The Boltzmann constant is the value that relates the average kinetic energy of a thermodynamic system or an object with its absolute temperature. Although they are often confusing, temperature and energy are not the same concept.
Temperature is a measure of energy, but not energy itself. With the Boltzmann constant, it is linked together as follows:
E c = (3/2) k B T
Boltzmann tombstone in Vienna. Source: Daderot on Wikipedia in English [CC BY-SA 3.0 (http://creativecommons.org/licenses/by-sa/3.0/)]
This equation is valid for an ideal monoatomic gas molecule of mass m , where E c is its kinetic energy given in Joules, k B is the Boltzmann constant and T is the absolute temperature in Kelvin.
In this way, as the temperature increases, the average kinetic energy per molecule of substance also increases, as expected. And the opposite happens when the temperature decreases, reaching the point where, if all movement ceases, the lowest possible temperature or absolute zero is reached.
When talking about average kinetic energy, it is necessary to remember that kinetic energy is associated with movement. And particles can move in a variety of ways, for example, moving, rotating or vibrating. Obviously, not everyone will do it the same way, and since there are countless, the average is used to characterize the system.
Some energy states are more likely than others. This concept is extremely important in thermodynamics. The energy considered in the previous equation is the kinetic energy of the translation. The probability of states and their relationship to the Boltzmann constant will be discussed a little later.
In 2018 he redefined Kelvin and with it the Boltzmann constant, which in the International System is about 1.380649 x 10 -23 JK -1 . It is possible to obtain much more precision for the Boltzmann constant, which has been determined in several laboratories around the world, by different methods.
Story
The famous constant owes its name to the Vienna-born physicist Ludwig Boltzmann (1844-1906), who dedicated his life as a scientist to the study of the statistical behavior of systems with many particles from the point of view of Newtonian mechanics.
Although nowadays the existence of the atom is universally accepted, in the 19th century the belief about whether the atom actually existed or was a device by which many physical phenomena were explained was in full debate.
Boltzmann was a strong supporter of the existence of the atom, and at the time he faced harsh criticism of his work by many colleagues, who considered it to contain insoluble paradoxes.
He stated that observable phenomena ‘at macroscopic levels could be explained by the statistical properties of constituent particles such as atoms and molecules.
Perhaps these criticisms are due to the deep episode of depression that led him to take his own life in early September 1906, when he still had a lot to do, because he was considered one of the great theoretical physicists of his time and there was very little left. . that other scientists contribute to corroborate the veracity of their theories.
It was not long before his demise, when new discoveries about the nature of the atom and its constituent particles were added to prove Boltzmann right.
Boltzmann’s constant and Planck’s work
Now the Boltzmann constant k B was introduced as known today some time after the work of the Austrian physicist. It was Max Planck, in his law on the emission of the black body, a work he presented in 1901, which at the time gave him the value of 1.34 x 10-23 J/K.
In 1933, a plaque defining entropy involving the famous constant was added to Boltzmann’s gravestone in Vienna as a posthumous tribute: S = k B log W , an equation that will be discussed later.
Today, Boltzmann’s constant is indispensable in the application of the laws of thermodynamics, statistical mechanics, and information theory, fields in which this sad-ending physicist pioneered.
Value and Equations
Gases can be described in macroscopic terms as well as in microscopic terms. For the first description, there are concepts like density, temperature and pressure.
However, it should be remembered that a gas is made up of many particles, which have an overall tendency to behave in a certain way. It is this trend that is measured macroscopically. One way to determine the Boltzmann constant is thanks to the well-known ideal gas equation:
pV = n. RT
Here p is the pressure of the gas, V is its volume, n is the number of moles present, R is the constant of the gas, and T is the temperature. In an ideal gas spring, the following relationship between the product pV is achieved and the kinetic energy of the K translation of the entire set is:
pV = (2/3). K
Therefore, kinetic energy is:
K = (3/2) nRT
By dividing by the total number of molecules present, which will be called N, the average kinetic energy of a single particle is obtained:
E c = K / N
E c = (3 / 2N) nRT
In a mole, there is the Avogadro number of particles N A and therefore the total number of particles is N = nN A, leaving:
E c = (3 / 2nN A ) nRT
Precisely the R / N A ratio is the Boltzmann constant, thus demonstrating that the average translational kinetic energy of a particle depends only on the absolute temperature T and not on other quantities, such as pressure, volume or even the type of molecule:
E c = (3/2) k B. T
Boltzmann constant and entropy
A gas has a certain temperature, but that temperature can correspond to different internal energy states. How to visualize this difference?
Consider the simultaneous toss of 4 coins and the ways in which they can fall:
Ways in which 4 coins can fall 4. Source: own elaboration
The set of coins can assume a total of 5 states, considered macroscopic , described in the figure. Which of these states would the reader say is the most likely?
The answer must be the state of 2 faces and 2 crossings, because it has a total of 6 possibilities, out of the 16 illustrated in the figure. E 2 4 = 16. These are equivalent to microscopic states .
What if 20 coins are played instead of 4? There would be a total of 2 20 possibilities or “microscopic states”. It’s a much larger number and more difficult to handle. To make it easier to handle large numbers, logarithms are very appropriate.
Now what seems obvious is that the state with the most disturbance is the most likely. More ordered states, such as 4 faces or 4 stamps, are slightly less likely.
The entropy of a macroscopic state S is defined as:
S = k B ln w
Where w is the number of possible microscopic states of the system and k B is the Boltzmann constant. Since ln w is dimensionless, entropy has the same units as k B : Joule / K.
This is the famous equation on Boltzmann’s headstone in Vienna. However, more than entropy, what matters is change:
ΔS = k B ln w 2 – k B ln w 1 = k B ln (w 2 / w 1 )
How is ok B calculated ?
The value of the Boltzmann constant is experimentally obtained very precisely, with measurements based on acoustic thermometry , which are performed using the property that establishes the dependence of the speed of sound in a gas on its temperature.
In effect, the speed of sound in a gas is given by:
Adiabatic B = γp
And ρ is the density of the gas. For the previous equation, p is the pressure of the gas in question and γ is the adiabatic coefficient, whose value for a given gas is found in the tables.
Metrology institutes are also experimenting with other ways to measure the constant, such as Johnson Noise Thermometry, which uses thermal fluctuations that occur randomly in materials, particularly conductors.
solved exercises
-Exercise 1
To locate:
a) The average kinetic energy of translation E c that has an ideal gas molecule at 25 °C
b) The kinetic energy of the K translation of molecules in 1 mol of this gas
c) The average velocity of an oxygen molecule at 25 °C
Data
m oxygen = 16 x 10 -3 kg / mol
Solution
a) E c = (3/2) k T = 1.5 x 1.380649 x 10 -23 J. K -1 x 298 K = 6.2 x 10 -21 J
b) K = (3/2) nRT = 5 x 1 mol x 8.314 J / mol .K x 298 K = 3716 J
c) E c = ½ mv 2 , considering that the oxygen molecule is diatomic and the molar mass must be multiplied by 2, it will have:
Find the change in entropy when 1 mole of gas that occupies a volume of 0.5 m 3 expands to occupy 1 m 3 .
Solution
ΔS = k B ln (w 2 / w 1 )
w 2 = 2 N w 1 (There were 2 4 microscopic states for the toss of the 4 coins, remember?)
Where N is the number of particles present in 0.5 mol of gas 0.5 x N A :
ΔS = k B ln (2 N w 1 / w 1 ) = k B ln 2 N = k B 0.5N A ln 2 = 2.88 J / K
