[MUSIC]
Lighthouse Scientific
Education presents a lecture
in the Gas series
The topic; Fundamentals of
Gas Law.
Material in this lecture relies
on an understanding of the
previous lecture The Basics
of Gas.
Gas law builds on the
'Basics of Gas' lecture and is
broken down into 2 lectures.
The lecture that follows this
one deals with
gas law that has only has 1
set of conditions.
That includes the
Ideal Gas Law.
Before preceding into gas
law some perspective of gas
basics will be helpful. The
gas laws that are covered
basics will be helpful. The
gas laws that are covered
in this lecture, that are based
on 2 conditions, are
Boyle's Law, Charles' Law,
Gay-Lussac's Law,
Avogadro's Law
and finally the Combined
Gas Law
Often these laws are
accompanied by problems
so an outline of problem
solving with gas law is
offered and put to the test
with 4 problems.
As for perspective, there are
4 basic properties
used to describe gases.
One way to consider
the properties is through their
effect on the number of
collisions that the
particles make.
Specifically, we can consider
how altering these properties
creates more collisions.
The first properly is amount,
number of particles.
Usually given in, or turned
into, moles. With amount,
more particles
mean more collisions.
Next is temperature. It is
given in, or turned into,
the temperature scale Kelvin.
Temperature is a measure
of heat and is related to the
speed of the particles.
Hotter particles move faster
and that causes
more collisions.
And then there is volume.
An example would be the
size of a container.
A small container has walls
that are closer together
and there is less available
space for the particles
to roam.
That leads to more collisions.
Finally, there is pressure. It
is defined as force per area.
In a gas, force is generated
by gas particles colliding
with the side of their
confinement (or really
with anything).
Force is proportional
to the number of collisions so
anything that increases
collisions increases
the pressure.
Considering gas from a
collision perspective can
help make sense of gas law.
While we are at the 4
properties
we might as well clear up
what 2 conditions
really means. It is simply
recognition that some
condition or gas property has
been changed.
There will be a set of values
for the 4 properties before
the change. A '1' subscript
indicates before.
And a set of values for the 4
properties after the change.
A subscript of '2'
indicates after.
Before and after or initial and
final are cues that
a change has occurred.
Not all the properties will
have different values
with the change.
T1 and T2 can be the same
temperature.
The gas laws covered in this
lecture relate the change
in values of 2 or more
properties that occur with
a change in condition.
In addition to a familiarity
with the 4 properties of a
gas, it is helpful to have a
familiarity with the Kinetic
Molecular Theory (KMT).
The purpose of the Kinetic
Molecular Theory is to
provide an understanding of
gas as particles.
It is premised on an ideal
gas and is essential to
building an understanding
that leads to gas law.
It was initially described in
the 'Basics of Gas' lecture
and is further developed (for
those students who will
need it) in the 'Gas:
Advanced Concepts' lecture.
Considerations of
KMT include:
gas consists of large number
of particles.
The particles move in
constant, random, straight-
-line motion. They speed up
when they are heated.
The gas particles are
separated by large distances
so the volume that a gas
occupies is mostly
empty space.
When dealing with gases
the term volume refers to the
space available for the
gas particles to roam around
and not the volume of
the particles.
Gas particles make up such
a tiny part of the available
volume that they can be
treated as if they essentially
take up none of the
available volume.
Their volume is neglected.
Since the particles are
zipping about at such
high speeds they can be
treated as if they have
no intermolecular interactions
upon collisions.
As long as there are no
chemical reactions occurring
this is a somewhat
reasonable assumption.
Any collisions between
particles is considered
as an elastic collision.
Elastic, in this context,
has the total energy of the
collision conserved.
We often treat the energy of
the cue ball in pool as
conserved. The energy can
be transferred
to the other balls but the total
energy remains constant.
With this understanding of
the KMT we can precede
into gas law under 2
conditions.
The first of those laws that
we will cover is named after
the 17th century natural
philosopher Robert Boyle;
Boyles Law.
The description of the law is
that pressure and volume
have an inverse relationship
if the properties of amount
and temperature are
held constant.
Mathematically, that is
P times V equals a constant.
The n and T in the subscript
indicate that
amount (in moles) and
temperature (T)
are held constant.
Pressure and volume will
change but not amount and
temperature.
What that equation is really
saying is that if one of these
two properties goes up, the
other one goes down
(and vice versa).
Raise the pressure and the
volume goes down.
Increase the volume and
the pressure goes down.
This law can be shown
graphically by plotting
volume on the x-axis and
pressure on the y axis.
As the volume is increased
the pressure is decreased.
This is what inversely
proportional looks like.
