### Gas Laws

```Gas Laws
Gas Pressure
Just means that gas is “pushing” on
something.
Gas Pressure
What’s going on inside?
Air:
Nitrogen 78%
Oxygen 21%
Argon ~1%
Carbon Dioxide <1%
Each of these particles
are constantly flying
around. Like a lotto ball!
Tire
They slam against
the container and
keep the tire “full”.
The particles press
against the walls.
Measuring Gas Pressure
Air:
Nitrogen 78%
Oxygen 21%
Argon ~1%
Carbon Dioxide <1%
Think of a giant ball pit
miles and miles up.
At the bottom of the ball pit, is like us walking around.
That’s the atmospheric pressure.
Measuring Gas Pressure
Vacuum
U-Tube
Can’t use it to
measure atmospheric
pressure, because
atmospheric pressure
presses on everything
equally.
So how
do we
measure
it?
Vacuum
It pushes down on
this side, and it
moves up on the
other side.
Measuring Gas Pressure
Vacuum
760 mmHg
We can measure that!
Take a ruler and
measure low to high in
milimeters!
The fluid that is contained
in this U tube, is mercury.
If we measure this at sea
level, we get. 760mmHg
between the bottom and
the top.
Measuring Gas Pressure
What if we go up a mountain or down into a mine?
Think about that ball pit again. If you’re at the bottom of
the ball pit will it weigh more or less than at the top?
Sea Level
More
Pressure
760mmHg
Less
Pressure
Measuring Gas Pressure of Containers
760 mmHg
800 mmHg
What if I snap off
the vacuum bulb?
40 mmHg
Because atmospheric
pressure is pushing down!
Measuring Gas Pressure
Barometer
Manometer
Gas Pressure Conversions
How do we measure things? Lots of ways!
Same goes with gas pressure.
Gas Pressure Units
mmHg
atmosphere
Torr
atm
Conversions
760 mmHg = 1 atm = 101.3kpa
kilopascal
kPa
Gas Pressure Conversions
The pressure inside a car tire is 225 kPa.
Express this value in both atm and mmHg.
760 mmHg = 1 atm = 101.3 kPa
225 kPa x 1 atm
=2.22 atm
101.3 kPa
225 kPa x 760 mmHg
=1688 mmHg
101.3 kPa
Boyle’s Law
If we keep the temperature the
same, we can predict what
pressure and volume will do.
Boyle’s Law
Pressure and Volume
Gas particles have a
bunch of room.
P= Low
V=High
Gas particles are squeezed
into smaller space.
P=High
V=Low
As pressure goes up, volume goes down. That means
inverse relationship.
Boyle’s Teeter Totter
Volume
Pressure
• When volume is high, pressure is low
• When the volume is low, pressure is high
• An Inverse relationship.
Boyle’s Law
Boyle’s law is explained by the equation P1V1=P2V2
Let’s get right to it!
At 1.70 atm, a sample of gas takes up 4.35 L. If the
pressure on the gas is increased to 2.40 atm, what
will the new volume be?
What do you know?
P1V1 = P2V2
(before) (after)
P1 (before pressure) =1.70 atm
(1.70 atm)(4.35L)=(2.40 atm)V2
V1 (before volume)= 4.35 L
P2 (after pressure) = 2.4 atm
7.40atm/L = (2.40atm)V2
V2 = ??
V2 =3.01L
Boyle’s Law
At 1.70 atm, a sample of gas takes up 4.35 L.
If the pressure on the gas is increased to 2.40
atm, what will the new volume be?
We increased the pressure, so we pushed down that
piston. We squeezed the molecules into a smaller
space. So the volume should go down!
Boyle’s Law
If I have 5.6 liters of gas in a piston at a pressure of
1.5 atm and compress the gas until its volume is 4.8 L,
what will the new pressure inside the piston be?
P1V1 = P2V2
(before) (after)
P1 (before pressure) = 1.5 atm
V1 (before volume)= 5.6 L
(1.5atm)(5.6L) = (P2)(4.8L)
?
P2 (after pressure) =
8.4 atm/L = (4.8L)P2
V2 = 4.8L
1.8 atm = P2
Charles’ Law
Charles’ law relates
volume and
temperature, while
keeping pressure the
same
V1 = V2
T1 T2
Charles’ Law
How could we test the theory that temperature and
volume are related?
Think about kinetic theory and molecules.
Charles’ Law
COLD
HOT
What’s going on
with the temp?
T= High
V= High
T = Low
V = Low
Charles’ law says that as the temp increases, so does
volume.
A direct relationship.
Charles’ Law
So now we can relate volume and temperature.
V1 = V2
T1
T2
MUST ALWAYS USE KELVIN TEMPERATURE in gas laws
A balloon takes up 625 L at 0°C. If it is heated to
80°C, what will its new volume be?
