The units in equation 4.18-4.25 are to be shown consistent on both the sides of each equation. Concept introduction: The thermodynamic properties of the system can be determined using natural variable equations. The natural variables such as temperature, volume, pressure and entropy can be used to describe the thermodynamic properties such as internal energy, Gibbs energy, enthalpy, Helmholtz energy.

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Answer 1

Question

Chapter 4, Problem 4.31E

Interpretation Introduction

Interpretation:

The units in equation 4.18-4.25 are to be shown consistent on both the sides of each equation.

Concept introduction:

The thermodynamic properties of the system can be determined using natural variable equations. The natural variables such as temperature, volume, pressure and entropy can be used to describe the thermodynamic properties such as internal energy, Gibbs energy, enthalpy, Helmholtz energy.

Expert Solution & Answer

Answer to Problem 4.31E

The units in equation 4.18-4.25 have been shown consistent on both the sides of each equation.

Explanation of Solution

The equation 4.18 is shown below.

$(∂U∂S)V=T$

Where,

• $U$ is the internal energy.

• $T$ is the temperature.

• $S$ is the entropy.

• $V$ is the volume.

The units of internal energy, temperature, entropy, pressure and volume is $J/mol$ , $K$ , $J/mol⋅K$ , $Pa$ and $m3$ respectively.

Substitute the units of internal energy and entropy in the left hand side expression as shown below.

$(∂U∂S)V=(J/molJ/mol⋅K)=(11/K)=K$

Thus, the unit obtained is Kelvin which is the unit of temperature. Thus, the units are consistent on the either side of the equation.

The equation 4.19 is shown below.

$(∂U∂V)S=−p$

Where,

• $U$ is the internal energy.

• $S$ is the entropy.

• $p$ is the pressure.

• $V$ is the volume.

The units of internal energy and entropy can also be expressed in per molecule, the units of internal energy and entropy is $J$ and $J/K$ respectively.

Substitute the units of internal energy and volume in the left hand side expression as shown below.

$(∂U∂V)S=(Jm3)$

The units of internal energy joule can be substituted as $Pa⋅m3$ as shown below.

$(∂U∂V)S=(Pa⋅m3m3)=Pa$

Thus, the unit obtained is Pascal which is the unit of pressure. Thus, the units are consistent on the either side of the equation.

The equation 4.20 is shown below.

$(∂H∂S)p=T$

Where,

• $H$ is the enthalpy.

• $T$ is the temperature.

• $S$ is the entropy.

• $p$ is the pressure.

The units of enthalpy, temperature, entropy, pressure and volume is $J/mol$ , $K$ , $J/mol⋅K$ , $Pa$ and $m3$ respectively.

Substitute the units of enthalpy and entropy in the left hand side expression as shown below.

$(∂H∂S)V=(J/molJ/mol⋅K)=(11/K)=K$

Thus, the unit obtained is Kelvin which is the unit of temperature. Thus, the units are consistent on the either side of the equation.

The equation 4.21 is shown below.

$(∂H∂p)S=V$

Where,

• $H$ is the enthalpy and its unit is $J$ per molecule.

• $S$ is the entropy and its units are $J/K$ per molecule.

• $p$ is the pressure and its unit is $Pa$ .

• $V$ is the volume and its unit is $m3$ .

The units of enthalpy and entropy can also be expressed in per molecule, the units of enthalpy and entropy is $J$ and $J/K$ respectively.

Substitute the units of enthalpy and pressure in the left hand side expression as shown below.

$(∂H∂p)S=(JPa)$

The units of enthalpy joule can be substituted as $Pa⋅m3$ as shown below.

$(∂H∂p)S=(Pa⋅m3Pa)=m3$

Thus, the unit obtained is cubic meter which is the unit of volume. Thus, the units are consistent on the either side of the equation.

The equation 4.22 is shown below.

$(∂A∂T)V=−S$

Where,

• $A$ is the Helmholtz energy.

• $T$ is the temperature.

• $S$ is the entropy.

• $V$ is the volume.

The units of Helmholtz energy, temperature, entropy, pressure and volume is $J/mol$ , $K$ , $J/mol⋅K$ , $Pa$ and $m3$ respectively.

Substitute the units of Helmholtz energy and temperature in the left hand side expression as shown below.

$(∂A∂T)V=(J/molK)=(Jmol⋅K)$

Thus, the unit obtained is $Jmol⋅K$ Kelvin which is the unit of entropy. Thus, the units are consistent on the either side of the equation.

The equation 4.23 is shown below.

$(∂A∂V)T=−p$

Where,

• $A$ is the Helmholtz energy.

• $p$ is the pressure.

• $V$ is the volume.

• $T$ is the temperature.

The units of Helmholtz energy and entropy can also be expressed in per molecule, the units of Helmholtz energy and entropy is $J$ and $J/K$ respectively.

Substitute the units of Helmholtz energy and volume in the left hand side expression as shown below.

$(∂A∂V)T=(Jm3)$

The units of Helmholtz energy joule can be substituted as $Pa⋅m3$ as shown below.

$(∂A∂V)T=(Pa⋅m3m3)=Pa$

Thus, the unit obtained is Pascal which is the unit of pressure. Thus, the units are consistent on the either side of the equation.

The equation 4.24 is shown below.

$(∂G∂T)p=−S$

Where,

• $A$ is the Gibbs energy.

• $T$ is the temperature.

• $S$ is the entropy.

• $V$ is the volume.

The units of Gibbs energy and entropy can also be expressed in per molecule, the units of Gibbs energy and entropy is $J$ and $J/K$ respectively.

Substitute the units of Gibbs energy and temperature in the left hand side expression as shown below.

$(∂G∂T)p=(J/molK)=(Jmol⋅K)$

Thus, the unit obtained is $Jmol⋅K$ Kelvin which is the unit of entropy. Thus, the units are consistent on the either side of the equation.

