[MUSIC]
Lighthouse Scientific
Education presents a lecture
in the Gas series
The topic; Gas
Law: One Condition.
Material in this lecture
relies on an understanding
of the previous lectures;
The Basics of Gas and
Fundamentals of Gas Law.
Gas law, in this series, is
broken down into 2 lectures.
The lecture previous to this
one deals with gas law that
has a change in condition.
This lecture features gas law
at a single condition.
Like the first gas law lecture
this one begins with some
perspective on gas and
that includes a
review of an ideal gas.
Under the 1 condition
banner, there are two gas
laws covered in this lecture.
The first one is the
Ideal Gas law: PV=nRT.
There are assumptions
with ideal gases that
do not always hold.
For those students who will
need that exposure,
deviation to the Ideal Gas
Law is covered.
The second law is
Dalton's Law of
Partial Pressure.
Like the 2 condition gas law
lecture, this one also spends
time considering problem
solving with gas law and
uses a couple 1 condition
gas law problems to firm
up the practice.
Before proceeding into
1 condition gas law a brief
revisit of gas basics will
keep us all on the same page.
There are 4 basic properties
used to describe gases and
they can be explored through
the number of collisions
that gas particles make.
The first property is
amount, number of particles.
Usually given or turned
into moles. With amount,
more particles mean
more collisions.
And then there 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.
Next there is volume.
And this refers to the volume
available for the gas
particle to roam about.
An example would be
the size of a container.
The number of collisions
a gas particle makes with
the walls of a container
is proportional to distance
between the walls.
The final property is pressure.
It is defined as force per area.
In a gas, the force is
generated by gas particles
colliding with the sides of its
confinement (or anything).
Since force is proportional to
the number of collisions,
anything that increases
the number of collisions
increases the pressure.
In the previous lecture we
covered 2 condition gas law.
These are used when the
values of two or more
of the 4 properties are
changed due to a change
in conditions.
The laws relate property
values before the change
to those after the change
and include the relationship
between pressure & volume;
Boyle's Law; the
relationship between
volume and temperature;
Charles' Law;
the relationship between
pressure and temperature;
Gay-Lussac's Law and the
relationship between volume
and amount; Avogadro's Law.
These laws can be
combined into an
inclusive 4 property
relationship;
the Combined Gas Law.
In this law we see that
the 4 properties of a gas
combine into a ratio that,
regardless of conditions,
comes to a single
constant value.
Otherwise these ratios
would not be equal
to each other.
This law bridges the 2
condition gas laws with the
Ideal Gas Law which we will
get to as soon as we review
what is meant by ideal gas.
An ideal gas is a theoretical,
and not completely
accurate, portrayal of a gas.
In many instances this model
stands in quite well for a real
gas and its overall simplicity
is useful for developing a
basic understanding of gases.
The model has 3
basic assumptions.
The first is that a gas
is composed of particles.
These particles can
be compounds or atoms.
They are thought of as
small hard spheres that have
no volume and do
not participate in
intermolecular interactions.
An analogy for lacking the
attractive and repulsive
forces of intermolecular
interactions is billiard balls on
a pool table. The balls collide
with each other but there is
no sticking together
or alteration to their
size or shape.
The second assumption is
that particles in a gas move
rapidly and are in
constant random motion.
They move in straight
lines and only change their
direction upon collisions.
After collision they will move
in a new straight line
until their next collision.
Collisions of gas
particles are said to be
perfectly elastic. That is,
upon collision, the
energy of motion is
transferred without loss.
In the billiard ball analogy,
when a 'break' occurs at the
start of the game the energy
of the cue ball is
transferred into the motion of
the other balls.
No energy is lost.
It is only communicated
to the other balls.
These 3 assumptions of an
ideal gas form the basis of
the Kinetic Molecular
Theory (KMT) which was used
extensively in the previous
gas law lecture to build
an understanding of
the behavior of gases.
OK, returning to the
Combined Gas law equation
from our list of gas laws.
For that law to hold,
pressure times volume
over amount times
the temperature
at one condition
is equal to PV over nT
at another condition.
The two fractions must be
equal to the same value.
The combined gas law
equation says that the
4 properties of a gas
are related to a constant.
There can be changes to the
values of the properties but
not a change to the ratio of
the values of the properties.
With a bit of algebra, this
equation can be rearranged
so that at any 1 condition the
4 properties of an ideal gas
are related as PV equals nRT.
That is the Ideal
Gas Law equation.
A quick peak at that algebra
starts with the Combined
Gas law having both sides
of the equation being
multiplied by the
denominator of the ratio, nT.
