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We saw previously how Sadi Carnot’s attempts to optimize the steam engine led to the concept of the Carnot engine, which is the heat engine that has the highest efficiency that is theoretically possible. When Rudolf Clausius analyzed the Carnot cycle mathematically, he came to the insight that there are different forms of energy, and that energy can be converted from one type to another. For example, heat can be converted to internal energy, which raises
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the temperature of a gas or object. Or, heat can be converted to useful mechanical work, as is done in heat engines. To quantitatively describe spontaneous, irreversible processes, such as the flow of heat from hot to cold, he introduced the concept of entropy. In the following, we aim to understand some of these results by modeling gases as a large amount of molecules moving around according to the laws of Newtonian mechanics. This is
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called the kinetic theory, and it aims to unify thermodynamics with the well-established Newtonian mechanics. It states that the internal energy of a gas is determined by the kinetic energy of the molecules, which explains why internal energy can be converted to mechanical work and vice versa. Moreover, it aims to explain entropy in terms of statistics: systems tend to evolve from statistically unlikely states to statistically likely states, so
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entropy should be a measure of the statistical likelihood of a state. Let’s see how we can derive the ideal gas law by modeling a gas as a large number of molecules moving around. The ideal gas law says that the pressure times the volume of a gas, is proportional to the amount of gas and its temperature. Here, n denotes the amount of gas in moles, and R denotes the constant of proportionality called the gas constant.
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But because we want to explain the ideal gas law in terms of the motion of individual molecules, let’s define capital N to be the number of gas molecules, and then the constant of proportionality becomes k_B, which is called Boltzmann’s constant. We can rewrite this into an expression for the pressure P. We will see how this pressure is caused by molecules moving around and bumping into the walls of the container. We can identify
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three factors that determine how much pressure the molecules exert on the walls. The first factor is the momentum of a molecule. Consider a single molecule moving towards a wall with a certain speed v, and a momentum given by its mass times its speed. Intuitively, we can expect that when the molecule bounces of the wall, the pressure it exerts is proportional to its momentum: so the faster the molecule moves, and the heavier the molecules are,
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the higher the pressure of the gas. The second factor that determines the pressure that molecules exert on the walls of the container, is the return time. A molecule doesn’t bounce of the wall only once, but rather it bounces back and forth between the walls. The more often the molecule bounces against the wall, the more pressure it exerts. If the molecule has a speed v, and the distance between the walls is L, then the time it takes for the
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particle to go back and forth is proportional to v divided by L. Since the length L is directly proportional to the volume of the container, we find that the pressure is proportional to the speed of the molecule, divided by the volume of the container. The third factor that determines the pressure that molecules exert on the walls of the container, is the number of molecules. If a single particle exerts a certain amount of pressure on the
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wall, then N molecules exert N times as much pressure. So the pressure P is proportional to the number of molecules N. Putting these three results together, tells us that the pressure is proportional to the number of molecules N, times the mass of a single molecule m, times the speed of a molecule squared, divided by the volume of the container V. When we compare this result with the ideal gas law, we find that they have the factor
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N/V in common. It follows then, that the remaining factors must be proportional to each other. That is, Boltzmann’s constant times the temperature must be proportional to the kinetic energy of a single molecule. So temperature is a measure of the kinetic energy of molecules in a gas. Now let’s analyze the result more quantitatively. Consider for now the motion of a molecule only in the x direction. If a particle moves
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with speed v_x at the wall, then after an elastic collision with the wall, the velocity is -v_x. Conservation of momentum then requires that an amount of momentum of 2mv_x has been transferred to the wall. So the momentum that is transferred per collision is 2mv_x. Now let’s find out how many times a single molecule collides with the wall during some time Delta t. During a time Delta t, the molecule travels a distance v_x Delta t. The distance of one
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round trip is 2L_x. The number of times that a single molecule collides with a wall in a time Delta t, is the distance it travels during that time, divided by the distance of a single round trip. The number of molecules is given by N. Now let’s calculate the total momentum that N molecules transfer to the wall during a given time Delta t. The total transferred momentum equals the amount of momentum transferred
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per collision, times the number of collisions that a single molecule undergoes during a time Delta t, summed over all molecules. We can plug in the values that we found previously. The 2’s cancel out, and the factor Delta t over L_x can be taken outside the summation. Now we must find how the total transferred momentum relates to the total pressure of the gas. Pressure is given by force divided by area. Because we consider the x-component of the velocities of the particles, the wall that
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the molecules collide with is in the yz-plane, so its area equals L_y times L_z. The average force is given by the change in momentum per unit time. We can plug in the expression we found for the transfer of momentum during a time Delta t. The Delta t cancels out, and the product L_x L_y L_z equals the volume of the box. We find in the final expression for the pressure P, a summation over all molecules. Let’s see how we can rewrite this expression
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more conveniently. First, we can take out a factor N, the number of molecules, so that the summation represents an average over all N molecules. Then we can take out a factor 2, so that we sum over something that resembles the kinetic energy. However, note that we only took into consideration the x-component of the molecule velocity. The total kinetic energy should also take into account the y- and z-components. If we
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assume a system that is symmetric in the x, y, and z directions, taking only into account the x-component of the velocity amounts to computing one third of the average kinetic energy. By comparing this result to the ideal gas law, we find that Boltzmann’s constant times the temperature, equals 2/3 the average kinetic energy of all molecules. We now know how the average kinetic energy relates to the temperature. The total internal
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energy is simply the average kinetic energy per molecule, times the number of molecules. We find that the total internal energy is given by 3/2 times N k_B T. This result can be interpreted using the equipartition theorem. The equipartition theorem states that the total internal energy is equally divided over the degrees of freedom of motion. In our case we have a total internal energy that is 3/2 N k_B T. This can be interpreted as the result
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of having three degrees of freedom for motion, where each degree of freedom carries ½ N k_B T units of energy. The three degrees of freedom come from the fact that a molecule has three velocity components: x, y and z. This result holds if the molecules consist of a single atom, so it holds for a monatomic gas. But what if the molecules consist of two atoms, so we have a diatomic gas? In that case, we have more degrees of freedom over
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which the energy can be distributed. Not only can molecules spend energy on moving around in the x- y- or z-direction, but they can also rotate. In the depicted orientation of the molecule, the molecule can rotate around its z-axis, or around its y-axis. Rotation along the x-axis, so along the length of the molecule, contains a negligible amount of energy because the moment of inertia is much smaller. These two extra degrees of freedom each contain additional factors of 1/2 N k_B T. The total
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internal energy now consists of 5 portions of N k_B T/2, so the total internal energy of a diatomic gas is given by 5/2 N k_B T, whereas for a monatomic gas it was given by 3/2 N k_B T. At higher temperatures, the molecule can also store energy in its vibrational modes. This adds one degree of freedom for motion, but there are two extra contributions to the energy: namely the kinetic energy due to the motion of the vibration, and the potential
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energy that is stored in the imaginary spring that binds the two atoms. Therefore, there are two extra contributions of N k_B T/2 to the total energy, so the total internal energy for a diatomic gas at high temperatures is given by 7/2 N k_B T. So we see that depending on the degrees of freedom the molecules in a gas have, the relation between the internal energy and the temperature changes. How can we verify this experimentally?