The equation 4.25 is shown below.

$(∂G∂p)T=V$

Where,

• $G$ is the Gibbs energy

• $T$ is the temperature.

• $p$ is the pressure.

• $V$ is the volume.

The units of Gibbs energy and entropy can also be expressed in per molecule, the units of Gibbs energy and entropy is $J$ and $J/K$ respectively.

Substitute the units of Gibbs energy and pressure in the left hand side expression as shown below.

$(∂G∂p)T=(JPa)$

The units of Gibbs energy joule can be substituted as $Pa⋅m3$ as shown below.

$(∂G∂p)T=(Pa⋅m3Pa)=m3$

Thus, the unit obtained is cubic meter which is the unit of volume. Thus, the units are consistent on the either side of the equation.

Conclusion

The units in equation 4.18-4.25 have been shown consistent on both the sides of each equation.

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A small artisanal cheesemaker is testing the acidity of their milk before it coagulates. During fermentation, bacteria produce lactic acid (K₁ = 1.4 x 104), a weak acid that helps to curdle the milk and develop flavor. The cheesemaker has measured that the developing mixture contains lactic acid at an initial concentration of 0.025 M. Your task is to calculate the pH of this mixture and determine whether it meets the required acidity for proper cheese development. To achieve the best flavor, texture and reduce/control microbial growth, the pH range needs to be between pH 4.6 and 5.0. Assumptions: Lactic acid is a monoprotic acid H H :0:0: H-C-C H :0: O-H Figure 1: Lewis Structure for Lactic Acid For simplicity, you can use the generic formula HA to represent the acid You can assume lactic acid dissociation is in water as milk is mostly water. Temperature is 25°C 1. Write the K, expression for the dissociation of lactic acid in the space provided. Do not forget to include state symbols.…

Curved arrows are used to illustrate the flow of electrons. Using the provided starting and product structures, draw the curved electron-pushing arrows for the following reaction or mechanistic step(s). Be sure to account for all bond-breaking and bond-making steps. :0: :0 H. 0:0 :0: :6: S: :0: Select to Edit Arrows ::0 Select to Edit Arrows H :0: H :CI: Rotation Select to Edit Arrows H. < :0: :0: :0: S:

Answer 2

Question

Chapter 4, Problem 4.31E

Interpretation Introduction

Interpretation:

The units in equation 4.18-4.25 are to be shown consistent on both the sides of each equation.

Concept introduction:

The thermodynamic properties of the system can be determined using natural variable equations. The natural variables such as temperature, volume, pressure and entropy can be used to describe the thermodynamic properties such as internal energy, Gibbs energy, enthalpy, Helmholtz energy.

Expert Solution & Answer

Answer to Problem 4.31E

The units in equation 4.18-4.25 have been shown consistent on both the sides of each equation.

Explanation of Solution

The equation 4.18 is shown below.

$(∂U∂S)V=T$

Where,

• $U$ is the internal energy.

• $T$ is the temperature.

• $S$ is the entropy.

• $V$ is the volume.

The units of internal energy, temperature, entropy, pressure and volume is $J/mol$ , $K$ , $J/mol⋅K$ , $Pa$ and $m3$ respectively.

Substitute the units of internal energy and entropy in the left hand side expression as shown below.

$(∂U∂S)V=(J/molJ/mol⋅K)=(11/K)=K$

Thus, the unit obtained is Kelvin which is the unit of temperature. Thus, the units are consistent on the either side of the equation.

The equation 4.19 is shown below.

$(∂U∂V)S=−p$

Where,

• $U$ is the internal energy.

• $S$ is the entropy.

• $p$ is the pressure.

• $V$ is the volume.

The units of internal energy and entropy can also be expressed in per molecule, the units of internal energy and entropy is $J$ and $J/K$ respectively.

Substitute the units of internal energy and volume in the left hand side expression as shown below.

$(∂U∂V)S=(Jm3)$

The units of internal energy joule can be substituted as $Pa⋅m3$ as shown below.

$(∂U∂V)S=(Pa⋅m3m3)=Pa$

Thus, the unit obtained is Pascal which is the unit of pressure. Thus, the units are consistent on the either side of the equation.

The equation 4.20 is shown below.

$(∂H∂S)p=T$

Where,

• $H$ is the enthalpy.

• $T$ is the temperature.

• $S$ is the entropy.

• $p$ is the pressure.

The units of enthalpy, temperature, entropy, pressure and volume is $J/mol$ , $K$ , $J/mol⋅K$ , $Pa$ and $m3$ respectively.

Substitute the units of enthalpy and entropy in the left hand side expression as shown below.

$(∂H∂S)V=(J/molJ/mol⋅K)=(11/K)=K$

Thus, the unit obtained is Kelvin which is the unit of temperature. Thus, the units are consistent on the either side of the equation.

The equation 4.21 is shown below.

$(∂H∂p)S=V$

Where,

• $H$ is the enthalpy and its unit is $J$ per molecule.

• $S$ is the entropy and its units are $J/K$ per molecule.

• $p$ is the pressure and its unit is $Pa$ .

• $V$ is the volume and its unit is $m3$ .

The units of enthalpy and entropy can also be expressed in per molecule, the units of enthalpy and entropy is $J$ and $J/K$ respectively.

Substitute the units of enthalpy and pressure in the left hand side expression as shown below.

$(∂H∂p)S=(JPa)$

The units of enthalpy joule can be substituted as $Pa⋅m3$ as shown below.

$(∂H∂p)S=(Pa⋅m3Pa)=m3$

Thus, the unit obtained is cubic meter which is the unit of volume. Thus, the units are consistent on the either side of the equation.

The equation 4.22 is shown below.

$(∂A∂T)V=−S$

Where,

• $A$ is the Helmholtz energy.