The strength of Bolye's law is
that it can be used to predict
pressure-volume behavior in
a before and after scenario.
Boyle's law says that
pressure times volume
equals a constant.
The 1 subscript indicates
that this pressure and
volume are for condition 1.
Usually it means a condition
before a change.
When a change does occur
(with constant amount
and temperature) there will
be a new pressure
and volume.
The 2 subscript indicates
the after change condition.
Since both sets of pressure
times volume are equal to
the same constant
they can be set equal to
each other.
This 4 variable equation can
be used to solve for any
one of the variables if the
other 3 are given.
For example, say we had a 2
liter container
under a 1 atm pressure.
According to Boyle's law,
the only way the volume can
be increased to 4 liters
(with constant amount and
temperature) is for
the new pressure to go
down. Specifically to 0.5 atm.
The actual procedure for
solving for the value is
covered later in the lecture.
Right now we are interested
in predicting the direction of
the change based on
the equation of the law. Now,
what direction would the
pressure need to go for that
starting volume of 2 liters
to decrease to 1 liter?
Boyle's law would argue
for an increase in presume.
2 atm in this case.
This is how Boyle's law is
applied.
But why does the law take
this form?
The law can be rationalized
using the Kinetic
Molecular Theory at 2
conditions.
Often the best way to
consider gas behavior
is to focus on pressure.
It is force per area.
Another factor to consider is
what is meant by keeping
certain properties constant.
For constant amount and
temperature it means that in
both conditions
(before and after) the same
numbers of particles
are moving at the
same speed.
Also keep in mind what we
learned in the
previous lecture:  pressure is
proportional to number of
collisions with the walls of
the container (this is force).
With these considerations
Boyle's law
can be rationalized.
If the volume of
the container is increased
(volume goes up),
the walls of the container are
farther away from each.
The gas particles have
longer distanced to travel
to strike the wall.  That
results in fewer collisions
which means less force on
the walls which  means
pressure goes down.
Fewer collisions-lower force-
less pressure.
If, however, the volume of
the container is decreased
(volume goes down),
the walls are closer together.
The gas particles have a
shorter distance to travel
before striking the wall.
That results in more
collisions which means more
force on the walls which means the pressure goes up.
There are rational
descriptions like this for each
gas law including the next
one and
it is named Charles' law.
it is named Charles' law.
This law says that volume
and temperature have
a direct relationship (at
constant amount
and pressure).
The equation is that volume
value over the temperature
value is a constant.
Amount and pressure cannot
change for this law to hold.
The equation tells us that if
one of these two properties
goes up, the other one also
goes up.
If one goes down the other
also goes down.
Graphically, that can be
shown with temperature
plotted on the x axis and
volume on the y axis.
As temperature increase so
does volume;
directly proportional.
The equation used in
problem solving starts with
volume over temperature
at a first condition being
equal to a constant.
With a change of condition a
new volume over
temperature equals that
same constant.
Of amount and pressure did
not change.
The before and after V over T
ratios are set equal to
each other and the 4 variable
2 condition Charles' law
is ready for problem solving.
For instance, a gas is found
at a volume of 2.0 l and a
temperature of 255 K. Under
Charles' Law conditions
the volume is increased to
2.2 l. Does temperature
increase or decrease?
Well, according to the
law if one property goes up
so does the other.
Temperature (T2) will also
have to go up;
275 K to be exact.
Actual problem solving
using gas law will be
discussed in detail later
in the lecture. Returning to
the original condition;
what happens to the volume
if the temperature
is decreased?
Right, the
volume is decreased.
Part of the student's
responsibility with gas law
is predicting whether a
property's value will go up
or down when a condition
is changed.
The gas law equation will
provide the logic.
Charles' law  can be
rationalized through the
Kinetic Molecular Theory at
2 conditions.
Even though the law focuses
on volume and temperature it
is easier to understand
through pressure.
Pressure is force per area.
Force is about collisions.
For this demonstration we
need to keep in mind that
at constant amount and
pressure, the same number
of particles exert the same
force on the container.
Another important factor to
consider is that hotter gas
particles (higher
temperature) are
moving faster.  They collide
(with the side of the
container) harder and
more often.
They produce a great force.
Cooler gases
(lower temperatures) has the
particles moving slower.
They collide (with the side of
the container) softer and
less often.
They produce a lesser force.
Taking this understand to the
volume. If the volume of the
container is increased the
walls of the container will be
farther apart.  If nothing else
changes there will be
less collisions with the wall
producing a lower pressure.
To prevent that the gas
particles will need to speed up
and to strike harder to
maintain the same force
and constant pressure (that's
a requirement of this law).