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The relation between the internal energy and the temperature determines the heat capacity of the gas. The heat capacity tells you how much energy you should put into the gas to raise its temperature by 1 degree. The heat capacity at constant volume is given by the change in internal energy divided by the change in the temperature. It is important to keep the volume fixed, because if the gas were allowed to expand, not all the heat energy
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that you put into the gas goes into raising the temperature of the gas. Part of the heat energy that you put in, would be spent on the mechanical work that is performed when the gas expands. Now let’s think about which part of the internal energy actually contributes to raising the temperature of the gas. We saw previously that the pressure of the gas is caused by the molecules bumping into the walls of the container. This is due to the translational motion of the molecules, not their rotations
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or vibrations. Since the ideal gas law tells us that temperature is directly proportional to the pressure, we can conclude that only the energy that is put into the translational motion of the molecules, contributes to the temperature of the gas. The energy that is put into rotations and vibrations don’t increase the temperature. Therefore, the more degrees of freedom for motion the molecules of a gas have, the more energy it takes to
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raise the temperature, so the higher the heat capacity. This is indeed what we see from the formulas that we derived for the internal energies for monatomic and diatomic gases. We can also use the equipartition theorem to find the heat capacity of solids, instead of ideal gases. In a solid, atoms do not move freely like molecules do in a gas, but rather they are bound to a lattice. When an atom vibrates, it has kinetic energy, and there
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is potential energy in the bond between atoms, which can be thought of as springs with a certain spring constant. According to the equipartition theorem, the kinetic energy and the potential energy each account for an energy of ½ N k_B T. Since the lattice is three-dimensional, there are three such contributions, leading to a total internal energy of 3N k_B T. The heat capacity per atom is therefore three times Boltzmann’s
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constant. This law is known as the Dulong-Petit law. For many materials, experimental observations agree well with theoretical predictions. But for very low temperatures, approaching absolute zero, the model fails for all materials, and the heat capacity goes to zero. Classical theory fails to explain this behavior, and it took the development of quantum mechanics to explain this phenomenon. We’ve seen how the average kinetic energy
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of molecules is related to the temperature of an ideal monatomic gas. But not all molecules have the same kinetic energy, and not all molecules have the same velocity. So we can ask: what is the velocity distribution of the molecules? Consider a monatomic ideal gas in a spherical container. Each molecule has a certain velocity vector, with some x-, y-, and z-component. We want to know the probability that a molecule has a velocity vector within
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a certain interval around v_x, v_y, v_z. This probability is given by a probability distribution function f, multiplied by the size of the interval. The question is: what is the formula for the probability distribution function f? The system is completely symmetric in the x-, y- and z-directions, so the probability that a molecule has a certain velocity in the x-direction, must be the same as the probability that the molecule has that velocity in the y- or z-direction. Moreover, the probability
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that a molecule has a certain velocity in the x-direction, should not depend on its velocity in the y- or z- direction. Therefore, we expect that the probability distribution function f of the velocity vector can be written as the product of the probability distribution functions of the separate components, which all follow the same probability distribution g. Moreover, the probability distribution f should be independent of any rotation of
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the coordinate system, because the system is fully rotationally symmetric. So the probability distribution can only depend on the magnitude of the velocity vector, not its direction. Combining these two requirements leads to the conclusion that the probability distribution function is an exponential function of the squared length of the velocity vector. As we required, it is a function of only the length of the velocity vector, and it can
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be factorized intro three identical probability distribution functions for the separate velocity components. We have now found the form of the probability distribution function, but it still depends on two unknown parameters A and B. How do we find their values? The first constraint we use to find the values of A and B, is the fact that the probability distribution must integrate to 1. To work with this constraint more easily, let’s
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define v as the length of the velocity vector. We can write the probability distribution function as a function of v. If we plug this equation into the integral, then we also want to change the integration variable to v. Note that one value of v, corresponds to the entire surface of a sphere in velocity space: multiple combinations of v_x, v_y, v_z, give the same value for v. Therefore, when we change the integration variable to
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v, we must include a factor of 4 pi v^2. We write the explicit expression for A. The integral can be solved analytically, which gives a value that depends on B. We have now found the relation between A and B. To find the value of B, we must use a second constraint. We know that the average kinetic energy of the molecules is given by 3/2 k_B T. The average kinetic energy is found by multiplying the kinetic energy, with the probability
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of the molecule having that kinetic energy, and integrating over all possible values of the kinetic energy. We equate this expression to the average kinetic energy of 3/2 k_B T. We can rearrange some factors to simplify the expression, and then we can solve the integral analytically. Plugging in the value we found for A, gives an expression where only B is unknown. Solving this equation for B gives that B equals
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½ m / k_B T. We have now used the two constraints to find the values of A and B. We can plug this into our expression for the probability distribution function. If we’re only interested in the probability that a molecule has a certain speed, and we don’t care about its direction, we multiply the probability distribution function with 4 pi v^2. This yields the Maxwell-Boltzmann distribution, which describes the velocity distribution for molecules in a monatomic
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ideal gas. If we plot this function for different temperatures, we see that for low temperatures the function has a narrow peak centered around low velocities, while for higher temperatures the function has a broader peak centered around higher velocities. So we’ve seen how Maxwell derived the velocity distribution of molecules using arguments of symmetry. We will now see how Boltzmann tried to achieve the same result, but using
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statistical arguments. To illustrate the line of thinking that he presented in his 1877 paper, consider a box containing seven molecules. The different molecules move around with different velocities. Let’s say the system contains a total internal energy of U, and that it can be divided into 7 discrete units of energy dE. This discretization of energy is purely for mathematical convenience, it doesn’t try to describe anything physical. The 7 units
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of energy are randomly distributed over the molecules. The idea is that through many random collisions, the molecules randomly exchange kinetic energy, and therefore, at any given point in time, the total kinetic energy is randomly distributed over the molecules. To describe the statistical distribution of the energy over the molecules, we can simply count how many molecules have a certain amount of energy. For example, there are three molecules
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that have 0 energy. There are two molecules that have 1 unit of energy. There is one molecule that has 2 units of energy, and there is one molecule that has 3 units of energy. The numbers n_0, n_1, n_2, n_3 therefore describe how the energy is distributed over the molecules. They must add up to the total number of molecules. Moreover, adding up all the energies should give the total number of units of energy in the system. Any distribution that satisfies
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these two constraints corresponds to a physically feasible distribution of the system. For example, we could put all the energy into one molecule, while all the other molecules have none. In this case, n_7 equals 1, and n_0 equals 6. Or, we could give all molecules exactly 1 unit of energy, so n_1 equals 7. So there are many possible ways in which we can distribute the energy over the molecules. Now the question is: what will we see in practice?
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If we were to measure the energy distribution over the molecules of a physical system, what would we measure, and why? If we think about the molecules moving around randomly, and randomly exchanging kinetic energy with each other, we should intuitively feel that it is statistically unlikely that by pure chance all the energy ends up in a single molecule. It should also feel unlikely that by pure chance all molecules end up having the exact
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same amount of energy. What is more likely, is that through random collisions and exchange of kinetic energy, some molecules happen to end up with more energy, and some with less. But how can we make this argument more rigorous? Can we quantitatively calculate the probability that the system randomly ends up with one distribution or the other, and then conclude which distribution is statistically most likely? Boltzmann proposed that you can simply count
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the number of ways in which you can achieve a certain distribution. For example, since there are seven molecules, there are seven ways to put all the energy into one single molecule. There is only one way to distribute the energy equally over all molecules. In how many ways can we achieve the distribution that we thought of as more likely? With this distribution, we have four groups. This first group contains the molecules that
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have 0 units of energy. With this distribution, three molecules fit in this group. The second group contains the molecules that have 1 unit of energy, and two molecules fit in this group. The third group contains only one molecule that has 2 units of energy, and the fourth group contains only one molecule that has 3 units of energy. Our task is now to figure out in how many ways we can distribute the seven molecules over these four groups. In
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the first slot, we can put any of the seven molecules, so we have seven possible choices. Let’s say we put molecule number 5 in that slot. For the next slot, we can choose any of the molecules, except molecule 5, so we have six possible choices. For example, we can choose molecule number one. For the next slot we can choose any molecule except molecules 5 or 1, so we have five possible choices, and we can continue this until we’ve
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filled all the slots. So how many ways are there in total to fill the slots? For the first slot, we had 7 possible choices. For each choice that we could pick, there would be 6 possible choices to fill the second slot. So in total there were 7 times 6 equals 42 possible ways to fill the first two slots. For each of those 42 choices, there are 5 ways to fill the third slot, so there are 7 times 6 times 5 ways to fills the first
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three slots. If we repeat this argument for all seven slots, we find that there is a total of 7 times 6 times 5 times 4 times 3 times 2 times 1 ways to fill all seven slots. This is more commonly expressed as seven factorial. So there are seven factorial ways to fill the slots. But the problem is that some ways of filling the slots correspond to the same physical configuration. So we’ve counted several configurations multiple times, whereas
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we should only count them once. For example, let’s say we swap the numbers in the first and the second slot. It means that they swap the amount of energy they have. But they have the same amount of energy, so it doesn’t change the physical configuration, so we should not count them as two separate ways to achieve the same energy distribution. More generally, we can move around the slots in the group in any way that we want, and we still end
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up with the same physical configuration. So how do we correct for this double counting? To find the number of ways in which we can move around the slots in the first group, we apply the same argument as we did before. There are three choices for the first slot, two choices for the second slot, and one choice for the third slot, so there are three factorial ways to move around the slots in the first group. Similarly, for the second group, which
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consists of two slots, there are two factorial ways to move around the slots. More generally, there are n_0 factorial times n_1 factorial times n_2 factorial etc. ways to achieve the same physical configuration. This product can be taken over all groups, even when the groups contain zero molecules, because 0 factorial is defined to be 1. So to find the number of physically different configurations that yield the same energy
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distribution over the molecules, we must divide the number of ways to fill the slots, by the number of ways that yield the same physical configuration. For our particular case, we find that there are 420 ways to achieve this particular energy distribution. We can verify this formula by observing that there are indeed 7 ways to put all the energy into a single molecule, and there is only 1 way to distribute the energy equally over all molecules. Indeed,
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there are many more configurations that yield the distribution that we identified as more likely. Now let’s introduce some terminology. The distribution that describes how many molecules have a certain amount of energy is called a ‘macrostate’. The specific configuration that specifies exactly which molecule has how much energy is called a ‘microstate’. A certain macrostate is more likely than another, because more microstates yield that same macrostate. The number of microstates that correspond
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to the same macrostate is called the multiplicity of the macrostate. So macrostates with a higher multiplicity are more likely than macrostates with a low multiplicity. So now we know what determines whether one macrostate is statistically more likely than another macrostate. But what is the exact probability to find a certain macrostate? To answer that question, we can list all the possible macrostates, and calculate their
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corresponding multiplicities. We saw that if all the energy is put into a single molecule, there are seven ways to achieve this distribution. If all molecules have exactly one unit of energy, there is only one way to achieve it. We can write down all the other distributions that have seven molecules, and seven units of energy. Using the formula that we found to calculate the multiplicity for a given distribution, we can complete the table. Adding
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all the multiplicities together, we find there are 1716 possible microstates for the system. Assuming that all microstates have an equal probability of occurring, we find that the most likely macrostate has a probability of 420/1716 of occurring. It is not very convenient to first make a list of all possible distributions to find the total number of microstates. Is there an easier way to compute the total number
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of microstates? Let’s represent the problem as follows. We have seven units of energy in the system. To distribute them over seven molecules, we insert six partitions. Each partition corresponds to a certain molecule. The number of units of energy that are to the left of each partition, indicates the energy of the molecule. The remaining units of energy are assigned to the seventh molecule. So in this example, molecule 1 has 0 units of energy, molecule 4 has 3 units of energy, etc.
