The concept of an ideal gas follows naturally from a particle-based view of matter. Some textbook authors prefer the term ‘perfect gas’ arguing that the word ‘ideal’ is usually used with a different meaning in physical chemistry [8, 14, 15], but others seem to have understandable reservations against calling any material ‘perfect’ [5,6,7]. In any case, this is a matter of language preference rather than a true scientific issue. This text uses the term ‘ideal gas,’ which reflects the personal choice of the authors.

In an ideal gas, particles move randomly and independently of each other; the only interaction between them is that they occasionally collide [8, 16,17,18]. These collisions are elastic, which means that the laws of momentum conservation and energy conservation can be stated for them in a simple manner.

Momentum is the product of the mass (m) and velocity (v) of an object. Mass is a scalar quantity in physics, but velocity is a vector that can often be conveniently stated with three independent components in a Cartesian coordinate system (vx, vy, vz, all three can be both positive and negative). Therefore, momentum is also a vector, and the conservation law directly applies to each of its components. Some confusion may arise from the fact that the quantities velocity and speed are not always distinguished clearly. In this text, we will strictly stick to the convention that speed (s) is the size of the velocity vector. Therefore, it is a scalar quantity without direction, and it cannot be negative. Speed can be calculated from the velocity vector using a simple geometric argument (Pythagorean theorem):

$$s = \sqrt ^ + v_^ + v_^ } .$$

(4)

Note that it is impossible to obtain the velocity from the speed alone, and confusion often arises from the fact the symbol v is used for speed as well.

The law of momentum conservation is straightforward in physics and can be applied to non-elastic collisions as well. The law of energy conservation is somewhat different. There are many different kinds of energy (e.g., electric, rotational, vibrational), but for an ideal gas, the only thing we care about is kinetic energy. This can be calculated by multiplying the mass with the square of the speed and then dividing by two, or also with the components of velocity:

$$E = \frac }} = \frac\left( ^ + v_^ + v_^ } \right).$$

(5)

Therefore, kinetic energy is a scalar quantity (this is of course valid for all kinds of energy). The assumption behind an elastic collision is that the particles do not change any of the other kinds of energy in it: they do not get irreversibly deformed, and if they rotate or vibrate, they do so in the same way after the collision. This means that the sum of the kinetic energies of the two colliding particles is the same before and after the encounter.

Two particles that are involved in a collision have six velocity components combined. Even if we know all of them before the collision, they will take new values after the event, so six unknowns arise. There will be three equations from the conservation of momentum (one for each component) and one additional equation from the conservation of energy. This means four equations altogether, which is not enough to define six unknowns. It is probably good to remember that considerations beyond the conservation laws are needed to calculate the outcome of an elastic collision, but fortunately, these are not necessary in dealing with the properties of ideal gases.

Many statements can be made about an ideal gas [5,6,7,8, 16,17,18], but the following is already sufficient as a physical definition: the only possible interaction between the particles is the elastic collision. Note that the concept is a scientific abstraction, and a substance in itself cannot be called an ideal gas. For example, hydrogen and helium are not ideal gases in a strict sense. The precise statement is always that a certain substance behaves like an ideal gas under certain conditions. For hydrogen and helium, the certain conditions include room temperature and atmospheric pressure. Even gold behaves like an ideal gas at 3500 K and 1 Pa.

Notably, the definition given above does not require identical particles in the gas. It can be a mixture of different substances. Most of the intensive physical properties of an ideal gas depend on the chemical composition. An obvious such property is molar mass (M); additional examples include density (ρ) and molar heat capacities (CV,m and Cp,m). There are few exemptions: for example, pressure, temperature, molar volume and concentration. These are the ones used in Eqs. 2 and 3, and this is a pivotal point in ideal gas behavior: one does not need to know the identity of the gas to work with these selected properties. This is the very reason why the forms given in Eqs. 2 and 3 are highly preferred to all other possibilities. Also, this was the essential discovery in Avogadro’s law [19, 20].

It should be noted that the elastic nature of the collisions between particles is emphasized in most textbooks (p. 11 in [8]). However, from a strict physical point of view, this is only true if the particles are structureless (e.g., noble gas atoms). In multiatomic molecules, there are additional internal forms of molecular motion that can store energy, most importantly rotations and vibrations. In a collision, it may happen that these are increased at the expense of the overall kinetic energy, or their decrease provides extra kinetic energy. Fluctuations of this kind are possible, but the equipartition theorem guarantees that the two conservation laws behind elasticity will be valid statistically over a large number of collisions.

