Chapter 4, Problem 4C.5E

(a)

Interpretation Introduction

Interpretation:

The contribution of each type of molecular motion to the heat capacity $C_{V,\text{m}}$ and their total contribution in the molecule $\text{HCN}$ have to be predicted.

Concept Introduction:

The energy of a molecule due to its motion in the three dimensional space is referred to as the translational energy.  The energy of a molecule due to its rotational motion in the three dimensional space is referred to as the rotational energy.  The total molar internal energy of a molecule at temperature $T$ is the sum of the contribution of translational and rotational kinetic energy.

Answer to Problem 4C.5E

The contribution of translational motion and rotational motion to the heat capacity $C_{V,\text{m}}$ of $\text{HCN}$ is $\underset{\_}{\frac{3}{2}RT}$ and $\underset{\_}{RT}$ respectively and their total contribution in the molecule $\text{HCN}$ is $\underset{\_}{\frac{5}{2}R}$.

Explanation of Solution

The $\text{HCN}$ molecule is a triatomic molecule as it is composed of one hydrogen atom, one carbon atom and one nitrogen atom.

The total degree of freedom for $\text{HCN}$ is $\text{9}$.  As $\text{HCN}$ molecule is linear, therefore, out of $\text{9}$ there are three translational and two rotational degree of freedom.  The contribution of each translational and rotational degree of freedom for a linear geometry is $\frac{1}{2}RT$.

The contribution of translational kinetic energy to the molar internal energy of $\text{HCN}$ molecule along $x$, $y$ and $z$ direction is shown below.

  $\begin{array}{c}{U}_{\text{m}}\left(\text{translation}\right)=3\times \left(\frac{1}{2}RT\right)\\ =\frac{3}{2}RT\end{array}$

Where,

• $U_{\text{m}}\left(\text{translation}\right)$ is translational kinetic energy.
• $R$ is the gas constant.
• $T$ is the temperature.

The contribution of rotational kinetic energy to the molar internal energy of $\text{HCN}$ molecule which has a linear geometry along $x$, $y$ and $z$ direction is shown below.

  $\begin{array}{c}{U}_{\text{m}}\left(\text{rotation,}\text{linear}\right)=2\times \left(\frac{1}{2}RT\right)\\ =RT\end{array}$

Where,

• $U_{\text{m}}\left(\text{rotation,}\text{linear}\right)$ is rotational kinetic energy.
• $R$ is the gas constant.
• $T$ is the temperature.

The total molar internal energy of $\text{HCN}$ is the sum of the contribution of translational and rotational kinetic energy.  The expression to calculate the total molar internal energy is shown below.

  $\Delta U_{\text{m}}=U_{\text{m}}\left(\text{translation}\right)+U_{\text{m}}\left(\text{rotation,}\text{linear}\right)$        (1)

Where,

• $\Delta U_{\text{m}}$ is the molar internal energy.
• $U_{\text{m}}\left(\text{translation}\right)$ is translational kinetic energy.
• $U_{\text{m}}\left(\text{rotation,}\text{linear}\right)$ is rotational kinetic energy.

The value of $U_{\text{m}}\left(\text{translation}\right)$ is $\frac{3}{2}RT$.

The value of $U_{\text{m}}\left(\text{rotation,}\text{linear}\right)$ is $RT$.

Substitute the value of $U_{\text{m}}\left(\text{translation}\right)$ and $U_{\text{m}}\left(\text{rotation,}\text{linear}\right)$ in equation (1).

    $\begin{array}{c}\Delta U_{\text{m}}=\frac{3}{2}RT+RT\\ =\frac{3RT+2RT}{2}\\ =\frac{5}{2}RT\end{array}$

The contribution of motion to the heat capacity $C_{V,\text{m}}$ of $\text{HCN}$ is calculated by the relation shown below.

  $C_{V,\text{m}}={\left(\frac{\partial U_{\text{m}}}{\partial T}\right)}_V$        (2)

Where,

• $C_{V,\text{m}}$ is the heat capacity at constant volume.
• $U_{\text{m}}$ is total molar internal energy
• $T$ is temperature.

The value of $U_{\text{m}}$ is $\frac{5}{2}RT$.

Substitute the value of $U_{\text{m}}$ in equation (2).

    $\begin{array}{c}{C}_{V,\text{m}}={\left(\frac{\partial \left(\frac{5}{2}RT\right)}{\partial T}\right)}_V\\ =\frac{5}{2}R{\left(\frac{\partial \left(T\right)}{\partial T}\right)}_V\\ =\frac{5}{2}R\end{array}$

Thus, the contribution of motion to the heat capacity $C_{V,\text{m}}$ of $\text{HCN}$ is $\underset{\_}{\frac{5}{2}R}$.