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Because there are 7 units of energy, and 6 partitions, there are a total 7+6=13 slots where we can place either a molecule or a partition. In how many ways can we place the six partitions? For the first partition there are 13 possible choices, for the second partition there are 12, for the third partition there are 11, and so forth. So in total there are 13 times 12 times 11 etc. ways to place the partitions. This expression can be written
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in terms of factorials, and then simplified. But, as before, different placements of the partitions may correspond to the same physical configuration. Since each physical configuration should be counted only once, we must correct for double counting. To correct for double counting, let’s see how many placements of the partitions correspond to the same physical configuration. Consider each partition and the units of energy corresponding
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to it. If we shuffle these blocks around, then each molecule still possesses the same amount of energy, so the physical configuration remains unchanged. So in how many ways can we shuffle these blocks around? Since there are 6 blocks, there are 6 factorial ways to shuffle them around. So to find the total number of microstates, we calculate the total number of ways to place the partitions, and divide by the number of ways we can rearrange the blocks. We can verify that in our case, the formula
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says there are 1716 microstates, which is indeed the number we found previously. So we found that the probability to find the system in a certain macrostate is given by the number of microstates corresponding to that macrostate, divided by the total number of microstates. In practice, when there are very many molecules, and the energy can be divided into arbitrarily small units, we will only observe the macrostate with the highest
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probability. So given a system with N molecules, and a fixed total internal energy U, what is the most likely macrostate? The probability to find the system in a certain macrostate, is given by the multiplicity of that macrostate, divided by the total number of microstates. So to find the most likely macrostate, we need to find the macrostate with the highest multiplicity omega. We already found an expression for the multiplicity of any macrostate. Since the total number of
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molecules N is fixed, maximizing omega means minimizing the denominator. Now consider the logarithmic function ln(x). If we have two inputs x, one larger than the other, then the larger input gives the larger output. Therefore, if the input x is minimized, then ln(x) is also minimized. So instead of minimizing the product of factorials, we can also minimize its logarithm. This is useful, because the logarithm of a product can be written as a
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sum of logarithms. Now let’s see how we can approximate the logarithm of a factorial. The factorial is the product of n, n-1, n-2, etc., and taking the logarithm allows us to write it as a sum of logarithms. If we represent this sum visually, we see that it can be approximated as the surface under the curve of the logarithm, which can be calculated using an integral. This integral can be calculated straightforwardly,
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and it yields us a result that is known as Stirling’s approximation. By using Stirling’s approximation, we can rewrite our minimization problem. We want to find the n_k that optimize this expression. Adding or subtracting a constant value from this expression doesn’t change the n_k for which this expression is minimal, so we can subtract the constant term. Moreover, we know that all n_k must sum to the total number of molecules, which is a fixed number
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N. So subtracting this number also doesn’t change the n_k for which this expression is minimal. We have now found the simplest form of the expression that we want to minimize. But when we minimize it, we must take care that the following constraints hold. First, the total number of molecules is fixed. Second, the total amount of energy must be fixed. So now we have formulated our constrained minimization problem. The standard way to solve such a problem is by using Lagrange multipliers.
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This method works as follows. We define a Lagrangian, which consists of the function that we want to minimize, and then we add the constraint functions, multiplied by the Lagrange multipliers alpha and beta. Now, instead of having a constrained optimization problem, we can find the global minimum of the Lagrangian with respect to the optimization variables and the Lagrange multipliers. This is so, because if we take the derivative of
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the Lagrangian with respect to the Lagrange multipliers, and set this derivative to 0 to find the extrema, the constraint corresponding to that Lagrange multiplier is automatically satisfied. When we take the derivative of the Lagrangian with respect to an optimization variable n_j, and set it equal 0 to find the extrema, then we find an expression for n_j that depends on the Lagrange multipliers. Recall that n_j
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represents the number of molecules that have j units of energy. We see that the number of molecules that have a certain energy, decreases exponentially with energy. The exponential e^{-(1+alpha)} merely serves to make sure that all n_j add up to the total number of molecules N. We now know how many molecules have a certain amount of energy in the most likely macrostate. Dividing by the total number of molecules
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N, we obtain the fraction of molecules that have a certain amount of energy. Recall that the discrete energy levels were a consequence of discretizing the energy into discrete units, which we did for mathematical convenience, so that we can count microstates. But in reality, kinetic energy is a continuous variable. So we can turn the ratios for the discrete energy levels into a continuous probability distribution function. The ratio of molecules that have
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an energy within an interval dE, is then given by the probability distribution function times the interval. The probabilities must integrate to 1, and the expectation value of the probability distribution equals the average energy per molecule. Now let’s compare the energy probability distribution function we found using Boltzmann’s method, to the probability distribution function we found using Maxwell’s method. Maxwell
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gave the probability that a molecule has a certain velocity. To convert this to the probability that a molecule has a certain energy, we substitute ½ mv^2 with E, and to substitute dv with dE, we compute the derivative of energy with respect to velocity, express it in terms of energy, and solve for dv. Now we can write the energy distribution, by taking Maxwell’s velocity distribution, and substituting E, and dE.
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We find that the energy probability function is proportional to sqrt(E) times e^{-E/k_B T}. Compared to the result we found using Boltzmann’s method, we see that they differ by a factor sqrt(E). Why is there a mismatch? To understand where the mismatch is coming from, let’s recall how we found Maxwell’s distribution. We started by assuming that the probability function for the velocity vector must be rotationally symmetric, and
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factorizable into identical probability functions for the separate velocity components. If we only care about the magnitude of the velocity vector, we must multiply with a factor of 4 pi v^2, because multiple velocity vectors have the same magnitude. When we substitute the expressions for E and dE, we find the probability function with a factor of sqrt(E). Now let’s see what happens if instead of a three-dimensional system, we consider a
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two-dimensional system, where the velocity vector only has two components. The same arguments of symmetry apply, but when we turn this function into a probability function for the magnitude of the velocity vector, we have to multiply with 2 pi v, because in two dimensions, it is a circle that corresponds to vectors with a constant magnitude, whereas in three dimensions it is the surface of a sphere. When we now write this probability function in terms of
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energy, we see that the two factors of v and 1/sqrt(E) cancel out, and the final result is consistent with Boltzmann’s result. So what exactly goes wrong in Boltzmann’s approach? The reason why Boltzmann’s method fails is because it assumes energy is exchanged in packages of equal size dE. In Maxwell’s method on the other hand, we would have to assume that each velocity component comes in packages of equal size dv. If the velocity
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components come in packages of equal size dv, then energy does not come in packages of equal size dE, because energy depends on the square of the velocity. If we assume that the energy states are separated equidistantly, we therefore overcount the higher energy states. On the other hand, in higher dimensions, there are more high energy states, because more velocity vectors correspond to the same level of energy. Therefore, in higher dimensions
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Boltzmann’s method undercounts the higher energy states. In two dimensions these two errors happen to cancel each other out exactly, so that Boltzmann’s method gives the same result as Maxwell’s method. Nonetheless, Maxwell’s method and Boltzmann’s method both predict the exponential dependence on energy. This factor is called the Boltzmann factor. The extra factor that is needed to properly count the energy states, is called
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the density of states. Now let’s see how we can make Boltzmann’s method work by assuming equidistant velocity components, as opposed to equidistant energy states. We have a system with a fixed number of molecules N. Instead of units of energy, we now distribute units of velocity components over the molecules. For each particular distribution, we can count how many molecules have a certain velocity vector. The number of ways to achieve a certain distribution is given by the multiplicity omega.
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To find the most likely macrostate, we must maximize the multiplicity omega, which is equivalent to minimizing its denominator, which is the same as minimizing its logarithm. The logarithm of a factorial is approximated using Stirling’s approximation. We have to minimize this function under the constraint that the total energy of the system is fixed, and the total number of molecules is fixed. This constrained optimization problem is solved
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by introducing Lagrange multipliers. We take the derivative of the Lagrangian with respect to the optimization variables, and set it equal to 0. We find that the optimization variable depends exponentially on the square of the velocity vector. This is exactly the starting point of Maxwell’s method, so the final results will be the same as well. In the final result, we find that the Lagrange multiplier beta is equal to 1/k_B T. This
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Lagrange multiplier beta is sometimes also called the thermodynamic beta. We have modeled an ideal gas as molecules randomly moving around and bumping into the walls of the container. This way, we explained the ideal gas law, we explained how internal energy is related to kinetic energy, and we explained using the equipartition theorem how the heat capacity depends on the degrees of freedom of each molecule. Using Boltzmann’s
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method of counting states, we can reproduce the molecular velocity distribution that Maxwell first found through arguments of symmetry. We will now see how we can interpret entropy in terms molecules and microstates. First recall how Clausius originally defined entropy. Based on his analysis of Carnot’s reversible heat engine, which makes it the most efficient heat engine theoretically possible, Clausius defined a change in entropy of a system, as the heat Q that is transferred
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into a system, divided by the temperature of that system. Clausius stated that for spontaneous processes, the entropy increases. According to Boltzmann’s model, systems evolve to the macrostate that has the highest multiplicity omega, because it is statistically the most likely macrostate. So according to Clausius, the entropy S spontaneously tends to a maximum, and according to Boltzmann, the multiplicity omega spontaneously tends to a maximum. Can
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we find a connection between the entropy S and the multiplicity omega? Ultimately, we want to define the entropy S in such a way that it adequately describes spontaneous, irreversible processes. So let’s have a look at a some spontaneous processes in a closed system that we want to describe. One spontaneous process is the change in the velocity distribution of the molecules. Suppose we start with a state where all the molecules
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have the same velocity vector. As the molecules move around and bounce into each other, eventually they will end up following a Maxwellian velocity distribution that we derived previously. So the molecules in a system spontaneously evolve from a non-Maxwellian distribution to a Maxwellian distribution. Another example of a spontaneous process is the change in position distribution of molecules, also known as diffusion. Suppose that all
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molecules start off lumped together in a corner. As they randomly move around, they end up spread out through the entire container. So the molecules in a system spontaneously evolve from a non-uniform distribution to a uniform distribution. We should be able to describe both these processes in terms of entropy and multiplicities of macrostates. We already saw how to describe the velocity distribution in terms of macrostates and multiplicities.
