{) =100}}^{\text{0}}\text{C}$. So,

  $T_0{\text{=100}}^{\text{0}}\text{C}\Rightarrow T_0{\text{=100}}^{\text{0}}\text{C+273}\text{.15=373}\text{.15}\text{K}$

  $T={?}^{\text{o}}\text{C=}T\text{+273}\text{.15}\text{K}$

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

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ C{O}_{2}5.4571.0450-1.157\end{array}$

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

Put values in equation (2)

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=5.457\times 373.15\times (\tau -1)+\frac{1.045\times {10}^{-3}}{2}{373.15}^2({\tau }^2-1)\\ +\frac{0}{3}\times {373.15}^3({\tau }^3-1)+\frac{-1.157\times {10}^5}{373.15}\times (\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=2036.28(\tau -1)+72.75({\tau }^2-1)-310.06(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}2036.28\tau -2036.28+72.75{\tau }^2-72.75-310.06+\frac{310.06}{\tau }\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}72.75{\tau }^2+2036.28\tau -2419.09+\frac{310.06}{\tau }\end{array}$

Given that

  $\begin{array}{l}\text{Q=12}\frac{\text{kJ}}{\text{mol}}\\ \Rightarrow Q=12\frac{\text{kJ}}{\text{mol}}\times \frac{1000\text{ J}}{\text{1}\text{kJ}}=12000\frac{\text{J}}{\text{mol}}\end{array}$

Now from equation (1),

  $\begin{array}{l}12000\frac{\text{J}}{\text{mol}}=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 12000\frac{\text{J}}{\text{mol}}=\text{8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}\text{×}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 1443.349\text{K}={\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\end{array}$

Put in abovE

  $\begin{array}{l}1443.349\text{K=}72.75{\tau }^2+2036.28\tau -2419.09+\frac{310.06}{\tau }\\ \\ 72.75{\tau }^2+2036.28\tau -3862.44+\frac{310.06}{\tau }=0\\ \\ \frac{72.75{\tau }^3+2036.28{\tau }^2-3862.44\tau +310.06}{\tau }=0\end{array}$

On solving,

  $\tau =1.7,0.084,-29.78$

Considering positive value of temperature $\tau =1.7$

Hence,

  $\begin{array}{l}\tau =\frac{T}{T_0}=\frac{T}{100+273.15}=1.7\\ \\ T=\text{634}\text{.355}\text{K}\end{array}$

(s)

Interpretation Introduction

Interpretation:

Final temperature ${\text{(T}}_2\text{)}$ of the sulphur dioxide is to be calculated when some amount of heat is given to the to the sulphur dioxide.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

  $Q=n\Delta H=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}$.     ...(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 4.3P

  $T=\text{626}\text{.1457}\text{K}$

Explanation of Solution

Given information:

  $\text{Q= 12 kJ/mol}$

  ${\text{initial temperature(T}}_{\text{1}}{\text{) =100}}^{\text{0}}\text{C}$. So,

  $T_0{\text{=100}}^{\text{0}}\text{C}\Rightarrow T_0{\text{=100}}^{\text{0}}\text{C+273}\text{.15=373}\text{.15}\text{K}$

  $T={?}^{\text{o}}\text{C=}T\text{+273}\text{.15}\text{K}$

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

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

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

Put values in equation (2)

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=5.699\times 373.15\times (\tau -1)+\frac{0.801\times {10}^{-3}}{2}{373.15}^2({\tau }^2-1)\\ +\frac{0}{3}\times {373.15}^3({\tau }^3-1)+\frac{-1.015\times {10}^5}{373.15}\times (\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=2126.58(\tau -1)+55.77({\tau }^2-1)-272(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}2126.58\tau -2126.58+55.77{\tau }^2-55.77-272+\frac{272}{\tau }\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}55.77{\tau }^2+2126.58\tau -2454.36+\frac{272}{\tau }\end{array}$

Given that

  $\begin{array}{l}\text{Q=12}\frac{\text{kJ}}{\text{mol}}\\ \Rightarrow Q=12\frac{\text{kJ}}{\text{mol}}\times \frac{1000\text{ J}}{\text{1}\text{kJ}}=12000\frac{\text{J}}{\text{mol}}\end{array}$

Now from equation (1),

  $\begin{array}{l}12000\frac{\text{J}}{\text{mol}}=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 12000\frac{\text{J}}{\text{mol}}=\text{8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}\text{×}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 1443.349\text{K}={\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\end{array}$

Put in abovE

  $\begin{array}{l}1443.349\text{K=}55.77{\tau }^2+2126.58\tau -2454.36+\frac{272}{\tau }\\ \\ 55.77{\tau }^2+2126.58\tau -3887.71+\frac{272}{\tau }=0\\ \\ \frac{55.77{\tau }^3+2126.58{\tau }^2-3887.71\tau +272}{\tau }=0\end{array}$

