Chapter 3, Problem 3A.4AE

(i)

Interpretation Introduction

Interpretation: The change in entropies of the system, surroundings and the total change in entropy, when a sample of nitrogen gas of mass $14\text{g}$ at $298\text{K}$ doubles its volume in an reversible isothermal expansion has to be calculated.

Concept introduction: Entropy is a thermodynamic quantity which represents the disorder or randomness of a system.  It is the thermal energy per unit temperature which is unavailable to do the useful work.  Entropy change depends only on the initial and final states.  In an isothermal process, the change in temperature is zero while in an adiabatic process the change in heat absorbed or given out is zero.

Answer to Problem 3A.4AE

The entropy change of the given the given system is $\underset{\_}{2.88J{K}^{-1}}$, the entropy change of the surroundings is $\underset{\_}{-2.88J{K}^{-1}}$, and the total entropy change is $\underset{\_}{0J{K}^{-1}}$.

Explanation of Solution

The relation between entropy change of the system, surroundings and the total entropy change is given by the equation

    $\Delta S_{\text{total}}=\Delta S_{\text{sys}}+\Delta S_{\text{surr}}$                                                                                     (1)

Where,

• • $\Delta S_{\text{total}}$ is the total change in entropy.
• • $\Delta S_{\text{sys}}$ is the change in entropy for system.
• • $\Delta S_{\text{surr}}$ is the change in entropy for surroundings.

For a reversible isothermal expansion, the total entropy change is zero.  Therefore  Therefore,

    $\Delta S_{\text{total}}=0$

Substitute the value of  $\Delta S_{\text{total}}$ in equation (1).Substitute the value of $\Delta S_{\text{total}}$ in equation (1).

  $\begin{array}{c}\Delta S_{\text{total}}=\Delta S_{\text{sys}}+\Delta S_{\text{surr}}\\ 0=\Delta S_{\text{sys}}+\Delta S_{\text{surr}}\end{array}$

Hence,

  $\Delta S_{\text{surr}}=-\Delta S_{\text{sys}}$                                                                                               (2)

The mass of nitrogen gas is $14\text{g}$.

The molar mass of nitrogen gas is $28\text{g}{\text{mol}}^{\text{-1}}$

The number of moles of nitrogen gas is calculated by the formula,

      $\text{Number}\text{of}\text{moles}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}$

Substitute the values of given mass and molar mass in the above equation.

  $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}\\ =\frac{14\text{g}}{28\text{g}{\text{mol}}^{\text{-1}}}\end{array}$

The expression for entropy change is given as

  $\Delta S_{\text{sys}}=2.303nR\log \frac{V_{\text{f}}}{V_{\text{i}}}$                                                                                    (3)

Where,

• • $R$ is the gas constant $\left(8.31{\text{4 J K}}^{-1}{\text{ mol}}^{-1}\right)$.
• • $n$ is the number of moles.
• • $V_{\text{f}}$ is the final volume.
• • $V_{\text{i}}$ is the initial volume.

The gas undergoes isothermal reversible expansion and doubles its volume. Hence,

    $V_{\text{f}}=2V_{\text{i}}$

Substitute the values of $n$, $\text{R}$, $V_{\text{i}}$ and $V_{\text{f}}$ in equation (3) to calculate the change in entropy of system of nitrogen gas as

  $\begin{array}{c}\Delta {\text{S}}_{\text{sys}}=2.303\times \frac{14\text{g}}{28\text{g}{\text{mol}}^{-1}}\times 8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\times \log \left(\frac{2V_{\text{i}}}{V_{\text{i}}}\right)\\ =\underset{\_}{2.88J{K}^{-1}}\end{array}$

Substitute the value of $\Delta {\text{S}}_{\text{sys}}$ in equation (2).

  $\begin{array}{c}\Delta S_{\text{surr}}=-\Delta S_{\text{sys}}\\ =-\underset{\_}{2.88J{K}^{-1}}\end{array}$

Hence, the change in entropy of system for nitrogen gas is $\underset{\_}{2.88J{K}^{-1}}$, the change in entropy for surroundings is $\underset{\_}{-2.88J{K}^{-1}}$ and the total entropy change is $\underset{\_}{0J{K}^{-1}}$

(ii)

Interpretation Introduction

Interpretation: The change in entropies of the system, surroundings and the total change in entropy when a sample of nitrogen of mass $14\text{g}$ at $298\text{K}$ doubles its volume in an isothermal irreversible expansion against $p_{\text{ex}}=0$ has to be calculated.

Concept introduction: Entropy is a thermodynamic quantity which represents the disorder or randomness of a system.  It is the thermal energy per unit temperature which is unavailable to do the useful work.  Entropy change depends only on the initial and final states.  In an isothermal process, the change in temperature is zero while in an adiabatic process the change in heat absorbed or given out is zero.

Answer to Problem 3A.4AE

The entropy change of the given system is $\underset{\_}{2.88J{K}^{-1}}$, the entropy change of the surroundings is $\underset{\_}{0J{K}^{-1}}$ and the total entropy change is $\underset{\_}{2.88J{K}^{-1}}$.

