Chapter 18, Problem 18.21QP

(a)

Interpretation Introduction

Interpretation:

For the given water sample various changes in enthalpy, entropy and free energy with temperature has to be explained.

Concept introduction:

Free energy:

Free energy is measured by subtracting the product of temperature and entropy from the enthalpy of a system.

$\begin{array}{l}\text{G}\text{=}\text{H}\text{-}\text{TS}\\ \text{where,}\\ \text{G}\text{-}\text{free}\text{energy;}\text{H}\text{-}\text{enthalpy}\\ \text{S}\text{-}\text{entropy and T}\text{-temperature}\text{.}\end{array}$

Relationship between ${\text{ΔG}}^{\text{o}}\text{,}{\text{ΔH}}^{\text{o}}\text{and}{\text{ΔS}}^{\text{o}}$ is given by

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{where,}\\ {\text{ΔG}}^{\text{o}}\text{-}\text{standard}\text{free}\text{energy}\text{change;}{\text{ΔH}}^{\text{o}}\text{-}\text{standard}\text{enthalpy}\text{change}\\ {\text{ΔS}}^{\text{o}}\text{-}\text{standard}\text{entropy}\text{change and T}\text{-}\text{temperature}\end{array}$

Answer to Problem 18.21QP

The temperature of the given water sample will increase gradually, and water will evaporate in the room.  This is a spontaneous process.

Explanation of Solution

To give: The change in the water sample

Given information,

$\begin{array}{l}\text{temperature}\text{of}\text{the}\text{water}\text{is}\text{25}{}^{\text{o}}\text{C}\text{.}\\ \text{temperature}\text{of}\text{the}\text{room}\text{is}\text{50}{}^{\text{o}}\text{C}\text{.}\end{array}$

The temperature of the given water sample will increase gradually, and then water sample will evaporate in the room.  This is a spontaneous process.

(b)

Interpretation Introduction

Interpretation:

For the given water sample various changes in enthalpy, entropy and free energy with temperature has to be explained.

Concept introduction:

Free energy:

Free energy is measured by subtracting the product of temperature and entropy from the enthalpy of a system.

$\begin{array}{l}\text{G}\text{=}\text{H}\text{-}\text{TS}\\ \text{where,}\\ \text{G}\text{-}\text{free}\text{energy;}\text{H}\text{-}\text{enthalpy}\\ \text{S}\text{-}\text{entropy and T}\text{-temperature}\text{.}\end{array}$

Relationship between ${\text{ΔG}}^{\text{o}}\text{,}{\text{ΔH}}^{\text{o}}\text{and}{\text{ΔS}}^{\text{o}}$ is given by

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{where,}\\ {\text{ΔG}}^{\text{o}}\text{-}\text{standard}\text{free}\text{energy}\text{change;}{\text{ΔH}}^{\text{o}}\text{-}\text{standard}\text{enthalpy}\text{change}\\ {\text{ΔS}}^{\text{o}}\text{-}\text{standard}\text{entropy}\text{change and T}\text{-}\text{temperature}\end{array}$

Answer to Problem 18.21QP

The sign of enthalpy is positive, since heat is absorbed for evaporation process.  The sign of entropy change is positive because water molecules changed their state from liquid to gas.

Explanation of Solution

To give: The enthalpy change of the water sample

Given information,

$\begin{array}{l}\text{temperature}\text{of}\text{the}\text{water}\text{is}\text{25}{}^{\text{o}}\text{C}\text{.}\\ \text{temperature}\text{of}\text{the}\text{room}\text{is}\text{50}{}^{\text{o}}\text{C}\text{.}\end{array}$

The sign of enthalpy is positive, since heat is absorbed by the water sample for evaporation process.  The sign of entropy is positive because water molecules changed their state from liquid to gas (free movement of molecules).

(c)

Interpretation Introduction

Interpretation:

For the given water sample various changes in enthalpy, entropy and free energy with temperature has to be explained.

Concept introduction:

Free energy:

Free energy is measured by subtracting the product of temperature and entropy from the enthalpy of a system.