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To define the macrostate, we count how many molecules have a certain velocity vector. To find the most likely macrostate, we maximize the expression for the multiplicity, which is equivalent to maximizing the negative logarithm of its denominator. This analysis requires that we discretize the velocity vectors that a molecule can have, but in reality the velocity of a molecule is a continuous variable. We then obtain a
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continuous probability distribution function, and the number of molecules that have a velocity within a certain interval is given by the total number of molecules, times the probability function, times the velocity interval. We can plug this into the expression for the logarithm of the multiplicity. That takes care of the velocity distribution of the molecules, but what about the position distribution? We then require a probability distribution
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function that specifies the probability that a molecule has a certain velocity vector, while it is also at a certain location in space. If the system has evolved to its equilibrium state, the velocities satisfy a Maxwellian distribution, while the molecules are spread out uniformly throughout space. So the probability distribution function for both velocity and position is given by Maxwell’s distribution times 1 divided by the volume of the container.
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How are we now going to relate this quantity to Clausius’ definition of entropy? Let’s say we have a box of gas with a fixed volume, and put in some heat. Because we assume fixed volume, Delta V is 0, the box performs no work through expansion, which means according to the first law of thermodynamics that all the heat energy that is put in goes into raising the internal energy of the gas. So according to Clausius’ definition of entropy, the
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change in entropy is given by the change in internal energy, divided by the temperature of the gas. Dividing both sides by Delta U tells us that the derivative of the entropy with respect to the internal energy is given by 1 divided by the temperature. We know from the equipartition theorem that for an ideal monatomic gas the internal energy is given by 3/2 N k_B T. By using the chain rule, we express dS/dU
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in terms of dS/dT. If we equate this expression to 1/T, we find that according to Clausius’ definition of entropy, for an ideal monatomic gas in a box of fixed volume, the derivative of entropy with respect to temperature is 3/2 N k_B / T. To find the relation between entropy and the logarithm of multiplicity, let’s see how their derivatives with respect to temperature compare to each other. To find the expression for ln omega, let’s first
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write down the expression for Nf ln Nf. We can plug in the expression for Maxwell’s distribution, and then we can calculate the integral. Integrating over the volume gives a factor V, so the factor N/V becomes N. The kinetic energy times Maxwell’s probability distribution, integrated over all velocities, gives the average kinetic of a molecule. Dividing it by k_B T gives 3/2. The other terms do not depend on velocity, so integrating Maxwell’s
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probability distribution over all velocities simply reduces to 1. If we lump all constant terms together in a single constant C, we can simplify the expression for ln omega. We can use this expression to calculate the derivative of ln omega with respect to temperature. This yields a value of 3/2 N/T, which differs only by a factor of k_B from the value for dS/dT. So if we multiply ln omega with k_B, the two expressions match. So we find that
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if we define entropy as Boltzmann’s constant times the logarithm of the multiplicity of the equilibrium macrostate, the definition is consistent with Clausius’ definition of entropy. With this definition, we see that if a system tends to the macrostate with the highest multiplicity, so with the highest statistical likelihood, it tends to the state with highest entropy, as Clausius postulated. Note however, that
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this is not the only definition of entropy that is consistent with Clausius’ definition. We made the definitions consistent with each other by considering the derivative with respect to temperature, so if we redefine entropy by adding any function that is independent of temperature, the new definition is still consistent with Clausius’ definition. This is an important point to keep in mind for later discussions, where we consider the expression
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for entropy for systems that can exchange particles. From the preceding analysis, we can also learn about the relevance of quasi-static processes. Our analysis heavily relied on the use of Maxwell’s velocity distribution, which is only valid for a gas that has settled into an equilibrium state. We used this equilibrium distribution to define entropy and have it match with Clausius’ definition of entropy, which involves a derivative with respect to internal energy. This means that as we change
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the energy of the system, the system always remains in an equilibrium state. Technically this is a contradiction: how can a system be in equilibrium, yet at the same time changing? Such a process is called quasi-static, and in practice it means that the system is changing so slowly that it never deviates too much from equilibrium. For example, if we have a system at a certain temperature, and we want to heat it up quasi-statically,
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it means we have to change the temperature in such small steps, that the system always remains close to equilibrium. Only in this case can we compute the total change in entropy using Clausius’ definition, so by dividing the transferred heat by the temperature. By contrast, consider the case where we heat up the gas in a non quasi-static manner. That is, we change the temperature of the environment suddenly, and then wait for the gas to settle
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into equilibrium. In this case Clausius’ definition cannot be used to calculate the total change in entropy of the gas. So we have these two different processes to heat up a gas. In one process you can use Clausius’ definition to calculate the total change in entropy, in the other you cannot. Nonetheless, in both cases the system starts and ends in the same states. Since the entropy is defined by the state of the system, the
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total entropy change in the system must be the same for these two processes. The difference is that in the non quasi-static process, entropy is generated by heat transfer, but then additional entropy is generated by the system settling into equilibrium, as it evolves from a macrostate with low multiplicity, to a state with high multiplicity. The combination of these two contributions yields the total entropy change of the system.
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So generally speaking, the change in entropy is larger than or equal to Q/T as defined by Clausius, and equality only holds for quasi static processes. So the total change in entropy of the system is eventually the same for quasi-static processes and non quasi-static processes. However, if we look at the total entropy change of both the system and the environment, we find that in the non quasi-static case the environment loses less entropy, because it
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transfers heat at a higher temperature. Therefore, a non quasi-static process generates more entropy in total: in the system and the environment combined. Another difference between the two processes is that the quasi-static process is internally reversible. If we start with the final state and cool it back down to the initial state quasi-statically, then the system goes through the same states as it did heating up, so it’s
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internally reversible. But if we cool down the system non quasi-statically, it doesn’t go through the same states as it did when we heated it up non quasi-statically. So the non quasi-static process is not internally reversible. So if we want to use Clausius’ definition to calculate the entropy change of a system, we have to specify that we consider the heat transfer for an internally reversible process, even if in practice the process was not quasi-static or internally reversible.
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Let’s now take some time to have a look at a few sidenotes, the first one being about exponential distributions. When we randomly distributed the total energy over a fixed number of molecules, we divided the total energy into discrete units, and then distributed those over the molecules by randomly placing partitions. The distance between two adjacent partitions indicates the energy that one molecule possesses. We found that the probability that
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a molecule possesses a certain amount of energy follows an exponential distribution. This is a very important result in statistical mechanics, but it is also used more broadly. For example, suppose that instead of units of energy, we distribute units of time. The partitions now represent random, uncorrelated events that are separated in time. The time interval between two consecutive events is then by the same logic also exponentially
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distributed. In physics, this result explains why particle lifetimes follow exponential distributions. Outside of physics, the exponential distribution is used to model the lifetime of a product. In queuing theory the result is used to model the time interval between two customers entering, or the time it takes to serve a customer. The second sidenote is about the physical relevance of Lagrange multipliers. Recall
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that according to Clausius’ definition, dS/dU equals 1/T if the volume of the system is fixed. If we plug in Boltzmann’s expression that relates entropy to the multiplicity of a macrostate, we find that d ln omega / dU equals 1/k_B T. This value of 1/k_B T is significant, because we had encountered it earlier as a Lagrange multiplier: we had a constrained optimization problem that we solved with Lagrange multipliers, and we found that the Lagrange
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multiplier for the energy constraint has the value 1/k_B T. Is this a mere coincidence? It turns out that this not a coincidence, but rather it is a general property of Lagrange multipliers. The physical meaning of the value of a Lagrange multiplier, is that it indicates by how much you can change the objective function, if you change the value of a constraint. This insight is particularly useful in for example economics. Let’s say you have a limited
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budget that you can invest to make profit. Of course you’d want to maximize your revenue, for a given budget: this is a constrained optimization problem that can be solved using Lagrange multipliers. Now you might be interested in the following question: suppose I had a slightly larger budget. How much more revenue would I be able to make? The answer to this question, is given by the value of the Lagrange multiplier. Let’s see why this is the case. Let’s say x and y are resources that you buy at
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a price f(x,y) to make revenue R(x,y). With a limited budget B you can buy a limited amount of resources, and within those limits you want to maximize your revenue. To solve this optimization problem, we introduce a Lagrange multiplier lambda. By setting the derivatives to 0 we find the x, y, and lambda that give the maximal revenue R for a given budget B. Note that the Lagrangian evaluated in the optimal point gives the optimal value for
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the revenue, because the constraint is satisfied, so the constraint term vanishes. So to see by how much our maximal revenue changes if we change the budget, we can take the derivative of the Lagrangian with respect to the budget. To calculate this derivative, we must apply the chain rule, because the optimal x, y, and lambda depend on the budget B. However, because x, y and lambda are such that the Lagrangian is at an optimum, their derivatives
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must be 0. We are left with the partial derivative of the Lagrangian with respect to B, which is equal to lambda. This is the result we were looking for, the Lagrange multiplier lambda indicates by how much the optimal value of the objective function changes if you change the value of the constraint. We will later see how other thermodynamic variables can be interpreted as Lagrange multipliers. Another important sidenote is one about distinguishability.