The temperature will have to
go up to accomplish that.
If, however, the volume of the
container is decreased
then the walls of the
container
will be closer together.
The particles will not have as
far to travel
to strike the walls of
the container.
If nothing else changes
there will be
more collisions with the wall
producing a higher pressure.
To prevent that the particles
will need to slow down and
strike softer to maintain the
same force
and that constant pressure.
Temperature will have
to go down
to accomplish this. Again
understanding gas from
the perspective of collisions
provides a rational for
gas behavior law.
The third gas law
is named for named for
Joseph Gay-Lussac.
It is that pressure and
temperature have a direct
relationship (at constant
amount and volume)
Its equation is pressure over
temperature being equal
to a constant.
The properties held constant
with this law are amount
and volume.
This equation is of the same
form as Charles' law and
like that law it implies that if
one of these 2 properties
goes up, the other one also
goes up.
If one goes down the other
goes down.
Graphically, it too is similar to
Charles' law.
As temperature increase the
pressure increases.
Another example of directly
proportional.
The useful version of this law
is obtained by noting that for
any single condition pressure
over temperature is a
constant.That is with a
constant amount and volume.
It is also true for a second
set of conditions that follow
Gay-Lussac's law. Setting
the two ratios of pressure
over temperature equal to
each other generates
the 4 variable equation.
Given 3 of these variables
the 4th can be solved for.
For example, under
conditions of Gay-Lussac's
law a gas is found to have a
pressure of 1 atm and
a temperature of 255 K.
What happens to the
temperature if the pressure is
increased to 1.2 atm?
The law is one of direct
proportion.
If pressure goes up then so
does the temperature.
Conversely, returning to the
original values,
the temperature is
decreased. We should then
expect the pressure to also
decrease.
Reasoning out Gay-Lussac's
law under KMT
at 2 conditions starts
with pressure,
force per area and a
recognition
that at constant amount and
volume, the same number
of particles are in the same
size container.
As argued with Charles' law
hotter gases
(higher temperature) has the
particles moving faster.
They collided (with the side
of the container) harder
and more often.   They
produce a greater force.
Cooler gas particles (lower
temperature)
has the particles moving
slower.  They collided
(with the side of the
container) softer
and less often.   They
produce a lesser force.
Applying these constraints to
pressure; to increase pressure
 gas particles need
to speed up and strike the
side of the container harder
and more often.
This occurs with a hotter
gas. Temperature goes up.
To decrease pressure gas
particles need to slow down
and strike the sides of
the container
softer and less often. That
occurs with a cooler gas.
Temperature goes down.
We have covered
3 gas laws so far.
The last of the 2 property 2
condition gas laws is
named for Amedeo
Avogadro.
We know this name from
Avogadro's number and the
mole. It's the same guy.
Avogadro's law has volume
and amount in
a direct relationship.
That is, when temperature
and pressure are held
constant.
The equation is volume
divided by amount (in moles)
equals a constant.
As with Gay-Lussac's
and Charles laws, if one of
the two properties goes up,
the other property goes up
and vice versa.
There is a common set
of conditions
that use Avogadro's law to
produce a
very useful relationship.
It takes place at standard
temperature and pressure,
STP.
That is a temperature of 0
degrees Celsius
and a pressure of 760 mm
Hg or 1 atm.
At STP, 1 mole of gas
occupies a volume of 22.4 l.
That mole-volume
relationship is referred to as
molar volume and can be
used in Avogadro's law.
Graphically, Avogadro's law
says that as
amount goes up so
does volume.
When considering the law
from 2 conditions, before
and after a change, volume
over amount under
the first condition is equal to
a constant as is volume
over amount at a
second condition.
Setting the two conditions
equal to each other has
the 4 variable equation used
in problem solving and
in trend prediction.  Say
there is 0.10 moles of gas in
a cylinder under a piston.
It initially has a volume
of 2 liters. If more matter is
added to the container
at the same temperature and
pressure the volume
will have to change. We can
use Avogadro's Law
to predict how the
volume changes.
If the amount of gas is
increased to 0.2 moles
there must be a matched
increase in the volume.
If the moles are doubled then
so is the volume.
They either both go up or
they both go down.
Using the KMT to rationalize
Avogadro's Law
begins with the definition of
pressure; pressure
equals force over area.
The limitations of the
Avogadro's Law, constant
temperature and pressure
has the gas particles having
the same speed
affecting the same force
per area.
So, If volume is increased:
the walls are farther apart.
That means the particles
have to travel further to strike
the wall which would lead to
less collisions resulting in
a lower force per area.
A lower pressure.
Since that is a violation of
this law
more gas must be added.