The nT on the left hand side
of the equation cancels out
leaving PV equal RnT or nRT.
The nice thing about this
equation is that given
any 3 of the 4 properties
of a gas, at a single
condition, the fourth
property can be solved for.
That is why it is such
an important equation.
Before proceeding,
we should build
a deeper understanding
of the gas constant R.
It is a very important
constant in chemistry.
Looking at the original form
of the equation it can be
seen that R must have units
from all 4 gas properties.
Its units must match the
units on the left had side of
the equation.
Furthermore, it must
have a ratio with
volume and pressure units
in the numerator
(such as liters and
atmospheres) and amount
and temperature units in the
denominator (such
as moles and Kelvin).
These are not the only
choice of unit denominations
that can be used.
For instance
presume unit of torr can
be used instead of atm.
Volume unit can also
vary but not so for moles
and Kelvin.
Amount and temperature
will always be
given in these units.
Since there are numerical
values attached to the
units there will need
to be a numerical value
attached to R.
When units of l-atm
over moles-K are used the
numerical value is 0.0821.
Together they are
the gas constant R.
When units of l-torr
over moles-K are used the
numerical value is 62.4.
This too is the gas constant R.
Since R is a constant.
These two versions of
R must be equal to each other.
That is statement that can
be backed-up with some math.
Consider the
first version of R.
It has atmospheres
for its pressure unit.
The other version of R has
torr as its pressure unit.
All of the other
units are identical.
In essence, these two forms
of the R constant only vary
in their numerical value
and the pressure unit.
In the 'Basics of Gas'
lecture we discussed converting
between pressure units.
We can do that here too even
though it feels
more complicated.
What is the conversion
between atm and torr?
It is 760 torr over 1 atm.
That is a unit factor.
It is oriented in this equation
so that atm units cancel
leaving torr units.
0.0821 times 760 is 62.4.
Does this mean that every
time a different set of units
are used that there will
need to be a conversion to
get the correct R?
Fortunately not.
Most text books provide a
table of commonly used R's
with the unit type listed
along with its appropriate
numerical value.
All these versions of R
are equal to each other.
Note that they all
have moles and Kelvin
in the denominator.
That minus 1 superscript
indicates that these are
in the denominator
of the unit ratio.
The different R's
reflect how pressure and
volume are expressed.
Most of the time people will
be treating gases as ideal
but some students will be
asked to consider where the
Ideal Gas law breaks down.
An opening question is
"Where do real gases differ
from ideal gases?" and the
parameters of interest are
within the Kinetic
Molecular Theory.
Remember that theory is
based on the ideal gas and it
has as a premise that gas
particles have no volume.
Their volume is neglected.
KMT also argues that
particles have elastic
collisions and that these
collisions occur without
intermolecular interactions.
In 1773 Johannes Van
der Waals proposed
2 modifications to
the Ideal Gas equation.
First he argued that at high
pressure the volume of the
gas cannot be neglected
which reduces total available
volume that the
particles have to roam.
That challenges
this premise of KMT.
We can consider why there
is a reduction of volume from
the particle's perspective.
Say a gas particle can be
modeled as a sphere
with a radius 'r'.
It can approach another gas
particle with a similar radius.
But the center of that first
particle cannot be found
in a circle centered
around the second particle.
If the center of the first
particle was found inside
that circle some part of
the first particle would be
inside the second particle
and that is not allowed.
For each particle there is
an expanded volume that the
other particles cannot
occupy and this effectively
reduced the available
volume to roam.
Van deer Waals came
up with a modification
to the volume term in
the Ideal Gas equation.
He says that a better
representation of the
ideal volume (from
the Ideal Gas law)
would be to take the real
volume of the container
and subtract out the
volume of the gas.
His subtraction term has
the amount as moles (n)
as well as a gas dependent
constant he called 'b'.
This term is an experimentally
determined value and it is
found in a look up table.
The second modification
Van der Waals made was
to the pressure.
He argued that
at high pressure
particles do interact which
modifies the real pressure.
This premise of KMT
is being challenged.
There is not an easy
cartoon to demonstrate how
Van der Waals came up with
his second modification
because, in reality, the
modification is a mix of
different factors
and is sensitive
to the amount of gas.
For us, we will say
that ideal pressure
(from the Ideal Gas
law) is a modification that
has the measured pressure
(real pressure) adjusted
by a term that has
an amount dependency
(moles); more particles
more interactions.
And a volume dependency;
bigger volume, more space
and less interactions.
It also has an experimentally
determined gas constant,
which, like 'b', is
found in a look up table.
Taken together, the
Ideal Gas Law equation
was adjusted by Van der Waals
with a modified pressure
times a modified volume
being equal to nRT.
