Introduction to entropy | Applications of thermodynamics | AP Chemistry | Khan Academy

The concept of entropy is related to the idea of microstates. To think about microstates, let's consider one mole of an ideal gas. Remember, n represents moles at a specific pressure, volume, and temperature. If the system of gas particles is at equilibrium, then the pressure, the volume, the number of moles, and the temperature all remain the same.

From a macroscopic point of view, nothing seems to be changing. However, from a microscopic point of view, things are changing all of the time. Looking at our gas particles here in the first box, imagine these gas particles slamming into the sides of the container and maybe slamming into each other and transferring energy from one particle to another.

If we think about the particles in our system at one moment in time in box one, and if we think about them at a different moment in time in box two, the particles might be in slightly different positions, and the velocities might have changed. Remember that the speed of a particle tells us how fast the particle is traveling. However, when we put an arrow on each particle, that also gives us the direction and the magnitude. The direction gives a velocity, so these arrows on these particles are meant to represent the velocities of the particles.

The kinetic energy of each particle is equal to (1/2) mv², where m is the mass of each particle and v is the velocity. A microstate refers to a microscopic arrangement of all of the positions and energies of the gas particles. Since we're dealing with an ideal gas here, by energies we're referring to the kinetic energies of the particles.

Going back to our boxes—box one, box two, and box three—each box shows a different microscopic arrangement of positions and energies of the gas particles. Therefore, each box represents one microstate. A good way to think about a microstate would be like taking a picture of our system of gas particles. From a macroscopic point of view, nothing seems to change, but taking a picture at the microscopic level, we see that the system is moving from one microstate into another into another.

The number of microstates available to this system of one mole of gas particles is a number that's too high for us to even comprehend. Now that we understand the concept of microstates, let's look at an equation developed by Boltzmann that relates entropy to the number of microstates. According to this equation, entropy, symbolized by S, is equal to Boltzmann's constant K times the natural log of W, and W represents the number of microstates in a system.

If the number of microstates of a system increases, that represents an increase in entropy. If the number of microstates decreases, that represents a decrease in the entropy of the system. Sometimes instead of using the word microstates, people will describe an increase in entropy as an increase in disorder or an increase in the dispersal of either matter or energy.

Sometimes using these terms helps us think about entropy. However, when we're using the equation developed by Boltzmann, we should think about these terms as meaning an increase in the number of microstates and therefore an increase in the entropy of the system. If we think about a decrease in disorder of the system or an increase in the order or a decrease in the dispersal of either matter or energy, that really relates to a decrease in the number of available microstates, which means a decrease in the entropy of the system.

Next, let's think about the change in entropy for a number of different situations. In our first situation, we're starting off with one mole of an ideal gas. Here are the gas particles in a container, and the container has a removable divider separating the container into two compartments. Let's say we go ahead and remove the divider, and now our gas particles are free to travel around in a larger volume.

If the initial volume is V1, let's say we have twice the volume for the final volume; therefore, V2, or the final volume, is equal to 2 * V1. The number of moles remains the same, so N1 is equal to 1 mole of our ideal gas, and N2 is also equal to 1 mole, so N2 is equal to N1. During the expansion of the gas, the temperature is kept constant; therefore, the initial temperature T1 is equal to the final temperature T2.

Next, let's think about what happens to the number of available microstates when the volume is doubled. When the volume is doubled, that increases the number of possible positions for the gas particles. Therefore, the number of microstates increases. Looking at our equation, if the number of microstates increases, then so does the entropy; therefore, we can say that S2 is greater than S1.

Thinking about the change in the entropy (∆S), if S2 is greater than S1, then S2 minus S1 would be a positive value. Therefore, the change in entropy for the free expansion of a gas when the temperature is constant is positive.

For our next situation, we're starting once again with one mole of an ideal gas; however, this time the volume will be held constant, and the temperature will be increased. So we start with 1 mole of our ideal gas at a certain volume (N1 and V1), and we are increasing the temperature but we're not changing the number of moles or the volume. Therefore the final number of moles N2 is equal to the initial number N1, which is 1 mole, and the final volume V2 is equal to the initial volume V1.

Since we are increasing the temperature, the final temperature T2 is greater than the initial temperature T1. Next, let's use a Maxwell-Boltzmann distribution over here on the right to explain what we see in our particulate diagrams on the left.