V1 = 625 L
T1 = 0 °C
T2 = 80 °C
V2 = ??
Must convert to Kelvin.
0 °C + 273 = 273K
80 °C + 273 = 353K
Charles’ Law
A balloon takes up 625 L at 0°C. If it is heated to 80°C,
what will its new volume be?
V1 = V2
T1
T2
V1 = 625 L
T1 = 273K
T2 = 353K
V2 = ??L
625L = V2
273K 353K
2.29L/K= V2
353K
808L = V2
Charles’ Law
At 27.00 °C a gas has a volume of 6.00 L. What
will the volume be at 150.0 °C?
V1 = V2
What’s the equation?
T1
T2
V1= 6.00 L
T1= 27 °C
V2= ??
T2= 150.0 °C
Must convert to Kelvin.
27 °C + 273 = 300K
150°C + 273 = 423K
Charles’ Law
At 27.00 °C a gas has a volume of 6.00 L. What
will the volume be at 150.0 °C?
V1 = V2
T1
T2
V1= 6.00 L
T1= 300K
V2= ??
T2= 423K
6.00L = V2
300K
423K
0.02L/K = V2
423K
8.46L = V2
Relationship between:
Amount of gas (n) and the Volume.
What happens to one, when I change the other?
air into it…will the volume increase?
Yes, a direct relationship
As the amount (in moles) goes up, so does the volume.
If we double the amount, it doubles the volume.
We only changed TWO things.
The volume and the amount of particles.
We didn’t mess with the pressure or the temperature,
they were held constant.
V1 = V2
n1
n2
V1 = V2
n1
n2
n1= 50g
V1 = 40L
Let’s try!
In a sample of gas, 50.0 g of oxygen gas (O2)
take up 48L of volume. Keeping the pressure
constant, the amount of gas is changed until
the volume is 79 L. How many mols of gas
are now in the container?
n2 = mol?
V2 = 79L
When doing Avogadro's law, “n” MUST be in moles!
V1 = V2
n1
n2
Before
1.6mol
n1=50g
V1 = 48L
After
n2 = g?
V2 = 79L
When doing Avogadro's law, “n” MUST be in moles!
50g O2 x 1 mol O2
= 1.6 mol O2
32g O2
1.6 mol O2 = n2
48L
79L
0.03 = n2
79L
2.6 mol = n2
Gay-Lussac’s Law
The pressure and Kelvin
temperature of a gas are
directly proportional, when
the volume remains
constant.
Gay Lussac’s Law
This law only applies to gases held at a constant volume.
Only the pressure and temperature will change.
P 1 = P2
T1
T2
Pi =initial pressure
Pf = final pressure
Ti = initial temperature (kelvin)
Tf = final temperature (kelvin)
The pressure in a sealed can of gas is 235 kPa when it
sits at room temperature (20C). If the can is warmed to
48C, what will the new pressure inside the can be?
Gay Lussac’s Law
The pressure in a sealed can of gas is 235 kPa when it sits at
room temperature (20°C). If the can is warmed to 48°C, what
will the new pressure inside the can be?
P1 = P2
T1
T2
P1 = 235 kPa
P2 = ?
T1 = 20°C
T2 = 48°C
Must convert to Kelvin
20°C + 273 = 293K
48°C + 273 = 321K
The pressure in a sealed can of gas is 235 kPa when it sits at
room temperature (20°C). If the can is warmed to 48°C, what
will the new pressure inside the can be?
P1 = P2
T1
T2
P1 = 235 kPa
P2 = ?
T1 = 293K
T2 = 321K
235 = Pf
293
321
0.80 = Pf
321
257.5 kPa = Pf
How to use these formulas
Charle’s Law
V1 = V2
T1
T2
V1 = V2
n1
n2
Gay Lussac’s Law
P1 = P2
T1
T2
They are all pretty much the
same equation, just different
variables!
Combined Gas Law
Charle’s Law
V1 = V2
T1
T2
Boyle’s Law
(P1)(V1) = (P2)(V2)
Gay Lussac’s Law
P1 = P2
T1
T2
What if I had a balloon. I
wanted to increase the pressure
and cool it down. What is the
volume? Do we have an
equation for that? P, T, V.
We can combine the laws!
Combined Gas Law
(P1)(V1) = (P2)(V2)
T1
T2
Combined Gas Law
A 40.0L balloon is filled with air at sea level (1.00
atm, 25.0 °C). It's tied to a rock and thrown in a a
cold body of water, and it sinks to the point where
the temperature is 4.0 ° C and the pressure is 11.00
atm. What will its new volume be?
Convert to Kelvin
(P1)(V1) = (P2)(V2)
25°C + 273 = 298K
T1
T2
4°C + 273 = 277K
P1P== 11 atm
atm
(1)(40) = (11)(V2)
1
P2P=
11 atm
298K
277K
2=
V1V=
40 L
0.13 = (11)(V2)
1= 40 L
V2V=
277K
2= ??