Explanation of Solution

The relation between entropy change of the system, surroundings and the total entropy change is given by the equation

    $\Delta S_{\text{total}}=\Delta S_{\text{sys}}+\Delta S_{\text{surr}}$                                                                                    (1)

Where,

• • $\Delta S_{\text{total}}$ is the total change in entropy
• • $\Delta S_{\text{sys}}$ is the change in entropy for system
• • $\Delta S_{\text{surr}}$ is the change in entropy for surroundings

For an irreversible isothermal expansion against $p_{\text{ex}}=0$  the entropy change of the surroundings is zero.  Therefore

  $\Delta S_{\text{surr}}=0$

Substitute the value of $\Delta S_{surr}=0$ in equation (1)

    $\begin{array}{l}\Delta S_{\text{total}}=\Delta S_{\text{sys}}+\Delta S_{\text{surr}}\\ \Delta S_{\text{total}}=\Delta S_{\text{sys}}+0\end{array}$

Hence,

    $\Delta S_{\text{total}}=\Delta S_{\text{sys}}$                                                                                                 (2)

The mass of nitrogen gas is $14\text{g}$.

The molar mass of nitrogen gas is $28\text{g}{\text{mol}}^{\text{-1}}$.

The number of moles of nitrogen gas is calculated by the formula,

      $\text{Number}\text{of}\text{moles}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}$

Substitute the values of given mass and molar mass in the above equation

  $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}\\ =\frac{14\text{g}}{28\text{g}{\text{mol}}^{\text{-1}}}\end{array}$

The expression for entropy change is given as

  $\Delta S_{\text{sys}}=2.303nR\log \frac{V_{\text{f}}}{V_{\text{i}}}$                                                                                    (3)

Where,

• • $R$ is the gas constant $\left(8.31{\text{4 J K}}^{-1}{\text{ mol}}^{-1}\right)$.
• • $n$ is the number of moles.
• • $V_{\text{f}}$ is the final volume.
• • $V_{\text{i}}$ is the initial volume.

The gas undergoes isothermal reversible expansion and doubles its volume Hence

    $V_{\text{f}}=2V_{\text{i}}$

Substitute the values of $n$, $\text{R}$, $V_{\text{i}}$ and $V_{\text{f}}$  in equation (3) to calculate the change in entropy of system of nitrogen gas as,

  $\begin{array}{c}\Delta {\text{S}}_{\text{sys}}=2.303\times \frac{14\text{g}}{28\text{g}{\text{mol}}^{-1}}\times 8.314\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\times \log \left(\frac{2V_{\text{i}}}{V_{\text{i}}}\right)\\ =\underset{\_}{2.88J{K}^{-1}}\end{array}$

Substitute the value of $\Delta {\text{S}}_{\text{sys}}$ in equation (2).

  $\begin{array}{c}\Delta S_{\text{total}}=\Delta S_{\text{sys}}\\ =\underset{\_}{2.88J{K}^{-1}}\end{array}$

Hence, the change in entropy of system for nitrogen gas is $\underset{\_}{2.88J{K}^{-1}}$, the change in entropy for surroundings is $\underset{\_}{0J{K}^{-1}}$ and the total entropy change is $\underset{\_}{2.88J{K}^{-1}}$.

(iii)

Interpretation Introduction

Interpretation: The change in entropies of the system, surroundings and the total change in entropy when a sample of nitrogen of mass $14\text{g}$ at $298\text{K}$ doubles its volume in an adiabatic reversible expansion has to be calculated.

Concept introduction: Entropy is a thermodynamic quantity which represents the disorder or randomness of a system.  It is the thermal energy per unit temperature which is unavailable to do the useful work.  Entropy change depends only on the initial and final states.  In an isothermal process, the change in temperature is zero while in an adiabatic process, the change in heat absorbed or given out is zero.

Answer to Problem 3A.4AE

The entropy change of the given system is $\underset{\_}{2.88J{K}^{-1}}$, the entropy change of the surroundings is $\underset{\_}{0}$ and the total entropy change is $\underset{\_}{2.88J{K}^{-1}}$.

Explanation of Solution

The relation between entropy change of the system, surroundings and the total entropy change is given by the equation,

    $\Delta S_{\text{total}}=\Delta S_{\text{sys}}+\Delta S_{\text{surr}}$                                                                                     (1)

Where,

• • $\Delta S_{\text{total}}$ is the total change in entropy
• • $\Delta S_{\text{sys}}$ is the change in entropy for system
• • $\Delta S_{\text{surr}}$ is the change in entropy for surroundings

For adiabatic reversible expansion the total entropy change is zero.  Therefore

    $\Delta S_{\text{total}}=0$

Substitute the value of $\Delta S_{\text{total}}$ in equation (1).

    $\begin{array}{c}\Delta S_{\text{total}}=\Delta S_{\text{sys}}+\Delta S_{\text{surr}}\\ 0=\Delta S_{\text{sys}}+\Delta S_{\text{surr}}\end{array}$

Hence,

    $\Delta S_{\text{surr}}=-\Delta S_{\text{sys}}$                                                                                                (2)

The expression for entropy change of surroundings is given as, as,

    $\Delta S_{\text{surr}}=\frac{\partial q}{T}$                                                                                                       (3)

For a reversible adiabatic process,,

  $\partial q=0$

Substitute the value of $\partial q$ in equation (3).

    $\begin{array}{c}\Delta S_{\text{surr}}=\frac{\partial q}{T}\\ =\frac{0}{T}\\ =0\end{array}$

Therefore

  ${\text{ΔS}}_{\text{surr}}=0$

Substitute the value of ${\text{ΔS}}_{\text{surr}}$ in equation (2).

$\begin{array}{c}\Delta S_{\text{sys}}=-\Delta S_{\text{surr}}\\ =0\underset{\_}{J{K}^{-1}}\end{array}$

Hence, the change in entropy of system for nitrogen gas is $\underset{\_}{0J{K}^{-1}}$, the change in entropy for surroundings is $\underset{\_}{0J{K}^{-1}}$ and the total entropy change is $\underset{\_}{0J{K}^{-1}}$.