(b)

Interpretation Introduction

Interpretation:

The contribution of each type of molecular motion to the heat capacity $C_{V,\text{m}}$ and their total contribution in the molecule ${\text{C}}_2{\text{H}}_6$ have to be predicted.

Concept Introduction:

Same as part (a).

Answer to Problem 4C.5E

The contribution of translational motion and rotational motion to the heat capacity $C_{V,\text{m}}$ of ${\text{C}}_2{\text{H}}_6$ is same for both that is $\underset{\_}{\frac{3}{2}RT}$ and their total contribution in the molecule ${\text{C}}_2{\text{H}}_6$ is $\underset{\_}{3R}$.

Explanation of Solution

The ${\text{C}}_2{\text{H}}_6$ molecule is composed of two carbon atoms and eight hydrogen atoms.

The total degree of freedom for ${\text{C}}_2{\text{H}}_6$ is $24$.  As ${\text{C}}_2{\text{H}}_6$ molecule is non-linear, therefore, out of $24$ there are three translational and three rotational degree of freedom.  The contribution of each translational and rotational degree of freedom for a non-linear geometry is $\frac{1}{2}RT$.

The contribution of translational kinetic energy to the molar internal energy of ${\text{C}}_2{\text{H}}_6$ molecule along $x$, $y$ and $z$ direction is shown below.

  $\begin{array}{c}{U}_{\text{m}}\left(\text{translation}\right)=3\times \left(\frac{1}{2}RT\right)\\ =\frac{3}{2}RT\end{array}$

Where,

• $U_{\text{m}}\left(\text{translation}\right)$ is translational kinetic energy.
• $R$ is the gas constant.
• $T$ is the temperature.

The contribution of rotational kinetic energy to the molar internal energy of ${\text{C}}_2{\text{H}}_6$ molecule which has a non-linear geometry along $x$, $y$ and $z$ direction is shown below.

  $\begin{array}{c}{U}_{\text{m}}\left(\text{rotation,}\text{nonlinear}\right)=3\times \left(\frac{1}{2}RT\right)\\ =\frac{3}{2}RT\end{array}$

Where,

• $U_{\text{m}}\left(\text{rotation,}\text{nonlinear}\right)$ is rotational kinetic energy.
• $R$ is the gas constant.
• $T$ is the temperature.

The value of $U_{\text{m}}\left(\text{translation}\right)$ is $\frac{3}{2}RT$.

The value of $U_{\text{m}}\left(\text{rotation,}\text{nonlinear}\right)$ is $\frac{3}{2}RT$.

Substitute the value of $U_{\text{m}}\left(\text{translation}\right)$ and $U_{\text{m}}\left(\text{rotation,}\text{nonlinear}\right)$ in equation (1).

    $\begin{array}{c}\Delta U_{\text{m}}=\frac{3}{2}RT+\frac{3}{2}RT\\ =\frac{3}{2}\left(RT+RT\right)\\ =\frac{3}{2}\left(2RT\right)\\ =3RT\end{array}$

The value of $U_{\text{m}}$ is $3RT$.

Substitute the value of $U_{\text{m}}$ in equation (2).

    $\begin{array}{c}{C}_{V,\text{m}}={\left(\frac{\partial \left(3RT\right)}{\partial T}\right)}_V\\ =3R{\left(\frac{\partial \left(T\right)}{\partial T}\right)}_V\\ =3R\end{array}$

Thus, the contribution of motion to the heat capacity $C_{V,\text{m}}$ of ${\text{C}}_2{\text{H}}_6$ is $\underset{\_}{3R}$.

(c)

Interpretation Introduction

Interpretation:

The contribution of each type of molecular motion to the heat capacity $C_{V,\text{m}}$ and their total contribution in the atom $\text{Ar}$ have to be predicted.

Concept Introduction:

Same as part (a).

Answer to Problem 4C.5E

The contribution of translational motion to the heat capacity $C_{V,\text{m}}$ of $\text{Ar}$ is $\underset{\_}{\frac{3}{2}RT}$ and their total contribution in the atom $\text{Ar}$ is $\underset{\_}{\frac{3}{2}R}$.

Explanation of Solution

The $\text{Ar}$ atom is a free atom.  It has $3$ total degree of freedom. 

There is only translational degree of freedom in this case and the contribution of each translational degree of freedom for $\text{Ar}$ atom is $\frac{1}{2}RT$.

The contribution of translational kinetic energy to the molar internal energy of $\text{Ar}$ atom along $x$, $y$ and $z$ direction is shown below.

