Chapter 2, Problem 2B.3AE

(i)

Interpretation Introduction

Interpretation: The values of $q\text{, }w\text{, }\Delta U$ and $\Delta H$ have to be calculated at constant pressure when the temperature is raised from ${\text{25 }}^{\text{ο}}\text{C}$ to ${\text{100 }}^{\text{ο}}\text{C}$.

Concept Introduction: The system is said to have gone through a process when the initial state is changed to the final state.  The properties of the system such as temperature, pressure, enthalpy, entropy are changes in thermodynamic process.

Thermodynamics address temperature and heat and their relationship to work, energy and matter.  Work is defined as the form of mechanical energy that is transferred from system to another.

Answer to Problem 2B.3AE

The value of value of $q$ and $\Delta H$ at constant pressure is $\underset{\_}{10.73kJ}$.

The value of $w$ at constant pressure is $\underset{\_}{-0.62355kJ}$.

The value of $\Delta U$ at constant pressure is $\underset{\_}{10.10kJ}$.

Explanation of Solution

It is given that the constant-pressure heat capacity of a sample of a perfect gas varies with temperature according to the expression shown below.

    $C_p/\left({\text{J K}}^{-1}\right)=20.17+0.3665\left(T/K\right)$                                                             (1)

The temperature is raised from ${\text{25 }}^{\text{ο}}\text{C}$ to ${\text{100 }}^{\text{ο}}\text{C}$.

The conversion of Celsius to Kelvin is done as,

    ${\text{0 }}^{\text{ο}}\text{C}=2\text{73 K}$

Therefore, the conversion of ${\text{25 }}^{\text{ο}}\text{C}$ to Kelvin is done as,

    $\begin{array}{c}{\text{25 }}^{\text{ο}}\text{C}=25+2\text{73 K}\\ =2\text{98 K}\end{array}$

Similarly, the conversion of ${\text{100 }}^{\text{ο}}\text{C}$ to Kelvin is done as,

    $\begin{array}{c}{\text{100 }}^{\text{ο}}\text{C}=100+2\text{73 K}\\ =373\text{ K}\end{array}$

At constant pressure,

    $q=\Delta H={\displaystyle \underset{T_1}{\overset{T_2}{\int }}{C}_p}dT$                                                                                          (2)

Substitute the value of heat capacity at constant pressure from equation (1) to equation (2).

  $\begin{array}{c}\Delta H={\displaystyle \underset{T_1}{\overset{T_2}{\int }}{C}_p}dT\\ ={\displaystyle \underset{T_1}{\overset{T_2}{\int }}20.17dT+{\displaystyle \underset{T_1}{\overset{T_2}{\int }}0.366\text{5 }TdT}}\\ =20.17\left(T_2-T_1\right)+0.366\text{5}\left(\frac{T_2^2}{2}-\frac{T_1^2}{2}\right)\end{array}$

Now, substitute the values of initial and final temperature to calculate value of $q$ and $\Delta H$.

    $\begin{array}{c}\Delta H=20.17\left(373-298\right)+0.366\text{5}\left(\frac{{\left(373\right)}^2}{2}-\frac{{\left(298\right)}^2}{2}\right)\\ =20.17\times 75+0.18325\left(139129-88804\right)\\ =10734.8\text{0 J}\\ =\underset{\_}{10.73kJ}\end{array}$

Therefore, the value of value of $q$ and $\Delta H$ at constant pressure is $\underset{\_}{10.73kJ}$ respectively.

The work done for a perfect gas is calculated by the formula,

    $\begin{array}{c}w=-pdV\\ =-nRdT\\ =-(1)RdT\\ =-RdT\end{array}$                                                                                                 (3)

Substitute the values of universal gas constant and temperature change in the above equation to calculate the work done.

    $\begin{array}{c}w=-8.31{\text{4 J mol}}^{-1}{\text{ K}}^{-1}\times \left(37\text{3 K}-29\text{8 K}\right)\\ =-8.31{\text{4 J mol}}^{-1}{\text{ K}}^{-1}\times 7\text{5 K}\\ =-623.5\text{5 J}\\ =\underset{\_}{-0.62355kJ}\end{array}$

Therefore, the work done at constant pressure is $\underset{\_}{-0.62355kJ}$.

The first law of thermodynamics is given by the expression,

    $\Delta U=q+w$                                                                                                    (4)

Substitute the values of $q$ and $w$ in the above formula to calculate the internal energy.

    $\begin{array}{c}\Delta U=10.7\text{3 kJ}+\left(-0.6235\text{5 kJ}\right)\\ =10.7\text{3 kJ}-0.6235\text{5 kJ}\\ =\underset{\_}{10.10kJ}\end{array}$

Hence, the value of $\Delta U$ is $\underset{\_}{10.10kJ}$.