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When we distributed units of energy, we did so by inserting partitions. But why would this be the correct way to randomly distribute the energy? For example, instead of modeling molecules as partitions, let’s model them as boxes. Each molecule is a box, and one by one we’re going to randomly distribute the units of energy among the boxes. This may seem like another reasonable way to randomly distribute energy over the molecules. But
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this method results in a different probability distribution, so it gives different physical predictions. So only one of the two methods can be correct, but not both. So what is the difference between these two methods, and why is one correct and one incorrect? The difference is that in the second approach, you assign the units of energy a certain order. Each unit of energy gets a label that says whether they are the first to be assigned
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to a molecule, or second, or third, etc. In other words, the particles are distinguishable. When you distribute the units of energy over the boxes, both may end up in the first box, or both may end up in the second box. But if each box ends up having one unit of energy, it matters whether the first unit of energy went to box 1 and the second to box 2, or whether the first unit of energy went to box 2 and the second to box 1.
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In the first method however, the particles are indistinguishable. All the units of energy lie along the energy axis, and when we place the partitions, we don’t consider all the different ways we can order the units of energy along the axis. In this case, either both units of energy end up in box 1, or in box 2, or each box has 1 unit of energy, but we don’t care about which unit of energy ends up on which side of the partition. As a result,
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in the case of distinguishable particles, there is a 1 in 2 chance that both molecules have one unit of energy. But in the case of indistinguishable particles, there is a 1 in 3 chance that both molecules have one unit of energy. So now we understand why it matters whether we choose the packages of energy to be distinguishable or indistinguishable. But how do we know which choice is correct? Keep in mind that in the end, we want to model
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the fact that molecules randomly collide with each other and transfer kinetic energy during the collision. But as the kinetic energy is transferred, it doesn’t make sense to ask ‘which part’ of the kinetic has been transferred. So if, for mathematical convenience, we model kinetic energy as discrete packages of energy, we also shouldn’t care which package in particular has been transferred during the collision. So the energy packets are indistinguishable.
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The key point here is that the discretization of energy was merely a mathematical trick, with no physical relevance. However, several decades later, when quantum physics was being developed, Einstein proposed that there is in fact a physical reality to discrete packets of energy. In order to still obtain the correct physical results, we have to assume that there are discrete particles, which are also fundamentally indistinguishable. As we saw, whether particles
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are distinguishable or indistinguishable affects the statistical properties of the system. So the notion of physical indistinguishable discrete particles led to new statistics such as Bose-Einstein statistics. A final sidenote is to be made about the role of phase space. We saw previously that Boltzmann’s first attempt to divide the energy into equal parts led to results that were inconsistent with Maxwell’s velocity distribution. Rather
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than energy, we had to divide velocity and space into equal parts. Therefore, we defined a probability distribution function that depends on position and velocity. Specifying both the position and velocity, or momentum, of a particle is typically done with the help of phase space. In phase space, position is plotted on one axis, and momentum is plotted on the other axis. If we’re going to count the number of molecules
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that have a certain position and momentum, we must define a finite interval of position and momentum. Because if we treat position and momentum as a continuous variable, then the probability that a particle occupies one particular point with perfect precision is 0. Choosing an interval of finite size is called coarse graining. Let’s see how the resolution with which we apply coarse graining, affects the way we calculate entropy. Suppose
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we have 10 molecules, and they all lie within the same region of phase space. In that case the multiplicity is 1, because there is only one way to distribute 10 molecules over one box. But now let’s increase the resolution of phase space. Let’s say that upon closer inspection, 4 of the molecules are in the first quadrant of the phase space region, 1 is in the second quadrant, 3 are in the third quadrant, and 2 are in the fourth quadrant.
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Using the formula for calculating the multiplicity of a macrostate, we find that there are many more ways to distribute the 10 molecules in this way over the four quadrants. So simply by increasing our resolution of phase space, we have increased the value of entropy. Now let’s increase the resolution so much, that there is only one molecule per box in phase space. The multiplicity of the state will then always be the same, because even as the
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system evolves in time, there will always be only 1 molecule per box in phase space. But this would mean that entropy can never increase, whereas the second law of thermodynamics says entropy should tend to a maximum. Coarse-grained entropy, however, can increase over time. Suppose that in the initial state the molecules are lumped together in phase space, and after some time they spread out. In the initial state the molecules may cover
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only one coarse-grained region, but in the final state they may cover multiple regions. So coarse graining is necessary to describe an increase in entropy over time. But then the question is: what should be the size of the coarse-grained region? The answer to that question didn’t come until quantum mechanics was developed: Heisenberg’s uncertainty principle states that the region a molecule occupies in phase space cannot be determined
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with arbitrary precision. This puts a limit on the resolution we can use in phase space. If we look at how the system evolves in coarse-grained phase space, we see that as the molecules cover a larger region in phase space, the multiplicity and entropy become higher. So we can try to express entropy in terms of the phase space volume that is accessible by the system of molecules. Recall that the multiplicity omega denotes the number of microstates
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that correspond to the same macrostate. To find the equilibrium state, we must find the macrostate with the highest multiplicity. To find the probability to observe that macrostate, we must calculate the total number of microstates J, and then divide Omega_max by J. For systems with a large number of molecules, the probability that the system tends to the equilibrium macrostate is practically 1. This means that instead of calculating the multiplicity of the most
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likely macrostate, we can also calculate the total number of microstates. This is why it is justified to calculate entropy by computing the total accessible phase space volume. First, as an example, let’s calculate the accessible phase space volume for a single particle. Its position in real space is restricted to a volume V, and its momentum is restricted to the shell of a sphere in momentum space, whose radius depends on the total energy U.
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To count the states in this phase space volume, we coarse grain the phase space into regions with a finite size h^3. If we assume knowledge of quantum mechanics, we can assume h to be Planck’s constant, otherwise it’s just an arbitrary constant to make sure we end up with the correct units. Then the total number of accessible states in phase space is the real-space volume times the momentum-space volume (because the position and momentum
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of the particle can be chosen independently), divided by the volume of a single coarse-grained region. We can write the momentum-space volume in terms of the total energy U, by using the formula for the surface of a sphere. Now let’s see what happens if we don’t just have a single particle, but instead we have a large number of particles N. Each particle lies somewhere in the real-space volume V. Because the positions of the N particles can
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be chosen independently, the accessible volume in 6N-dimensional phase space, is V^N. The momenta of the N particles are such that the total energy is a fixed value U. In the 1-particle case, we saw that this corresponds to the surface of a sphere in three-dimensional momentum space. For the N-particle case, it corresponds to the surface of a hypersphere in 3N-dimensional momentum space, which can be calculated using a formula that depends on the radius r and
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the dimension d. In our case, the radius is sqrt(2mU), where U is the total energy of the system, and the dimension is 3N. Now we divide the phase space volume into coarse-grained regions, each with volume h^3N. The total number of accessible microstates is given by the real-space volume times the momentum-space volume divided by the volume of one coarse-grained region. We can plug in the expression for the surface of the momentum
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hyper-sphere, and simplify the expression by applying Stirling’s approximation. To calculate the entropy, we take the logarithm and multiply with Boltzmann’s constant. This gives an expression for the entropy of a monatomic ideal gas with N molecules and a fixed total energy U. We now have calculated the entropy of a monatomic ideal gas twice: once using Boltzmann’s method of combinatorics, and now using coarse-grained
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phase space volume. Let’s compare the two to each other. If we put together all the constant terms into some constant C, we see that the two formulas are the same, except for an extra occurrence of the number of molecules N. Even though the expressions are different, they are both consistent with Clausius’ definition of entropy, because both expressions have the same derivative with respect to the temperature. But it leaves the question: which
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of these expressions is correct? In what way should entropy depend on the number of molecules N? To answer this question, we consider the Gibbs paradox. The paradox is as follows. Suppose we have two identical containers of gas, with the same number of molecules, volume, temperature, and therefore also the same entropy S_0. The two containers are separated by a partition. The total multiplicity of the two systems
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together is given by the product of the separate multiplicities, because for each one of the Omega_1 microstates of the first system, the second system can be in any one of the Omega_2 microstates of the second system, because the two systems are fully independent. The entropy is given by the logarithm of the multiplicities, so a product of multiplicities is a sum of entropies. Now let’s see what happens if we remove the partition.
