>> narrator: In this video,
we'll take a closer look
at Gas Law calculations.
The first gas law we'll
talk about is Boyle's Law.
Boyle's Law tells us
that as the pressure
increases on a gas,
the volume of that gas
will decrease.
Pressure and volume
are inversely related,
according to Boyle's Law.
Let's work a sample problem
using Boyle's Law.
We have 20.5 liters
of nitrogen gas,
held at room temperature,
which is 25 degrees Celsius,
and 742 torrs of pressure.
That sample's pressure is
increased to 9.8 atmospheres.
Assume that temperature
is held constant.
What would be the new volume
at this pressure
of 9.8 atmospheres?
Remember Boyle's Law is P1 V1,
that is the pressure times
volume at one set of conditions,
equals P2 times V2,
that is the pressure times
volume at a second set
of conditions.
Let's plug in the values
that we're given for pressure
and volume and solve
for the unknown.
What we're trying to
solve for is a new volume,
in other words, V2.
Since we want to solve for V2,
we would like to isolate it
on one side of the equation,
and we're going to do that
by dividing both sides
of the equation by P2.
Doing so, we get
that P1 times V1
divided by P2 equals V2.
Now all we need to do
is plug in the values
that were given to us
in the problem
for P1V1 and P2.
You may have noticed
that the initial pressure
was given in units of torr.
And that the final pressure,
P2, was given in units
of atmosphere.
Both the torr and the atmosphere
are non-SI units of pressure.
One atmosphere is equal
to 760 torr.
When using any of these
gas law equations,
units must match
so that they cancel out,
so let's go ahead
and convert 742 torr
to atmosphere, so that
our pressure units match.
The conversion factor
we need is that there are
760 torr in one atmosphere,
so we can convert
from 742 torr to atmospheres
by placing what we start with,
the 742 torr in
the upper left hand corner
of a conversion chart,
then bringing down
the units of torr.
Place 760 torr in the
denominator and 1 atmosphere
in the numerator.
Then, when we multiply
742 x 1. We get 742,
and when we divide that by 760.
Our answer is 0.976,
and since the units
of torr canceled,
we are left with the units
of atmosphere.
And that tells us
that 742 torr is equal
to 0.976 atmosphere.
Plugging all our values
into the rearranged Boyle's Law,
we get that .976 atmosphere
times 20.5 liters
divided by 9.8 atmosphere
equals 2.0 liters.
You give it a try now.
Here's a real life problem
that operates on the
principles of Boyle's Law.
On October 14th 2012
an Australian skydiver
and daredevil,
Felix Baumgartner,
jumped to earth from
a helium balloon
in the stratosphere.
He set multiple records
on that date including being
the first person to break
the sound barrier without
vehicular power.
On the ground
before his takeoff,
Felix's team filled
his balloon with
5.10 x 10 to the 6th liters
of helium as he prepared
to take off.
The atmospheric pressure at
the launch was 1 atmospheres.
When he jumped from
approximately 39 km
from the earth's surface,
the volume of his balloon was
8.50 x 10 to the 8th liters.
What was the atmospheric
pressure on his balloon
the instant Felix began
his 4 minute and 19 second
free fall towards Earth?
When you're ready,
press pause, work the problem,
and then press play to see
if you got it right.
As you saw in the first
example, we rearranged
Boyle's Law to solve for
the desired variable.
In the last problem,
it was the new volume.
In this problem we are solving
for the new pressure,
or according to
the equation, P2.
P1, or the original pressure,
was 1 atmosphere, the initial
volume of the balloon was
5.10 x 10 to the 6th liters.
The new volume of the balloon
was 8.50 x 10 to the 8th liters
right before Felix jumped.
We needed to solve for
the final pressure,
so we rearranged Boyle's Law
to isolate P2
by dividing both sides
of the equation by V2.
Once we plugged in the values
that we had for P1, V1, and V2,
and solved, we found
that the external
atmospheric pressure
on Felix's balloon was
only 0.006 atmosphere.
This makes sense,
because at that altitude,
there aren't many gaseous
molecules that can apply
an external pressure
on his balloon.
The second gas law we're going
to look at is Charles' Law.
Charles's Law examines
the relationship between
volume and the absolute
temperature of a gas.
In this case, volume
and absolute temperature
are directly proportional
to one another.