Also, an individual collision with the wall seems to violate the conservation of momentum as the particle bounces back (the direction of the velocity changes) without any counterpart, as the wall is stationary. This picture is still useful as the wall is considered to have such a large mass that gaining a little bit of momentum from a molecular collision will not change its state of motion. In addition, similarly to the intermolecular case, the overall effect of a large number of molecular collisions with all parts of the container wall will give an average of zero change in the overall momentum.

A sequence of thought presented in many textbooks will be reproduced here (p. 11 in [8]). Imagine a sample of ideal gas composed of only one substance, and attempt to calculate the pressure using the scheme shown in Fig. 1. Notice that this shows three different two-dimensional projections of a three-dimensional system, which are not always easy to understand intuitively. In this scheme, A is a flat surface whose plane will sometimes be referred to as the wall; M1, M2, M3, M4, M5 and M6 are particles; the arrows drawn represent velocity vectors of the particles. The middle view is a perspective one, the left view is taken in a direction parallel to surface A, whereas the right view is from a direction perpendicular to A. As all particles are identical, the arrows are also proportional to the momenta of the particles. Consider A (notice the italics, meaning that it is a physical property now rather than a geometrical element), the area of the flat surface A. For simplicity, a square is drawn as A in Fig. 1, but the following arguments are valid for all shapes, even ones with curved boundaries. Try to count the particles colliding to this surface in a time interval of Δt. (Symbol Δ is a capital Greek delta, the fourth letter in the Greek alphabet. This notation is very common, but not particularly didactic as the combination of one Greek and one Latin letter results in a single scientific symbol. This could be difficult to distinguish from the product of two symbols. Similar notation customs are used for differentials with the letter d and for partial differentials with the symbol ∂.) Some knowledge of the velocities of the particles is necessary to do this, but only in the direction that is perpendicular to the chosen surface A. Let the symbol for this direction be x. Consider particle M1, whose velocity component in the direction x is vx. If vx is the same as the average of this property for all particles, then half of the articles in the rectangular box of base A and depth vxΔt collide with A in this interval. The reason why only half of the particles do this is that in truly random motion, the other half moves away from A and will not collide within time Δt. The volume of the box is AvxΔt, so using the concentration (c) of the gas, the amount of substance for the colliding particles is cAvxΔt/2. The number of such particles is obtained by multiplication with the Avogadro constant NA (= R/k) to yield NAcAvxΔt/2.

Particles and velocities considered for the derivation of pressure in an ideal gas (three different views). A: Flat surface upon which the pressure is calculated; M1, M2, M3, M4, M5, M6: individual particles; arrows: velocities of particles; v: velocity of particle P; vx: velocity of particle M1 in direction x

Our main interest is the pressure, which is defined in physics as force divided by the area upon which it acts. Furthermore, force is described in Newton’s laws as the overall momentum change divided by the time interval in which it occurs. When particle M1 collides with the wall, it is a very special kind of collision: a bounce (meaning that the mass of the wall is so much higher than the mass of the molecule that the wall does not move at all). The physical laws of bouncing require that the velocity of the particle in direction x changes from vx to − vx, while other components remain unchanged. This is shown by a dotted black arrow in Fig. 1. So, the overall change of velocity in direction x is |− vx− vx |= 2vx (the | | signs represent absolute value). If the mass of the particle is m, this means a momentum change of 2mvx. Therefore, the pressure is obtained by multiplying this momentum change (2mvx) with the number of particles colliding (NAcAvxΔt/2) and then dividing by the time interval (Δt) and the surface area (A):

$$p = 2mv_ \frac}} v_ \Delta t}} = cN_}} mv_^ .$$

(6)

It can be seen that the surface area and the time interval conveniently disappear during simplification, which is readily rationalized by the physical observation that the pressure of a sample of gas is independent of where and when it is measured.

