6.1 Atomic Model

6.1.1 Protons, Neutrons, and Electrons

Atom. The name atom is derived from the Greek word ατομοϛ (atomos) that literally translates to “indivisible” or “not to be cut”. The concept of matter being ultimately composed of elements that may not be further divided was first brought forth by Dalton.¹

Proton and Neutron. Atoms are composed of a very dense nucleus that consists of neutrally charged neutrons and positively charged protons. The positive charge of a proton amounts to one elementary charge that is equal to 1.602 176 565 × 10⁻¹⁹ C. The mass of protons is 1.672 621 777 × 10⁻²⁷ kg. The mass of neutrons is 1.674 927 351 × 10⁻²⁷ kg and they are usually considered to have approximately the same mass. The nucleus is held together by electromagnetic forces.

Electron. The electrons are arranged in shells around the nucleus. Electrons are three orders of magnitude lighter than both protons and neutrons and carry one negative charge. Its weight is 9.109 382 91 × 10⁻³¹ kg. An atom is classified by the number of protons, in electrically neutral states. If the number of electrons is equal to the number of protons (which is the default state), the atom is referred to as being electrically neutral or uncharged.

Ion. If the number of electrons and protons is not equal, the atom becomes charged. It is then referred to as an ion. As an example, Na⁺ indicates that the sodium atom has acquired one positive elementary charge or, to be correct, it lost one electron. The atom is then referred to as a cation. If the atom gained one elementary charge, it is referred to as an anion. An example is the chlorine anion Cl⁻. Higher charges may, of course, also occur, e.g., in copper sulfate CuSO₄ where the copper ion is double-positively charged Cu²⁺ and the sulfate is double-negatively charged ${\text{SO}}_4^{2-}$. Uncharged atoms can be transformed into ions by removing or adding electrons. Removal of electrons requires a certain amount of energy that is referred to as ionization energy. Thus the process of removing an electron from an atom is referred to as ionization, whereas the inverse process is referred to as recombination.

6.1.2 Periodic Table

Element. The number of protons in the nucleus defines the chemical element. Per definition, an element is a substance that cannot be decomposed any further. The number of protons is referred to as the atomic number that identifies the respective chemical element. The elements are arranged in the periodic table of the elements (see Fig. 6.1) in order of increasing atomic number. For each element, the number of protons and the number of neutrons are the same, with the exception of hydrogen, which has one proton, but no neutrons. The first composition of all known chemical elements was provided by the French chemist Antoine de Lavoisier [2]. It contained 33 substances. Today, we know 118 elements, 88 of which occur naturally.

Nuclide. If an atom is characterized by its number of protons and neutrons, it is referred to as a nuclide. Some elements have more than one nuclide, i.e., they occur with different neutron count. Such nuclides are referred to as isotopes of the pure element, i.e., as versions of the atom with varying number of neutrons, but the same number of protons. Isotopes are denoted with two preceding numeric values. The upper integer denotes the sum of neutrons and protons in the nucleus, the lower integer, the number of protons: ${}_1^1\text{H}$ is protium (the naturally occurring form of hydrogen with one proton only), ${}_1^2\text{H}$ is deuterium (commonly used hydrogen substitute in chemical spectroscopy with one proton and one neutron), ${}_1^3\text{H}$ is tritium (the radioactive isotope of hydrogen with one proton and two neutrons).

Many elements occur as mixtures of their isotopes naturally. Examples include carbon, which occurs besides the normal ${}_6^{12}\text{C}$ (99 %) also as ${}_6^{13}\text{C}$ (1 %), that is important, e.g., in nuclear magnetic resonance (NMR) spectroscopy. Only a few elements, e.g., fluorine or sodium occur as only a single isotope. These elements are referred to as mononuclidic element.

The periodic table is arranged in periods and groups. Elements that belong to the same group in the periodic table have similar physical and chemical properties, as they share the same number of so-called valence electrons, i.e., electrons in the outmost shell. Elements from the same period vary with respect to their physical and chemical behavior but within periods, certain trends in their behavior can be observed.

Period. Within the same period (going from left to right):

• atom radii decrease, due to the fact that a higher number of protons in the atom nucleus exerts higher attractive forces on the electrons, thus pulling them closer to the nucleus

• ionization energy increases due to the increasing interaction forces between the nucleus and the electrons

• electronegativity (which is the ability to attract electrons) increases due to the fact that an increasingly positively charged nucleus exerts higher attraction forces

Group. Within the same group (going from top to bottom):

• atom radii increase due to the fact that the atom has more electrons

• ionization energy decreases as the attractive forces from the nucleus decrease, thus making it easier to remove electrons

• electronegativity decreases as the positively charged nucleus is further apart from electrons in close proximity due to the fact that the atom radius is bigger

By convention, the groups in the periodic tables are assigned certain names:

• group 1 - “alkali metals”

• group 2 - “alkaline earth metals”

• group 3 to 12 - “transition metals”

• group 13 - “boron group elements”

• group 14 - “carbon group elements”

• group 15 - “pnictogens”

• group 16 - “chalcogens”

• group 17 - “halogens”

• group 18 - “noble gases”

Please note, that sometimes periodic tables may additionally subdivide atoms into categories, which comprise atoms from different periods and groups with similar physical or chemical properties. These categories include, e.g., post-transition metals or metalloids. In the periodic table in Fig. 6.1 these categories are color-coded. As this categorization is not standardized, periodic tables from different sources may vary.

6.4 Ideal Gases

6.4.1 Definition

Most of the theoretical assumptions on gases will be made using the ideal gas equations. The concept of ideal gases is a simplification of the dynamics of real gases, but in most cases it is sufficient for describing a number of properties of real gases. Two basic assumptions are made for ideal gases.

• There are no attractive forces between the gas molecules - the only interaction of the gas molecules are collisions among the molecules and with the container wall.

• The total volume of all gas molecules is small compared to the total volume of the container in which the gas is stored - the gas molecules can thus be assumed as infinitesimal small points.