• $T$ is the temperature.

• $S$ is the entropy.

• $V$ is the volume.

The units of Helmholtz energy, temperature, entropy, pressure and volume is $J/mol$ , $K$ , $J/mol⋅K$ , $Pa$ and $m3$ respectively.

Substitute the units of Helmholtz energy and temperature in the left hand side expression as shown below.

$(∂A∂T)V=(J/molK)=(Jmol⋅K)$

Thus, the unit obtained is $Jmol⋅K$ Kelvin which is the unit of entropy. Thus, the units are consistent on the either side of the equation.

The equation 4.23 is shown below.

$(∂A∂V)T=−p$

Where,

• $A$ is the Helmholtz energy.

• $p$ is the pressure.

• $V$ is the volume.

• $T$ is the temperature.

The units of Helmholtz energy and entropy can also be expressed in per molecule, the units of Helmholtz energy and entropy is $J$ and $J/K$ respectively.

Substitute the units of Helmholtz energy and volume in the left hand side expression as shown below.

$(∂A∂V)T=(Jm3)$

The units of Helmholtz energy joule can be substituted as $Pa⋅m3$ as shown below.

$(∂A∂V)T=(Pa⋅m3m3)=Pa$

Thus, the unit obtained is Pascal which is the unit of pressure. Thus, the units are consistent on the either side of the equation.

The equation 4.24 is shown below.

$(∂G∂T)p=−S$

Where,

• $A$ is the Gibbs energy.

• $T$ is the temperature.

• $S$ is the entropy.

• $V$ is the volume.

The units of Gibbs energy and entropy can also be expressed in per molecule, the units of Gibbs energy and entropy is $J$ and $J/K$ respectively.

Substitute the units of Gibbs energy and temperature in the left hand side expression as shown below.

$(∂G∂T)p=(J/molK)=(Jmol⋅K)$

Thus, the unit obtained is $Jmol⋅K$ Kelvin which is the unit of entropy. Thus, the units are consistent on the either side of the equation.

The equation 4.25 is shown below.

$(∂G∂p)T=V$

Where,

• $G$ is the Gibbs energy

• $T$ is the temperature.

• $p$ is the pressure.

• $V$ is the volume.

The units of Gibbs energy and entropy can also be expressed in per molecule, the units of Gibbs energy and entropy is $J$ and $J/K$ respectively.

Substitute the units of Gibbs energy and pressure in the left hand side expression as shown below.

$(∂G∂p)T=(JPa)$

The units of Gibbs energy joule can be substituted as $Pa⋅m3$ as shown below.

$(∂G∂p)T=(Pa⋅m3Pa)=m3$

Thus, the unit obtained is cubic meter which is the unit of volume. Thus, the units are consistent on the either side of the equation.

Conclusion

The units in equation 4.18-4.25 have been shown consistent on both the sides of each equation.

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Answer 3

Question

Chapter 4, Problem 4.31E

Interpretation Introduction

Interpretation:

The units in equation 4.18-4.25 are to be shown consistent on both the sides of each equation.

Concept introduction:

The thermodynamic properties of the system can be determined using natural variable equations. The natural variables such as temperature, volume, pressure and entropy can be used to describe the thermodynamic properties such as internal energy, Gibbs energy, enthalpy, Helmholtz energy.

Expert Solution & Answer

Answer to Problem 4.31E

The units in equation 4.18-4.25 have been shown consistent on both the sides of each equation.

Explanation of Solution

The equation 4.18 is shown below.

$(∂U∂S)V=T$

Where,

• $U$ is the internal energy.

• $T$ is the temperature.

• $S$ is the entropy.

• $V$ is the volume.

The units of internal energy, temperature, entropy, pressure and volume is $J/mol$ , $K$ , $J/mol⋅K$ , $Pa$ and $m3$ respectively.

Substitute the units of internal energy and entropy in the left hand side expression as shown below.

$(∂U∂S)V=(J/molJ/mol⋅K)=(11/K)=K$

Thus, the unit obtained is Kelvin which is the unit of temperature. Thus, the units are consistent on the either side of the equation.

The equation 4.19 is shown below.

$(∂U∂V)S=−p$

Where,

• $U$ is the internal energy.

• $S$ is the entropy.

• $p$ is the pressure.

• $V$ is the volume.

The units of internal energy and entropy can also be expressed in per molecule, the units of internal energy and entropy is $J$ and $J/K$ respectively.

Substitute the units of internal energy and volume in the left hand side expression as shown below.

$(∂U∂V)S=(Jm3)$

The units of internal energy joule can be substituted as $Pa⋅m3$ as shown below.

$(∂U∂V)S=(Pa⋅m3m3)=Pa$

Thus, the unit obtained is Pascal which is the unit of pressure. Thus, the units are consistent on the either side of the equation.

The equation 4.20 is shown below.

$(∂H∂S)p=T$

Where,

• $H$ is the enthalpy.

• $T$ is the temperature.

• $S$ is the entropy.

• $p$ is the pressure.

The units of enthalpy, temperature, entropy, pressure and volume is $J/mol$ , $K$ , $J/mol⋅K$ , $Pa$ and $m3$ respectively.

Substitute the units of enthalpy and entropy in the left hand side expression as shown below.

$(∂H∂S)V=(J/molJ/mol⋅K)=(11/K)=K$

Thus, the unit obtained is Kelvin which is the unit of temperature. Thus, the units are consistent on the either side of the equation.

The equation 4.21 is shown below.

$(∂H∂p)S=V$

Where,

• $H$ is the enthalpy and its unit is $J$ per molecule.

• $S$ is the entropy and its units are $J/K$ per molecule.

• $p$ is the pressure and its unit is $Pa$ .

• $V$ is the volume and its unit is $m3$ .

