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Current time:0:00Total duration:11:54

AP.Chem:

ENE‑4 (EU)

, ENE‑4.A (LO)

, ENE‑4.A.1 (EK)

, ENE‑4.A.2 (EK)

- [Instructor] The concept
of entropy is related to the idea of microstates. And to think about microstates, let's consider one mole of an ideal gas. So 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. So 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. So 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. So if we think about the
particles in our system at one moment in time, in box 1, if we think about them at
a different moment in time, in box 2, 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 and the
direction give a velocity. So these arrows on these
particles are meant to represent the velocities of the particles. And the kinetic energy
of each particle is equal to 1/2 mv squared, 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. So going back to our boxes,
box 1, box 2 and box 3, 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. So 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, into another. So 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. So if the number of microstates
of a system increases, that represents an increase in entropy, and 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. And if we think about a
decrease in the 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, so 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. So 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 times V1. The number of moles remains the same. So n1 is equal to one
mole of our ideal gas and n2 is also equal to one 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. And 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. And thinking about the change
in the entropy, delta 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 one mole of our ideal gas at a certain volume, so 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 one 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. And that's what the
Maxwell-Boltzmann distribution tells us here. So 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. So 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. But it also means, if we look at the
Maxwell-Boltzmann distribution, there are greater range
of speeds available to the particles. And 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. Therefore, the final entropy, S2, is greater than the initial entropy, S1. So when we think about
the change in entropy, delta S for the system,
if S2 is greater than S1, the change in entropy is positive. So for a system of gas particles with constant volume and
constant number of moles, if we increased the temperature, there's 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 particulate diagram on the right. So the final number of moles, n2, is greater than the initial number, n1. But we're gonna keep the
temperature the same. So the final temperature is equal to the initial temperature, and we're gonna 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. Therefore, the final entropy, S2, is greater than the initial entropy, S1. And once again, the change in entropy, if S2 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 the gaseous state, we assume there are no
more intermolecular forces between the gas particles. So we're assuming this is an ideal gas. And 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. And if we increase the
number of microstates, we increase the entropy. Therefore, thinking
about the entropy change, this would be the final entropy, S2, minus the initial entropy, S1. And 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 gasses 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. But remember, really, disorder is really just a way to describe an increased number of
available microstates. And for our final situation, let's look at a reaction
that involves only gases. On the left side, we can see there's 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, delta S. That'd be equal to the final entropy
minus the initial entropy. So 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, delta S, will be negative. So 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.

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