One of the known consequences of these assumptions is the fact that ideal gases fill the entire volume with which they are provided. If the volume is decreased, the pressure of the gas molecules increases as they are more tightly packed. If the temperature is increased without increasing the volume, the pressure increases. From these basic correlations, the ideal gas law is defined as

$pV=nRT$

$pV=n{N}_A\frac{R}{N_A}T$

$pV=N{k}_{B}T$

$p=n_D{k}_{B}T$

R is the universal gas constant and has a value of 8.314 462 1 J mol⁻¹ K⁻¹ according to National Institute of Standards and Technology (NIST). Dividing R by N_A results in the Boltzmann constant k_B. Dividing the number of molecules by the volume results in the number density n_D, i.e., the number of molecules per volume.

The ideal gas law (see Eq. 6.4) is often used in slightly modified form using Eq. 6.2 and Eq. 6.3 as

$\begin{array}{l}pV=\frac{m}{M}RT\hfill \\ p\frac{V}{m}=\frac{R}{M}T\hfill \\ \frac{p}{\rho }=R_{\text{S}}T\hfill \\ p=\rho R_{\text{S}}T\hfill \end{array}$

Here we have introduced the specific gas constant R_S normalized to the molecular weight of the gas used. This is also a constant, although it is (obviously) different for every gas. A selection of commonly used gases is given in Tab. 6.1.

Molar Volume. The volume any gas at ambient pressure will occupy can be calculated simply from the ideal gas law in Eq. 6.4. Using p = 1 bar = 10⁵ Pa, n = 1 mol, R = 8.314 462 1 J mol⁻¹ K⁻¹ and a temperature of T = 0 C = 273.1499 K will result in a total volume of V_M,0 = 0.022 414 m³ or 22.414 L. This volume L is commonly referred to as the molar volume. It is the volume occupied by 1 mol of a gas at standard conditions. It is equal for all gases. The number of molecules contained in this volume is referred to as the Loschmidt number N_L. It is often confused with the Avogadro number N_A, which defines the number of molecules per mole. The two numbers are interconnected by the molar volume

$\text{L}=\frac{N_{\text{A}}}{N_{\text{L}}}$

The Committee on Data for Science and Technology (CODATA) recommends a value of 2.687 × 10²⁵ for the Loschmidt constant. In general most gases, especially at atmospheric (or lower) pressure as well as elevated temperatures (room temperature or above), will be described sufficiently precise with the ideal gas law.

6.4.2 Mean Free Path

The most important interaction mechanism among the molecules of gases are collision events. The average length a gas molecule can travel before it encounters a collision event is referred to as the mean free path. We assume a particle of diameter d traveling at a velocity of v. In the time span t it will have traveled the distance

$l_t=vt$

Its cross-section area is given by πd². Therefore it will have covered a total interaction volume V_v of

$V_v=l_t\pi d^2=vt\pi d^2$

in the time span t. The shape of this volume is cylindrical. Now, we assume that the particle is not alone in the control volume. We assume a total number of n_D particles per unit volume (referred to as number density) to be in the control volume. The number of particles in the volume through which our particle traveled is given by V_v n_D. Our particle would have collided with each of them. Therefore, the average mean path, i.e., the path traveled per collision would be

$\begin{array}{l}\lambda =\frac{\text{distance}\text{traveled}}{\text{number}\text{of}\text{collisions}}=\frac{l_t}{V_v{n}_D}\hfill \\ =\frac{vt}{vt\pi d^2{n}_D}\hfill \\ =\frac{1}{\pi d^2{n}_D}\hfill \end{array}$

Now this is only an approximation as it assumes the particles we collide with to be static. This is obviously not true. If our particle is moving, the other particles will move as well. Therefore, we need to take into account their relative velocity ${\overrightarrow{v}}_{\text{diff}}$ instead of the absolute velocity. This velocity is given as

$\begin{array}{ll}{\overrightarrow{v}}_{\text{diff}}\hfill & =\sqrt{{\left({\overrightarrow{v}}_1-{\overrightarrow{v}}_2\right)}^2}\hfill \\ \hfill & =\sqrt{\overline{{\overrightarrow{v}}_1{\overrightarrow{v}}_1}-2\overline{{\overrightarrow{v}}_1{\overrightarrow{v}}_2}+\overline{{\overrightarrow{v}}_2{\overrightarrow{v}}_2}}\hfill \end{array}$

where we consider the average velocities only. Now as ${\overrightarrow{v}}_1$ and ${\overrightarrow{v}}_2$ are uncorrelated and random, the average product $\overline{{\overrightarrow{v}}_1{\overrightarrow{v}}_2}$ will be zero as we do not expect the particles to move in a predominantly parallel or coordinated fashion. Furthermore they are of the same magnitude. Therefore we find

$\begin{array}{ll}{\overrightarrow{v}}_{\text{diff}}\hfill & =\sqrt{{\left({\overrightarrow{v}}_1\right)}^2+{\left({\overrightarrow{v}}_2\right)}^2}\hfill \\ \hfill & =\sqrt{{\left(2{\overrightarrow{v}}_1\right)}^2}\hfill \\ \hfill & =\sqrt{2}{\overrightarrow{v}}_1\hfill \end{array}$

which tells us that the average velocity of the particles will be higher by a factor of $\sqrt{2}$ when considering collision with moving particles instead of collision with static particles. The average mean free path in Eq. 6.9 will therefore be smaller by this factor resulting in

$\lambda =\frac{1}{\sqrt{2}\pi d^2{n}_D}$

Now the number of particles per volume n_D can be calculated from Eq. 6.7 as

$n_D=\frac{p}{k_{B}T}$

in which we find Eq. 6.10 to be

$\lambda =\frac{k_{B}T}{\sqrt{2}\pi d^{2}p}$

For an example of how to approximate the diameter of a molecule see section 9.3.4. Some experimental values for commonly used gases are given in Tab. 6.2.