$\begin{array}{l}\text{G}\text{=}\text{H}\text{-}\text{TS}\\ \text{where,}\\ \text{G}\text{-}\text{free}\text{energy;}\text{H}\text{-}\text{enthalpy}\\ \text{S}\text{-}\text{entropy and T}\text{-temperature}\text{.}\end{array}$

Relationship between ${\text{ΔG}}^{\text{o}}\text{,}{\text{ΔH}}^{\text{o}}\text{and}{\text{ΔS}}^{\text{o}}$ is given by

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{where,}\\ {\text{ΔG}}^{\text{o}}\text{-}\text{standard}\text{free}\text{energy}\text{change;}{\text{ΔH}}^{\text{o}}\text{-}\text{standard}\text{enthalpy}\text{change}\\ {\text{ΔS}}^{\text{o}}\text{-}\text{standard}\text{entropy}\text{change and T}\text{-}\text{temperature}\end{array}$

Answer to Problem 18.21QP

The water molecules are easily moving and in higher temperature.  Hence, the sign of entropy change is positive.  The sign of enthalpy change is already positive, so the only way to have a spontaneous process is when the sign of entropy change is also positive.

Explanation of Solution

To give: The entropy change of the water sample

Given information,

$\begin{array}{l}\text{temperature}\text{of}\text{the}\text{water}\text{is}\text{25}{}^{\text{o}}\text{C}\text{.}\\ \text{temperature}\text{of}\text{the}\text{room}\text{is}\text{50}{}^{\text{o}}\text{C}\text{.}\end{array}$

The water molecules are in easily moving state and in higher temperature.  Hence, the sign of entropy change is positive.

Given process is a spontaneous process, the sign of enthalpy change is already positive, so the sign of entropy change is should be positive.

(d)

Interpretation Introduction

Interpretation:

For the given water sample various changes in enthalpy, entropy and free energy with temperature has to be explained.

Concept introduction:

Free energy:

Free energy is measured by subtracting the product of temperature and entropy from the enthalpy of a system.

$\begin{array}{l}\text{G}\text{=}\text{H}\text{-}\text{TS}\\ \text{where,}\\ \text{G}\text{-}\text{free}\text{energy;}\text{H}\text{-}\text{enthalpy}\\ \text{S}\text{-}\text{entropy and T}\text{-temperature}\text{.}\end{array}$

Relationship between ${\text{ΔG}}^{\text{o}}\text{,}{\text{ΔH}}^{\text{o}}\text{and}{\text{ΔS}}^{\text{o}}$ is given by

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{where,}\\ {\text{ΔG}}^{\text{o}}\text{-}\text{standard}\text{free}\text{energy}\text{change;}{\text{ΔH}}^{\text{o}}\text{-}\text{standard}\text{enthalpy}\text{change}\\ {\text{ΔS}}^{\text{o}}\text{-}\text{standard}\text{entropy}\text{change and T}\text{-}\text{temperature}\end{array}$

Answer to Problem 18.21QP

Given process is a spontaneous process.  Hence, the sign of free energy change is negative.

Explanation of Solution

To give: The free energy change of the water sample

Given information,

$\begin{array}{l}\text{temperature}\text{of}\text{the}\text{water}\text{is}\text{25}{}^{\text{o}}\text{C}\text{.}\\ \text{temperature}\text{of}\text{the}\text{room}\text{is}\text{50}{}^{\text{o}}\text{C}\text{.}\end{array}$

Given process is a spontaneous process, hence, the sign of free energy should be negative.

For a spontaneous process $\text{ΔG}\text{<}\text{0}\text{.}$

(e)

Interpretation Introduction

Interpretation:

For the given water sample various changes in enthalpy, entropy and free energy with temperature has to be explained.

Concept introduction:

Free energy:

Free energy is measured by subtracting the product of temperature and entropy from the enthalpy of a system.

$\begin{array}{l}\text{G}\text{=}\text{H}\text{-}\text{TS}\\ \text{where,}\\ \text{G}\text{-}\text{free}\text{energy;}\text{H}\text{-}\text{enthalpy}\\ \text{S}\text{-}\text{entropy and T}\text{-temperature}\text{.}\end{array}$

Relationship between ${\text{ΔG}}^{\text{o}}\text{,}{\text{ΔH}}^{\text{o}}\text{and}{\text{ΔS}}^{\text{o}}$ is given by

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{where,}\\ {\text{ΔG}}^{\text{o}}\text{-}\text{standard}\text{free}\text{energy}\text{change;}{\text{ΔH}}^{\text{o}}\text{-}\text{standard}\text{enthalpy}\text{change}\\ {\text{ΔS}}^{\text{o}}\text{-}\text{standard}\text{entropy}\text{change and T}\text{-}\text{temperature}\end{array}$

Answer to Problem 18.21QP

The temperature of the given water sample will decrease gradually, and then water will evaporate in the room.  This is spontaneous process.