Amount has to go up to
maintain the same number
of collisions with the further
away walls.
Each individual particle will
strike the wall less often
but there will be more
particles.
On the other hand, if volume
is decreased:
then walls are closer
together
which means that there
should be
more collisions with the side
of the container
and that is an increase in
force and pressure.
Since this is not allowed
under these conditions,
gas must be removed.
Amount goes down.
Each individual gas particle
will strike the wall more often
but there will
be less of them. This will
maintain the pressure.
An understanding of gas law
from the perspective of the
law equation and KMT
provide a deeper foundation.
Let's take stock of what we
have covered so far,
four gas laws.
They each relate two of the 4
properties of a gas while
holding the other two
properties constant.
In reality these four gas laws
actually spring
out of, or can be combined
into,
a single equation that
relates pressure,
volume, amount and
temperature at 2 conditions.
This is the Combined
Gas Law.
The equation is pressure
times volume over amount
times temperature at one
condition being equal to
the same ratio in another
condition.
Now, if one or more
quantities are constant they
can be removed from the
equation. They still need
to be noted as constants
though.
Such as is the case for
constant amount.
The combine gas law in this
circumstance will have
3 of the 4 properties as
variables.
The 4th property is place as
a subscript to indicate that
it is a constant.
A similar construction
of the Combined Gas law is
found with constant
temperature as well as
constant pressure
and constant volume.
All of these equations can
be used like the other
gas laws.
What happens to this
Combined Gas law
if 2 properties are held
constant?
What do we get if amount
and temperature are
constant and only volume
and pressure can vary?
Why that's Boyle's law.
Remove n and T from the
Combined Gas law and out
pops Boyle's Law.
What if pressure and amount
are held constant
and only volume and
temperature can vary?
That's Charles' law.  Keeping
volume and amount constant
and letting pressure and
temperature change
produces Gay-Lussac's law.
Holding pressure and
temperature fixed and
just letting amount and
volume vary
produces Avogadro's law.
So gas law at two conditions
is really just a manifestation
of holding some properties
constant and letting
other adjust to circumstance.
While we have all the gas
laws up we can show
which sets of variables trend
together or trend in
opposite directions.
If the property variables
are multiplied by each other,
like in Boyle's law, then the
only way for that
multiplication to remain
constant is that
if 1 property goes up the
other one goes down.
Conversely if 1 property goes
down the other goes up.
Inversely proportional.
If, however, one of the
property variables is
divided by another, as in
these 3 gas laws
then to have the math
produce a constant
both properties have to
increase (in the same
proportion) or both decrease
in the same proportion.
This analysis gives the trend
of the change which is a
handy tool in addressing
gas behavior.
To actually get the value of a
changed property using
gas law some problem
solving is in order and that is
the second part of the
lecture.
We will use our standard
problem solving
approach when dealing with
gas law.
That approach will be
discussed in detail
before proceeding to 4
practice problems.
The most important step in
problem solving is to
identifying the known and
unknown variables.
We'll call that the set-up.
Gas problems usually come
in one or two varieties.
Does the problem have 1 set
of condition or 2 sets?
Does the problem have 1 set
of condition or 2 sets?
1 set of conditions is where
any, and/or all, of the
4 gas properties are given
with only 1 value.
As we will see in the
next lecture,
'Gas Law:  One Condition'
such problems  provide
information of all but one of
the 4 gas properties.
The student is asked to
determine the value for
the 1 unknown property.
A new constant R
will be introduced there.
In this lecture we will focus
on problems that give more
than one value for some of
the properties.
A change has occurred in the
problem and two or more
properties will have before
and after the change values.
In this set up 2 or more
properties of the gas will
have variables for 2
conditions.
Subscript's 1 and 2 indicate
before and after the change.
Properties that are unaltered
or remain the same
in both conditions generally
can be
omitted from the set-up in
this type of problem.
A couple points of order
concerning the set-up need
to be addressed stating with
temperature.
For gas law all temperatures
that are involved in the
calculation must be
converted into Kelvin.
Also, there should only be
one unknown.
That is all of the other
variables needed to solve
the equation have to be
given or readily deduced.
The second step is to find a
relationship between the
known and the unknowns.
The unknown variable in the
equation will also need to be
solved for or isolated on one
side of the equation.  With
gas law covered in this
lecture the unknown will
never be isolated by itself
on one side of the equation.
The equation will always
need to be solved
algebraically for that.
In this particular example
Charles' law is
the relationship and the
unknown variable, T2,
is in the denominator
of a fraction.
The math looks like it might
be of the unpleasant variety.