Certainly a more complicated
version of the Ideal Gas Law
but one that more accurately
models real gas behavior.
The second major topic
of 1 condition gas law is
Dalton's law of Partial Pressure.
It was recognized
by the great early
chemist John Dalton
and it is a recognition
that different gases are
sometimes mixed together.
Dalton's law deals
with the total pressure, P
subtotal, of a mix of gases.
It is a straightforward law
that claims that the total
pressure of a mix of gases in
a container is the sum of the
pressure that each gas
would exert if it were alone.
That is if each gas were
in the same volume container
and at the same temperature.
Otherwise we
would be comparing
apples and oranges.
Mathematically, that has
the total pressure of all the
gases in a mix being equal
to the sum of the pressure
of the individual gases.
It doesn't matter how
many gases there are,
just add up their
individual contributions.
There is one more
piece of terminology.
Each gas contributes to only
part of the total pressure
So any one gas is said to
contribute a partial pressure.
Dalton's Law of Partial
Pressure has these partial
pressures summing
to a total pressure.
We can show this with a
carton of a container that
contains two gases;
red sphere gas particle
and yellow sphere
gas particles.
Both sets of gas particles
are colliding with the sides
of the container
creating a total pressure.
Dalton argued that this
total pressure is the same
as the pressure in an equal
volume container that just
contains the red sphere
gas particles plus
the pressure in an equal
volume container that
just contains the
yellow sphere particles.
Alone the two gases
each have a pressure.
Together they each
have a partial pressure.
If partial pressure problems
come up at all in a student's
homework they generally
come up in one of two ways.
First is where the
student is asked to find
partial pressure
of just one gas.
The solution to the problem
has the student algebraically
isolating the unknown
partial pressure on one side
of the equation and then
solving by inserting known
pressure values.
If the partial pressure of
gas '3' is asked for then
isolating that variable
begins by subtracting all of
the other partial pressures
from both sides
of the equation.
The resulting math lets
the student enter known
values and solve.
The other type of question
ties into the previous topic;
ideal gas.
The Ideal Gas equation
can be solved for pressure
by diving both sides
of the equation by its
neighbor, volume.
Volume cancels out on
the left hand side and we have
another equation for pressure.
Bringing up the
original equation of
Dalton's Law of Partial
Pressure, we can merge the
two equations and have
another way to express
the total pressure.
This one requires the moles
of all the gases in the mix to
be summed together in
a total number of moles.
One might call these the
partial moles and treat them
just like the
partial pressures.
It is now time to look
at problem solving and
this is somewhat of a repeat
of the process outlined
in the 'Gas Law: 2
Condition lecture'.
This review will highlight
the differences in
solving the two
types of problems
and highlight solving 1
condition gas law problems.
As always, problem solving
begins with a careful reading
of the problem and an
identification of the known
and unknowns in the problem.
While it is not a necessary
step, it is helpful to ask
whether the problem
is dealing with 1
or 2 conditions.
It is not always clear
which type it is but a
telling sign is whether there
is a property, or
properties, with 2 values?
2 value properties will
use the 1 and 2 subscripts.
This type of problem
solving is covered in detail
in the previous lecture.
In this lecture we are going
to be concerned with 1
condition problems that have
the gas properties
with only 1 value.
For these problems the set-up
will be populated with the
variables in the Ideal
Gas law P, V, n, R, and T.
When boiled down, problems
using this set-up are finding
a single unknown property
when the other 3 properties
are provided for or
readily determined.
2 condition problems
want to know how
one property changes when
another property changes.
If it is still unclear
what type of problem
your dealing with just
populate the set-up
with the known and
unknowns and look for
gas law that relates them.
One note concerning
the gas constant R.
Choose an R that has
the same units as do
the properties in the set-up.
In the end, all but one of the
units in R will
have to cancel out.
And as with other gas related
calculations, all temperature
values need to be
converted into Kelvin.
Once the problem has been
scoured for knowns and
unknowns it is time to find
a mathematical relationship
that unites them.
Additionally, the mathematical
relationship will need to
be solved for the unknown.
With these terms in the set-up
the relationship should
be the Ideal Gas law.
3 of the 4 properties
are known; 1 is not.
Using the number of moles as
the unknown for
demonstration purposes,
the equation will need to be
modified to get 'n' isolated on
one side of the equation.
Fortunately, the Ideal Gas
Law does not have
a denominator,
like most of the 2 condition
gas law equations do,
and therefore solves
in a single step.
Divide both sides of the
equation by the neighbor or
neighbors of the unknown.