The units of enthalpy and entropy can also be expressed in per molecule, the units of enthalpy and entropy is $J$ and $J/K$ respectively.

Substitute the units of enthalpy and pressure in the left hand side expression as shown below.

$(∂H∂p)S=(JPa)$

The units of enthalpy joule can be substituted as $Pa⋅m3$ as shown below.

$(∂H∂p)S=(Pa⋅m3Pa)=m3$

Thus, the unit obtained is cubic meter which is the unit of volume. Thus, the units are consistent on the either side of the equation.

The equation 4.22 is shown below.

$(∂A∂T)V=−S$

Where,

• $A$ is the Helmholtz energy.

• $T$ is the temperature.

• $S$ is the entropy.

• $V$ is the volume.

The units of Helmholtz energy, temperature, entropy, pressure and volume is $J/mol$ , $K$ , $J/mol⋅K$ , $Pa$ and $m3$ respectively.

Substitute the units of Helmholtz energy and temperature in the left hand side expression as shown below.

$(∂A∂T)V=(J/molK)=(Jmol⋅K)$

Thus, the unit obtained is $Jmol⋅K$ Kelvin which is the unit of entropy. Thus, the units are consistent on the either side of the equation.

The equation 4.23 is shown below.

$(∂A∂V)T=−p$

Where,

• $A$ is the Helmholtz energy.

• $p$ is the pressure.

• $V$ is the volume.

• $T$ is the temperature.

The units of Helmholtz energy and entropy can also be expressed in per molecule, the units of Helmholtz energy and entropy is $J$ and $J/K$ respectively.

Substitute the units of Helmholtz energy and volume in the left hand side expression as shown below.

$(∂A∂V)T=(Jm3)$

The units of Helmholtz energy joule can be substituted as $Pa⋅m3$ as shown below.

$(∂A∂V)T=(Pa⋅m3m3)=Pa$

Thus, the unit obtained is Pascal which is the unit of pressure. Thus, the units are consistent on the either side of the equation.

The equation 4.24 is shown below.

$(∂G∂T)p=−S$

Where,

• $A$ is the Gibbs energy.

• $T$ is the temperature.

• $S$ is the entropy.

• $V$ is the volume.

The units of Gibbs energy and entropy can also be expressed in per molecule, the units of Gibbs energy and entropy is $J$ and $J/K$ respectively.

Substitute the units of Gibbs energy and temperature in the left hand side expression as shown below.

$(∂G∂T)p=(J/molK)=(Jmol⋅K)$

Thus, the unit obtained is $Jmol⋅K$ Kelvin which is the unit of entropy. Thus, the units are consistent on the either side of the equation.

The equation 4.25 is shown below.

$(∂G∂p)T=V$

Where,

• $G$ is the Gibbs energy

• $T$ is the temperature.

• $p$ is the pressure.

• $V$ is the volume.

The units of Gibbs energy and entropy can also be expressed in per molecule, the units of Gibbs energy and entropy is $J$ and $J/K$ respectively.

Substitute the units of Gibbs energy and pressure in the left hand side expression as shown below.

$(∂G∂p)T=(JPa)$

The units of Gibbs energy joule can be substituted as $Pa⋅m3$ as shown below.

$(∂G∂p)T=(Pa⋅m3Pa)=m3$

Thus, the unit obtained is cubic meter which is the unit of volume. Thus, the units are consistent on the either side of the equation.

Conclusion

The units in equation 4.18-4.25 have been shown consistent on both the sides of each equation.

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Answer 4

Question

Chapter 4, Problem 4.31E

Interpretation Introduction

Interpretation:

The units in equation 4.18-4.25 are to be shown consistent on both the sides of each equation.

Concept introduction:

The thermodynamic properties of the system can be determined using natural variable equations. The natural variables such as temperature, volume, pressure and entropy can be used to describe the thermodynamic properties such as internal energy, Gibbs energy, enthalpy, Helmholtz energy.

Expert Solution & Answer

Answer to Problem 4.31E

The units in equation 4.18-4.25 have been shown consistent on both the sides of each equation.

Explanation of Solution

The equation 4.18 is shown below.

$(∂U∂S)V=T$

Where,

• $U$ is the internal energy.

• $T$ is the temperature.

• $S$ is the entropy.

• $V$ is the volume.

The units of internal energy, temperature, entropy, pressure and volume is $J/mol$ , $K$ , $J/mol⋅K$ , $Pa$ and $m3$ respectively.

Substitute the units of internal energy and entropy in the left hand side expression as shown below.

$(∂U∂S)V=(J/molJ/mol⋅K)=(11/K)=K$

Thus, the unit obtained is Kelvin which is the unit of temperature. Thus, the units are consistent on the either side of the equation.

The equation 4.19 is shown below.

$(∂U∂V)S=−p$

Where,

• $U$ is the internal energy.

• $S$ is the entropy.

• $p$ is the pressure.

• $V$ is the volume.

The units of internal energy and entropy can also be expressed in per molecule, the units of internal energy and entropy is $J$ and $J/K$ respectively.

Substitute the units of internal energy and volume in the left hand side expression as shown below.

$(∂U∂V)S=(Jm3)$

The units of internal energy joule can be substituted as $Pa⋅m3$ as shown below.

$(∂U∂V)S=(Pa⋅m3m3)=Pa$

Thus, the unit obtained is Pascal which is the unit of pressure. Thus, the units are consistent on the either side of the equation.

The equation 4.20 is shown below.

$(∂H∂S)p=T$

Where,

• $H$ is the enthalpy.

• $T$ is the temperature.

• $S$ is the entropy.

• $p$ is the pressure.

The units of enthalpy, temperature, entropy, pressure and volume is $J/mol$ , $K$ , $J/mol⋅K$ , $Pa$ and $m3$ respectively.

Substitute the units of enthalpy and entropy in the left hand side expression as shown below.

