### Video Transcript

In this video, weβre going to be
talking about the bulk properties of an ideal gas. In particular, weβll be looking at
a very useful equation, known as the ideal gas equation, which links together three
important properties of a gas β β volume, temperature, and pressure. But to start off with, letβs define
exactly whatβs meant by the bulk properties of an ideal gas.

Firstly, what do we mean by bulk
properties? Well, bulk properties describe the
average or collective behavior of a large number of particles. We know that gases consist of gas
molecules. And even a small amount of gas
contains a large number of molecules. This means that in most situations,
if we tried to describe the individual properties of each molecule in a gas, for
example, their velocities and their positions, we can see that things would get
incredibly complicated very quickly. Especially considering that weβd
have to measure these quantities in three dimensions and that constantly
changing. So instead of all this, itβs much
easier and more useful to just consider the bulk properties of a gas.

In the case of an ideal gas, the
bulk properties weβre interested in are the volume, pressure, and temperature of the
gas. These properties group the behavior
of all of the molecules in a gas together. So we can think of a gas as a
single cohesive thing, rather than as billions of individual molecules. The second thing we need to define
is what we mean by an ideal gas. When we use the word ideal in
day-to-day life, we use it to describe something which is really good or
convenient. When we use the word ideal in
science, it has a similar meaning. We use it to refer to a convenient,
simplified description of something.

So when we talk about an ideal gas,
weβre talking about a simplified approximation of how our gas behaves. Specifically, we define an ideal
gas as consisting of particles which donβt interact with each other and which have
negligible size. In other words, weβre simplifying
our description of a gas by saying that there are no forces between the gas
molecules such as electrostatic repulsion and that the volume of the gas molecules
is negligible compared to the volume of the entire gas. The reason that we simplify our
description in this way is that it makes the maths we need to describe that gas much
simpler while still being accurate enough to be useful in many situations.

Physicists have used these two
simplifying assumptions to come up with a simple and useful equation that relates
these three bulk properties of a gas. This equation is known as the ideal
gas equation, and it looks like this. The ideal gas equation tells us
that the pressure of an ideal gas multiplied by its volume is equal to a constant of
proportionality multiplied by the temperature of the gas. Now, pressure is equivalent to
force divided by area. When we talk about the pressure of
a gas in a container, weβre talking about the pressure that the gas molecules exert
on the walls of the container by colliding with them. The pressure of a gas is equal to
the total force that the gas molecules exert on the inside of that container divided
by the total internal area of the walls of the container.

Because all actions have an equal
and opposite reaction, we can also think of this the other way around. The pressure that the gas exerts on
its container is exactly the same as the pressure that the container exerts on the
gas. Depending on the context that weβre
looking at, it might sometimes be easier to think of the pressure of a gas as being
an outward, expansive pressure that acts on its container. But at other times, it might be
easier to think of pressure as being an inward squeezing pressure that the container
exerts on a gas. But thinking about it in these
different ways doesnβt change how we calculate it or its value.

When we think about the pressure of
a gas in a container, which will tend to be a reasonably small amount of gas, the
pressure can be measured at any point throughout the gas. And it takes the same values at
each of these points. The units of pressure are pascals
represented by a capital βπβ and a lowercase βπ.β And because pressure is equal to
force divided by area, one pascal is equivalent to one newton per meter squared. The next variable in this equation,
volume, is much easier to conceptualize. The volume of a gas is just the
amount of 3D space that it occupies, which we measure in meters cubed.

The other variable in this equation
is temperature. Like with most equations in physics
that involve temperature, itβs really important that the value that we use here is
measured in kelvin rather than Celsius. We can recall that in the Celsius
temperature scale, zero degrees Celsius is arbitrarily defined as being the freezing
point of water. However, the lowest temperature
that an object can theoretically have is actually negative 273.15 degrees
Celsius. So this temperature, known as
absolute zero, is where we define the zero point of the kelvin scale.

Now, other than the positions of
the zero points, the kelvin scale and the degrees Celsius scale are actually the
same. That is, a temperature increase of
one degree Celsius is exactly the same as a temperature increase of one kelvin,
which means that the freezing point of water is 273.15 kelvin. In other words, the temperature
measured in kelvin, which we could call π π, is equal to the temperature measured
in Celsius, which we could call π π plus 273.15, although weβll often round this
to three significant figures, in which case itβs 273. Whenever we use the ideal gas
equation, we need to make sure that any temperatures given in Celsius are first
converted into kelvin. Otherwise, our calculation will
give us an incorrect result.

The final part of this equation to
look at is the constant of proportionality π. This constant, like most constants
in physical equations, just make sure that when we use the ideal gas equation, we
get an answer which is in the correct units. For example, if we substituted in a
pressure in pascals and a volume in meters cubed, then we could obtain a temperature
in kelvin. One important thing to know about
this constant of proportionality is that it depends on the amount of gas that we
have.