The good news is that this
equation, and the other gas
law equations, solve
algebraically for the unknown
in same 1 or 2 steps.
The 2 step kinds have
relationship with fractions.
There are denominators.
Several of the covered gas
laws fall under this
description. The first step in
isolating the
unknown in these laws is to
get rid of the fraction and
lose the denominator.
When that is done the
equations will take on a form
like Boyle's law in which
there's only
multiplication (no division).
The second step deals with
this form of the equation but
before we get
to that step we need to
explore the first step;
getting rid of the
denominator.
Turning equations with
division into equations
with only multiplication is
accomplished by
cross multiplying.
And what does that entail?
Take Charles' law for
instance. The 'cross' means
across the equal sign.
Multiply V1 by T2
and set that equal to the
multiplication of T1 times V2.
V1 times T2 equal V2 times
T1. The new equation does
not have a denominator and
can be solved with an
additional step.
Returning to the list
of gas law under the
denominator heading,
these laws can be converted
into modified gas laws with
no denominator by cross
multiplying.
Starting with the law we just
solved; V1 times T2
equals V2 times T1.
Gay-Lussac's law solves
in a similar manner; P1
times T2 equal P2 times T1.
and so does Avogadro's law:
V1 times n2 equals
V2 times n1.
In each case there is a
pairing of a property from
before the change with
a property from the after
the change.
All laws that start with
denominators will be mixing
before and after variables
when cross multiplying.
Sneaking Boyle's law back in
the list we are now ready for
the final algebra step in
isolating an unknown, and
that is to divide both sides of
the equation by the
neighbor or neighbors of the
unknown variable.
Bringing back this version of
Charle's law,
and saying that the unknown
variable is T2,
the neighbor of T2 is V1.
Divide both sides of
the equation by V1. On the
left hand side of equation
the V1's cancel out leaving
our unknown
variable isolated by itself on
one side of the equation.
At this point we would
normally jump
to step 3 but it might be
helpful to check out what
would be needed to solve a
gas law equation if the setup
had different variables and
the equation in the
relationship did not have
a denominator. In that case
there would be
no need for cross multiplying
and we could just jump to
the final step in the algebra
and noting that the unknown
in the problem is V2 and
the neighbor of V2 is P2,
divide both sides of
the equation by P2.
Canceling P2's out on the
right-hand side produces the
solved algebraic equation.
Take the equation with the
unknown isolated on
one side of the equation to
the solution step;  step 3.
Add values from the set-up
into the equation, cancel
units and do the math.
Follow these steps closely
when first attempting gas
law problems.
They greatly reduce the
confusion that sometime
accompanies new material.
On to 4 practice problems to
see the steps in work.
Problem 1: a gas has a
volume of 300 ml
at 40  degrees Celsius.
If the volume
changes to 400 ml, what is
the new temperature
(assume constant pressure
and amount)?
An early question that one
might ask of this problem
is if it is a 1 or 2 condition
problem.
The surest way to answer
that question is to see if
there is a property that
has 2 given values.
Looking at the problem it can
be seen
that volume, given in ml, has
1,....2 condition values.
The variables in the set up
will have
subscripts  1 (before) and 2
(after).
On to the set-up.
As mentioned
there are two volumes
in the problem.
Which one is from before
the change?
Which one gets the
subscript of 1.
Which one is from after
 the change?
Which one gets the
subscript of 2?
The problems says that the
volume is changed to 400 ml
V2 is 400 ml. That means
300 ml must be V1.
The other property it
mentions is temperature.
It gives one value as 40
degrees Celsius and asks
for a new temperature.
That tells us that there two
temperatures in this problem.
Which one is the initial or the
before the
change temperature?
A question that can be asked
is whether the 40 degrees Celsius is associated
with the volume V1 of 300 ml
or the volume V2 of 400 ml?
It is connected to V1 so it
must be T1.
These two are at the
same  condition.
What does that mean
for T2?
Does it remain constant?
If it did the problem would
not ask for a new
temperature.
It changes but to what.
It is the unknown.
What about the other 2 gas
properties pressure
and amount? Pressure and
amount do not change
and do not need to be
included in the set-up.
Are we ready for step 2?
Actually no.
Degrees Celsius needs to be
converted into Kelvin.
Gas problems are solved in
the absolute
temperature scale.
Fortunately, the equation is
pretty straight forward.
Take the temperature in
Celsius and add 273.15.
40 degrees plus 273.15.
Remember that we line up
significant figures
when adding so the 0 in the
1's place of the 40 is the
digit of last significant. T1
should be given as 313 K.
Actually, we can wait until
the end of the problem to
round but we've done it here
for demonstration purposes.