$(∂H∂S)V=(J/molJ/mol⋅K)=(11/K)=K$

Thus, the unit obtained is Kelvin which is the unit of temperature. Thus, the units are consistent on the either side of the equation.

The equation 4.21 is shown below.

$(∂H∂p)S=V$

Where,

• $H$ is the enthalpy and its unit is $J$ per molecule.

• $S$ is the entropy and its units are $J/K$ per molecule.

• $p$ is the pressure and its unit is $Pa$ .

• $V$ is the volume and its unit is $m3$ .

The units of enthalpy and entropy can also be expressed in per molecule, the units of enthalpy and entropy is $J$ and $J/K$ respectively.

Substitute the units of enthalpy and pressure in the left hand side expression as shown below.

$(∂H∂p)S=(JPa)$

The units of enthalpy joule can be substituted as $Pa⋅m3$ as shown below.

$(∂H∂p)S=(Pa⋅m3Pa)=m3$

Thus, the unit obtained is cubic meter which is the unit of volume. Thus, the units are consistent on the either side of the equation.

The equation 4.22 is shown below.

$(∂A∂T)V=−S$

Where,

• $A$ is the Helmholtz energy.

• $T$ is the temperature.

• $S$ is the entropy.

• $V$ is the volume.

The units of Helmholtz energy, temperature, entropy, pressure and volume is $J/mol$ , $K$ , $J/mol⋅K$ , $Pa$ and $m3$ respectively.

Substitute the units of Helmholtz energy and temperature in the left hand side expression as shown below.

$(∂A∂T)V=(J/molK)=(Jmol⋅K)$

Thus, the unit obtained is $Jmol⋅K$ Kelvin which is the unit of entropy. Thus, the units are consistent on the either side of the equation.

The equation 4.23 is shown below.

$(∂A∂V)T=−p$

Where,

• $A$ is the Helmholtz energy.

• $p$ is the pressure.

• $V$ is the volume.

• $T$ is the temperature.

The units of Helmholtz energy and entropy can also be expressed in per molecule, the units of Helmholtz energy and entropy is $J$ and $J/K$ respectively.

Substitute the units of Helmholtz energy and volume in the left hand side expression as shown below.

$(∂A∂V)T=(Jm3)$

The units of Helmholtz energy joule can be substituted as $Pa⋅m3$ as shown below.

$(∂A∂V)T=(Pa⋅m3m3)=Pa$

Thus, the unit obtained is Pascal which is the unit of pressure. Thus, the units are consistent on the either side of the equation.

The equation 4.24 is shown below.

$(∂G∂T)p=−S$

Where,

• $A$ is the Gibbs energy.

• $T$ is the temperature.

• $S$ is the entropy.

• $V$ is the volume.

The units of Gibbs energy and entropy can also be expressed in per molecule, the units of Gibbs energy and entropy is $J$ and $J/K$ respectively.

Substitute the units of Gibbs energy and temperature in the left hand side expression as shown below.

$(∂G∂T)p=(J/molK)=(Jmol⋅K)$

Thus, the unit obtained is $Jmol⋅K$ Kelvin which is the unit of entropy. Thus, the units are consistent on the either side of the equation.

The equation 4.25 is shown below.

$(∂G∂p)T=V$

Where,

• $G$ is the Gibbs energy

• $T$ is the temperature.

• $p$ is the pressure.

• $V$ is the volume.

The units of Gibbs energy and entropy can also be expressed in per molecule, the units of Gibbs energy and entropy is $J$ and $J/K$ respectively.

Substitute the units of Gibbs energy and pressure in the left hand side expression as shown below.

$(∂G∂p)T=(JPa)$

The units of Gibbs energy joule can be substituted as $Pa⋅m3$ as shown below.

$(∂G∂p)T=(Pa⋅m3Pa)=m3$

Thus, the unit obtained is cubic meter which is the unit of volume. Thus, the units are consistent on the either side of the equation.

Conclusion

The units in equation 4.18-4.25 have been shown consistent on both the sides of each equation.

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Answer 5

Chapter 4, Problem 4.31E

Interpretation Introduction

Interpretation:

The units in equation 4.18-4.25 are to be shown consistent on both the sides of each equation.

Concept introduction:

The thermodynamic properties of the system can be determined using natural variable equations. The natural variables such as temperature, volume, pressure and entropy can be used to describe the thermodynamic properties such as internal energy, Gibbs energy, enthalpy, Helmholtz energy.

Expert Solution & Answer

Answer to Problem 4.31E

The units in equation 4.18-4.25 have been shown consistent on both the sides of each equation.

Explanation of Solution

The equation 4.18 is shown below.

$(∂U∂S)V=T$

Where,

• $U$ is the internal energy.

• $T$ is the temperature.

• $S$ is the entropy.

• $V$ is the volume.

The units of internal energy, temperature, entropy, pressure and volume is $J/mol$ , $K$ , $J/mol⋅K$ , $Pa$ and $m3$ respectively.

Substitute the units of internal energy and entropy in the left hand side expression as shown below.

$(∂U∂S)V=(J/molJ/mol⋅K)=(11/K)=K$

Thus, the unit obtained is Kelvin which is the unit of temperature. Thus, the units are consistent on the either side of the equation.

The equation 4.19 is shown below.

$(∂U∂V)S=−p$

Where,

• $U$ is the internal energy.

• $S$ is the entropy.

• $p$ is the pressure.

• $V$ is the volume.

The units of internal energy and entropy can also be expressed in per molecule, the units of internal energy and entropy is $J$ and $J/K$ respectively.

Substitute the units of internal energy and volume in the left hand side expression as shown below.

$(∂U∂V)S=(Jm3)$

The units of internal energy joule can be substituted as $Pa⋅m3$ as shown below.