On solving,

  $\tau =1.678,0.073,-39.88$

Considering positive value of temperature $\tau =1.678$

Hence,

  $\begin{array}{l}\tau =\frac{T}{T_0}=\frac{T}{100+273.15}=1.678\\ \\ T=\text{626}\text{.1457}\text{K}\end{array}$

(t)

Interpretation Introduction

Interpretation:

Final temperature ${\text{(T}}_2\text{)}$ of the wateris to be calculated when some amount of heat is given to the to the water.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

  $Q=n\Delta H=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}$.     ...(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 4.3P

  $T=\text{706}\text{.37}\text{K}$

Explanation of Solution

Given information:

  $\text{Q= 12 kJ/mol}$

  ${\text{initial temperature(T}}_{\text{1}}{\text{) =100}}^{\text{0}}\text{C}$. So,

  $T_0{\text{=100}}^{\text{0}}\text{C}\Rightarrow T_0{\text{=100}}^{\text{0}}\text{C+273}\text{.15=373}\text{.15}\text{K}$

  $T={?}^{\text{o}}\text{C=}T\text{+273}\text{.15}\text{K}$

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

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ H_{2}O3.471.45000.121\end{array}$

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

Put values in equation (2)

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=3.470\times 373.15\times (\tau -1)+\frac{1.450\times {10}^{-3}}{2}{373.15}^2({\tau }^2-1)\\ +\frac{0}{3}\times {373.15}^3({\tau }^3-1)+\frac{0.121\times {10}^5}{373.15}\times (\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=1294.83(\tau -1)+100.95({\tau }^2-1)+32.43(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}1294.83\tau -1294.83+100.95{\tau }^2-100.95+32.43-\frac{32.43}{\tau }\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}100.95{\tau }^2+1294.83\tau -1363.35-\frac{32.43}{\tau }\end{array}$

Given that

  $\begin{array}{l}\text{Q=12}\frac{\text{kJ}}{\text{mol}}\\ \Rightarrow Q=12\frac{\text{kJ}}{\text{mol}}\times \frac{1000\text{ J}}{\text{1}\text{kJ}}=12000\frac{\text{J}}{\text{mol}}\end{array}$

Now from equation (1),

  $\begin{array}{l}12000\frac{\text{J}}{\text{mol}}=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 12000\frac{\text{J}}{\text{mol}}=\text{8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}\text{×}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 1443.349\text{K}={\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\end{array}$

Put in abovE

  $\begin{array}{l}1443.349\text{K=}100.95{\tau }^2+1294.83\tau -1363.35-\frac{32.43}{\tau }\\ \\ 100.95{\tau }^2+1294.83\tau -2796.7-\frac{32.43}{\tau }=0\\ \\ \frac{100.95{\tau }^3+1294.83{\tau }^2-2796.7\tau -32.43}{\tau }=0\end{array}$

On solving,

  $\tau =1.893,-0.011,-14.708$

Considering positive value of temperature $\tau =1.893$

Hence,

  $\begin{array}{l}\tau =\frac{T}{T_0}=\frac{T}{100+273.15}=1.893\\ \\ T=\text{706}\text{.37}\text{K}\end{array}$

(u)

Interpretation Introduction

Interpretation:

Final temperature ${\text{(T}}_2\text{)}$ of the nitrogen is to be calculated when some amount of heat is given to the to the nitrogen.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

  $Q=n\Delta H=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}$.     ...(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 4.3P

  $T=\text{770}\text{.55}\text{K}$

Explanation of Solution

Given information:

  $\text{Q= 12 kJ/mol}$

  ${\text{initial temperature(T}}_{\text{1}}{\text{) =100}}^{\text{0}}\text{C}$. So,

  $T_0{\text{=100}}^{\text{0}}\text{C}\Rightarrow T_0{\text{=100}}^{\text{0}}\text{C+273}\text{.15=373}\text{.15}\text{K}$

  $T={?}^{\text{o}}\text{C=}T\text{+273}\text{.15}\text{K}$

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

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ N_{2}3.2800.59300.040\end{array}$

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

Put values in equation (2)

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=3.280\times 373.15\times (\tau -1)+\frac{0.593\times {10}^{-3}}{2}{373.15}^2({\tau }^2-1)\\ +\frac{0}{3}\times {373.15}^3({\tau }^3-1)+\frac{0.040\times {10}^5}{373.15}\times (\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=1223.93(\tau -1)+41.29({\tau }^2-1)+10.72(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}1223.93\tau -1223.93+41.29{\tau }^2-41.29+10.72-\frac{10.72}{\tau }\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}41.29{\tau }^2+1223.93\tau -1254.5-\frac{10.72}{\tau }\end{array}$

