Gas Pressure Just means that gas is pushing on something Gas Pressure Tire Whats going on inside Air Nitrogen 78 Oxygen 21 Argon 1 Carbon Dioxide lt1 Each of these particles are constantly flying around Like a lotto ball ID: 621950
Download Presentation The PPT/PDF document "Gas Laws" is the property of its rightful owner. Permission is granted to download and print the materials on this web site for personal, non-commercial use only, and to display it on your personal computer provided you do not modify the materials and that you retain all copyright notices contained in the materials. By downloading content from our website, you accept the terms of this agreement.
Slide1
Gas LawsSlide2
Gas Pressure
Just means that gas is “pushing” on something.Slide3
Gas Pressure
Tire
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!
They slam against the container and keep the tire “full”. The particles press against the walls.Slide4
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. Slide5
Measuring Gas Pressure
U-Tube
Can’t use it to measure atmospheric pressure, because atmospheric pressure presses on everything equally.
Vacuum
So how do we measure it?
Vacuum
It pushes down on this side, and it moves up on the other side.Slide6
Measuring Gas Pressure
Vacuum
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.
760 mmHgSlide7
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 PressureSlide8
Measuring Gas Pressure of Containers
800 mmHg
40 mmHg
What if I snap off the vacuum bulb?
Because atmospheric pressure is pushing down!
760 mmHgSlide9
Measuring Gas Pressure
Barometer
ManometerSlide10
Gas Pressure Conversions
How do we measure things? Lots of ways! Same goes with gas pressure.
Gas Pressure Units
mmHg atmosphere kilopascalTorr
atm
kPaConversions760 mmHg = 1
atm = 101.3kpaSlide11
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
kPa225 kPa
x 1 atm
101.3 kPa
=2.22
atm
225 kPa x 760 mmHg 101.3 kPa
=1688 mmHgSlide12
Boyle’s Law
If we keep the temperature the same, we can predict what pressure and volume wil
l do.Slide13
Boyle’s Law
Pressure and Volume
Gas particles have a bunch of room.
Gas particles are squeezed into smaller
space.
What about volume?
V=High
V=Low
As pressure goes up, volume goes down. That means
inverse
relationship.
P= Low
P=HighSlide14
Boyle’s
Teeter Totter
When volume is high, pressure is low
When the volume is low, pressure is
high
An Inverse relationship. Like when I buy clothes
Pressure
VolumeSlide15
Boyle’s Law
Boyle’s law is explained by the equation P
1
V1=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?
P
1V1 = P2
V2(before) (after)What do you know?
P1 (before pressure) = V1 (before volume
)=P2 (after pressure) =
V2 = ??(
1.70 atm)(4.35L)=(2.40 atm)V2
7.40atm/L = (2.40atm)V2V2 =3.01L
1.70 atm
4.35 L2.4 atmSlide16
Boyle’s Law
Does that answer make sense?
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!Slide17
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?
P1V
1
= P
2
V2(before) (after)P
1 (before pressure) =
V1 (before volume
)=P2 (after pressure) =
V2 =(
1.5atm)(5.6L) = (P2)(4.8L)8.4
atm/L = (4.8L)P21.8 atm
= P21.5 atm
5.6 L?
4.8LSlide18
Charles’ Law
Charles’ law relates volume and
temperature
, while keeping pressure the same
V
1
=
V
2
T
1
T2Slide19
Charles’ Law
How could we test the theory that temperature and volume are related?
Think about
kinetic theory and molecules.Slide20
Charles’ Law
HOT
COLD
T= High
T = Low
V= High
V = Low
Charles’ law says that as the
temp increases, so does volume.
A
direct relationship.
What’s going on with the temp?Slide21
Charles’ Law
So now we can relate volume and temperature.
V
1 = V2T
1
T
2
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?