In the particulate diagram on the left, we can see that the particles are moving with different velocities or different magnitudes or speeds, and we can see that by the length of the arrows in the box. That's what the Maxwell-Boltzmann distribution tells us here. For the one in blue, we know that there are a range of speeds available to the particles.

So, the area under the curve represents all of the particles in the gas sample. Some are moving at a slower speed, some are moving at a higher speed, but most of them are moving at a speed close to the center of this peak. Increasing the temperature means that, on average, the particles are moving faster. So we can see that on average, the length of these arrows is longer than the length of the arrows in the particulate diagram on the left.

It also means if we look at the Maxwell-Boltzmann distribution, there is a greater range of speeds available to the particles. If there's a greater range of speeds or greater range of velocities, that means there's a greater range of kinetic energies, which means there are more possible microstates available to the system of gas particles. Therefore, increasing the temperature causes an increase in the number of possible microstates, and an increase in the number of microstates means an increase in the entropy. Thus, the final entropy S2 is greater than the initial entropy S1.

So when we think about the change in entropy (∆S) for the system, if S2 is greater than S1, the change in entropy is positive. For a system of gas particles with constant volume and constant number of moles, if we increase the temperature, there is an increase in the entropy.

For our next situation, we're once again starting with one mole of an ideal gas at a certain temperature T1 and a certain volume V1. This time we're going to increase the number of moles from one mole of an ideal gas to two moles of an ideal gas. So if we had four particles on the left, now we have eight particles in the diagram.

In the particulate diagram on the right, the final number of moles N2 is greater than the initial number N1, but we're going to keep the temperature the same. The final temperature is equal to the initial temperature, and we're going to keep the volume constant too. So V2, or the final volume, is equal to the initial volume V1.

Because we have increased the number of particles, there are more possible arrangements of particles and also more ways to distribute the energy. Therefore, when we increase the number of moles, there's an increase in the number of possible microstates, and increasing the number of microstates in a system increases the entropy. Thus, the final entropy S2 is greater than the initial entropy S1.

Once again, the change in entropy (∆S) if S2 is greater than S1 would be positive. So increasing the number of moles of gas particles increases the entropy of the system.

For our next situation, let's consider the evaporation of liquid water into gaseous water. Water molecules in the liquid state are held together by intermolecular forces, with hydrogen bonding being the most important intermolecular force holding the particles together. Water molecules in the liquid state are still free to move; however, when water molecules in the liquid state get converted into water molecules in a gaseous state, we assume there are no more intermolecular forces between the gas particles.

So we're assuming this is an ideal gas. If there are no intermolecular forces between the gas particles, we've increased the freedom of movement of the water molecules. We've increased the number of possible positions, and therefore we've increased the number of microstates available. If we increase the number of microstates, we increase the entropy.

Considering the entropy change, this would be the final entropy S2 minus the initial entropy S1. Since the entropy increased going from liquid water to gaseous water, the change in entropy would be positive. This is also one example where thinking about entropy in terms of disorder can be helpful because gases are more disordered than liquids, you could say, and therefore going from a liquid to a gas increases the amount of disorder, which increases the entropy. However, remember, really disorder is just a way to describe an increased number of available microstates.

For our final situation, let's look at a reaction that involves only gases. On the left side, we can see there are two moles of SO₂ and one mole of O₂. So there are three moles of gas on the reactant side, and on the product side, there are two moles of gas. Next, let's think about the change in entropy for this reaction (∆S), which would be equal to the final entropy minus the initial entropy.

Thinking about the initial entropy S1 and the final entropy S2, we went from three moles of gas to two moles of gas. Remember that a decrease in the number of moles of gas means a decrease in the number of possible microstates. That means a decrease in the entropy; therefore, S2 is less than S1, or we could say that S1 is greater than S2.

Therefore, for the change in the entropy, if S1 is greater than S2, we're subtracting a larger number from a smaller number, which means the change in entropy for this reaction (∆S) will be negative.

As a quick summary, when trying to figure out the change in entropy, we need to consider the number of available microstates. If the number of available microstates increases, then the change in entropy is positive. If the number of available microstates decreases, then the change in entropy is negative.