Explanation of Solution

To give: The change in the water sample

Given information,

$\begin{array}{l}\text{temperature}\text{of}\text{the}\text{water}\text{is}\text{75}{}^{\text{o}}\text{C}\text{.}\\ \text{temperature}\text{of}\text{the}\text{room}\text{is}\text{50}{}^{\text{o}}\text{C}\text{.}\end{array}$

The temperature of the given water sample will decrease gradually, and then water will evaporate in the room.  This is a spontaneous process.

(f)

Interpretation Introduction

Interpretation:

For the given water sample various changes in enthalpy, entropy and free energy with temperature has to be explained.

Concept introduction:

Free energy:

Free energy is measured by subtracting the product of temperature and entropy from the enthalpy of a system.

$\begin{array}{l}\text{G}\text{=}\text{H}\text{-}\text{TS}\\ \text{where,}\\ \text{G}\text{-}\text{free}\text{energy;}\text{H}\text{-}\text{enthalpy}\\ \text{S}\text{-}\text{entropy and T}\text{-temperature}\text{.}\end{array}$

Relationship between ${\text{ΔG}}^{\text{o}}\text{,}{\text{ΔH}}^{\text{o}}\text{and}{\text{ΔS}}^{\text{o}}$ is given by

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{where,}\\ {\text{ΔG}}^{\text{o}}\text{-}\text{standard}\text{free}\text{energy}\text{change;}{\text{ΔH}}^{\text{o}}\text{-}\text{standard}\text{enthalpy}\text{change}\\ {\text{ΔS}}^{\text{o}}\text{-}\text{standard}\text{entropy}\text{change and T}\text{-}\text{temperature}\end{array}$

Answer to Problem 18.21QP

The sign of enthalpy is negative because heat is released from water sample.  The entropy change is positive because water molecules are escaping into the gaseous state.

Explanation of Solution

To give: The enthalpy change of the water sample

Given information,

$\begin{array}{l}\text{temperature}\text{of}\text{the}\text{water}\text{is}\text{75}{}^{\text{o}}\text{C}\text{.}\\ \text{temperature}\text{of}\text{the}\text{room}\text{is}\text{50}{}^{\text{o}}\text{C}\text{.}\end{array}$

The sign of enthalpy is negative $\left(\text{ΔH}\text{<}\text{0}\right)$ because heat is released from water sample.

The entropy change is positive $\left(\text{ΔS}\text{>}\text{0}\right)$ because water molecules are escaping into the gaseous state.

(g)

Interpretation Introduction

Interpretation:

For the given water sample various changes in enthalpy, entropy and free energy with temperature has to be explained.

Concept introduction:

Free energy:

Free energy is measured by subtracting the product of temperature and entropy from the enthalpy of a system.

$\begin{array}{l}\text{G}\text{=}\text{H}\text{-}\text{TS}\\ \text{where,}\\ \text{G}\text{-}\text{free}\text{energy;}\text{H}\text{-}\text{enthalpy}\\ \text{S}\text{-}\text{entropy and T}\text{-temperature}\text{.}\end{array}$

Relationship between ${\text{ΔG}}^{\text{o}}\text{,}{\text{ΔH}}^{\text{o}}\text{and}{\text{ΔS}}^{\text{o}}$ is given by

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{where,}\\ {\text{ΔG}}^{\text{o}}\text{-}\text{standard}\text{free}\text{energy}\text{change;}{\text{ΔH}}^{\text{o}}\text{-}\text{standard}\text{enthalpy}\text{change}\\ {\text{ΔS}}^{\text{o}}\text{-}\text{standard}\text{entropy}\text{change and T}\text{-}\text{temperature}\end{array}$

Answer to Problem 18.21QP

The sign of entropy change is negative because water molecules are at a lower temperature and their movement is slow.  But, some water molecules evaporates and becomes vapor state this process will increases the entropy.

Explanation of Solution

To give: The entropy change of the water sample

Given information,

$\begin{array}{l}\text{temperature}\text{of}\text{the}\text{water}\text{is}\text{75}{}^{\text{o}}\text{C}\text{.}\\ \text{temperature}\text{of}\text{the}\text{room}\text{is}\text{50}{}^{\text{o}}\text{C}\text{.}\end{array}$

The sign of entropy change is negative because water molecules are at a lower temperature and their movement is slow.

But, some water molecules evaporates and becomes vapor state this process will increases the entropy.

(h)

Interpretation Introduction

Interpretation:

For the given water sample various changes in enthalpy, entropy and free energy with temperature has to be explained.

Concept introduction:

Free energy:

Free energy is measured by subtracting the product of temperature and entropy from the enthalpy of a system.