Tab. 6.2

Gas molecule diameters of some commonly used gases [4]

Gas	Diameter d m	Average velocity${\overline{\overrightarrow{v}}}_{\text{mean}}$ m s−1
air	0.36	467
Ar	0.36	182
CO2	0.45	379
H2	0.27	1 769
He	0.22	1 256
Kr	0.41	274
N2	0.37	475
O2	0.36	444
Xe	0.48	219

6.4.3 Gas Pressure

We can estimate the gas pressure from kinetic theory by a similar simple model. Assume a particle being enclosed in a volume of width L along the x-axis. The particle moves in this container and collides with the side wall transferring the momentum

$\Delta p=p_x-\left(p_x\right)=2p_x=2m{v}_x$

The collisions on one specific side of the container occurs every

$\Delta t=\frac{2L}{v_x}$

The force during this impact is given by

$F=\frac{\Delta p}{\Delta t}=\frac{m{v}_x^2}{L}$

If we assume a total of N number of gas molecules to be present in the container, we take the average of their velocities along the x-axis $\overline{v_x}$. The forces exerted by all of these molecules on one side of the container will then be

$F_N=\frac{Nm{\overline{v}}_x^2}{L}$

If we consider the average speed of all molecules in the container ${\overline{v}}^2$, we can again assume that their movement is random and not preferably in one direction. Therefore we can assume that the total fraction of the velocity along the x-axis will be a third of ${\overline{v}}^2$ in which case we find

$F_N=\frac{Nm{\overline{v}}^2}{3L}$

Referring to the total force exerted onto the area L² we find for the pressure

$\begin{array}{l}p=\frac{F_N}{L^2}=\frac{Nm{\overline{v}}^2}{3L^3}\hfill \\ =\frac{Nm{\overline{v}}^2}{3V}\hfill \\ =\frac{n_{D}m{\overline{v}}^2}{3}\hfill \end{array}$

where we have used V = L³ to express the volume and n_D as the number density, i.e., the number of molecules per volume.

6.4.4 Kinetic Energy

Next we will derive the kinetic energy, i.e., the correlation between the temperature and the movement of the particles. Applying the ideal gas law Eq. 6.6 to Eq. 6.12 we find

$\begin{array}{ll}\frac{n_{D}m{\overline{v}}^{2}V}{3}\hfill & =N{k}_{B}T=pV\hfill \\ \frac{m{\overline{v}}^2}{3}\hfill & =k_{B}T\hfill \end{array}$

Using the kinetic energy relation $E_{\text{kin}}=\frac{1}{2}m{v}^2$ for a single particle we can rewrite Eq. 6.13 as

${\text{E}}_{\text{kin}}=\frac{1}{2}m{v}^2=\frac{3}{2}{k}_{B}T$

Obviously the total kinetic energy of the system is simply

${\text{E}}_{\text{kin,}\text{total}}=N\frac{1}{2}m{\overrightarrow{v}}^2=N\frac{3}{2}{k}_{B}T$

where we need to use to root mean square velocity, i.e., the average velocity of the particles.

Mean Square Velocity. Extending these models we can also derive the average speed of one molecule or particle usually as the root mean square velocity. It can be directly derived from Eq. 6.14 as

$v=\sqrt{\frac{3k_{B}T}{m}}$

It gives the average velocity of all molecules in the control volume. Obviously, not all particles have the same speed. The speed follows the Maxwell¹speed distribution. In the following section we will shortly derive this distribution.

6.4.5 One-Dimensional Velocity Probability Distribution

We begin by studying the velocity distribution along one axis. As we will see in a moment, this distribution is, mathematically speaking, a Boltzmann¹distribution. We will derive the velocity probability distribution F (v_x, v_y, v_z) which will give us the likeliness that a particle will travel at a given set of values for the independent variables v_x, v_y, and v_z. So for each values of v_x, v_y, and v_z, this distribution will give us the likeliness that a particle will travel at exactly these speeds. This value is 0 ≤ F (v_x, v_y, v_z) ≤ 1 with the overall integral amounting to 1. This is due to the fact that each particle must travel at one set of velocities and if we integrate over all of them we will eventually catch all particles. As the movement of the particles in uncorrelated, i.e., it is random, we can surely assume that this combined three-dimensional distribution can be split into individual distributions along each axis that are independent. We therefore note

$F\left(v_x,v_y,v_z\right)=f\left(v_x\right)g\left(v_y\right)h\left(v_z\right)$

with f (v_x), g (v_y), and h (v_z) being the velocity probability distributions along each axis. Again, these functions will return the likeliness of a particle traveling at a given velocity. We can already note here that these distributions will most likely look identical. This is due to the fact that the particle movement is random and we do not expect to see any preferred movement direction, and therefore no predominant velocity distribution. We quickly transfer this to a sum by taking the natural logarithm of Eq. 6.17 resulting in

$\ln \left(F\left(v_x,v_y,v_z\right)\right)=\ln \left(f\left(v_x\right)\right)+\ln \left(f\left(v_y\right)\right)+\ln \left(f\left(v_z\right)\right)$

One may wonder why we do this. If we assume the distributions to look the same in all directions, we may just as well derive only the distribution in one axis and simply repeat this process with the other two. Therefore, we now take the partial derivative with respect to v_x (we could also take one of the other two) which results in

$\begin{array}{ll}\frac{\partial }{\partial v_x}\ln \left(F\left(v_x,v_y,v_z\right)\right)\hfill & =\frac{\partial }{\partial v_x}\ln \left(f\left(v_x\right)\right)+\frac{\partial }{\partial v_x}\ln \left(f\left(v_y\right)\right)+\frac{\partial }{\partial v_x}\ln \left(f\left(v_z\right)\right)\hfill \\ \hfill & =\frac{d}{d{v}_x}\ln f\hfill \end{array}$

Now it becomes obvious why the natural logarithm helped: The distributions f (v_y) and f (v_z) have disappeared entirely because we converted the product of Eq. 6.4 to a sum (Eq. 6.18). We know that $v^2=v_x^2+v_y^2+v_z^2$ in which case we can rewrite Eq. 6.19 as