As already mentioned, the simple sequence of thought leading to Eq. 6 can be found in many textbooks as a derivation of the ideal gas law, although the work used as a reference here [8] was notably improved compared to its original version with the publication of new editions. However, this derivation is seldom taken to completion. Instead, Eq. 6 is typically simply compared to the ideal gas law (whose validity is assumed before the end of the derivation!) and a formula is obtained for vx. This typical approach is most unfortunate as there is a simple line of thought that can be used to complete the derivation easily. This relies on the equipartition principle: first, the term mvx2/2 is recognized as the contribution of the motion in direction x (Ex) to the average kinetic energy of the particles; then, the equipartition principle is applied to replace this term with kT/2:

$$p = cN_}} mv_^ = cN_}} 2E_ = cN_}} 2\frackT = cN_}} kT = cRT.$$

(7)

It is seen that we arrived at the form of the ideal gas law written in Eq. 3, so the derivation is complete now. Historically, this equation could be used to define the universal constants k and R.

As already pointed out, textbooks typically do not bring the derivation to completion. In the absence of the use of equipartition, this is not even a proper derivation of the ideal gas law. However, there are other problems with the presented sequence of thought as well.

The reader is asked to stop for a moment at this point and try to find lapses of logic in the sequence of thought presented above in Sect. "A simplistic view of pressure." Normally, few people are inclined to do this as textbook knowledge is assumed to be beyond any doubt. Unfortunately, this is far from the truth. Similarly to all other documents, textbooks also contain errors, which may range from minor typos or regrettable omissions to outright fallacies.

The next subsection will list a (not necessarily complete) collection of such lapses of logic in this sequence of thought. None of them are entirely beyond repair, but they necessitate additional remarks at the very least. If the reader was successful in identifying one or more of them, it is a favorable sign of the ability to think scientifically.

Consider particle M4 in Fig. 1. It is within the rectangular box we are interested in, but is moving away from surface A, so its velocity component in direction x is negative and will not hit A. This is why we divided the number of particles in the rectangular box with 2, right? However, we took vx as the average for all particles. Since the velocity components can be both positive and negative, and the motions are totally random, the average will certainly be 0. One might be inclined to update the definition by declaring that vx should be the average for those particles that move toward A, but this still leads to problems that are described in Sect. "Averages."

Consider particle M3 in Fig. 1. It is not within the rectangular box for which we have made the count, but it does collide with A within the given time interval. Or consider particle M2. It is within the box; it moves toward A, but collides with the wall outside A. So, M3 is not counted but contributes to the pressure, whereas M2 is counted but does not in fact contribute. Refuge from this problem can be found in the fact that there is a very large number of molecules in the system and their motion is completely random, which gives a good reason to hope that the effects of entering and leaving particles will cancel each other exactly.

Consider particles M5 and M6 in Fig. 1. They are both within the rectangular box of interest, and they are both moving toward A. However, they collide with each other before hitting the wall (only the left view of Fig. 1 shows this fact clearly) and change the direction of their motion. There are several good remedies for this problem. The first is noticing that the conservation of momentum law is valid separately for the different components of the vector. As the collision cannot change the sum of the momenta of the two particles in direction x, this momentum will reach surface A and exert a force despite the intermolecular collision. The second is based on the fact that all appearances of the time interval Δt are cancelled by the operations in Eq. 6. That means we are free to select a Δt that is short enough to avoid any such collisions. The third way out is a highly counterintuitive one: we did not clearly specify this earlier, but we imagine the particles to be mass points at the first approach [21]. Roughly speaking, points do not have measurable sizes in any direction, so they do not collide. To satisfy even the most pugnacious mathematician: the probability of two mass points colliding in this model is zero in any finite time interval. A simultaneously amusing and amazing further fact in probability theory is that an event with zero probability is still not impossible [22]. This line will not be pursued here any further: understanding the quantitative description and the consequences of collisions between particles in an ideal gas needs such scientific depth that a separate article will be devoted to it.

The subtitle is honest here but may still be misleading. No, we will not argue that Einstein’s relativity theory should be brought into the discussion in any way. What we simply mean is that reference points are needed to measure velocity or speed. These very often take the form of a coordinate system. The derivation presented to calculate the pressure tacitly assumed that the wall itself does not move; only the particles change their positions. So, the coordinate system is fixed to the container of the gas sample. Still, we were notably free to choose a direction to be designated x. There will be some consequences of this freedom, which we will discover in a later section. Also, at this point, it may be useful to state that most of the derivations presented in this article are only valid if the center of mass of the gas sample does not move relative to the container. In other words, transferring the results to the case of flowing gases would need additional mental work.