Now it is time to move to
step 2. Here we ask what
relates the known variables
to the unknown variable?
We should also take the
constant properties
of pressure and amount into
consideration.
Taken all together the
relationship is Charle's law.
It is not however, solved for
the unknown variable T2.
T2 is in the denominator and
the algebra needed to isolate
this variable begins....with
cross multiplying.
Cross means across the
equal sign.
That is V1 times T2 equals
V2 times T1.
With the denominator
cleared the second step
is to divide both sides of the
equation by the
variable or variables next to
the unknown variable.
The unknown is variable T2.
The variable next to it is V1.
Divide both sides of the
equation by V1.
Cancel the V1's from the left
hand side of the equation
and T2 is isolated by itself on
one side of the equation.
Now it is time to solve that
equation using values
from the set-up. V2, (400 ml)
times T1 (313 K) divided by
V1 (300 ml).
The ml units cancel out
leaving the temperature
unit of Kelvin. The math 400
times 313 divided by 300
gives a solution to the
equation of 417 Kelvin.
That should be added to the
set-up because it is not the
final answer to the problem.
While gas problems are
solved in the Kelvin scale,
temperatures are almost
never given or measured
in Kelvin. Thermometers
either read Celsius
or Fahrenheit. This Kelvin
values should be converted
back into units of the starting
temperature.
The equation for converting
Kelvin into Celsius has
273.15 being subtracted from
the Kelvin value.
417 - 273.15 (actually the
same rules for significant
figures is applied here)
gives a final temperature and
a solution to the problem of
144 degrees Celsius.
A quick note. We rounded
temperature values
twice in this problem.
It is probably best to only
round once at the end of the
problem.
On to problem 2. A sample of
gas with a pressure of
75 atm is initially in a
1.0 liter container.
It is allowed to expand into a
second 1.0 liter container.
What is the new pressure if
temperature
is held constant? Perhaps a
cartoon of this transition
would help clarify the
change.
Starting with two 1 litter
containers that are initially
separated from each other.
All the gas is in the 1.0 liter
container on the left.
The pressure in that
container is 75 atm.
If the stopper between
the containers is removed
the gas can sweep into that
second container effectively
diluting the gas.
The gas has more volume to
occupy with the same
amount of particles.
The gas cannot exert the
same amount of pressure as
it did in the smaller volume.
The pressure has to go
down. But by how much?
We again begin the process
of problem solving
by asking if this a 1  or 2
condition problem.
Does a search of the
problem show one of the 4
gas properties with 2 different
values or
at 2 different conditions?
Yes, volume changes in
this problem.
This is a 2 condition problem
and properties will be
given subscripts 1 or 2.
To the set-up of known
and unknown variables.
Since volume is highlighted
we can ask what is the initial
volume in this scenario?
What is V1? The problem
clearly states that the initial
volume is 1.0 l. Initial is a
term that belongs to
the 'before the change'.
What is the 'after the change'
V2 value.
Well the problem says the
second volume is also 1.0
liters so is V2 1 liter?
It is not. If V1 and V2 were
both 1 liter then volume
would be constant and that is
not the case.
The key term here is
expanded.
V1 is expanded.
A volume is added to V1.
1.0 l is added to V1.
V2 is V1 + 1.0 l.
V2 is 2 liters.
What else is changing in
this problem?
Does temperature change?
No it's a constant.
What about amount? Well it
doesn't specifically say
that it is a constant but it
implies that there is a set
amount of gas particles that
will be allowed to flow into an
additional volume.
Amount is implied to be a
constant.
Returning to the problem we
can find
the fourth property pressure.
There is also a reference
to a new pressure, an after
the change pressure.
Pressure varies in this
problem.
There will be a P1 and a P2.
Which one gets to claim
that 75 atm?  Well, is 75
attached to the condition
of V1 or V2? It is attached to
the initial volume.
The new pressure is
therefore attached
to the new volume.
P2 is our unknown.
We have our values in the
set up and are ready to move
into the relation step.
If we had any temperature
values they would need to be
converted in
to Kelvin here but we do not.
In the relationship step we
ask if there is an equation
that relates the known and
unknowns in the set-up.
We also keep in mind the
constants (temperature
and amount).
What relates change in
volume and pressure?
Boyle's law. P1V1 equals
P2V2.
Is the equation solved for the
unknown P2?
No, it is not.
Since there is no
denominator in the equation
there will be no need to cross
multiple so this equation is
solved with one step.
Isolation of the unknown
is achieved by dividing both
sides of the equation by the
neighbor of the unknown.
P2 is the unknown which
makes V2 the neighbor.
Divide both sides by V2.
