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 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.

${\left(\frac{\partial U}{\partial S}\right)}_V=T$

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 $\text{J}/\text{mol}$, $\text{K}$, $\text{J}/\text{mol}\cdot \text{K}$, $\text{Pa}$ and ${\text{m}}^3$ respectively.

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

$\begin{array}{rll}{\left(\frac{\partial U}{\partial S}\right)}_V& =& \left(\frac{\text{J}/\text{mol}}{\text{J}/\text{mol}\cdot \text{K}}\right)\\ & =& \left(\frac{1}{1/\text{K}}\right)\\ & =& \text{K}\end{array}$

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.

${\left(\frac{\partial U}{\partial V}\right)}_S=-p$

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 $\text{J}$ and $\text{J}/\text{K}$ respectively.

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

${\left(\frac{\partial U}{\partial V}\right)}_S=\left(\frac{\text{J}}{{\text{m}}^3}\right)$

The units of internal energy joule can be substituted as $\text{Pa}\cdot {\text{m}}^3$ as shown below.

$\begin{array}{rll}{\left(\frac{\partial U}{\partial V}\right)}_S& =& \left(\frac{\text{Pa}\cdot {\text{m}}^3}{{\text{m}}^3}\right)\\ & =& \text{Pa}\end{array}$

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.

${\left(\frac{\partial H}{\partial S}\right)}_p=T$

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 $\text{J}/\text{mol}$, $\text{K}$, $\text{J}/\text{mol}\cdot \text{K}$, $\text{Pa}$ and ${\text{m}}^3$ respectively.

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

$\begin{array}{rll}{\left(\frac{\partial H}{\partial S}\right)}_V& =& \left(\frac{\text{J}/\text{mol}}{\text{J}/\text{mol}\cdot \text{K}}\right)\\ & =& \left(\frac{1}{1/\text{K}}\right)\\ & =& \text{K}\end{array}$

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.

${\left(\frac{\partial H}{\partial p}\right)}_S=V$

Where,

• $H$ is the enthalpy and its unit is $\text{J}$ per molecule.

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

• $p$ is the pressure and its unit is $\text{Pa}$.

• $V$ is the volume and its unit is ${\text{m}}^3$.

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

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

${\left(\frac{\partial H}{\partial p}\right)}_S=\left(\frac{\text{J}}{\text{Pa}}\right)$

The units of enthalpy joule can be substituted as $\text{Pa}\cdot {\text{m}}^3$ as shown below.

$\begin{array}{rll}{\left(\frac{\partial H}{\partial p}\right)}_S& =& \left(\frac{\text{Pa}\cdot {\text{m}}^3}{\text{Pa}}\right)\\ & =& {\text{m}}^3\end{array}$

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.

${\left(\frac{\partial A}{\partial T}\right)}_V=-S$

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 $\text{J}/\text{mol}$, $\text{K}$, $\text{J}/\text{mol}\cdot \text{K}$, $\text{Pa}$ and ${\text{m}}^3$ respectively.

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

$\begin{array}{rll}{\left(\frac{\partial A}{\partial T}\right)}_V& =& \left(\frac{\text{J}/\text{mol}}{\text{K}}\right)\\ & =& \left(\frac{\text{J}}{\text{mol}\cdot \text{K}}\right)\end{array}$

Thus, the unit obtained is $\frac{\text{J}}{\text{mol}\cdot \text{K}}$ 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.

${\left(\frac{\partial A}{\partial V}\right)}_T=-p$

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 $\text{J}$ and $\text{J}/\text{K}$ respectively.

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

${\left(\frac{\partial A}{\partial V}\right)}_T=\left(\frac{\text{J}}{{\text{m}}^3}\right)$

The units of Helmholtz energy joule can be substituted as $\text{Pa}\cdot {\text{m}}^3$ as shown below.

$\begin{array}{rll}{\left(\frac{\partial A}{\partial V}\right)}_T& =& \left(\frac{\text{Pa}\cdot {\text{m}}^3}{{\text{m}}^3}\right)\\ & =& \text{Pa}\end{array}$

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.

${\left(\frac{\partial G}{\partial T}\right)}_p=-S$

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 $\text{J}$ and $\text{J}/\text{K}$ respectively.

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

$\begin{array}{rll}{\left(\frac{\partial G}{\partial T}\right)}_p& =& \left(\frac{\text{J}/\text{mol}}{\text{K}}\right)\\ & =& \left(\frac{\text{J}}{\text{mol}\cdot \text{K}}\right)\end{array}$

Thus, the unit obtained is $\frac{\text{J}}{\text{mol}\cdot \text{K}}$ 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.

${\left(\frac{\partial G}{\partial p}\right)}_T=V$

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 $\text{J}$ and $\text{J}/\text{K}$ respectively.

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

${\left(\frac{\partial G}{\partial p}\right)}_T=\left(\frac{\text{J}}{\text{Pa}}\right)$

The units of Gibbs energy joule can be substituted as $\text{Pa}\cdot {\text{m}}^3$ as shown below.

$\begin{array}{rll}{\left(\frac{\partial G}{\partial p}\right)}_T& =& \left(\frac{\text{Pa}\cdot {\text{m}}^3}{\text{Pa}}\right)\\ & =& {\text{m}}^3\end{array}$

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.