That means as volume goes up,
temperature must also go up.
Let's work an example
using Charles's Law.
A gas that occupies 247 mL
at 22 degrees Celsius is
heated to 98 degrees Celsius.
Since pressure is
not mentioned in the problem,
we will assume pressure
is held constant.
What would be the new volume
at 98 degrees Celsius?
Looking at our rearranged
equation, that would be V2.
Let's write down Charles's Law
to remind us what we're
working with.
V1 over T1 equals V2 over T2.
Since we're trying to solve
for the new volume, or V2,
we'd like to isolate V2
on one side of the equation.
We're going to do that
by multiplying both sides
of the equation by T2
and that will give us
the rearranged Charles's Law
shown here.
Now we just plug in the values
that were given to us
in the problem.
Remember that temperature
must be in Kelvin.
The initial volume was 247 mL,
and the initial temperature
was 22 degrees Celsius.
We need to convert that
to Kelvin, which you do
by adding 273 to
the Celsius temperature.
Twenty two degrees Celsius
is the same as 295 Kelvin.
The final temperature
was 98 degrees Celsius,
again we need to convert
that to Kelvin, so we add 273
to the Celsius temperature
to convert it.
98 degrees Celsius is
the same as 371 Kelvin.
Now we have the equation
247 mL times 371
divided by 295.
And now we carry out
this calculation, and we
will get the new volume.
V1 is 247 mL.
T2 is 371 Kelvin.
Our final volume, therefore,
is 311 milliliters.
Just as we expected
from Charles' Law,
the volume of the gas
increased due to the increase
in temperature.
Here's a problem
for you to try.
A helium filled balloon
has a volume of 2.9 liters
at 33 degrees C.
The volume of the balloon
decreases to 2.48 liters
when it's placed outside
on a cold day.
What does the temperature
outside have to be in order
to cause the volume
of the balloon to decrease
to 2.48 liters?
Make sure you report
that temperature in Kelvin.
Pause the video
while you work this out
and when you're ready,
press play and check your work.
It's usually simpler to work
out these gas law calculations
by listing the given
information in the problem.
One of the most common
mistakes students make is not
matching the correct volume
and temperatures together
so be careful with that.
Your original volume
was 2.9 liters,
The new volume was 2.48 liters,
so volume has decreased.
What do we expect to happen
to the temperature according
to Charles's Law?
Remember they are
directly related so we would
expect the temperature
to also decrease.
So we're expecting T2
to be less than 306 Kelvin.
When you rearrange
Charles's Law to solve
for the new temperature
and plug in the values, you get
that T2 equals 261.7 Kelvin.
In fact, temperature did
go down from 306 Kelvin.
The final gas law we're
going to work with
in this Closer Look is
called the Gay-Lussac law.
We've examined all combinations
of the four variables
that describe the properties
of a gas except for pressure
and temperature together.
And that's exactly
what Gay-Lussac law does.
It turns out that pressure
and absolute temperature
are also directly related.
We've done enough
of these gas law problems
so that you can probably
work a sample problem
by yourself.
Let's give it a try.
According to the
National Football League,
NFL footballs must have
an internal pressure
of 0.85 to 0.92 atmospheres
to be considered legal.
Footballs at pressures
under 0.85 atmospheres
are considered illegal
because they are easier
for quarterbacks to grip
and throw.
If the temperature inside
the field house
in Foxboro Massachusetts,
where footballs are inflated
during pre-game,
is 294 K before kickoff
and the initial pressure
of the football is
0.88 atmospheres,
will the footballs be at
a legal pressure at halftime
of the game?
Assume the temperature
on the field is 275K.
What will the pressure of
the footballs be at halftime?
Go ahead and press pause,
and when you think you have
the answer, press play
and check your work.
Did you find that the final
pressure was 0.82 atmospheres?
If so, good work.
If not, let's take a look at
how we arrived at this answer.
The first thing we did
was rearrange Gay-Lussac's Law
so that P2 was isolated
by itself.
Next, we plugged in the values
given to us in the problem
for P1, T2, and T1,
and then we solved for
the final pressure, P2,
and found the pressure
inside the football was
0.82 atmospheres at halftime.
So we have evidence that
the football would have
a lower pressure due to
a lower temperature.
And that's a closer look
at Gas Laws Calculations
For more Chemistry Matters,
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available to you.