Sticking with the unknown
in the problem as the number
of moles divide both side
of the equation by
R T, (n's neighbors).
Canceling out the R and T
on the right hand side of the
equation produces the
modified Ideal Gas law
equation with the unknown
isolated on one side
of the equation.
It doesn't matter which side.
The final step is to
take that new equation and
insert values from the set-up.
Be sure to cancel units to
make sure the
orientation of the known
values are correct.
Let's do a couple problems.
What is the pressure (in atm)
inside a 10.0 liter container
that holds 4.00 moles of
gas at a temperature of
1000 degrees C?
Starting with the set-up.
Is this a 1 condition problem
or a 2 condition problem?
Sometimes it's easier to see
if there is a property that has
2 values which would
indicate that this just might
be a 2 condition problem.
Well, the problem has
a single value for volume,
a single value for amount,
and a single value
for temperature.
No value is given for the
fourth property pressure.
All single values. 3 of the 4
gas properties are given
and the fourth is asked for.
It is a one condition
problem and the set-up is
going to be populated
with the variables of the
Ideal Gas law.
As for the knowns;
starting with volume.
It is given as 10.0 l.
Amount is 4.00 moles.
Temperature is 1000
degrees C. That leaves the
fourth property pressure
as the unknown and units of
atmospheres are called for.
It's a bit tempting to take
on the gas constant
here but there is a
bit of business with
temperature to handle first.
The Celsius value needs to
be converted into Kelvin.
The equation for the
conversion has the 273.15
being added to the
Celsius temperature.
That will be 1000 + 273.
The point 15 is dropped
because the 1000 does not
have significant figures
in those positions. Rounding
doesn't need to occur here.
It can be dealt with in
the final rounding step.
An easy math step has
the temperature as 1273
which is added to the set-up.
To complete the set-up an
appropriate R has to be found.
Usually that involves
a search of a look up table.
We need to find an R that
has atm for pressure, liters
for volume and moles and K
for amount and temperature.
That one will do.
Add it in and the
set-up is complete.
Now to the relationship.
What law or equation relates
the 4 properties of a gas?
It is the Ideal Gas law
equation; PV equals nRT.
It will need to be modified
to have the unknown (pressure)
isolated on one
side of the equation.
There is no denominator
in the equation so there is
no need to do any
cross multiplying.
We can just jump to the
step that has us divided both
sides of the equation by
the neighbor of the unknown.
P is the unknown and
V is its neighbor.
Divide both sides
of the equation by V.
Cancel V's out on the left
hand side of the equation
and pressure isolated on
one side of the equation.
Use that equation
in the solve step.
From the set-up we
have an n of 4.00 moles,
times the gas constant R
times the temperature of 1273 K
all divided by the
volume of 10.0 liters.
One of the more confusing
aspects of dealing the
Ideal Gas law equation
is cancelling units with
the gas constant R. In
this gas law equation there
is a fraction or ratio of
units in the numerator of
the equation. Since
the R constant ratio is
in the numerator of the gas
law equation we can treat
the units in the numerator
ratio like they are in the
numerator of the gas
equation and the units in the
denominator of the R
ratio like they are in the
denominator of the
gas law equation.
With that liters cancel
out as do moles and K.
The only unit left in R is atm.
Multiplying 4.00 by
0.0821 by 1273 and then
dividing by 10.0 gives a
pressure value of 41.8 atm.
Given 3 of the 4 gas
properties, find the fourth.
While we have this problem
up we can investigate
how changing the unit
requirement for the unknown
variable pressure into torr
will alter the solution.
Adjusting the set-up for a
new pressure unit is a start
but the real change
needs to come from
the gas constant R.
The current version has units
of atm for pressure so this
version or R will not work.
Returning to the
look up table we see
an R that has torr as its
pressure unit without altering
the units of volume,
amount and temperature.
Place this new R in the
set-up and replace the old
R in the equation
in the solve step.
Notice that l, mole
and K still cancel out.
The new math comes
to 31,000 torr.
This is not a different
pressure from the one
calculated for
atmosphere units.
It is the same pressure
in different units
and it highlights the
importance of using
the appropriate R.
Onto second problem.
How many grams of
CO2 are present at STP
in a 5.0 l container?
There doesn't seem to
be much information here
but we should head on over to
the set-up and see what can
be teased out of the problem.
A starting question can be
whether this is a 1 condition
or a 2 condition problem.
Looking at the problem we
see a single temperature
and pressure presented in
the Standard Temperature
and Pressure (STP).
There is also a single volume.
There is no instance
where a gas property is given
two different values so this
is a 1 condition gas
problem and the set-up
will have the variables from
the Ideal Gas law equation.