Must convert to Kelvin.0 °C + 273 = 273K80 °C + 273 = 353K
625 L0 °C
??V1 =
T1 =
T2 = V2 =
80 °CSlide22
Charles’ Law
V
1 =
V2T1 T
2
A balloon takes up 625 L at 0°C. If it is heated to 80°C, what will its new volume be?
V
1
= 625 LT1 =
273KT2 = 353K
V2 = ??L625L = V
2273K 353K2.29L/K= V2
353K808L = V2Slide23
Charles’ Law
At 27.00 °C a gas has a volume of 6.00 L. What will the volume be at 150.0 °C?
What’s the equation?
V1
=
V
2
T1 T2
V1=
T1=
V2=
T2=6.00 L
27 °C??
150.0 °CMust convert to Kelvin.27
°C + 273 = 300K150°C + 273 = 423KSlide24
Charles’ Law
At 27.00 °C a gas has a volume of 6.00 L. What will the volume be at 150.0 °C?
V
1 = V2
T
1
T
2V1=
T1
=
V2=T2=
6.00 L??
300K423K
6.00L = V2300K 423K
0.02L/K = V2 423K
8.46L = V2Slide25
Avogadro’s Law
Relationship between:
Amount of gas (n) and the Volume.
What happens to one, when I change the other?
I start with the first balloon, and then blow more air into it…will the volume increase?
Yes, a direct relationshipSlide26
Avogadro’s Law
As the amount (in moles) goes up, so does the volume.
If we double the amount, it doubles the volume.Slide27
Avogadro’s Law
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. n1
=
n
2
V1 V2Slide28
Avogadro’s Law
n
1 =
n2 V1 V
2
Let’s try!
In a sample of gas, 50.0 g of oxygen gas (O
2) 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?
n1=
n2 = V1 = V
2 = When doing Avogadro's law, “n” MUST be in moles!
50g40L
mol?79LSlide29
Avogadro’s Law
n
1 =
n2 V1 V
2
Before
Aftern1=50g n
2 = g?V1
= 48L V2 = 79L
When doing Avogadro's law, “n” MUST be in moles!50g O2 x
1 mol O2 32g O2
= 1.6 mol O2
1.6mol1.6 mol O2
48L= n2
79L0.03 = n2
79L2.6 mol = n2Slide30
Gay
Lussac’s
Law
This law only applies to gases held at a constant volume. Only the pressure and temperature
will change.
P
i
= Pf
Ti T
fP
i =initial pressurePf = final pressureTi = initial temperature (kelvin)T
f = 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?Slide31
Gay
Lussac’s
LawThe 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?P
i
=
PfTi T
fP
i= 235 kPa
Pf= ?Ti= 20°CTf= 48°C
Must convert to Kelvin20°C + 273 = 293K48°C + 273 = 321K
Pi= 235 kPaPf= ?Ti
= 293KTf= 321K235293
= Pf 321
0.80 = Pf 321
257.5 kPa = PfSlide32
How to use these formulas
Charle’s
Law
V1 = V2T
1
T
2
Avogadro’s LawV1 =
V2n1
n2
Gay Lussac’s LawP1 = P2
T1 T2They are all pretty much the same equation, just different variables!
Write the equations for each variable and for each law.Slide33
Combined Gas Law
Charle’s
Law
V1 = V2T
1
T
2
Boyle’s Law(P1)(V
1) = (P2)(V
2)
Gay Lussac’s LawP1 = P2T
1 T2What 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.
I can combine the laws!Combined Gas Law
(P1)(V1) = (P2)(V
2) T1 T2Slide34
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?(P
1
)(V
1
) = (P2)(V2
) T1 T
2
P1= 1 atmP2= 11 atm
V1= 40 LV2= ??T1
= 25 °CT2= 4 °CConvert to Kelvin25°C + 273 = 298K
4°C + 273 = 277KP1= 1 atm
P2= 11 atmV1= 40 LV2
= ??T1= 298KT2= 277K
(1)(40) = (11)(V2) 298K 277K
0.13 = (11)(V2) 277K
36.01 = (11)(V2)3.27 L = V2