  $\begin{array}{c}{U}_{\text{m}}\left(\text{translation}\right)=3\times \left(\frac{1}{2}RT\right)\\ =\frac{3}{2}RT\end{array}$

Where,

• $U_{\text{m}}\left(\text{translation}\right)$ is translational kinetic energy.
• $R$ is the gas constant.
• $T$ is the temperature.

The value of $U_{\text{m}}\left(\text{translation}\right)$ is $\frac{3}{2}RT$.

Substitute the value of $U_{\text{m}}\left(\text{translation}\right)$ in equation (1).

    $\Delta U_{\text{m}}=\frac{3}{2}RT$

The value of $U_{\text{m}}$ is $\frac{3}{2}RT$.

Substitute the value of $U_{\text{m}}$ in equation (2).

    $\begin{array}{c}{C}_{V,\text{m}}={\left(\frac{\partial \left(\frac{3}{2}RT\right)}{\partial T}\right)}_V\\ =\frac{3}{2}R{\left(\frac{\partial \left(T\right)}{\partial T}\right)}_V\\ =\frac{3}{2}R\end{array}$

Thus, the contribution of motion to the heat capacity $C_{V,\text{m}}$ of $\text{Ar}$ is $\underset{\_}{\frac{3}{2}R}$.

(d)

Interpretation Introduction

Interpretation:

The contribution of each type of molecular motion to the heat capacity $C_{V,\text{m}}$ and their total contribution in the molecule $\text{HBr}$ have to be predicted.

Concept Introduction:

Same as part (a).

Answer to Problem 4C.5E

The contribution of translational motion and rotational motion to the heat capacity $C_{V,\text{m}}$ of $\text{HBr}$ is $\underset{\_}{\frac{3}{2}RT}$ and $\underset{\_}{RT}$ respectively and their total contribution in the molecule $\text{HBr}$ is $\underset{\_}{\frac{5}{2}R}$.

Explanation of Solution

The $\text{HBr}$ molecule is a diatomic molecule as it is composed of one hydrogen atom and one bromine atom.

The total degree of freedom for $\text{HBr}$ is $6$.  As $\text{HBr}$ molecule is linear, therefore, out of $6$ there are three translational and two rotational degree of freedom.  The contribution of each translational and rotational degree of freedom for a linear geometry is $\frac{1}{2}RT$.

The contribution of translational kinetic energy to the molar internal energy of $\text{HBr}$ molecule along $x$, $y$ and $z$ direction is shown below.

  $\begin{array}{c}{U}_{\text{m}}\left(\text{translation}\right)=3\times \left(\frac{1}{2}RT\right)\\ =\frac{3}{2}RT\end{array}$

Where,

• $U_{\text{m}}\left(\text{translation}\right)$ is translational kinetic energy.
• $R$ is the gas constant.
• $T$ is the temperature.

The contribution of rotational kinetic energy to the molar internal energy of $\text{HBr}$ molecule which has a linear geometry along $x$, $y$ and $z$ direction is shown below.

  $\begin{array}{c}{U}_{\text{m}}\left(\text{rotation,}\text{linear}\right)=2\times \left(\frac{1}{2}RT\right)\\ =RT\end{array}$

Where,

• $U_{\text{m}}\left(\text{rotation,}\text{linear}\right)$ is rotational kinetic energy.
• $R$ is the gas constant.
• $T$ is the temperature.

The value of $U_{\text{m}}\left(\text{translation}\right)$ is $\frac{3}{2}RT$.

The value of $U_{\text{m}}\left(\text{rotation,}\text{linear}\right)$ is $RT$.

Substitute the value of $U_{\text{m}}\left(\text{translation}\right)$ and $U_{\text{m}}\left(\text{rotation,}\text{linear}\right)$ in equation (1).

    $\begin{array}{c}\Delta U_{\text{m}}=\frac{3}{2}RT+RT\\ =\frac{3RT+2RT}{2}\\ =\frac{5}{2}RT\end{array}$

The value of $U_{\text{m}}$ is $\frac{5}{2}RT$.

Substitute the value of $U_{\text{m}}$ in equation (2).

    $\begin{array}{c}{C}_{V,\text{m}}={\left(\frac{\partial \left(\frac{5}{2}RT\right)}{\partial T}\right)}_V\\ =\frac{5}{2}R{\left(\frac{\partial \left(T\right)}{\partial T}\right)}_V\\ =\frac{5}{2}R\end{array}$

Thus, the contribution of motion to the heat capacity $C_{V,\text{m}}$ of $\text{HBr}$ is $\underset{\_}{\frac{5}{2}R}$.