Filling in those
values starts with STP.
These are set values
and are either remembered
or looked up.
Standard temperature is
0 degrees Celsius and
standard pressure is 1 atm.
An equivalent pressure
value is 101.3 kilopascals.
And then there is volume.
It's at 5.0 l.
3 of the 4 gas
properties are known.
The forth property,
amount in moles, is not.
Furthermore, amount
is asked for in grams
and not moles so
there is going to be an
additional unknown
and an additional step
in this problem.
Once we solve for moles
we will need to
convert that into grams.
This problem is more
complicated because
it has 2 unknowns.
There are still a few items
to deal with in the set-up.
Let's not forget to convert
the temperature into Kelvin
0.0 degrees plus 273.2.
There's a rounding to
2 in the last
significant digit.
The temperature
in Kelvin is 273.2.
The set-up will be completed
with the addition of the
appropriate gas constant
R. From the look up table,
which version of R has
pressure in kPa, volume in l
and moles and K for
amount and temperature?
It's this one right here.
Notice that none
of these R values
have amount in grams.
Everything is going
to go through the mole.
With known and unknowns in
hand a relationship is sought
and that is going to be
the Ideal Gas Law equation.
The unknown variable
moles of gas will need to
be isolated. Since there
is no denominator in
this equation we jump
to step 2 and divide both
sides of the equation by the
neighbors of the
unknown variable.
The unknown is n and
the neighbors are RT.
Dividing both sides by RT and
canceling RT out on
the right hand side
gives an equation with
our unknown isolated on
one side of the equation.
Take that equation
to the solve step
and insert know
values from the set-up.
Pressure, 101.3 kPa,
time volume, 5.0 l
divided by the
appropriate gas constant R
and temperature in K.
There may be an issue
with canceling units since
the fraction, or ratio,
of units in the R constant
are in the denominator
of the gas law equation.
In fraction math, a fraction
in the denominator can be
cleared or removed by
multiplying the equation by
the inverse of that fraction.
In other words
multiple the equation
by the flip of the units.
Now moles and Kelvin are
in the numerator and liters
and kPa are in the denominator.
That allows
canceling of liters,
kPa and K leaving just moles.
This does seem like a
fair amount of work to deal
with units so let's just
place the units, back where
they belong and recognize
that when the Ideal Gas law
equation is solve
correctly that units will
cancel properly.
With the unit set it is time to
take out the calculator and
do the math. 101.3 times 5.0
divided 8.3145 divided by 273.2
There are 0.223 moles
of CO2 in this container.
Add that to the set-up
because the problem is
not actually finished.
The problem asks for
grams of CO2.
Do we have enough
information in the set-up
to tackle that conversion?
How do we convert moles
of CO2 into grams of CO2?
This is a skill developed in
the 'Mole' series of lectures.
It says to use the
molecular weight of CO2
as a conversion factor.
Make sure the unit moles are
in the denominator so that they
cancel out leaving grams.
0.223 moles is 9.8 g of CO2.
When dealing with moles in
gas law the ability to move
between weight and
moles will be required.
And that ends the
material of the lecture.
As a recap, the Ideal Gas
law relates the 4 properties
of a gas, at 1 condition, to
a constant. Most of the time
that constant is
given as a capital R
It is called the gas constant.
It uses the units of the
4 properties and a numerical
value that relates the units.
If different units are used
for the gas properties then the
numerical value
could be different.
Make sure that the
appropriate R is used in a
calculation by
canceling out units.
The second topic of the
lecture was Dalton's Law of
Partial Pressure. It says
that the total pressure of gas
(P-total) is the sum
of the pressure of the
individual gases. Those
individual pressures are
called partial pressures.
Dalton's law and the
Ideal Gas equation can be
combined to give a total
pressure that is a reflection
of the sum of the moles of
all of the gases in the mix.
The Ideal Gas equation is
used to solve the gas law
problems in which the
properties of a gas
are at 1 condition.
In the set-up part of
the process, the known and
unknown values are gathered
and there should only be
1 value per
variable or property.
The set-up will contain the
variables from the Ideal Gas
Law equation with 3 of
the 4 properties having
known values; 1 unknown.
Be sure to use
appropriate R constant.
As for relating known and
unknown variables in these
problems it will most
likely be the Ideal Gas law.
The unknown variable is
solved in the equation by
dividing both sides of the
equation by the variables
next to, or the neighbors
of, the unknown variable.
The unknown property is
solved for by inputting
values from set-up.
Cancel units to verify
correct input of values.
And that completes our lecture.
Staying with that last remark.
Check units when using gas
law. It helps catch errors.
[MUSIC]