$\begin{array}{ll}\frac{\partial }{\partial v_x}\ln F\hfill & =\frac{d}{d{v}_x}\ln f\hfill \\ \frac{\partial }{\partial v}\ln F\frac{\partial v}{\partial v_x}\hfill & =\frac{d}{d{v}_x}\ln f\hfill \end{array}$

We have to find $\frac{\partial v}{\partial v_x}$ which we find as

$\begin{array}{ll}\frac{\partial v}{\partial v_x}v\hfill & =\frac{\partial }{\partial v_x}\sqrt{v_x^2+v_y^2+v_z^2}=\frac{1}{2\sqrt{v_x^2+v_y^2+v_z^2}}2v_x\hfill \\ \hfill & =\frac{v_x}{v}\hfill \end{array}$

that after inserting into Eq. 6.20 yields

$\frac{\partial }{\partial v}\ln F\frac{1}{v}=\frac{1}{v_x}\frac{d\ln f}{d{v}_x}$

As already stated, we could repeat the same process for the other two velocity components and this would result in a similar equation as Eq. 6.21 with only the right-hand side being different

$\frac{\partial }{\partial v}\ln F\frac{1}{v}=\frac{1}{vy}\frac{d\ln g}{d{v}_y}$

$\frac{\partial }{\partial v}\ln F\frac{1}{v}=\frac{1}{v_z}\frac{d\ln h}{d{v}_z}$

Here we employ a trick we use intensively during the derivation of solutions to partial differential equations: If several differential equations all amount to the same value (in this case the left-hand side of Eq. 6.21), they must be constants (see section 8.3.3 for details on this neat little trick). We call this constant −λ and proceed with solving Eq. 6.21. Please note that the other two equations (Eq. 6.22 and Eq. 6.23) are solved likewise.

$\begin{array}{ll}\frac{1}{v_x}\frac{d\ln f}{d{v}_x}\hfill & =-\lambda \hfill \\ \frac{d\ln f}{d{v}_x}\hfill & =-\lambda v_x\hfill \\ d\ln f\hfill & =-\lambda v_{x}d{v}_x\hfill \\ \ln f\hfill & =-\frac{1}{2}\lambda v_x^2+c\hfill \\ f\left(v_x\right)\hfill & =c_1{\text{e}}^{-\frac{1}{2}\lambda v_x^2}\hfill \end{array}$

where we still need to find the two constants c₁ and λ.

Determining c₁. The c₁ can easily be found by taking into account that the integral of the distribution must be 1, i.e., if we sum up all probabilities for all velocities, we must inevitably collect all particles. Therefore

$\begin{array}{ll}{\int }_{-\infty }^{+\infty }f\left(v_x\right)d{v}_x\hfill & ={\int }_{-\infty }^{+\infty }{c}_1{\text{e}}^{-\frac{1}{2}\lambda v_x^2}d{v}_x\hfill \\ \hfill & =c_1{\int }_{-\infty }^{+\infty }{\text{e}}^{-\frac{1}{2}\lambda v_x^2}d{v}_x\hfill \\ \hfill & \stackrel{!}{=}1\hfill \end{array}$

The integral ${\int }_{-\infty }^{+\infty }{e}^{-a{x}^2}dx$ is one of the Gaussian integrals and it evaluates to $\sqrt{\frac{\pi }{a}}$. You can also use Maple to verify that this is correct. In this case we find

$c_1\sqrt{\frac{2\pi }{\lambda }}\stackrel{!}{=}1\to c_1=\sqrt{\frac{\lambda }{2\pi }}$

Determining λ. Next we need to determine λ. This we do by taking into account that all kinetic energies of all particles combined must be identical to the overall energy of the system. This energy is given by Eq. 6.15 as

$\frac{1}{3}N\frac{3}{2}{k}_{B}T=\frac{1}{2}N{k}_{B}T$

The reason for the $\frac{1}{3}$ prefactor arises from the fact that we only take one third of the energy contribution as we are only considering the velocity component v_x currently. Again, as the movement is random and uncorrelated, the energy contributions are identical. Now we know that the kinetic energy along the x-axis of one single particle is

$E_{\text{kin}}=\frac{1}{2}m{v}_x^2$

We use Eq. 6.25 and integrate over all velocities in which case we find

$\begin{array}{ll}{E}_{\text{kin,}\text{total}}\hfill & =N\frac{1}{2}m{\int }_{-\infty }^{+\infty }f\left(v_x\right)v_x^{2}d{v}_x\hfill \\ \hfill & =N\frac{1}{2}m\sqrt{\frac{\lambda }{2\pi }}{\int }_{-\infty }^{+\infty }{\text{e}}^{{-}^{\frac{\lambda }{2}}}{v}_x^{2}d{v}_x\hfill \end{array}$

where we have yet another Gaussian integral which evaluates to

${\int }_{-\infty }^{+\infty }{e}^{-a{x}^2}{x}^{2}dx=\frac{1}{2}\frac{\sqrt{\pi }}{a\frac{3}{2}}$

Therefore we can evaluate Eq. 6.27 and set it equal to Eq. 6.26 finding

$\begin{array}{ll}{E}_{\text{kin, total}}\hfill & =\frac{1}{4}Nm\sqrt{\frac{\lambda }{2\pi }}\frac{\sqrt{\pi }}{{\left(\frac{\lambda }{2}\right)}^{\frac{3}{2}}}=\frac{1}{2}\frac{Nm}{\lambda }\hfill \\ \hfill & \stackrel{!}{=}\frac{1}{2}N{k}_{B}T\hfill \\ \frac{m}{\lambda }\hfill & =k_{B}T\to \lambda =\frac{m}{k_{B}T}\hfill \end{array}$

One-Dimensional Energy Distribution. Now we have all constants and can rewrite Eq. 6.24 as

$f\left(v_x\right)=\sqrt{\frac{m}{2\pi k_{B}T}}{\text{e}}^{-\frac{m{v}_x^2}{2k_{B}T}}$

This is the one-dimensional energy distribution, i.e., the probability function that a given particle will travel at a given speed which is equivalent to stating that a particle will have a certain kinetic energy. As already stated, this function is a Boltzmann distribution. We can quickly transfer it to a three-dimensional case.