$\begin{array}{l}\text{G}\text{=}\text{H}\text{-}\text{TS}\\ \text{where,}\\ \text{G}\text{-}\text{free}\text{energy;}\text{H}\text{-}\text{enthalpy}\\ \text{S}\text{-}\text{entropy and T}\text{-temperature}\text{.}\end{array}$

Relationship between ${\text{ΔG}}^{\text{o}}\text{,}{\text{ΔH}}^{\text{o}}\text{and}{\text{ΔS}}^{\text{o}}$ is given by

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{where,}\\ {\text{ΔG}}^{\text{o}}\text{-}\text{standard}\text{free}\text{energy}\text{change;}{\text{ΔH}}^{\text{o}}\text{-}\text{standard}\text{enthalpy}\text{change}\\ {\text{ΔS}}^{\text{o}}\text{-}\text{standard}\text{entropy}\text{change and T}\text{-}\text{temperature}\end{array}$

Answer to Problem 18.21QP

Given process is a spontaneous process.  Hence, the sign of free energy change is negative.

Explanation of Solution

To give: The free energy change of the water sample

Given information,

$\begin{array}{l}\text{temperature}\text{of}\text{the}\text{water}\text{is}\text{75}{}^{\text{o}}\text{C}\text{.}\\ \text{temperature}\text{of}\text{the}\text{room}\text{is}\text{50}{}^{\text{o}}\text{C}\text{.}\end{array}$

Given process is a spontaneous process, hence, the sign of free energy should be negative.

For a spontaneous process $\text{ΔG}\text{<}\text{0}\text{.}$

(i)

Interpretation Introduction

Interpretation:

For the given water sample various changes in enthalpy, entropy and free energy with temperature has to be explained.

Concept introduction:

Free energy:

Free energy is measured by subtracting the product of temperature and entropy from the enthalpy of a system.

$\begin{array}{l}\text{G}\text{=}\text{H}\text{-}\text{TS}\\ \text{where,}\\ \text{G}\text{-}\text{free}\text{energy;}\text{H}\text{-}\text{enthalpy}\\ \text{S}\text{-}\text{entropy and T}\text{-temperature}\text{.}\end{array}$

Relationship between ${\text{ΔG}}^{\text{o}}\text{,}{\text{ΔH}}^{\text{o}}\text{and}{\text{ΔS}}^{\text{o}}$ is given by

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{where,}\\ {\text{ΔG}}^{\text{o}}\text{-}\text{standard}\text{free}\text{energy}\text{change;}{\text{ΔH}}^{\text{o}}\text{-}\text{standard}\text{enthalpy}\text{change}\\ {\text{ΔS}}^{\text{o}}\text{-}\text{standard}\text{entropy}\text{change and T}\text{-}\text{temperature}\end{array}$

Answer to Problem 18.21QP

The temperature of the water will not change, and water molecules will evaporate, this is a spontaneous process.

Explanation of Solution

To give: The change in the water sample

Given information,

$\begin{array}{l}\text{temperature}\text{of}\text{the}\text{water}\text{is}\text{50}{}^{\text{o}}\text{C}\text{.}\\ \text{temperature}\text{of}\text{the}\text{room}\text{is}\text{50}{}^{\text{o}}\text{C}\text{.}\end{array}$

The temperature of the water will not change, and water molecules will evaporate, this is a spontaneous process.

(j)

Interpretation Introduction

Interpretation:

For the given water sample various changes in enthalpy, entropy and free energy with temperature has to be explained.

Concept introduction:

Free energy:

Free energy is measured by subtracting the product of temperature and entropy from the enthalpy of a system.

$\begin{array}{l}\text{G}\text{=}\text{H}\text{-}\text{TS}\\ \text{where,}\\ \text{G}\text{-}\text{free}\text{energy;}\text{H}\text{-}\text{enthalpy}\\ \text{S}\text{-}\text{entropy and T}\text{-temperature}\text{.}\end{array}$

Relationship between ${\text{ΔG}}^{\text{o}}\text{,}{\text{ΔH}}^{\text{o}}\text{and}{\text{ΔS}}^{\text{o}}$ is given by

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{where,}\\ {\text{ΔG}}^{\text{o}}\text{-}\text{standard}\text{free}\text{energy}\text{change;}{\text{ΔH}}^{\text{o}}\text{-}\text{standard}\text{enthalpy}\text{change}\\ {\text{ΔS}}^{\text{o}}\text{-}\text{standard}\text{entropy}\text{change and T}\text{-}\text{temperature}\end{array}$

Answer to Problem 18.21QP

The sign of enthalpy is positive because heat is absorbed by water molecules for their evaporation.  The sign of entropy change is positive because the phase changes of water molecules from liquid to gas.