$(∂U∂V)S=(Pa⋅m3m3)=Pa$

Thus, the unit obtained is Pascal which is the unit of pressure. Thus, the units are consistent on the either side of the equation.

The equation 4.20 is shown below.

$(∂H∂S)p=T$

Where,

• $H$ is the enthalpy.

• $T$ is the temperature.

• $S$ is the entropy.

• $p$ is the pressure.

The units of enthalpy, temperature, entropy, pressure and volume is $J/mol$ , $K$ , $J/mol⋅K$ , $Pa$ and $m3$ respectively.

Substitute the units of enthalpy and entropy in the left hand side expression as shown below.

$(∂H∂S)V=(J/molJ/mol⋅K)=(11/K)=K$

Thus, the unit obtained is Kelvin which is the unit of temperature. Thus, the units are consistent on the either side of the equation.

The equation 4.21 is shown below.

$(∂H∂p)S=V$

Where,

• $H$ is the enthalpy and its unit is $J$ per molecule.

• $S$ is the entropy and its units are $J/K$ per molecule.

• $p$ is the pressure and its unit is $Pa$ .

• $V$ is the volume and its unit is $m3$ .

The units of enthalpy and entropy can also be expressed in per molecule, the units of enthalpy and entropy is $J$ and $J/K$ respectively.

Substitute the units of enthalpy and pressure in the left hand side expression as shown below.

$(∂H∂p)S=(JPa)$

The units of enthalpy joule can be substituted as $Pa⋅m3$ as shown below.

$(∂H∂p)S=(Pa⋅m3Pa)=m3$

Thus, the unit obtained is cubic meter which is the unit of volume. Thus, the units are consistent on the either side of the equation.

The equation 4.22 is shown below.

$(∂A∂T)V=−S$

Where,

• $A$ is the Helmholtz energy.

• $T$ is the temperature.

• $S$ is the entropy.

• $V$ is the volume.

The units of Helmholtz energy, temperature, entropy, pressure and volume is $J/mol$ , $K$ , $J/mol⋅K$ , $Pa$ and $m3$ respectively.

Substitute the units of Helmholtz energy and temperature in the left hand side expression as shown below.

$(∂A∂T)V=(J/molK)=(Jmol⋅K)$

Thus, the unit obtained is $Jmol⋅K$ Kelvin which is the unit of entropy. Thus, the units are consistent on the either side of the equation.

The equation 4.23 is shown below.

$(∂A∂V)T=−p$

Where,

• $A$ is the Helmholtz energy.

• $p$ is the pressure.

• $V$ is the volume.

• $T$ is the temperature.

The units of Helmholtz energy and entropy can also be expressed in per molecule, the units of Helmholtz energy and entropy is $J$ and $J/K$ respectively.

Substitute the units of Helmholtz energy and volume in the left hand side expression as shown below.

$(∂A∂V)T=(Jm3)$

The units of Helmholtz energy joule can be substituted as $Pa⋅m3$ as shown below.

$(∂A∂V)T=(Pa⋅m3m3)=Pa$

Thus, the unit obtained is Pascal which is the unit of pressure. Thus, the units are consistent on the either side of the equation.

The equation 4.24 is shown below.

$(∂G∂T)p=−S$

Where,

• $A$ is the Gibbs energy.

• $T$ is the temperature.

• $S$ is the entropy.

• $V$ is the volume.

The units of Gibbs energy and entropy can also be expressed in per molecule, the units of Gibbs energy and entropy is $J$ and $J/K$ respectively.

Substitute the units of Gibbs energy and temperature in the left hand side expression as shown below.

$(∂G∂T)p=(J/molK)=(Jmol⋅K)$

Thus, the unit obtained is $Jmol⋅K$ Kelvin which is the unit of entropy. Thus, the units are consistent on the either side of the equation.

The equation 4.25 is shown below.

$(∂G∂p)T=V$

Where,

• $G$ is the Gibbs energy

• $T$ is the temperature.

• $p$ is the pressure.

• $V$ is the volume.

The units of Gibbs energy and entropy can also be expressed in per molecule, the units of Gibbs energy and entropy is $J$ and $J/K$ respectively.

Substitute the units of Gibbs energy and pressure in the left hand side expression as shown below.

$(∂G∂p)T=(JPa)$

The units of Gibbs energy joule can be substituted as $Pa⋅m3$ as shown below.

$(∂G∂p)T=(Pa⋅m3Pa)=m3$

Thus, the unit obtained is cubic meter which is the unit of volume. Thus, the units are consistent on the either side of the equation.

Conclusion

The units in equation 4.18-4.25 have been shown consistent on both the sides of each equation.

Answer 6

Chapter 4, Problem 4.31E

Interpretation Introduction

Interpretation:

The units in equation 4.18-4.25 are to be shown consistent on both the sides of each equation.

Concept introduction:

The thermodynamic properties of the system can be determined using natural variable equations. The natural variables such as temperature, volume, pressure and entropy can be used to describe the thermodynamic properties such as internal energy, Gibbs energy, enthalpy, Helmholtz energy.

Expert Solution & Answer

Answer to Problem 4.31E

The units in equation 4.18-4.25 have been shown consistent on both the sides of each equation.

Explanation of Solution

The equation 4.18 is shown below.

$(∂U∂S)V=T$

Where,

• $U$ is the internal energy.

• $T$ is the temperature.

• $S$ is the entropy.

• $V$ is the volume.

The units of internal energy, temperature, entropy, pressure and volume is $J/mol$ , $K$ , $J/mol⋅K$ , $Pa$ and $m3$ respectively.

Substitute the units of internal energy and entropy in the left hand side expression as shown below.

$(∂U∂S)V=(J/molJ/mol⋅K)=(11/K)=K$

Thus, the unit obtained is Kelvin which is the unit of temperature. Thus, the units are consistent on the either side of the equation.

The equation 4.19 is shown below.