Three-Dimensional Energy Distribution. Returning to Eq. 6.17 where we stated that the three energy distributions for v_x, v_y, and v_z can be derived independently, we already stated that the functions we derive for each of them look identical (see Eq. 6.21, Eq. 6.22, and Eq. 6.23). Therefore we can reassemble the three-dimensional velocity probability distribution or energy probability distribution using Eq. 6.28 as a template resulting in

$\begin{array}{ll}F\left(v_x,v_y,v_z\right)\hfill & =f\left(v_x\right)g\left(v_y\right)h\left(v_z\right)={\left(\sqrt{\frac{m}{2\pi k_{B}T}}\right)}^3\left({\text{e}}^{-\frac{m{v}_x^2}{2k_{B}T}}{\text{e}}^{-\frac{m{v}_y^2}{2k_{B}T}}{\text{e}}^{-\frac{m{v}_z^2}{2k_{B}T}}\right)\hfill \\ \hfill & ={\left(\frac{m}{2\pi k_{B}T}\right)}^{\frac{3}{2}}{\text{e}}^{-\frac{m}{2k_{B}T}\left(v_x^2+v_y^2+v_z^2\right)}\hfill \\ \hfill & ={\left(\frac{m}{2\pi k_{B}T}\right)}^{\frac{3}{2}}{\text{e}}^{-\frac{m{v}^2}{2k_{B}T}}\hfill \end{array}$

Eq. 6.29 is the energy distribution in three dimensions.

6.4.6 Maxwell Speed Distribution

We now want to transfer the three-dimensional energy or velocity distribution into a function of only one variable. Instead of supplying the values v_x, v_y, and v_z we want to supply only one value v thereby reducing the number of independent variables to only one. This is actually very simple to do: Eq. 6.29 gives us the probability of a particle traveling at a given set of velocities for v_x, v_y, and v_z, we simply need to integrate over the individual independent variables. We therefore transfer F (v_x, v_y, v_z) to spherical coordinates using one of the unit vectors as the velocity component. We therefore end up with a function F (θ, ϕ, v) which we simply need to integrate over the first two independent variables to result in a function of only the velocity as independent variable. In transferring the function to spherical coordinates, we have to take into account the Jacobian determinant (see section 7.5) in which case our integral has to be transformed as

$\begin{array}{ll}d{v}_{x}d{v}_{y}d{v}_z\hfill & =v^2\sin \theta d\theta d\phi dv\hfill \\ F\left(v_x,v_y,v_z\right)d{v}_{x}d{v}_{y}d{v}_z\hfill & =F\left(\theta ,\phi ,v\right)v^2\sin \theta d\theta d\phi dv\hfill \end{array}$

As stated, we do not want to integrate over the velocity here, therefore we only integrate twice. As usual, we integrate 0 ≤ θ ≤ π and 0 ≤ ϕ ≤ 2π in spherical coordinates. Thus we find

$\begin{array}{ll}F\left(v\right)\hfill & ={\int }_0^{\pi }{\int }_0^{2\pi }F\left(\theta ,\phi ,v\right)v^2\sin \theta d\theta d\phi \hfill \\ \hfill & ={\int }_0^{\pi }\sin \theta d\theta {\int }_0^{2\pi }d\phi F\left(\theta ,\phi ,v\right)v^2\hfill \\ \hfill & ={\int }_0^{\pi }\sin \theta d\theta {\int }_0^{2\pi }d\phi F\left(\theta ,\phi ,v\right)v^2\hfill \\ \hfill & =4\pi F\left(\theta ,\phi ,v\right)v^2\hfill \\ \hfill & =4\pi v^2{\left(\frac{m}{2\pi k_{B}T}\right)}^{\frac{3}{2}}{\text{e}}^{-\frac{m{v}^2}{2k_{B}T}}\hfill \end{array}$

Do not be confused because we are inserting v² instead of r² - we are working with velocities as the unit vector of our coordinate systems, which is why we have to use v instead of the parameter r which usually only indicates a length. Eq. 6.30 is the Maxwell velocity or speed distribution. It will return the likelihood of a particle to travel at the given velocity v. The unit of the function is $\frac{1}{\text{m}{\text{s}}^{-1}}$.

The distribution shown in Fig. 6.3 is calculated for a water molecule with M = 18 u at various temperatures. As you can see the strong shoulder observed at lower temperatures flattens out as the temperature increases. The whole distribution shifts toward the right, allowing a wider range of velocities. This is a very characteristic feature of gases at higher temperature.

6.6.1 Work

The easiest component is work. It is the sum of all the volume forces acting on the control volume. This work is expressed as

$dW=-pdV$

We will see a good example of such volume work in Fig. 6.4c.

6.6.2 Heat

Heat is the energy transferred into and out of the system at a given temperature. Combustion processes are typical examples of heat being applied to a system.

6.6.3 Internal Energy

The first energy term as seen from section 6.6 is the internal energy U of a thermodynamic system, which is defined as

$U=U_{\text{pot}}+U_{\text{kin}}$

The energy is defined as the sum of all potential U_pot and kinetic energy U_kin required in order to create the system. U_kin comprises the energy required for the motion of the atoms or compounds within the system. However, U_kin does not include the kinetic energy of the lateral movement of the system as a whole. U_pot comprises the energy stored as mass (as well as by the electron configurations of the atoms), the chemical bonds (which may further react), and the physical forces (e.g., electromagnetic forces due to the presence of dipoles or internal stress/strain) within the system or among the atoms or compounds of the system.