Explanation of Solution

To give: The enthalpy change of the water sample

Given information,

$\begin{array}{l}\text{temperature}\text{of}\text{the}\text{water}\text{is}\text{50}{}^{\text{o}}\text{C}\text{.}\\ \text{temperature}\text{of}\text{the}\text{room}\text{is}\text{50}{}^{\text{o}}\text{C}\text{.}\end{array}$

The sign of enthalpy change is positive, since heat is absorbed by the water sample for evaporation process.

The sign of entropy change is positive because water molecules changed their state from liquid to gas (free movement of molecules).

(k)

Interpretation Introduction

Interpretation:

For the given water sample various changes in enthalpy, entropy and free energy with temperature has to be explained.

Concept introduction:

Free energy:

Free energy is measured by subtracting the product of temperature and entropy from the enthalpy of a system.

$\begin{array}{l}\text{G}\text{=}\text{H}\text{-}\text{TS}\\ \text{where,}\\ \text{G}\text{-}\text{free}\text{energy;}\text{H}\text{-}\text{enthalpy}\\ \text{S}\text{-}\text{entropy and T}\text{-temperature}\text{.}\end{array}$

Relationship between ${\text{ΔG}}^{\text{o}}\text{,}{\text{ΔH}}^{\text{o}}\text{and}{\text{ΔS}}^{\text{o}}$ is given by

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{where,}\\ {\text{ΔG}}^{\text{o}}\text{-}\text{standard}\text{free}\text{energy}\text{change;}{\text{ΔH}}^{\text{o}}\text{-}\text{standard}\text{enthalpy}\text{change}\\ {\text{ΔS}}^{\text{o}}\text{-}\text{standard}\text{entropy}\text{change and T}\text{-}\text{temperature}\end{array}$

Answer to Problem 18.21QP

For a spontaneous process the sign of free energy change should be positive, here the sign of enthalpy change is already positive so the sign of entropy change should be positive.

Explanation of Solution

To give: The entropy change of the water sample

Given information,

$\begin{array}{l}\text{temperature}\text{of}\text{the}\text{water}\text{is}\text{50}{}^{\text{o}}\text{C}\text{.}\\ \text{temperature}\text{of}\text{the}\text{room}\text{is}\text{50}{}^{\text{o}}\text{C}\text{.}\end{array}$

For a spontaneous process $\text{ΔG}\text{<}\text{0}\text{.}$

Given process is a spontaneous process, the sign of enthalpy change is already positive, so the sign of entropy change should be positive.

(l)

Interpretation Introduction

Interpretation:

For the given water sample various changes in enthalpy, entropy and free energy with temperature has to be explained.

Concept introduction:

Free energy:

Free energy is measured by subtracting the product of temperature and entropy from the enthalpy of a system.

$\begin{array}{l}\text{G}\text{=}\text{H}\text{-}\text{TS}\\ \text{where,}\\ \text{G}\text{-}\text{free}\text{energy;}\text{H}\text{-}\text{enthalpy}\\ \text{S}\text{-}\text{entropy and T}\text{-temperature}\text{.}\end{array}$

Relationship between ${\text{ΔG}}^{\text{o}}\text{,}{\text{ΔH}}^{\text{o}}\text{and}{\text{ΔS}}^{\text{o}}$ is given by

$\begin{array}{l}{\text{ΔG}}^{\text{o}}\text{=}{\text{ΔH}}^{\text{o}}\text{-}{\text{TΔS}}^{\text{o}}\\ \text{where,}\\ {\text{ΔG}}^{\text{o}}\text{-}\text{standard}\text{free}\text{energy}\text{change;}{\text{ΔH}}^{\text{o}}\text{-}\text{standard}\text{enthalpy}\text{change}\\ {\text{ΔS}}^{\text{o}}\text{-}\text{standard}\text{entropy}\text{change and T}\text{-}\text{temperature}\end{array}$

Answer to Problem 18.21QP

Given process is a spontaneous process.  Hence, the sign of free energy change is negative.

Explanation of Solution

To give: The free energy change of the water sample

Given information,

$\begin{array}{l}\text{temperature}\text{of}\text{the}\text{water}\text{is}\text{50}{}^{\text{o}}\text{C}\text{.}\\ \text{temperature}\text{of}\text{the}\text{room}\text{is}\text{50}{}^{\text{o}}\text{C}\text{.}\end{array}$

Given process is a spontaneous process, hence, the sign of free energy change should be negative.

For a spontaneous process $\text{ΔG}\text{<}\text{0}\text{.}$