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First consider the formula obtained using the phase space volume. If the partition is in place, the total entropy is twice the entropy of a single container. If we remove the partition, N becomes 2N, and V becomes 2V. If we plug this into our formula for entropy, we see that by removing the partition, the entropy has increased by a certain amount. Now we can do the same thing for Boltzmann’s expression for the entropy. If the partition is in place,
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the total entropy is twice the entropy of a single container. If we remove the partition, N becomes 2N, and V becomes 2V. We see that now, the total entropy does not change when we remove the partition. So to determine which expression for entropy is correct, we must ask: should the entropy increase when removing the partition, or not? To answer that question, let’s recall Clausius’ original motivation to introduce the concept
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of entropy. He introduced entropy to quantify the degree to which a system has irreversibly transformed. So we must ask: does our system undergo an irreversible transformation when we remove the partition? If so, the entropy must increase, otherwise the entropy must remain the same. Whether an irreversible transformation has taken place, depends on the level of detail with which we observe our system. Let’s say we keep track of each molecule individually.
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After removing the partition, each individual molecule moves around randomly, and if we insert the partition again it is highly unlikely that by pure random chance each individual molecule happens to be on the same side of partition as it originally was. In that sense, removing the partition causes an irreversible transformation, so the entropy should increase, as predicted by the phase space volume formula. However, practically we cannot keep track
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of each molecule individually, and we typically wouldn’t care either. All we’re interested in are the macroscopic properties of the system, and these don’t change if you remove or re-insert the partition. So according to this view, no irreversible change has taken place when removing the partition, so the entropy shouldn’t increase. This is what Boltzmann’s formula predicts. But let’s say that the two gases on either side of the partition
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are different gases, which we can easily distinguish practically, and whose properties are relevant to us. When we remove the partition, the two gases mix, which is an irreversible transformation because the two gases don’t spontaneously unmix when we re-insert the partition. So in this case, the entropy should increase when removing the partition. Using Boltzmann’s expression for the entropy, an increase in entropy is achieved by calculating the entropy
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for the two gases separately, both before and after the partition is removed. After the partition is removed, the only thing that changes for each gas is that the volume V becomes 2V. If we plug this in, we see that indeed removing the partition causes the entropy to increase. So we see that the phase space volume approach gives an unwanted result, because it assumes the molecules are distinguishable, whereas
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in practice we cannot or don’t care to distinguish separate molecules. How can we change the derivation of the phase space volume formula so that it properly takes into account the practical indistinguishability of the molecules, and so that it gives the correct formula for entropy? In the original derivation, we counted the total number of microstates by calculating the accessible phase space volume and dividing by the volume of the coarse-grained regions.
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Making the molecules indistinguishable means saying that we don’t care which molecule occupies which coarse-grained region. That is, for a given microstate, changing the labels of the molecules gives the same microstate, and we should not count them double. We can change N labels in N factorial ways, so dividing by N factorial corrects for the double counting. To see how this affects the formula for entropy, we apply Stirling’s approximation and take
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the logarithm. Rearranging the terms gives the entropy for an ideal monatomic gas, which is also known as the Sackur-Tetrode equation. Now we can ask: why did Boltzmann’s formula for the entropy already seem to take into account the indistinguishability? We did not seem to have any such assumption explicitly during its derivation. Recall that to find the most likely macrostate using Boltzmann’s method, we maximized the multiplicity omega.
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But when we did so, we ignored the factor N factorial, because it is constant in a closed system. So we didn’t consciously take into account indistinguishability, but we just happened to have made a lucky simplification in the derivation. As a final remark, it’s important to realize the difference between particles that are practically indistinguishable, and particles that are fundamentally indistinguishable.
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The answer to the Gibbs paradox was that practically we don’t care about what happens to each individual molecule: it’s possible in principle to keep track of each molecule, but why should we? This is different from the reason why we considered the discrete energy packets to be indistinguishable. The energy packets are fundamentally indistinguishable, because if kinetic energy is transferred between molecules, it makes no sense to speak of ‘which part
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of the kinetic energy’ has been transferred. This fundamental indistinguishability is a part of quantum theory. Some sources may claim that this fundamental quantum indistinguishability was what solved the Gibbs paradox, but that claim is contested: ”Quantum theory did not resolve any paradox, because there was no paradox”. We have now seen what entropy means on a molecular level, and how we can calculate entropy for
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a monatomic ideal gas. What we will do now, is to see how thermodynamic quantities can be rederived using statistical mechanics as the fundamental starting point. More specifically, if we can calculate the entropy of a system, simply by counting the possible microstates, we want to be able to calculate properties like temperature and pressure from the entropy. To see how quantities like temperature and pressure can be derived from entropy, let’s
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consider the ideal monatomic as an example. We already know the expression for the entropy of a monatomic ideal gas. The fundamental assumption that we’re going to use, is that in a closed system, entropy tends to increase to a maximum. This increase in entropy can for example be achieved by heat flowing from a high temperature to a low temperature. The total entropy is given by the sum of entropies of the two separate
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systems. The exchange of heat energy must be such that the total entropy will be maximized. So let’s see how the total entropy changes as heat flows from A to B. The change in total entropy is due to the change in entropy of system A, and the change in entropy of system B. We know that whatever energy system A gains, system B must lose, so dU_B equals minus dU_A. Dividing both sides by dU_A gives the derivative of the total entropy with respect to the energy
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in system A. For the entropy to be maximal, the derivative must be 0. So for the system to be in equilibrium, and to have no further heat exchange take place, both systems must have the same value for dS/dU. So dS/dU appears to be an important quantity to describe heat flow between systems. More specifically, it is the derivative of entropy with respect to energy at constant volume and number of particles. Using the expression
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for the entropy of an monatomic ideal gas, and using the equipartition theorem, we find that dS/dU is one divided by temperature. Since systems want to maximize their entropy, the derivative of entropy with respect to energy can be interpreted as the eagerness to gain energy: a high derivative means that the system can increase its entropy a lot when it takes in energy, so it is very eager to take in energy. So a system with a low temperature is very eager to take in energy, and a system with
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a high temperature is less eager to take in energy. So cold systems tend to take in heat from hot systems, and the transfer of heat stops when both systems have equal temperature. Note that this is basically Clausius’ definition of entropy, so we were already familiar with this expression. But whereas Clausius used this expression to define entropy, we now consider entropy to be the more fundamental quantity on the basis of statistical mechanics,
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and we use it to define temperature. Now let’s consider two gases that have a movable partition between them, so they can exchange volume. Again we can ask what exchange of volume must take place in order to maximize the entropy, and if the volume is the only thing that is changing, we will find that the derivative of entropy with respect to volume must the same for both systems. The derivative is taken for a constant number
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of particles and constant energy. Note that the constraint of constant energy is not a trivial one, because if you for example compress a gas adiabatically, its energy can increase. Using the expression for the entropy of a monatomic ideal gas, and using the ideal gas law, we find that the derivative of entropy with respect to volume at constant energy and constant number of particles, is given by the pressure divided by temperature. So
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pressure divided by temperature can be interpreted as the eagerness to gain volume, given a constant number of particles and constant energy. Recall that another spontaneous irreversible process that we wanted entropy to describe, is the diffusion of particles. So if we have two compartments A and B that can exchange particles, we can again ask when an equilibrium is reached. An equilibrium is reached when the entropy is maximal, and dS/dN is the same
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for both containers. More specifically, we take the derivative of entropy with respect to the number of particles, assuming that the energy and volume stay fixed. Again this is not a trivial constraint, because if a particle is transferred from one compartment to another, it usually carries kinetic energy with it. This derivative is used to define a new quantity, called the chemical potential. In analogy with pressure, we define the chemical
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potential such that dS/dN is the chemical potential divided by temperature. The derivative dS/dN can be interpreted as the eagerness of a system to gain particles. We include a minus sign in the definition so that the chemical potential mu intuitively acts like a potential. That is, particles tend to flow from high chemical potentials to low chemical potentials. We have now used the derivatives of entropy to define temperature, pressure, and chemical potential. These partial derivatives can be
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used to express a total change in entropy. Multiplying both sides by the temperature T, and solving for the change in energy dU, yields the equation that is known as the fundamental thermodynamic relation. These partial derivatives also provide another valuable insight, namely that they are all Lagrange multipliers. Recall that the Lagrange multiplier of a certain constraint, indicates by how much the optimal value of the objective function changes when
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the value of the constraint is changed. So all these partial derivatives correspond to Lagrange multipliers of constraints under which the entropy is maximized. We already saw that 1/k_B T is the Lagrange multiplier for the energy constraint. The constraint tells us that there is a fixed amount of energy that is distributed over the molecules. If we neglect the density of states, we found that the number of molecules
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that have a certain energy follows an exponential distribution, where the exponent contains the Lagrange multiplier. Similarly then, pressure divided by temperature must be the Lagrange multiplier for the volume constraint. The molecules are distributed over a fixed amount of volume, so if we randomly distribute the molecules over space, then for a one-dimensional volume the number of particles that have a certain volume is exponentially distributed,
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where the exponent contains the Lagrange multiplier. The average of the exponential distribution is then given by one over the Lagrange multiplier, which according to the ideal gas law is the total volume divided by the number of molecules, as you’d intuitively expect. The chemical potential divided by the temperature is the Lagrange multiplier for the number of particles constraint. Different types of particles have their own constraint and their
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own Lagrange multiplier, so their own chemical potential. Particles belong to a different type, whenever they are practically distinguishable, so for example if they are different substances, or if they are in different compartments. Having introduced the new concept of chemical potential, let’s see how it can be used in the context of chemical reactions. We start with the fundamental thermodynamic relation for two types of molecules, A and B. Let’s
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first consider the case of a system at a fixed temperature and fixed volume. We saw in a previous video that in such a system, the Helmholtz free energy determines whether a process occurs spontaneously: the system spontaneously tends to minimize its Helmholtz free energy. We can write the expression for a small change in the Helmholtz free energy dF, by applying the product rule to the term TS. Then we can substitute the expression
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for dU, where the term TdS will cancel. We have now found an expression for the change in the Helmholtz free energy. Note that dV and dT will be zero if we assume a system of fixed temperature and volume. Now let’s suppose molecules of type A can be converted to molecules of type B. For example, type A could be water molecules in the liquid phase, and type B could be molecules in the vapor or steam phase. In more complicated chemical
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reactions, more types of molecules or particles can be involved. We see that the chemical potentials of the two substances can be defined as the partial derivatives of the Helmholtz free energy. The system will tend to a state where the Helmholtz free energy is minimal, so its derivative vanishes. If type A gains dN particles, then type B must lose dN particles, so dN_B equals -dN_A. Substituting the definitions of chemical
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potential, we find that the systems tends towards an equilibrium state where the chemical potentials are equal. This makes sense intuitively: when there’s no variation in potential, there’s no driving force to change the system. The second example we consider is an open container, where the volume can change, and the temperature and pressure are fixed. We saw in a previous video, that in this case the Gibbs energy tends to a minimum. We can
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write the expression for a difference in the Gibbs energy by applying the product rule to PV, and to TS. If we plug in the expression for dU, we see that the term TdS cancels, and the term PdV cancels. We end up with an expression for the change in Gibbs energy, that contains a change in pressure and a change in temperature. But we assumed that pressure and temperature are constant, so these terms vanish.