Given that

  $\begin{array}{l}\text{Q=12}\frac{\text{kJ}}{\text{mol}}\\ \Rightarrow Q=12\frac{\text{kJ}}{\text{mol}}\times \frac{1000\text{ J}}{\text{1}\text{kJ}}=12000\frac{\text{J}}{\text{mol}}\end{array}$

Now from equation (1),

  $\begin{array}{l}12000\frac{\text{J}}{\text{mol}}=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 12000\frac{\text{J}}{\text{mol}}=\text{8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}\text{×}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 1443.349\text{K}={\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\end{array}$

Put in abovE

  $\begin{array}{l}1443.349\text{K=}41.29{\tau }^2+1223.93\tau -1254.5-\frac{10.72}{\tau }\\ \\ 41.29{\tau }^2+1223.93\tau -2697.85-\frac{10.72}{\tau }=0\\ \\ \frac{41.29{\tau }^3+1223.93{\tau }^2-2697.85\tau -10.72}{\tau }=0\end{array}$

On solving,

  $\tau =2.065,-0.003,-31.703$

Considering positive value of temperature $\tau =2.065$

Hence,

  $\begin{array}{l}\tau =\frac{T}{T_0}=\frac{T}{100+273.15}=2.065\\ \\ T=\text{770}\text{.55}\text{K}\end{array}$

(v)

Interpretation Introduction

Interpretation:

Final temperature ${\text{(T}}_2\text{)}$ of the hydrogen cyanide is to be calculated when some amount of heat is given to the to the hydrogen cyanide.

Concept Introduction:

The final temperature is calculated from below equation where amount of heat is given:

  $Q=n\Delta H=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}$.     ...(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 4.3P

  $T=\text{654}\text{.13}\text{K}$

Explanation of Solution

Given information:

  $\text{Q= 12 kJ/mol}$

  ${\text{initial temperature(T}}_{\text{1}}{\text{) =100}}^{\text{0}}\text{C}$. So,

  $T_0{\text{=100}}^{\text{0}}\text{C}\Rightarrow T_0{\text{=100}}^{\text{0}}\text{C+273}\text{.15=373}\text{.15}\text{K}$

  $T={?}^{\text{o}}\text{C=}T\text{+273}\text{.15}\text{K}$

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

  $\begin{array}{l}iA{10}^{3}B{10}^{6}C{10}^{-5}D\\ HCN4.7361.3590-0.725\end{array}$

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

Put values in equation (2)

  $\begin{array}{l}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=A{T}_0(\tau -1)+\frac{B}{2}{T}_0{}^2({\tau }^2-1)+\frac{C}{3}{T}_0{}^3({\tau }^3-1)+\frac{D}{T_0}(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=4.736\times 373.15\times (\tau -1)+\frac{1.359\times {10}^{-3}}{2}{373.15}^2({\tau }^2-1)\\ +\frac{0}{3}\times {373.15}^3({\tau }^3-1)+\frac{-0.725\times {10}^5}{373.15}\times (\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}=1767.24(\tau -1)+94.61({\tau }^2-1)-194.29(\frac{\tau -1}{\tau })\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}1767.24\tau -1767.24+94.61{\tau }^2-94.61-194.29+\frac{194.29}{\tau }\\ {\displaystyle \underset{373.15}{\overset{T}{\int }}\frac{\Delta C^{\circ }{}_P}{R}dT}\text{=}94.61{\tau }^2+1767.24\tau -2056.14+\frac{194.29}{\tau }\end{array}$

Given that

  $\begin{array}{l}\text{Q=12}\frac{\text{kJ}}{\text{mol}}\\ \Rightarrow Q=12\frac{\text{kJ}}{\text{mol}}\times \frac{1000\text{ J}}{\text{1}\text{kJ}}=12000\frac{\text{J}}{\text{mol}}\end{array}$

Now from equation (1),

  $\begin{array}{l}12000\frac{\text{J}}{\text{mol}}=R{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 12000\frac{\text{J}}{\text{mol}}=\text{8}\text{.314}\frac{\text{J}}{\text{mol}\text{K}}\text{×}{\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\\ 1443.349\text{K}={\displaystyle \underset{T_0}{\overset{T}{\int }}\frac{\Delta C_P}{R}dT}\end{array}$

Put in abovE

  $\begin{array}{l}1443.349\text{K=}94.61{\tau }^2+1767.24\tau -2056.14+\frac{194.29}{\tau }\\ \\ 94.61{\tau }^2+1767.24\tau -3499.49+\frac{194.29}{\tau }=0\\ \\ \frac{94.61{\tau }^2+1767.24\tau -3499.49\tau +194.29}{\tau }=0\end{array}$

On solving,

  $\tau =1.753,0.057,-20.49$

Considering positive value of temperature $\tau =1.753$

Hence,

  $\begin{array}{l}\tau =\frac{T}{T_0}=\frac{T}{100+273.15}=1.753\\ \\ T=\text{654}\text{.13}\text{K}\end{array}$