Now, compare the velocities of particles M1, M2, M3, M4, M5 and M6 in Fig. 1. Of course, they are different because the directions of movement are different. However, a closer look will reveal that even the speeds are different. This is a natural state of affairs in a gas sample: the hypothetical state in which all the particles move with the same speed is definitely outside the boundaries of classical thermodynamics.

The speeds and velocities of different particles vary then; yet, we used a single value for vx. First, we tried to define it as a bona fide average and then attempted to update the definition to require an average for those particles that move toward A but warned that this will create even more problems. Now, it is time to deliver on this promise. The problem is averaging itself, which can be demonstrated in an example that requires some numerical calculations.

Consider two gas particles moving at different speeds, for example, two helium atoms, one moving at 100 m s−1, the other one at 500 m s−1. It is a very natural thing to state that the average speed of the two molecules is 300 m s−1 in this case. Now, calculate the kinetic energy of the two particles (E = ms2/2). The mass of a helium atom is m = M/NA = 0.00400 kg mol−1/6.02 × 1023 mol−1 = 6.64 × 10−27 kg, so the kinetic energy of the one traveling at 100 m s−1 is 3.32 × 10−23 J = 33.2 yJ (y stands for yocto, a rarely used prefix for 10−24) or 20.0 J mol −1. The kinetic energy of the helium atom moving at 500 m s−1 is 830 yJ (or 500 J mol−1), so the average energy of the two atoms is 432 yJ or 260 J mol−1 (note that the decimal part is not given, as this would run against the established scientific use of significant digits). If one calculates the kinetic energy for the average speed of 300 m s−1, the result is 299 yJ (180 J mol−1). There is no mistake here: the average energy cannot be calculated from the average speed! Or at least not from this average. When discussing the description of ideal gases, many textbooks introduce the concept of root mean square speed, which is, as the name implies, the square root of the average of the squares of the two speeds, srms = ((1002 + 5002)/2)1/2 = 361 m s−1. If the average energy is calculated from srms, the correct value of 432 yJ (or 260 J mol−1) is achieved.

The origin of the problem is quite deep. When we consider a large collection of things (such as molecules) whose properties differ, there is no natural way of calculating ‘the average.’ There are several different averages [22]. The one that seems the most natural to many users is called the arithmetic mean: we add all the properties and then divide by the number of things considered. The root mean square has the mathematical name of quadratic mean, and there are others as well. When we try to characterize the properties of a bunch of particles with different individual properties, it is difficult to know which of the means is needed in a formula. It is often necessary to explore the probabilities at which certain values occur in a physical property. This idea will be elaborated in later sections.

A careful reader might have already spotted an apparent contradiction. The end of Sect. "The notion of an ideal gas" defined the ideal gas as a sample for which the only possible interaction between the particles is the elastic collision. On the other hand, Sect. "Collisions" said that if the particles of an ideal gas are thought to be mass points, they collide with zero probability. This may simply be a language issue, but by formal logic, the absence of collisions actually still satisfies the ‘only possible interaction…’ criterion. The easiest way to exclude the possibility of collisions is to have only one particle in a container.

Could one particle make sense as an ideal gas? It certainly collides with the walls of the container so it must exert some pressure. It is more difficult to imagine how we could attribute temperature to a single particle. Nevertheless, it is worth putting some effort into this as it is an exceptionally well characterized system in quantum mechanics for which a reasonably detailed mathematical description is given in textbooks as well. It is called particle in a box (p. 264 in [8]).

All the physical properties of a particle in a box in thermodynamic equilibrium can be obtained from its stationary wave function Ψstat, which is the solution of the stationary Schrödinger equation:

$$\begin - \frac }}^ m}}\frac \Psi_}}} \left( \right)}} }} - \frac }}^ m}}\frac \Psi_}}} \left( \right)}} }} - \frac }}^ m}}\frac \Psi_}}} \left( \right)}} }} \hfill \\ + V\left( \right)\Psi_}}} \left( \right) = E\Psi_}}} \left( \right). \hfill \\ \end$$

(8)