36.01 = (11)(V2)
T1T=
1= 298K
25°C
2= 277K
T2T=
3.27 L = V2
4°C
Ideal Gas Law
How can we describe what’s going on
in this container? What variables can
we think of?
Temperature (T)
313K
Pressure (P)
3.18 atm
95.2 L
Volume (V)
Amount of Gas (n) 7.5 mol
Did you know that if we know
3 of the 4 variables, we can
find the last one?
Ideal Gas Law
Ideal gas law: PV = nRT
Temperature (T)
313K
Pressure (P)
3.18
?? atm
Volume (V)
95.2 L
Amount of Gas (n) 7.5
?? mol
Whatwould
How
if we needed
we rearrange
the the
problem of
amount
to gas
find(n)?
P?
P =nRT
V
PV = n
RT
Ideal Gas Law
PV = nRT
So what is R?
R is a constant! For most cases,
R = 0.0821 L▪atm/mol ▪K
V=L
P = atm
T=K
n = mol
Those units
look familiar.
The units on “R”
MUST match the
units in the
problem!
Ideal Gas Law
“R” will come in many forms.
R = 62.4 L▪mmHg /K ▪mol
R = 8.31 L▪kPa /K ▪mol
NOT A BIG DEAL! The “R” constant will always be
given, just use the right constant.
Ideal Gas Law
2.3 moles of Helium gas are at a pressure of 1.70 atm,
and the temperature is 41°C. What is volume of the
gas?
PV = nRT
P = 1.70 atm
V = ??
n = 2.3 mol
R = 0.0821 L▪atm/K ▪mol
T = 41°C
Convert to Kelvin
41°C + 273 = 314K
Ideal Gas Law
2.3 moles of Helium gas are at a pressure of 1.70 atm,
and the temperature is 41°C. What is volume of the
gas?
PV = nRT
Rearrange the equation.
P = 1.70 atm
V = nRT
P
V = ??
V = (2.3 mol)(314K) x 0.0821 L ▪atm
n = 2.3 mol
1.70 atm
K ▪ mol
R = 0.0821 L▪atm/K ▪mol
V = 59.3
T = 314K
1.7
V = 34.9 L
Ideal Gas Law
At a certain temperature, 3.24 moles of CO2
gas at 2.15 atm takes up a volume of 35.28
L. What is this temperature (in Celsius)?
P = 2.15 atm
V = 35.28 L
T = ??
Do the units given match
the R?
n = 3.24 mol
R = 0.0821 L▪atm/K ▪mol
Ideal Gas Law
PV = nRT
At a certain temperature, 3.24 moles of CO2
gas at 2.15 atm takes up a volume of 35.28
L. What is this temperature (in Celsius)?
P=
V=
T=
2.15 atm
35.28 L
??
Rearrange the equation.
T = PV
nR
T = (2.15 atm)(35.28L) X K ▪ mol
n = 3.24 mol
(3.24 mol)
0.0821 L ▪ atm
R = 0.0821 L▪atm/K ▪mol
Ideal Gas Law
Charle’s Law
V1 = V2
T1
T2
V1 = V2
n1
n2
Who wants to memorize
all of these?!?!
Gay Lussac’s Law
P1 = P2
T1
T2
Ideal Gas Law
PV = nRT
Combined Law
(P1)(V1) = (P2)(V2)
T1
T2
You don’t have to!
Gas Law
Just memorize one!
Ideal Gas Law
PV = nRT
Can use it for any of the gas
law problems!
Warning:
If this blows your mind and you get totally confused,
just memorize the equations.
Gas Law
Before
After
P1 = 3 atm P2 = 7atm
T1 = ??
T2 = 150k
PV =nRT
PV = nRT
V
V
P = nRT
T
VT
P= nR
T V
Rearrange the ideal equation
so that the variables given are
on the same “side”
You’ve found the equation
you need to use. You don’t
need “n, R, or V”.
P1 = P2
T1
T2
Gas Law
PV = nRT
P1 = 1,217 mmHg
P2 = 732 mmHg
V1 = ??
V2 = 42L
Rearrange the
equation so the
variables you’re
looking for are on
the same side of the
equation.
Easy! PV is already on the same side. Now just double it.
P1V1 = P2V2
Gas Law
PV = nRT
V1 = 7.5L
V2 = 1.2L
n1= 32 mol
n2 = ?
Rearrange the equation so V
and n are on the same side.
PV = nRT
P
P
V = nRT
P
V = nRT
n
Pn
V1 = V2
n1 n2
Gas Law
Before
V1 = ?
P1 = 96 kPa
T1 = 12K
After
V2 = 54L
P2 = 112 kPa
T2 = 42K
PV = nRT
T
T
P1V1 = P2V2
T1
T2
PV = nRT
Rearrange so V, T, P are
on same side.
```