In fact, its value is proportional
to the number of gas molecules. This means that this constant of
proportionality is only a constant if the number of gas molecules is kept the
same. So itβs important to be aware of
this whenever we use the ideal gas equation. As long as the number of gas
molecules doesnβt change, the ideal gas equation tells us that the pressure of an
ideal gas multiplied by its volume is proportional to its temperature. So now that weβve talked about what
an ideal gas is and how its bulk properties can be related to each other using the
ideal gas equation, letβs take a look at some example questions.

Which of the following formulas
most correctly represents the relation between the pressure, volume, and absolute
temperature of an ideal gas? (A) π over π is proportional to
π, (B) π over π is proportional to π, (C) π times π is proportional to π, (D)
π over π is proportional to π, or (E) π over π is proportional to π.

So in this question, weβve been
given five different formulas. And we need to decide which one of
them most accurately represents the relation between these three quantities. We can start by recalling which
symbols we generally use to represent these quantities. Well, we usually use capital π for
pressure, capital π for volume, and a capital π for absolute temperature. Since each of our available answers
uses all three of these symbols, that means that they all represent some kind of
relation between the three quantities that weβre interested in.

The key phrase in our question
which can help us figure out which one of these options is most correct is ideal
gas. We can recall that an ideal gas is
a simplified description of a gas based on the assumptions that gas molecules have
negligible size and donβt interact with each other. Now thereβs one key equation that
we should always think of whenever we hear the phrase ideal gas. And that is the ideal gas equation,
ππ equals ππ. This equation gives us the
relationship between three of the main bulk properties of an ideal gas β β the
pressure π, the volume π, and the absolute temperature πβ β which are, of course,
the same as the three quantities weβre being asked about in this question.

But unfortunately, while this does
give us a correct relationship between the pressure, volume, and absolute
temperature of an ideal gas, none of our available answers look the same as this
equation. In fact, one of the first things we
might have noticed about the answer options weβve been given is that while they are
formulas, none of them are equations. That is, they donβt have equal
signs. Instead, each of the formulas weβve
been given has this symbol. This is the proportionality
sign. We use it to represent the fact
that both sides of the formula are proportional to each other.

So, for example, if two variables
π΄ and π΅ were proportional to each other, that could be represented by this
formula. What this means is that while the
two variables are not necessarily equal to each other, they do increase or decrease
proportionally to each other. Which means that, for example, if
we were to double the size of π΄, that is, multiply it by two, then weβd also find
that the size of π΅ would double as well. Or, for example, if we multiplied
π΅ by 0.25, then weβd find that π΄ is multiplied by 0.25 as well.

You might recall that itβs possible
to turn any proportionality statement into an equality statement. That is, we can turn a statement
like this involving a proportionality sign into an equation with an equal sign. We can do this by simply replacing
the proportionality sign with an equal sign and introducing a constant of
proportionality, generally represented by a πΎ, which is multiplied by one of the
variables. This gives us an equation which
expresses exactly the same relationship between π΄ and π΅, as this proportionality
statement does. We can see that in this equation,
if we were to multiply π΅ by two, for example, then because π΄ is equal to some
number times π΅, π΄ would also have to increase by two times.

In this case, the constant of
proportionality πΎ plays the role of scaling π΅ such that πΎ times π΅ is exactly
equal to π΄. In many physical equations, we also
find that the constant of proportionality plays an additional role of making sure
that the units on the left-hand side of the equation are equivalent to the units on
the right-hand side of the equation. Itβs also worth noting that there
are two ways that we can turn this proportionality statement into an equation. We can write the constant of
proportionality on the right-hand side of the equation like this such that itβs
multiplied by π΅. Or we could write our equation like
this, with the constant of proportionality multiplied by π΄.

Itβs really important to note that
these values of πΎ are not the same because we have effectively defined πΎ in a
different way in each equation. In fact, theyβll even have
different units. Theyβre different constants
entirely. So we should really call one of
them πΎ one and the other one πΎ two to distinguish between them. So now, both of these equations and
this statement of proportionality are exactly equivalent to each other. Itβs also important to note that
statements of proportionality can have more than one variable on each side of the
proportionality sign, which we can see is the case in all of the answer options
weβve been given.

For example, we could say that π΄
is proportional to π΅ times πΆ. In this case, if we were to
multiply the size of π΄ by two, then we would find that the product of π΅ and πΆ is
multiplied by two as well. Having more than one variable on
either side of a proportionality statement doesnβt change the way that we can turn
it into an equation. So the statement π΄ is proportional
to π΅ times πΆ is equivalent to π΄ equals πΎ one times π΅πΆ or πΎ two times π΄
equals π΅πΆ.