In this equation, x, y and z are the spatial coordinates in a Cartesian system, m is the mass of the particle, E is the energy of the particle, h is the Planck constant, π is the well-known constant equal to the ratio of the circumference and diameter of a circle (3.1415…), and V(x,y,z) is the potential function that characterizes the interaction of the particle with the outside world. The form of V(x,y,z) for the particle in the box model is such that it is 0 within the box, whereas it is infinitely large outside the box. This partial differential equation resembles a single algebraic equation for more than one unknown, in which there is no single solution; the best we can achieve is to find a connection between the unknowns. Similarly, in a partial differential equation, the “solution” is a connection between the independent variables. However, typically there are other, non-differential equations that must also be satisfied by the wave function in addition to the Schrödinger equation.

In the particle in a box model, these additional equations are called boundary conditions, and they reflect the physical fact that the particle cannot occur in or outside the walls of the container. For a rectangular box of dimensions a × b × c, these boundary conditions can be stated in a simple manner:

$$\Psi_}}} \left( \right) = \Psi_}}} \left( \right) = \Psi_}}} \left( \right) = \Psi_}}} \left( \right) = \Psi_}}} \left( \right) = \Psi_}}} \left( \right) = 0.$$

(9)

Finding the solution of a partial differential equation is usually not an easy process. One can certainly look for help from software such as Wolfram Alpha [13], but in the case of Eqs. 8–9, this is not necessary as the solution is given in many textbooks. It takes the following form (p. 269 in [8]):

$$\Psi_}}} = \sqrt} \sin \left( \sqrt m} }}x} \right)\sqrt} \sin \left( \sqrt m} }}y} \right)\sqrt} \sin \left( \sqrt m} }}z} \right).$$

(10)

The energy of a particle in a box is quantized. The expression giving the possible energy levels contains three quantum numbers (nx, ny, nz) that are independent of each other and characterize the movement in the three spatial directions (p. 270 in [8]):

$$E = E_ + E_ + E_ = \frac^ h^ }} }} + \frac^ h^ }} }} + \frac^ h^ }} }}.$$

(11)

As already stated, finding the solution shown in Eqs. 10–11 is quite a demanding task, but once known, it is easy to make sure that it satisfies the equations by simple substitutions.

As a next step, we can calculate the force acting on one of the walls of the box. Since force does not appear in any of Eqs. 8–11, it must be connected to the existing physical properties somehow. For this, it is useful to recall a piece of basic physical knowledge: energy is the ability to do work, and work can be calculated as the product of the force and the displacement in the direction of the force. So, when the wall of the rectangular box in direction x is moved a bit, the force can be calculated as the ratio of the energy change and the displacement in direction x. In more formal mathematical language, it is said that the force is the derivative of the energy regarding a change in the side length a. As the energy is already known as a function of a, this calculation can be brought to an easy conclusion:

$$F = \frac}E}}}a}} = \frac}}}a}}\left( ^ h^ }} }}} \right) = \frac^ h^ }}\frac}}}a}}\left( }}} \right) = \frac^ h^ }}\left( }}} \right) = - \frac^ h^ }} }} = - \frac }}.$$

(12)

In an ideal gas, the particles are independent of each other, and if they are considered mass points, they do not collide with each other. So, for N particles (N = nNA), the overall force is obtained by simply summing the contributions from the individual particles:

$$F = \sum\limits_^ \left( i \right)}}} = - \frac\sum\limits_^ \left( i \right)} .$$

(13)

At this point, we can use the equipartition theorem again. The fact that the average energy per degree of freedom for a single molecule is kT/2 can be put into a mathematical equation for our case as follows:

$$\frac\sum\limits_^ \left( i \right)} = \frackT.$$

(14)

The pressure is the absolute value of force divided by the area of the wall perpendicular to direction x, which is easily given by the product of the two sides (bc):

$$p = \frac} = \frac\sum\limits_^ \left( i \right)} }} = \frackTN}} = \frac = \frac.$$

(15)

So, the most common form of the ideal gas law (Eq. 1) can be derived directly from the quantum mechanical model of a particle in a box.

Thus far, this derivation was presented for a rectangular box. However, the ideal gas law in fact has no limitations on the shape of the box in which the gas is stored. The previous considerations can be somewhat extended into containers of different shapes, but these rely on solving the stationary Schrödinger equation for different boundary conditions. While particle in a (rectangular!) box is a typical textbook model in quantum mechanics, other cases are seldom handled. This article gives two more examples: the case of a cylindrical box and the case of a spherical box.