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We see that we can write the chemical potentials as partial derivatives of Gibbs energy, assuming fixed pressure and temperature. Because the system spontaneously tends to a state with minimal Gibbs energy, we can find the condition for the equilibrium state by setting the derivative of the Gibbs energy equal to 0. If type A gains dN particles, then type B must lose dN particles, so dN_B equals -dN_A. Substituting the definitions of chemical potential, we
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again find that the system tends towards an equilibrium state where the chemical potentials are equal. So far we have considered systems with large numbers of particles. An ideal gas consists of so many particles, that we can apply Stirling’s approximation and make meaningful statistical predictions. Moreover, when we derived the statistics for an ideal gas, we assumed that the system is isolated, so that its total energy is fixed. What we’re going to do
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now, is to consider systems that don’t necessarily contain lots of particles, and that can exchange energy and particles with an environment, or reservoir. We will see that we can still make statistical predictions, by considering the entropy change of the reservoir. Let’s say that we have a reservoir that contains oxygen molecules. In it, there’s a system, which is a red blood cell, to which an oxygen molecule can bind itself. Let’s
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say that if no oxygen molecule is bound to the blood cell, the energy of the system is 0, as a reference. If an oxygen molecule binds to the blood cell, the energy of the system becomes some negative value epsilon. We now want to find the probability that an oxygen molecule will bind to the blood cell, and to do that, we will consider the entropy change of the reservoir. First consider the state where no oxygen molecule is bound to the blood
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cell. Then the energy of the system is 0. The number of oxygen molecules in the system is also 0. In this state, the reservoir has a certain entropy S_0. Now consider the state where an oxygen molecule is bound to the red blood cell. The energy of the system becomes epsilon. The number of oxygen molecules in the system is 1. The system and the reservoir together form a closed system, so due to the conservation of energy,
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if the energy of the system has become epsilon, the energy of the reservoir must have changed by minus epsilon. Similarly, the total number of oxygen molecules is conserved, so if the system has gained an oxygen molecule, then the reservoir must have lost an oxygen molecule. We saw previously the formula for the change in entropy as a function of change in energy and change in number of particles. So if the change in energy is minus epsilon, and the
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change in number of particles is -1, then the change in entropy of the reservoir is its chemical potential, minus the binding energy epsilon, divided by the temperature. The new entropy of the reservoir we call S_1. Now let’s see what the probabilities for the two states are. The ratio of the probabilities, is the ratio of the multiplicities, because the fundamental assumption of statistical mechanics is that each microstate is equally
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likely. From Boltzmann’s definition of entropy, we know that the multiplicity is given by the exponential of the entropy divided by Boltzmann’s constant. Dividing these two exponentials gives the exponential of the difference between the two entropies. And since we just found the expression for the difference in entropies, we now know that the ratio of probabilities is given by e^(- {epsilon-mu} / k_B T). So this is the ratio of probabilities, but what are the individual probabilities? To
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find those, we use the fact that the two probabilities must add up to 1. Then we can solve for the probability that no oxygen molecule is bound to the blood cell, and the probability that an oxygen molecule is bound to the blood cell. Now let’s see if these expressions make sense intuitively. It says that if the binding energy is lower, the probability is higher that an oxygen molecule is bound to the blood cell. This makes sense, because binding then
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becomes energetically more favorable. It also says that if the chemical potential is higher, then it is more likely that an oxygen molecule binds to the blood cell. This also makes sense, because the more oxygen molecules there are, the more likely it is that one binds to the blood cell. Moreover, at very high temperatures, the probability that an oxygen molecule is bound to the blood cell becomes 50/50, so a higher temperature
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seems to correspond to more randomness in the system. This appears consistent with the expressions for Helmholtz and Gibbs free energy. These free energies spontaneously tend to a minimum, and for higher temperatures, it means maximizing entropy, even if it goes at the expense of increasing internal energy. We have seen what happens if a blood cell is surrounded by oxygen molecules. But what happens if there are also molecules of a different
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type, such as carbon monoxide? What will be the probabilities that the blood cell binds with oxygen, carbon monoxide, or nothing? Let state 0 be the state where the blood cell binds with nothing. In this state, the system has a reference energy of 0. In state 1 the blood cell binds with an oxygen molecule, causing the system to have energy epsilon_1. In state 2 the blood cell binds with a carbon monoxide molecule, causing the system to have
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energy epsilon_2. So in state 0 the system energy is 0, the number of molecules is 0, and the entropy of the reservoir has some reference value S_0. In state 1 the system energy is epsilon_1, there is one particle of type 1, oxygen, and zero particles of type two, carbon monoxide. By conservation of energy and particles, it means that the energy of the reservoir has changed by minus epsilon_1, and the number
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of particles of type 1 has decreased by 1 in the reservoir. Using the fundamental thermodynamic relation, we can calculate the change in entropy of the reservoir, and with that we can find entropy of the reservoir S_1. We can do the same exercise for state 2 of the system. To calculate the ratios in probability, we use the fact that the probability is proportional to the multiplicity of the macrostate, which is the exponential of the entropy. Therefore,
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the ratio in probabilities is the exponential of the difference in entropy, divided by Boltzmann’s constant. If we calculate the ratios for the two different states, and we require that all the probabilities should add up to 1, we can calculate the individual probability by defining the normalization constant Z, and dividing the exponential for each state by this fixed constant. The exponentials of the change in entropy for different states
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are called Gibbs factors. The normalization constant that is found by adding all Gibbs factors together, is called the Grand partition function. The probabilities for the different states are found by the dividing the Gibbs factors by the Grand partition function. More generally speaking, if you consider a system that can exchange particles with a reservoir, then for each microstate s of the system, you can calculate the Gibbs factor,
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which involves the system's energy, minus the number of particles in the system times the chemical potential of the reservoir, divided by the temperature of the reservoir. Summing these Gibbs factors over all microstates gives the Grand partition function. In case the system can only exchange energy, but no particles, with the reservoir, the Gibbs factors reduce to Boltzmann factors, and the Grand partition function is simply called the partition function.