$(∂U∂V)S=−p$

Where,

• $U$ is the internal energy.

• $S$ is the entropy.

• $p$ is the pressure.

• $V$ is the volume.

The units of internal energy and entropy can also be expressed in per molecule, the units of internal energy and entropy is $J$ and $J/K$ respectively.

Substitute the units of internal energy and volume in the left hand side expression as shown below.

$(∂U∂V)S=(Jm3)$

The units of internal energy joule can be substituted as $Pa⋅m3$ as shown below.

$(∂U∂V)S=(Pa⋅m3m3)=Pa$

Thus, the unit obtained is Pascal which is the unit of pressure. Thus, the units are consistent on the either side of the equation.

The equation 4.20 is shown below.

$(∂H∂S)p=T$

Where,

• $H$ is the enthalpy.

• $T$ is the temperature.

• $S$ is the entropy.

• $p$ is the pressure.

The units of enthalpy, temperature, entropy, pressure and volume is $J/mol$ , $K$ , $J/mol⋅K$ , $Pa$ and $m3$ respectively.

Substitute the units of enthalpy and entropy in the left hand side expression as shown below.

$(∂H∂S)V=(J/molJ/mol⋅K)=(11/K)=K$

Thus, the unit obtained is Kelvin which is the unit of temperature. Thus, the units are consistent on the either side of the equation.

The equation 4.21 is shown below.

$(∂H∂p)S=V$

Where,

• $H$ is the enthalpy and its unit is $J$ per molecule.

• $S$ is the entropy and its units are $J/K$ per molecule.

• $p$ is the pressure and its unit is $Pa$ .

• $V$ is the volume and its unit is $m3$ .

The units of enthalpy and entropy can also be expressed in per molecule, the units of enthalpy and entropy is $J$ and $J/K$ respectively.

Substitute the units of enthalpy and pressure in the left hand side expression as shown below.

$(∂H∂p)S=(JPa)$

The units of enthalpy joule can be substituted as $Pa⋅m3$ as shown below.

$(∂H∂p)S=(Pa⋅m3Pa)=m3$

Thus, the unit obtained is cubic meter which is the unit of volume. Thus, the units are consistent on the either side of the equation.

The equation 4.22 is shown below.

$(∂A∂T)V=−S$

Where,

• $A$ is the Helmholtz energy.

• $T$ is the temperature.

• $S$ is the entropy.

• $V$ is the volume.

The units of Helmholtz energy, temperature, entropy, pressure and volume is $J/mol$ , $K$ , $J/mol⋅K$ , $Pa$ and $m3$ respectively.

Substitute the units of Helmholtz energy and temperature in the left hand side expression as shown below.

$(∂A∂T)V=(J/molK)=(Jmol⋅K)$

Thus, the unit obtained is $Jmol⋅K$ Kelvin which is the unit of entropy. Thus, the units are consistent on the either side of the equation.

The equation 4.23 is shown below.

$(∂A∂V)T=−p$

Where,

• $A$ is the Helmholtz energy.

• $p$ is the pressure.

• $V$ is the volume.

• $T$ is the temperature.

The units of Helmholtz energy and entropy can also be expressed in per molecule, the units of Helmholtz energy and entropy is $J$ and $J/K$ respectively.

Substitute the units of Helmholtz energy and volume in the left hand side expression as shown below.

$(∂A∂V)T=(Jm3)$

The units of Helmholtz energy joule can be substituted as $Pa⋅m3$ as shown below.

$(∂A∂V)T=(Pa⋅m3m3)=Pa$

Thus, the unit obtained is Pascal which is the unit of pressure. Thus, the units are consistent on the either side of the equation.

The equation 4.24 is shown below.

$(∂G∂T)p=−S$

Where,

• $A$ is the Gibbs energy.

• $T$ is the temperature.

• $S$ is the entropy.

• $V$ is the volume.

The units of Gibbs energy and entropy can also be expressed in per molecule, the units of Gibbs energy and entropy is $J$ and $J/K$ respectively.

Substitute the units of Gibbs energy and temperature in the left hand side expression as shown below.

$(∂G∂T)p=(J/molK)=(Jmol⋅K)$

Thus, the unit obtained is $Jmol⋅K$ Kelvin which is the unit of entropy. Thus, the units are consistent on the either side of the equation.

The equation 4.25 is shown below.

$(∂G∂p)T=V$

Where,

• $G$ is the Gibbs energy

• $T$ is the temperature.

• $p$ is the pressure.

• $V$ is the volume.

The units of Gibbs energy and entropy can also be expressed in per molecule, the units of Gibbs energy and entropy is $J$ and $J/K$ respectively.

Substitute the units of Gibbs energy and pressure in the left hand side expression as shown below.

$(∂G∂p)T=(JPa)$

The units of Gibbs energy joule can be substituted as $Pa⋅m3$ as shown below.

$(∂G∂p)T=(Pa⋅m3Pa)=m3$

Thus, the unit obtained is cubic meter which is the unit of volume. Thus, the units are consistent on the either side of the equation.

Conclusion

The units in equation 4.18-4.25 have been shown consistent on both the sides of each equation.

Answer 7

Chapter 4, Problem 4.31E

Interpretation Introduction

Interpretation:

The units in equation 4.18-4.25 are to be shown consistent on both the sides of each equation.

Concept introduction:

The thermodynamic properties of the system can be determined using natural variable equations. The natural variables such as temperature, volume, pressure and entropy can be used to describe the thermodynamic properties such as internal energy, Gibbs energy, enthalpy, Helmholtz energy.

Answer 8

Expert Solution & Answer

Answer to Problem 4.31E

The units in equation 4.18-4.25 have been shown consistent on both the sides of each equation.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

The equation 4.18 is shown below.

$(∂U∂S)V=T$

Where,

• $U$ is the internal energy.