6.6.4 Enthalpy

The internal energy is the energy required for creating a thermodynamic system. However, this system must occupy a given space and the energy required for creating this space needs to be added to the total energy required for creating and placing a system. This combined energy is referred to as enthalpy and is defined as

$H=U+pV$

The first term U sums up the internal energy of the system (see Eq. 6.34). The second term pV is the volume work done at the system or performed by the system.

6.6.5 Changes of Internal Energy and Enthalpy

Now if internal energy and enthalpy change over time, the total differentials of Eq. 6.34 and Eq. 6.35 have to be found. We will begin with the enthalpy

$dH=dU+pdV+VdP$

for which the last term is often neglected if processes at constant pressure are being studied. Often, the enthalpy has to be formulated as a function of pressure and temperature. Therefore, the partial differential is formulated as

$dH=\frac{\partial H}{\partial T}dT+\frac{\partial H}{\partial p}dp\left(\text{real}\text{gas}\right)$

$\begin{array}{ll}\approx \frac{\partial H}{\partial T}dT\hfill & \left(\text{ideal}\text{gas}\right)\hfill \\ \approx c_{p}dT\hfill & \left(\text{perfact}\text{gas}\right)\hfill \end{array}$

Here we have already introduced two simplifications when working with fluids in thermodynamics, namely the ideal gas and the perfect gas. In general, the latter term is often used interchangeably with the first. An ideal gas is a gas for which the enthalpy is only a function of the temperature. Whenever there is a linear correlation between enthalpy and temperature change (given by the proportionality constant c_p) we refer to the gas as being a perfect gas. We have discussed the properties of ideal gases in section 6.4. In ideal gases, the enthalpy is only a function of the temperature, not of the pressure. In perfect gases (which are commonly used in thermodynamics), the dependence of the enthalpy on the temperature is a constant, the isobaric heat capacity c_p.

In an analogy, the dependence of the internal energy is usually formulated as a function of the volume and the temperature. It is given by

$\begin{array}{ll}dU=\frac{\partial U}{\partial T}dT+\frac{\partial U}{\partial V}dV\hfill & \left(\text{real}\text{gas}\right)\hfill \\ =\frac{\partial U}{\partial T}dT\hfill & \left(\text{ideal}\text{gas}\right)\hfill \\ =c_{v}dT\hfill & \left(\text{perfect}\text{gas}\right)\hfill \end{array}$

In ideal gases the change of the internal energy is only dependent on the temperature. For perfect gases this dependency is given by a constant, the isochoric heat capacity c_v. The unit of this constant is J kg⁻¹ K⁻¹, but sometimes it is normalized to the molecular weight. The unit is then J g⁻¹ K⁻¹.

Heat Capacity of Fluids. Please note that for liquids (being incompressible), both enthalpy and internal energy are linearly dependent on temperature and the values of c_p and c_v can be assumed to be almost identical. For water, c_v ≈ c_p = 75.2724 kJ mol⁻¹ = 4.1818 J g⁻¹ K⁻¹ if normalized to the molecular weight at standard ambient temperature and pressure (STP) conditions.

6.6.6 Reaction Enthalpy

The change of enthalpy is an important means of accessing chemical reactions. Consider Eq. 6.36 at constant pressure, which then simplifies to

$dH=dU+pdV$

The amount of energy emitted from or absorbed by a system during a change of state is referred to as heat of reaction. The heat of reaction is characteristic for a chemical reaction. It is given in kJ, although sometimes the unit used is kcal where 1 kcal = 4.187 kJ. As it is dependent on the amount of substance used, it is normalized to 1 mol. The heat of reaction originating from 1 mol is referred to as the reaction enthalpy ∆H.

Exothermic and Endothermic. Depending on whether or not a chemical reaction creates or absorbs heat, the reaction is either classified as being exothermic if heat is released or endothermic if heat is absorbed. By convention, for exothermic reactions ∆H < 0 whereas for endothermic reactions ∆H > 0.

A typical example of an exothermic reaction is the formation of ammonia from hydrogen gas and nitrogen gas

${\text{N}}_2+3{\text{H}}_2\to {\text{NH}}_3\Delta H=-92.3\text{kJ}{\text{mol}}^{-1}$

A typical example for an endothermic reaction is the formation of nitric oxide from oxygen gas and nitrogen gas

${\text{O}}_2+{\text{N}}_2\to 2\text{NO}\Delta H=+180.6\text{kJ}{\text{mol}}^{-1}$

Temperature and Pressure Dependency. As we have seen during the study of the thermodynamics of the enthalpy, this value is dependent both on pressure and temperature. It is therefore required to always state at which conditions the respective reaction was carried out. Usually, the reaction will be done at STP conditions in which case the symbol ∆H⁰ is used and the value referred to as standard reaction enthalpy. It is usually sufficient to only indicate one ∆H value and the respective pressure and temperature as the ∆H for other pressure and temperatures can be calculated using Eq. 6.37 and the respective constants.

Hess’s Law. It is important to keep in mind that the thermodynamic state variables do not depend on the process used to obtain the thermodynamic system. This is referred to as Hess’s¹law which states that by reacting educts to a given product each reaction path will add up to the same ∆H value. It is therefore not possible to find an “energetically cheaper” way to obtain a certain product.

As an example, we’ll study the reaction of carbon and oxygen gas to carbon dioxide. There are two distinct reaction pathways one can take. The first one is the direct oxidation given by

$\text{C}+{\text{O}}_2\to {\text{CO}}_2\Delta H^0=-393.8\text{kJ}{\text{mol}}^{-1}$

The second option is the oxidation to carbon monoxide and a subsequent oxidation of carbon monoxide to carbon dioxide

$\begin{array}{l}\text{C}+\frac{1}{2}{\text{O}}_2\to \text{CO}\Delta H^0=-110.6\text{kJ}{\text{mol}}^{-1}\hfill \\ \text{CO}+\frac{1}{2}{\text{O}}_2\to {\text{CO}}_2\Delta H^0=-283.2\text{kJ}{\text{mol}}^{-1}\hfill \end{array}$

If you add up the two ∆H values you will see that they are equal. Thus it is not important which reaction path is taken.