A cylindrical box is imagined in a way that the cross section of the box in the yz plane is a circle of radius R, whereas in the two other perpendicular planes it is a rectangle. The height of the cylinder is denoted a; its meaning is exactly the same as that of c for a rectangular box. The details and the solution of the Schrödinger equation are given in the Supplementary Information. What is important is the formula obtained for the possible quantized energy levels of a particle. This contains an nx quantum number analogous to that seen in Eq. 9. There are two additional quantum numbers that arise for motion in the direction perpendicular to the x axis, but they do not appear in the energy formula. Instead, a series of infinitely many isolated values denoted \(__}\) appear, which represent the independent variables at which the value of the Bessel function of the first kind is 0. This might sound like a frightening definition, but for our purposes, it is a constant whose values will not even be needed. The full energy formula is:

$$E = \frac Z_ }}^ }}^ mR^ }} + \frac n_^ }} }}.$$

(16)

Now, two different ways of calculating pressure arise. In the first, the collisions of the “side” of the cylinder must be considered. The force here is the derivative of the energy with respect to R, the radius of the cylinder:

$$F_ = \frac}E}}}R}} = \frac}}}R}}\left( Z_ }}^ }}^ mR^ }}} \right) = \frac Z_ }}^ }}^ m}}\frac}}}R}}\left( }}} \right) = \frac Z_ }}^ }}^ m}}\left( }}} \right) = - 2\frac Z_ }}^ }}^ mR^ }} = - \frac.$$

(17)

In calculating the pressure, it must be considered that the movement characterized by radius R is two dimensional now, which means that the average energy will be 2 × ½kT. The surface to be considered is the curved surface of the cylinder here (2πRa):

$$p = \fracRa}} = \frac\frackTN}}Ra}} = \fracR^ a}} = \frac = \frac.$$

(18)

The same can be done in the direction z, which would give the pressure acting on the bases of the cylinder. The force here is calculated by derivation with respect to a:

$$F_ = \frac}E}}}a}} = \frac}}}a}}\left( n_^ }} }}} \right) = \frac n_^ }}\frac}}}a}}\left( }}} \right) = \frac n_^ }}\left( }}} \right) = - 2\frac n_^ }} }} = - \frac$$

(19)

Since z is a single dimension here, ½kT is necessary for energy. The surface is the area of the circular base of the cylinder:

$$p = \fracR^ }} = \frac\frackTN}}R^ }} = \fracR^ a}} = \frac = \frac.$$

(20)

The case of the spherical box (with radius R) also brings some difficulties in solving the stationary Schrödinger equation; the solution is shown in the Supplementary Information. The quantized energy levels are given by a formula that includes a constant Zl, which represents the independent variables at which the value of the spherical Bessel function of the first kind (jl) is 0. Again, the actual values of Zl will not be needed:

$$E = \frac Z_^ }}^ R^ }}$$

(21)

The force acting upon the spherical wall is obtained from a derivation similar to the previous ones:

$$F = \frac}E}}}R}} = \frac}}}R}}\left( Z_^ }}^ mR^ }}} \right) = \frac Z_^ }}^ m}}\frac}}}R}}\left( }}} \right) = \frac Z_^ }}^ m}}\left( }}} \right) = - 2\frac Z_^ }}^ mR^ }} = - \frac.$$

(22)

Now, R characterizes the boundaries of a three-dimensional movement, so the average energy should be taken as 3 × ½kT. For calculating the pressure, the surface of the sphere is needed now:

$$p = \fracR^ }} = \frac\frackTN}}R^ }} = \fracR^ }} = \frac = \frac.$$

(23)

In all cases (Eqs. 15, 18, 20 and 23), the validity of the ideal gas law is confirmed for containers of different shapes. However, it must be recognized that only simple shapes can be handled by this method, and even then, the derivation is notably burdensome as it involves finding the solutions of the stationary Schrödinger equation with different boundary conditions. So, overall, this line of thought shows quite nicely the fundamental origins of the ideal gas law but is very impractical as a proof and becomes close to impossible for more complicated container shapes. So, in the next section, a line of thought will be sought where the container shape does not matter so much.