• $T$ is the temperature.

• $S$ is the entropy.

• $V$ is the volume.

The units of internal energy, temperature, entropy, pressure and volume is $J/mol$ , $K$ , $J/mol⋅K$ , $Pa$ and $m3$ respectively.

Substitute the units of internal energy and entropy in the left hand side expression as shown below.

$(∂U∂S)V=(J/molJ/mol⋅K)=(11/K)=K$

Thus, the unit obtained is Kelvin which is the unit of temperature. Thus, the units are consistent on the either side of the equation.

The equation 4.19 is shown below.

$(∂U∂V)S=−p$

Where,

• $U$ is the internal energy.

• $S$ is the entropy.

• $p$ is the pressure.

• $V$ is the volume.

The units of internal energy and entropy can also be expressed in per molecule, the units of internal energy and entropy is $J$ and $J/K$ respectively.

Substitute the units of internal energy and volume in the left hand side expression as shown below.

$(∂U∂V)S=(Jm3)$

The units of internal energy joule can be substituted as $Pa⋅m3$ as shown below.

$(∂U∂V)S=(Pa⋅m3m3)=Pa$

Thus, the unit obtained is Pascal which is the unit of pressure. Thus, the units are consistent on the either side of the equation.

The equation 4.20 is shown below.

$(∂H∂S)p=T$

Where,

• $H$ is the enthalpy.

• $T$ is the temperature.

• $S$ is the entropy.

• $p$ is the pressure.

The units of enthalpy, temperature, entropy, pressure and volume is $J/mol$ , $K$ , $J/mol⋅K$ , $Pa$ and $m3$ respectively.

Substitute the units of enthalpy and entropy in the left hand side expression as shown below.

$(∂H∂S)V=(J/molJ/mol⋅K)=(11/K)=K$

Thus, the unit obtained is Kelvin which is the unit of temperature. Thus, the units are consistent on the either side of the equation.

The equation 4.21 is shown below.

$(∂H∂p)S=V$

Where,

• $H$ is the enthalpy and its unit is $J$ per molecule.

• $S$ is the entropy and its units are $J/K$ per molecule.

• $p$ is the pressure and its unit is $Pa$ .

• $V$ is the volume and its unit is $m3$ .

The units of enthalpy and entropy can also be expressed in per molecule, the units of enthalpy and entropy is $J$ and $J/K$ respectively.

Substitute the units of enthalpy and pressure in the left hand side expression as shown below.

$(∂H∂p)S=(JPa)$

The units of enthalpy joule can be substituted as $Pa⋅m3$ as shown below.

$(∂H∂p)S=(Pa⋅m3Pa)=m3$

Thus, the unit obtained is cubic meter which is the unit of volume. Thus, the units are consistent on the either side of the equation.

The equation 4.22 is shown below.

$(∂A∂T)V=−S$

Where,

• $A$ is the Helmholtz energy.

• $T$ is the temperature.

• $S$ is the entropy.

• $V$ is the volume.

The units of Helmholtz energy, temperature, entropy, pressure and volume is $J/mol$ , $K$ , $J/mol⋅K$ , $Pa$ and $m3$ respectively.

Substitute the units of Helmholtz energy and temperature in the left hand side expression as shown below.

$(∂A∂T)V=(J/molK)=(Jmol⋅K)$

Thus, the unit obtained is $Jmol⋅K$ Kelvin which is the unit of entropy. Thus, the units are consistent on the either side of the equation.

The equation 4.23 is shown below.

$(∂A∂V)T=−p$

Where,

• $A$ is the Helmholtz energy.

• $p$ is the pressure.

• $V$ is the volume.

• $T$ is the temperature.

The units of Helmholtz energy and entropy can also be expressed in per molecule, the units of Helmholtz energy and entropy is $J$ and $J/K$ respectively.

Substitute the units of Helmholtz energy and volume in the left hand side expression as shown below.

$(∂A∂V)T=(Jm3)$

The units of Helmholtz energy joule can be substituted as $Pa⋅m3$ as shown below.

$(∂A∂V)T=(Pa⋅m3m3)=Pa$

Thus, the unit obtained is Pascal which is the unit of pressure. Thus, the units are consistent on the either side of the equation.

The equation 4.24 is shown below.

$(∂G∂T)p=−S$

Where,

• $A$ is the Gibbs energy.

• $T$ is the temperature.

• $S$ is the entropy.

• $V$ is the volume.

The units of Gibbs energy and entropy can also be expressed in per molecule, the units of Gibbs energy and entropy is $J$ and $J/K$ respectively.

Substitute the units of Gibbs energy and temperature in the left hand side expression as shown below.

$(∂G∂T)p=(J/molK)=(Jmol⋅K)$

Thus, the unit obtained is $Jmol⋅K$ Kelvin which is the unit of entropy. Thus, the units are consistent on the either side of the equation.

The equation 4.25 is shown below.

$(∂G∂p)T=V$

Where,

• $G$ is the Gibbs energy

• $T$ is the temperature.

• $p$ is the pressure.

• $V$ is the volume.

The units of Gibbs energy and entropy can also be expressed in per molecule, the units of Gibbs energy and entropy is $J$ and $J/K$ respectively.

Substitute the units of Gibbs energy and pressure in the left hand side expression as shown below.

$(∂G∂p)T=(JPa)$

The units of Gibbs energy joule can be substituted as $Pa⋅m3$ as shown below.

$(∂G∂p)T=(Pa⋅m3Pa)=m3$

Thus, the unit obtained is cubic meter which is the unit of volume. Thus, the units are consistent on the either side of the equation.

Answer 12

Conclusion

The units in equation 4.18-4.25 have been shown consistent on both the sides of each equation.

Answer 13

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Answer to Problem 4.31E

Explanation of Solution

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Chapter 4 Solutions