Cancel out the V2's
on the right hand side and
the P2 is isolated in an
equation which is ready
for step 3.
Solve the problem using
values from the set-up.
P1 (75 atm) times V1 (1.0 l)
divided by V2 (2.0 l).
Units of liters cancel leaving
atm units.
Expanding from volume 1
liter to 2 liters drops the
pressure from 75 atm to
37.5 atm.
Doubling the volume
halves the pressure under
these conditions.
One property goes up while
the other goes down.
To be completely accurate
the 37.5
can be rounded to the two
significant figure value of 38.
Example 3: A sample of gas
in a 10.0 l container has
observed values of 100 torr
and 30.0 degree C.
What is the pressure if the
container is heated
to 130 degree C? This
problem has a slightly
different look but it will be
solved in a similar manner.
What type of gas problem
is this?
Is it under 1 condition or is it
under 2 conditions?
Is there a before and an
after?
Is there a gas property with 2
two stated values?
According to the problem
there are two
different temperatures. Good
bet that this is a 2 condition
problem and the variables
will have subscripts of 1 and 2.
So what are the known and
unknowns in this problem?
We saw that there are two
temperatures but which one
is which? What is the initial
temperature (T1)
and which is the final (T2).
A careful reading of the
problem has the gas heated
to 130 degrees Celsius.
That must be the final or the
after temperature.
The initial temperature will
therefore be
30 degrees C.
It isn't necessary to
find the initial value first.
As long as the values are
correct the variables can be
added to the set-up in
any order. What else
changes in this problem?
What about volume?
It starts out at 10.0 liters.
Does it change?
It doesn't appear so.
All the problem says is that
this container is heated.
Volume will therefore be
treated as a constant.
Does amount change?  Is
any gas added, removed
or involved in a chemical
reaction?
No, nothing has is happening to the gas except for
it being heated. That leaves
the fourth property
pressure and we see that it
uses the unit torr.
One way to decide if the 100
torr given in the problem is
P1 or P2 is to ask whether
this pressure is associated
with the initial or final
temperature.
It is associated with the 30
degrees C which is T1.
The other way to approach
P1 and P2 is to is ask
whether the new pressure
called for in the problem is
associated with the initial or
final temperature.  It belongs
to the condition with
the gas heated to 130
degrees C and its value is
not given in the problem.
It is our unknown.
Before proceeding to the
relationship step there
is some cleanup work that
needs to be done.
The Celsius values will need
to be converted into Kelvin.
Bringing up the conversion
equation and
inserting 30 degrees in the
Celsius position.
The last position of
significance is noted for
demonstration purposes and
the addition gives a T1
of 303 K.
Add that value to the set-up.
For T2 put the 130 in the
Celsius position.
It too has a final significant
figure in the ones position.
The addition gives T2 as
403. Add that to the set-up
and we are ready to find an
equation that relates the
known and unknowns.
What law relates changes in
temperature and pressure?
Remember that volume and
amount are held constant. It
is Gay-Lussac's law and the
unknown P2 is not isolated
so some algebra is in order.
The first step in the process
is to cross multiple in order
to remove the denominators
in the equation.
Cross means across the
equal sign.
That produces the modified
Gay-Lussac's law equation
P1 time T2 equals
P2 times T1.
Now that the equation only
involves multiplication solve
for the unknown P2 by
dividing both sides of the
equation by its neighbor.
P2's neighbor is T1.
Divide both sides by T1 and
canceling the T1's from the
right hand side of the
equation
gives the isolated
P2 equation.
All that is left is to input
values from the set-up
and solve. P1 (100 torr),
times T2 (403 K)
divided by T1 (303 K)
with units K canceling out
yields a final pressure of
133 torr.  Temperature goes
up; pressure goes up.
Okay, one more.
At STP 1.0 moles of gas
have a volume of 22.4 l.
What will its volume be at
4.0 atm and 25 degree C?
There seems to be a lot
going on here.
Is this a 1  or 2 condition
problem?
In the previous problems we
looked into the problem for
gas properties that have
more than 1 value.
This one doesn't show that.
It implies it.
It expects the student to
know that STP
means standard temperature
and pressure.
 Atmospheres are used
here rather
 than mm of Hg
because they are the units
used in the problem.
Since the problem provides
a second pressure and
temperature we are
to reason that this is a 2
condition problem
with subscripts 1 and 2 will
be used.
It's going to be a busy
set-up.
We already noted that there
is a change in temperature
and in pressure from the
initial condition of STP.
T1 T2, P1 P2
The initial values of T1 and
P1 are gotten
from the definition of STP.
These values can be
memorized or looked up.