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At this point it is useful to introduce a different definition for the entropy, which is called the Gibbs entropy, and we will see how it relates to the Boltzmann definition of entropy that we defined earlier. We defined the Boltzmann entropy using the logarithm of the multiplicity of a certain macrostate. That is, the system consists of many individual particles, each of which can be in a particular state ‘a’. n_a denotes the number of particles
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that are in state ‘a’. The multiplicity omega denotes in how many ways we can distribute capital N particles in such a way that each state ‘a’ has n_a particles. Now let’s consider a single particle in a system that is in thermal contact with a reservoir at temperature T. In this case, there is a certain probability P(a) that this single particle is in state ‘a’. If that same system now has many particles that act independently
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from each other, then we expect that the number of particles in state ‘a’, is approximately the total number of particles times the probability for a single particle to be in state ‘a’. We can rewrite this expression by taking the N outside of the summation, and writing the logarithm of a product as the sum of logarithms. The sum of all probabilities equals 1, which is constant, so that term can be neglected and absorbed in the arbitrary constant. What
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we find is that the Boltzmann entropy can be calculated using the probability distribution for a single particle, and then multiply with the total number of particles. For the Gibbs entropy we’re going to do something slightly different: instead of looking at the probability distribution for a single particle, we’re going to consider the probability for the multi-particle system as a whole. This probability function describes the probability that the
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entire system finds itself in a certain microstate s, where each individual particle has certain state s_j, which could be any of the single-particle states ‘a’. The Gibbs entropy is then defined using the sum over all system probabilities, rather than the single-particle probabilities. So let’s see how these two definitions of entropy are related to each other. To see how the Gibbs definition and Boltzmann definition of entropy are related, let’s consider a
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simple system consisting of only two particles, and each particle can be in only one of two single-particle states. We denote the state that particle 1 is in as s_1, so s_1 can be either a_1 or a_2. The state that particle 2 is in is s_2, so s_2 can also be either a_1 or a_2. The probability that a particle is in state a_1 or state a_2, is given by P(a_1) and P(a_2) respectively. This means that there are four possible system states
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in total. Either both particles are in state a_1, both are in state a_2, or the two particles can be in different states, with either particle being in state a_1 or a_2. Now let’s calculate the probabilities for the corresponding system states, assuming that the probability that a particle is in a certain state does not depend on the state of the other particle. The probability for the system to be in the first system state is then the probability
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for a single particle to be in state a_1, squared. To compute the Gibbs entropy, we have to multiply this probability with the logarithm of itself. We can write the logarithm of the product as the sum of logarithms. We can do the same exercise for all four system states: multiply the two single-particle probabilities with each other to find the system probability, multiply this with its logarithm, and then write the logarithm of the product as the
1:41:29
sum of logarithms. To find the Gibbs entropy, we must add all these results together. Let’s first collect all terms containing P(a_1) ln P(a_1). This term is multiplied by P(a_1)+P(a_2), because if one particle of the system is in state a_1, the other particle can be either in state a_1 or a_2. There is a factor of 2, because either of the two particles can be in state a_1. The same argument can be repeated to collect all terms containing P(a_2)
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ln P(a_2). This result can be simplified, by observing that the sum of all probabilities gives 1. The factor 2 comes from the fact that there are two particles, so in the more general case of N particles, this factor can be written as N. We are left with an expression that is the same as Boltzmann’s definition of entropy. So it turns out that the Gibbs definition of entropy and the Boltzmann definition of
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entropy are equivalent. However, in deriving this result we had to assume that the particles are non-interacting, so the probability of a system state can be written as the product of probabilities for single-particle states. If that is not the case, so the particles are interacting, then the two definitions are not equivalent. It has been demonstrated that in that case, it is Gibbs’ definition of entropy that is consistent with Clausius’ definition of entropy; Boltzmann’s definition is not.
1:43:07
Now that we know how we can calculate the entropy of a system using probabilities for system states as opposed to single-particle states, we can now look at how the partition function can be used to compute thermodynamic quantities is a convenient manner. Recall that the partition function is the sum of Boltzmann factors over all possible microstates of the system. The probability for the system to be in a certain microstate is given by
1:43:34
its Boltzmann factor divided by the partition function. Keep in mind that the system that we are considering is one that is in thermal contact with a reservoir of a certain temperature. This means that the system can exchange energy with the reservoir, so the energy of the system is not fixed, but rather it can fluctuate. Therefore, we cannot unambiguously define the system’s internal energy as we did when applying Boltzmann’s method of combinatorics,
1:44:04
but instead we can define the system’s average energy. This average value, or its expectation value, is given by summing the energies of all possible microstates multiplied by their probabilities. The probabilities can be expressed using Boltzmann factors and the partition function. From the expression for the partition function, we see that if we take its derivative with respect to beta, we get this factor of the energy
1:44:33
E in the summation. Using the chain rule, we can rewrite this expression as the derivative of the logarithm of the partition function. Now let’s see how we can use the partition function to calculate the entropy of the system. Note that here we use the Gibbs definition for entropy, where we use the probabilities of entire system states, rather than probabilities of single-molecule states. We can rewrite this expression using Boltzmann
1:45:03
factors and the partition function. One term can be rewritten using the average energy that we calculated earlier. The second term can be simplified by realizing that the sum of all probabilities equals 1. By plugging in the definition of beta, we find that the entropy of the system is given by its average energy divided by the temperature, plus Boltzmann’s constant times the logarithm of the partition function. This expression for the entropy
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can be plugged into the definition for the Helmholtz free energy. We find that the Helmholtz free energy is given by minus k_B T times the logarithm of the partition function. As an example, let’s calculate the partition function of an ideal gas, and verify that we can use the partition function to calculate the average energy and the entropy. First let’s calculate the partition function for a single molecule. The system is in thermal
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contact with a reservoir at temperature T, so the total energy of the system is not fixed. This means that the range of possible momentum values for the molecule is not restricted, and the summation over all momentum values therefore goes from minus infinity to plus infinity. To coarse-grain the continuous phase space variables x and p, we divide dx dp by h^3. This integral can be calculated analytically. In case we consider multiple distinguishable
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particles, the energy of the total system state is given by the sum of the squared momenta of the individual particles. The exponentials and the integrals can be factorized, so the partition function of the N-particle system can be written as the Nth power of the single-particle partition function. But recall from our discussion of the Gibbs paradox, that we want to consider the particles as practically indistinguishable. This means
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that we must correct for the double-counting of identical system states. If the state consists of N particles, then the N particles can be permuted in N factorial ways, so we have to divide the partition function by N factorial to correct for the double counting of identical states. We can plug in the expression of the single-particle partition function, and then rewrite the expression using the thermodynamic beta. Using the expression we found for the
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average energy, we find that it is equal to 3N / 2beta, which is 3/2 N k_B T. This is consistent with what we found using the equipartition theorem. Now, we can calculate the entropy using the average energy and the partition function. By approximating the N factorial using Stirling’s approximation, we find an expression for the entropy of an ideal monatomic gas, which is consistent with the Sackur-Tetrode equation
1:48:05
that we found earlier. Let’s now summarize the different sorts of systems that we have considered. The first type of system was one that is isolated, so the total energy, the number of particles, and volume are fixed. We assume that each microstate that satisfies these constraints is equally likely. The set of all possible microstates that satisfy these constraints, and their probabilities which are all equal, are referred to as the microcanonical
1:48:35
ensemble. To find the most likely macrostate, we find the macrostate with the largest number of microstates. To do this, we assume that the system consists of many microscopic degrees of freedom. For example, each velocity component of each single particle constitutes a degree of freedom. By assuming a large number of degrees of freedom, we can apply Stirling’s approximation. And by using Lagrange multipliers we find that the probability that a certain
1:49:06
degree of freedom assumes a certain value, is proportional to the Boltzmann factor. We also considered a system which is in thermal contact with a reservoir of a fixed temperature. In this case the number of particles in the system is still conserved, but heat energy can be exchanged. So in this case the constant variables are T, N, V. The total energy E was fixed in the microcanonical ensemble, but in this case, it can fluctuate. For this
1:49:38
system, the probability of each microstate is given by the Boltzmann factor divided by the partition function. The set of all microstates and their corresponding probabilities are referred to as the canonical ensemble. Lastly, we saw systems where both the energy and number of particles can fluctuate due to contact with a reservoir of constant temperature and chemical potential. So the constant variables are T, mu, and V. The probability for each
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possible microstate of the system is given by the Gibbs factor divided by the grand partition function. The set of all possible microstates and their corresponding probabilities are referred to as the grand canonical ensemble. Now let’s summarize all that we have seen. Clausius stated that internal energy and mechanical work can be converted into each other. This is explained by kinetic theory: on a microscopic level, internal energy is the kinetic energy
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of molecules. With this model, we can derive the ideal gas law, and relate the temperature to the internal energy of a monatomic ideal gas. More generally, the equipartition theorem states that each microscopic degree of freedom carries ½ k_B T units of energy. Using the equipartition theorem we can calculate the average velocity of the molecules, but Maxwell went one step further, and derived the probability distribution for the velocities of the molecules
1:51:14
using arguments of symmetry. Boltzmann tried to reproduce the same result as Maxwell, by using counting arguments. He assumed that discrete packets of kinetic energy are randomly distributed over the molecules, and that each distribution at the microscopic level is equally likely. The equilibrium state that we observe at the macroscopic level, is the macrostate that has the most microstates. Boltzmann concluded that in order to reproduce
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Maxwell’s distribution, it is not the energy, but momentum that should be randomly distributed in packages of equal size. When we also consider the spatial distribution of the molecules, we must divide phase space in equal portions. Clausius’ definition of entropy can be interpreted in the context of statistical mechanics as the logarithm of the multiplicity of a state. Alternatively, it can be interpreted as the phase space volume that is accessible by the
1:52:15
system. In this way, we can derive the Sackur-Tetrode formula for the entropy of an ideal gas. We considered the Gibbs paradox to determine how the entropy should depend on the number of particles N. Clausius’ definition of entropy could not determine that, because it only considered exchange of heat, not exchange of particles. The conclusion was that we have to consider the particles as practically indistinguishable. Using the formula for the entropy of an ideal
1:52:47
monatomic gas as a starting point, we derived how differentials of thermodynamic quantities are related to each other, and we introduced the chemical potential mu. We saw that in chemical reactions, mu is given by the derivative of Helmholtz free energy or Gibbs free energy with respect to the number of particles. We found that if a system is in thermal contact with a reservoir, the probability for a certain microstate is proportional to the Boltzmann
1:53:16
factor. If the system can also exchange particles with the reservoir, the probability for a certain microstate is proportional to the Gibbs factor. In order to calculate the entropy, given the probabilities for system states, we introduced the Gibbs definition of entropy. To find the probability distributions for the system microstates, we must normalize the Boltzmann factors using the partition function. This partition function is also
1:53:45
a useful quantity to calculate other thermodynamic quantities. Finally, we introduced the terms: microcanonical ensemble, canonical ensemble, and grand canonical ensemble, to describe the different ways in which a system can interact with an environment, and the way that this affects its statistical properties.
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