(ii)

Interpretation Introduction

Interpretation: The values of $q\text{, }w\text{, }\Delta U$ and $\Delta H$ have to be calculated at constant volume when the temperature is raised from ${\text{25 }}^{\text{ο}}\text{C}$ to ${\text{100 }}^{\text{ο}}\text{C}$.

Concept Introduction: The system is said to have gone through a process when the initial state is changed to the final state.  The properties of the system such as temperature, pressure, enthalpy, entropy are changes in thermodynamic process.

Thermodynamics address temperature and heat and their relationship to work, energy and matter.  Work is defined as the form of mechanical energy that is transferred from system to another.

Answer to Problem 2B.3AE

The value of $\Delta U$ at constant volume is $\underset{\_}{10.1112kJ}$.

The value of value of $\Delta H$ at constant volume is $\underset{\_}{10.73475kJ}$.

At constant volume, $w=0$.

The value of $q$ at constant volume is $\underset{\_}{10.1112kJ}$.

Explanation of Solution

At constant volume,

    $C_p-C_v=nR$                                                                                                 (5)

For a perfect gas,

    $n=1$

Therefore,

    $\begin{array}{l}{C}_p-C_v=R\\ C_v=C_p-R\end{array}$

Substitute the value of heat capacity at constant pressure in the above formula to calculate the heat capacity at constant volume.

    $\begin{array}{c}{C}_v=C_p-R\\ C_v=20.17+0.3665\times T-8.314\\ C_v=11.856+0.3665\times T\end{array}$

Now,

    $\begin{array}{c}{C}_v={\left(\frac{\partial U}{\partial T}\right)}_v\\ \partial U={\displaystyle \int C_{v}dT}\end{array}$                                                                                                 (6)

Substitute the values of heat capacity at constant volume and integrate it.

    $\begin{array}{c}\partial U={\displaystyle \underset{T_1}{\overset{T_2}{\int }}\left(11.856+0.3665\times T\right)}dT\\ ={\displaystyle \underset{T_1}{\overset{T_2}{\int }}11.856dT}+{\displaystyle \underset{T_1}{\overset{T_2}{\int }}0.3665\times TdT}\\ =11.856\left(T_2-T_1\right)+0.366\text{5}\left(\frac{T_2^2}{2}-\frac{T_1^2}{2}\right)\\ =11.856\left(37\text{3 K}-29\text{8 K}\right)+0.366\text{5}\left(\frac{{\left(373\right)}^2}{2}-\frac{{\left(298\right)}^2}{2}\right)\end{array}$

Simplify the above equation,

    $\begin{array}{c}\Delta U=11.856\times 75+0.18325\left(139129-88804\right)\\ =889.2+9222.0\\ =10111.2\text{ J}\\ =\underset{\_}{10.1112kJ}\end{array}$

Hence, the value of $\Delta U$ is $\underset{\_}{10.1112kJ}$.

The relation between change in enthalpy, change in internal energy, and change in volume is,

    $\Delta H=\Delta U+\Delta \left(pV\right)$                                                                                       (7)

Where,

• Ø  $\Delta H$ is the change in enthalpy.
• Ø  $\Delta U$ is the change in internal energy.
• Ø  $\Delta V$ is the change in volume.
• Ø  $p$ is the pressure.

The equation (7) is also written as,

    $\begin{array}{c}\Delta H=\Delta U+\Delta \left(pV\right)\\ \Delta H=\Delta U+\Delta pV+p\Delta V\end{array}$

At constant volume, $p\Delta V=0$. Therefore,

    $\begin{array}{c}\Delta H=\Delta U+\Delta pV\\ \Delta H=\Delta U+V\left(\frac{nR{T}_2}{V}-\frac{nR{T}_1}{V}\right)\\ \Delta H=\Delta U+nR\left(T_2-T_1\right)\end{array}$

Substitute the values of $\Delta U$, universal gas constant and temperatures in the above formula to calculate the change in enthalpy.

    $\begin{array}{c}\Delta H=10111.2\text{ J}+1\times 8.31{\text{4 J mol}}^{-1}{\text{ K}}^{-1}\left(37\text{3 K}-29\text{8 K}\right)\\ =10111.2\text{ J}+8.31{\text{4 J mol}}^{-1}{\text{ K}}^{-1}\times 7\text{5 K}\\ =10734.7\text{5 J}\\ =\underset{\_}{10.73475kJ}\end{array}$

Hence, the value of value of $\Delta H$ is $\underset{\_}{10.73475kJ}$.

At constant volume, $w=0$. Now, the value of $q$ is calculated as,

    $\begin{array}{c}\Delta U=q+0\\ \Delta U=q\\ q=\underset{\_}{10.1112kJ}\end{array}$

Hence, the value of $q$ is $\underset{\_}{10.1112kJ}$.