It is 0 degrees Celsius
and 1 atm.
These are exact values and
have an infinite number of
significant figures.
As for T2 and P2, the
problem says they change to
25 degrees Celsius
and 4 atm.
So far so good but there isn't
an unknown yet so
clearly we're not done.
Returning to the problem
we see the mention of a third
property, volume.
It is said that at STP, by
definition, 1 mole of gas
occupies by 22.4 liters.
The problem then goes on
to ask for a new volume due
to an increase in
temperature and pressure.
There is a V1 and a V2.
We can see that the 22.4 l is
attached to the STP
and must be V1.  V2 is an
unknown in this problem.
We should also consider the
property of amount.
Is any gas added, removed
or involved in a
chemical reaction?
It doesn't appear so and
that makes amount a
constant in this problem.
It could be added to the set-
up as a constant
but it doesn't have to be.
We still need to convert
temperatures into Kelvin
before proceeding into
the relationship step.
To speed matters along
the steps of the conversion
will be omitted and 0 degrees
Celsius will be said to equal
a value of 273 K
and 25 degrees C  to 298 K.
Now we are ready to
consider the relationship
between the known and
unknown variables.
There are more variables in
this set up than in the
previous three problems but
there is a gas law that
can handle a change in 3
properties (and
constant amount). It is the
Combined Gas law
equation and it is not solve
for the unknown V2.
This might be a more
involved equation but the
unknown is isolated in the
same manner as it was in the
previous problems.
Since the equation has
denominators the first step is
to cross multiple.
Everything in the numerator
of the left side fraction
times the denominator of the
right side fraction is equal
to the denominator of the left
side fraction times
everything in the numerator
of the right side fraction.
The mathematically modified
Combined Gas law equation
is V1 x P1 x T2 =
V2 x P2 x T1
and is ready for the
isolation step.
Divide both sides of the
equation by the neighbors
of the unknown.
V2 is the unknown and its
neighbors are P2 and T1.
This is the first times that we
have more than 1 neighbor
but that doesn't change the
procedure one bit.
Divide both sides of the
equation by P2 times T1.
Since the fraction on the
right side has P2 times T1 in
both numerator and
denominator these values
can be canceled out leaving
V2 isolated by itself.
It is time to take this busy
equation to the
solution step and input
values from the set-up.
V1 (22.1 l) times  P1 (1 atm)
times T2 (298 K).
All  over P2 (4.0 atm)
time T1 (273 K).
Note that the units atm
cancel out as does unit K
leaving just the units liters
which is consistent
with an unknown volume.
Getting out the calculator
and multiplying 22.1 times 1
times 298 and then
diving by 4 and 273 will
get a V2
of 6.11 liters
Noticing that P2 only has 2
significant figures
V2 is rounded to 2 sig. figs,
6.1 l.
And that concludes the
material for this lecture.
As a recap. Gas law was
investigated in situations
that involved a change in
condition.  2 condition laws
The first law is Boyle's Law:
Pressure and volume have
an inverse relationship.
If one goes up the
other goes down.
P1V1 equals P2V2 when
amount and temperature are
held constant.
Charles' Law:  volume and
temperature have
a direct relationship. If one
property goes up so does
the other. V1 divided by T1
equals V2 divided by T2
when amount and pressure
are held constant.
Gay-Lussac's Law:  pressure
and temperature have
a direct relationship. If one
property goes down so does
the other. P1 divided by T1
equals P2 divided by T2
when amount and volume are
held constant.
Avogadro's Law:  volume and
amount
have a direct relationship.
If one property goes up so
does the other.
V1 divided by n1 equals V2
divided by n2
when temperature and
pressure are held constant.
These gas laws can be
combined into a law
of its own:It relates pressure,
volume, amount and
temperature at 2 conditions.
Any variable held constant is
removed from the equation.
We also looked at problem
solving as related to
gas law using our 3 step
problem solving technique.
In the set-up we asked if the
problem had
any of the 4 properties of a
gas given with 2 values.
When this is the case
subscripts 1 and 2 are used
to indicate before and after
the change in condition.
In the relationship step we
looked for a gas law based
on the variables in the set-up
and a recognition of what
properties were held
constant.
The unknown variable was
then isolated by 1 or 2 steps.
If it the gas law had a
denominator we
crossed multiplied in order to
remove denominators.
With only multiplication in the
equation we divided
both sides of the equation by
the neighbor or neighbors
of the unknown. With that
the unknown is isolated.
We solved by inputting
values from the set-up
and canceling units.  And
that concludes our lecture.
Understanding gas law is a
recognition that gas law is
based on the 4 properties of
a gas!
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