6.6.7 Standard Formation Enthalpy

Hess’s law gives rise to a very important concept in chemistry: The enthalpy change during a reaction can be calculated as the difference of “absolute” enthalpies of products and educts as

$\Delta H_j^0={\displaystyle \sum H_{\text{products}}-}{\displaystyle \sum H_{\text{educts}}}$

As enthalpies cannot be measured on an absolute scale, but only as changes of the enthalpy, the “absolute zero” value for the enthalpy is interpreted as the enthalpy that the most stable form, i.e., the most stable allotrope of an element would have at STP conditions. The entropy change that occurs while reacting this element to a given compound is referred to as the formation enthalpy. If the reaction is carried out at STP conditions, the enthalpy change is referred to as standard formation enthalpy$\Delta H_f^0$. Please note that this value refer to 1 mol of product. Taking our two earlier examples, we can give the standard formation enthalpy of ammonia to be $\Delta H_f^0\left({\text{NH}}_3\right)=-46.1\text{kJ}{\text{mol}}^{-1}$ and of carbon dioxide to be $\Delta H_f^0\left({\text{CO}}_2\right)=-393.8\text{kJ}{\text{mol}}^{-1}$.

As an example of a more complex reaction, let us consider the following reaction which is the important reaction if the blast furnace process by which iron oxide iron(III)oxide (commonly referred to as rust) is converted to iron

${\text{Fe}}_2{\text{O}}_3+3\text{CO}\left(\text{g}\right)\to 2\text{Fe}\left(\text{s}\right)+3{\text{CO}}_2\left(\text{g}\right)$

for which the values for $\Delta H_f^0$ can be read from Tab. 6.3 as $\Delta H_f^0\left({\text{CO}}_2\right)=-393.8\text{kJ}{\text{mol}}^{-1}$, $\Delta H_f^0\left({\text{Fe}}_2{\text{O}}_3\right)=-824.8\text{kJ}{\text{mol}}^{-1}$, $\Delta H_f^0\left(\text{CO}\right)=-110.6\text{kJ}{\text{mol}}^{-1}$. For iron we find $\Delta H_f^0\left(\text{Fe}\right)=0\text{kJ}{\text{mol}}^{-1}$ because it is an element. Therefore using Eq. 6.40 we obtain

Tab. 6.3

Standard formation enthalpy $\Delta H_f^0$ of some compounds. Taken from [8]

Compound	$\Delta H_f^0$, kJ mol−1	Compound	$\Delta H_f^0$, kJ mol−1
gases
O3	+142.8	H2O	-242.0
H2S	-20.6	CO	-110.6
CO2	-393.8	SO2	-297.0
liquids
HF	-271.3	HCl	-92.4
HBr	-36.4	HI	+26.5
H2O	-286.0	H2O2	-187.9
NH3	-46.1
solids
NaCl	-411.3	Fe2O3	-824.8
Cgraphite	0	Cdiamond	1.9
dissociation
H	+218.1	O	+249.3
F	+79.5	Cl	+121.8
N	+472.5

$\begin{array}{ll}\Delta H_f^0\left({\text{Fe}}_2{\text{O}}_3\to \text{Fe}\right)\hfill & =0\text{kJ}{\text{mol}}^{-1}-3393.8\text{kJ}{\text{mol}}^{-1}-\left(-824.8\text{kJ}{\text{mol}}^{-1}-3110.6\text{kJ}{\text{mol}}^{-1}\right)\hfill \\ \hfill & =-24.8\text{kJ}{\text{mol}}^{-1}\hfill \end{array}$

6.7 Second Law of Thermodynamics

6.7.1 Entropy

The first law of thermodynamics states that energy is conserved. This is the basis on which the enthalpy transitions of thermodynamic systems can be calculated. However, it does not state why and how physical and chemical transitions occur. This is what is described by the second law of thermodynamics. This law focuses on entropy, which was already introduced as a thermodynamic state variable.

The concept of entropy was originally introduced by Carnot.¹ Entropy is sometimes referred to as the state of disorder of a system. However, definitions based on terms such as “disorder” or “chaos” are often ambiguous as the terms may have distinctly different meanings. Scientifically speaking, a system in equilibrium is in perfect disorder, meaning that there is no local gradient in concentration, temperature, etc., and therefore no useful (chemical or physical) potential from which energy could be gained. The entropy of such a system is high. Entropy may therefore, more intuitively, be defined as the degree of dispersal of energy or the absence of any gradient or potential, i.e., the homogeneity of the energy distribution. A good example is a block of ice placed in a warm atmosphere. Initially there is a large gradient in temperature which may be used to gain useful energy from the system. One example of energy usage of such as system would be an energy generator such as a Seebeck generator that exploits temperature gradients to create electrical potential differences and thus, current. Over time, the ice will melt and the temperatures will equilibrate. The system has increased its entropy as the temperature gradient, i.e., the temperature potential has diminished.

The second law of thermodynamics states that only thermodynamic transitions which increase the entropy will effectively happen. This is very important, e.g., for predicting if a certain chemical reaction will take place or not. If the reaction would result in the increase of entropy, then this reaction can (in principle) take place. Similar to enthalpy values, which can only be given as relative values and for which therefore an arbitrary zero value has to be defined, entropy values are also referenced. By definition, an ideal crystalline substance or a perfect gas at absolute zero temperature 0K will have an entropy of zero. As we will see, it is sufficient to have a reference entropy value from which the entropy at any given temperature and pressure can be calculated. The most commonly used entropy value is the standard entropy S⁰ which is the entropy of 1 mol substance at STP conditions. This value is a material constant (see Tab. 6.4). Looking at Tab. 6.4 some interesting details can be gained

• The most chemically stable compounds (such as graphite or diamond) have low S⁰ values.

• Solids have lower S⁰ values than liquids or gases.

• Liquids have lower S⁰ values than gases.

• Solids with a high degree of order (as is the case in crystals) have very low S⁰ values.