Looking back at the question, we
can see that the ideal gas equation gives us the relation that weβre looking
for. However, the available answers are
all proportionality statements. This means that we can find our
answer by turning this equation into a proportionality statement. In the ideal gas equation, we have
two variables, π and π, on one side of the equation and one variable, π, on the
other. The presence of this constant of
proportionality means the equation tells us π times π is proportional to π. These two statements are equivalent
ways of expressing the same relation between the three variables. And we can also see that this
statement of proportionality is one of our answer options. So option (C) is the correct
answer. The formula that most correctly
represents the relation between the pressure, volume, and absolute temperature of an
ideal gas is π times π is proportional to π.

Now that weβve answered that, letβs
look at another example question.

A gas cylinder holds 3.25 meters
cubed of gas at a pressure of 520 kilopascals and a temperature of 300 kelvin. At what temperature would the gas
pressure in the cylinder become 865 kilopascals? Answer to three significant
figures.

So this question describes a
cylinder of gas with a volume of 3.25 meters cubed, a pressure of 520 kilopascals,
which is the same as 520,000 pascals, and a temperature of 300 kelvin. Weβre being asked to calculate the
temperature at which the gas pressure in the cylinder would become 865 kilopascals
or 865,000 pascals. So we could imagine the same
cylinder of gas at sometime later where the pressure has changed to 865 kilopascals
or 865,000 pascals and the temperature has taken some unknown value that we need to
work out. And because itβs the same cylinder,
we can assume that the volume would go unchanged.

At this point, because we have two
different sets of values for the quantities volume, pressure, and temperature, letβs
call the initial set of values on the left π one, π one, and π one and the
changed values on the right can be π two, π two, and π two. So our challenge in this question
is to find π two. So to start with, letβs think of
the variables that we have to deal with in this question and see if we can think of
any equations that might help us.

Well, this question asks us to
consider the volume, pressure, and temperature of the gas. One equation that gives us the
relationship between these quantities is the ideal gas equation. This tells us that the pressure of
an ideal gas multiplied by its volume is equal to some constant multiplied by its
absolute temperature. We can recall that the ideal gas
equation makes the assumptions that gas molecules have negligible size and donβt
interact with each other. Even though this question talks
about a real gas, itβs still reasonable to use the ideal gas equation as it can
still give us accurate answers without being too complicated.

Now, in this question, weβre trying
to find the value of π. So letβs rearrange the ideal gas
equation to make π the subject and do this by just dividing both sides of the ideal
gas equation by π, giving us ππ over π equals π. Since we need to find the
temperature of the gas after the pressure has increased, it seems reasonable that we
can just plug these values, π two and π two, into this equation. And it will tell us π two. However, while this is technically
true, if we tried doing this, weβll quickly realize that we donβt know what the
value of π is. This is because π actually takes
different values depending on the number of gas molecules that weβre dealing
with. So itβs not something we can just
look up.

So as it stands, we canβt
substitute π two and π two straight into the equation and just get a value of π
two. However, because we have a set of
initial conditions for the volume, pressure, and temperature of the gas, there is
something else we can do with the ideal gas equation to find the answer. If we start with ππ equals ππ
and divide both sides of the equation by π, we get ππ over π equals π. Because the amount of gas molecules
in this question is constant, this means that π is constant. So regardless of how we try to
change the pressure, volume, or temperature of a fixed amount of gas, we find that
the value of ππ over π is always constant.

In other words, π one times π one
divided by π one is equal to π two times π two divided by π two. Writing the ideal gas equation in
this way is actually equivalent to its more familiar form ππ equals ππ. However, it makes it easy to
calculate how variables change without needing to know π. And we can use this formulation for
this question. Since weβre trying to find π two,
letβs first rearrange this to make π two the subject. First, we can multiply both sides
of the equation by π two, then multiply both sides of the equation by π one, and
finally divide both sides of the equation by π one π one.

With the equation in this form
because we know the initial temperature, initial pressure, and initial volume before
the pressure increased and we know the volume and the pressure after the pressure
increased. We can just substitute all of these
values in and calculate π two. Letβs just give ourselves some more
space. And we know that π one is 300
kelvin, π two is 865,000 pascals, π two is 3.25 meters cubed, π one is 520,000
pascals, and π one is 3.25 meters cubed as well. Plugging all this into our
calculator, we get an answer of 499.04 which has units of kelvin since itβs a
temperature. And rounding our answer to three
significant figures gives us a final answer of 499 kelvin.

Now that weβve looked at some
example questions, letβs summarize what weβve talked about in this video. Firstly, we defined an ideal gas by
the assumptions that particles in a gas have negligible size and donβt interact with
each other. We also defined bulk properties as
properties that result from the average behavior of many particles. And the bulk properties of an ideal
gas are pressure, volume, and temperature. Weβve seen how these three
quantities are related by the ideal gas equation, ππ equals ππ, where π is
proportional to the number of gas molecules.

And finally, weβve seen how we can
use the ideal gas equation to obtain the formula π one π one over π one equals π
two π two over π two. And weβve shown how we can use this
equation to calculate changes in the pressure, volume, and temperature of the gas
without needing to know the value of π.