6.7.2 Entropy Changes due to Chemical Reaction

During a reaction, the entropy changes. Large changes of entropy occur if the state of a reagent is changed, e.g., if a liquid and a solid react to a gas

$\begin{array}{ll}\frac{1}{2}{\text{O}}_2\left(\text{g}\right)+\text{C}\left(\text{s}\right)\rightleftharpoons \text{CO}\left(\text{g}\right)\Delta S^0\hfill & =197.7-\left(\frac{205.2}{2}+5.7\right)\text{J}{\text{mol}}^{-1}\hfill \\ \hfill & =+89.4\text{J}{\text{mol}}^{-1}\hfill \end{array}$

Very small values are obtained if gases are converted to liquids or solids

$\begin{array}{ll}\text{Ca}\left(\text{s}\right)+\frac{1}{2}{\text{O}}_2\left(\text{g}\right)\rightleftharpoons \text{CaO}\left(\text{s}\right)\Delta S^0\hfill & =39.8-\left(41.7+\frac{205.2}{2}\right)\text{J}{\text{mol}}^{-1}\hfill \\ \hfill & =-104.5\text{J}{\text{mol}}^{-1}\hfill \end{array}$

In general, the entropy change during a phase transition can be calculated as

$\begin{array}{ll}\Delta S\hfill & =\frac{\Delta H_{\text{vap}}}{T}\hfill \\ \hfill & \approx 10.5R\approx 85\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\text{to}88\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\hfill \end{array}$

This is the case as phase transitions are isothermal processes, i.e., the temperature of a medium does not change during a phase transition. As stated, this value is interestingly constant for many liquids and can be approximated by Eq. 6.41 at STP conditions. This approximation is referred to as Trouton’s¹rule. Tab. 6.5 states some of the most relevant enthalpies of vaporization.

6.7.3 Mathematical Description

Mathematically, the entropy for a reversible isothermal process is described as

$dS=\frac{dQ}{T}$

It has to be noted that most thermodynamic processes are not isothermal and thus the entropy is usually expressed in diagrams as function of the temperature S (T). Therefore according to Eq. 6.42, the heat transferred is the integral Q = ∫ S (T) dT.

Using Eq. 6.42 and Fig. 6.2 the second law of thermodynamics can be expressed in its most general form as

$\frac{d{\sum }_i{S}_i}{dt}={\displaystyle \sum _k\frac{{\dot{Q}}_k}{T_k}+}{\displaystyle \sum _i{\dot{m}}_l{s}_l+}{\dot{S}}_{\text{produced}}$

where the individual terms describe the following contributions

• $\frac{d{\sum }_i{S}_i}{dt}$ describes the temporal changes of the entropy values S_i of all compounds i contained in the system

• ${\sum }_k\frac{{\dot{Q}}_k}{T_k}$ describes the entropy transferred into and out of the system by heat flows ${\dot{Q}}_k$ at constant temperatures T_k; this term is zero for adiabatic processes

• ${\sum }_l{\dot{m}}_l{s}_l$ describes the entropy brought into and out of the system by all mass flows ${\dot{m}}_l$; this term is zero for closed and contained systems

• ${\dot{S}}_{\text{produced}}$ describes the entropy produced inside of the system during the respective process; this term is zero for isentropic processes

Please note that the overall change of entropy can never be negative, otherwise the respective process or reaction would not occur at all. Now the question arises, how can some chemical reactions have negative S⁰ values? The answer to this is that the overall process must have a positive change of entropy. If a single reaction ${\dot{S}}_{\text{produced}}$ within the control volume of this process has a negative change in entropy, the system would need to compensate this by providing heat $\frac{{\dot{Q}}_k}{T_k}$ and thus raise the overall change of entropy to a positive value.

6.7.4 Reversible and Irreversible Processes

Let us study examples for reversible and irreversible processes (see Fig. 6.4). A gas volume is restricted inside of a vessel to a fraction V₁ if the total volume by a separator which can be removed or displaced. Removing the separator will result in expansion of the gas to the total volume of the vessel (Fig. 6.4a). If the gas is heated, it will expand and displace the separator (see Fig. 6.4b). In both cases, we assume an isothermal adiabatic process. The system is closed and contained and we assume ideal gas behavior.

We will start discussing the case displayed in Fig. 6.4b. Taking Eq. 6.32, we can therefore assume

$\begin{array}{ll}W+Q\hfill & =\frac{d\left(mu\right)}{dt}\hfill \\ \hfill & =m\frac{du}{dt}\hfill \end{array}$

As stated, we assume isothermal expansion and ideal gases therefore taking Eq. 6.39, we assume the overall change of internal energy to be zero. Therefore

$Q=-W$

Furthermore, the overall entropy change of the system must be zero if the process is (as assumed) reversible. Taking Eq. 6.43 we derive

$\begin{array}{l}\frac{dS}{dt}=\frac{\dot{Q}}{T}+{\dot{S}}_{\text{produced}}\hfill \\ =0\hfill \end{array}$

Therefore the system will need to consume the same amount of entropy as provided by means of heat transfer. As entropy cannot be consumed, the ambiance has to provide this entropy. In consequence, the entropy of the gas will be increased by heat transfer, whereby the entropy of the ambient will be reduced by the same amount. The important thing to note is that the entropy provided by the ambiance can be restored if the system would provide heat to the ambient, thereby reversing the process.

The expansion can be used to do work W (see Fig. 6.4c). This work can be expressed as

$\begin{array}{l}dW=-pAdx\hfill \\ =-pdV\hfill \end{array}$

Using the ideal gas law Eq. 6.4 we derive

$p=\frac{nRT}{V}$

and therefore

$\begin{array}{ll}dW\hfill & =-\frac{nRT}{V}dV\hfill \\ W\hfill & =-nRT\ln V+c_1\hfill \end{array}$

$W_{1\to 2}=nRT\ln \frac{V_1}{V_2}$

$=nRT\ln \frac{p_2}{p_1}$

$=T\Delta S_{1\to 2}$