- 98 Views
- Uploaded on
- Presentation posted in: General

Gas Laws

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Gas Laws

Just means that gas is “pushing” on something.

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.

Tire

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.

Vacuum

Vacuum

U-Tube

It pushes down on this side, and it moves up on the other side.

So how do we measure it?

Can’t use it to measure atmospheric pressure, because atmospheric pressure presses on everything equally.

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.

Vacuum

760 mmHg

We can measure that!

Take a ruler and measure low to high in milimeters!

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

760 mmHg

800 mmHg

40 mmHg

What if I snap off the vacuum bulb?

Because atmospheric pressure is pushing down!

Barometer

Manometer

How do we measure things? Lots of ways! Same goes with gas pressure.

Gas Pressure Units

mmHgatmospherekilopascal

Torr

atmkPa

Conversions

760 mmHg = 1 atm = 101.3kpa

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 kPax 1 atm

101.3 kPa

=2.22 atm

225 kPax 760 mmHg

101.3 kPa

=1688 mmHg

If we keep the temperature the same, we can predict what pressure and volume will do.

Pressure and Volume

What about volume?

Gas particles have a bunch of room.

Gas particles are squeezed into smaller space.

P= Low

V=High

P=High

V=Low

As pressure goes up, volume goes down. That means inverse relationship.

Volume

Pressure

- When volume is high, pressure is low
- When the volume is low, pressure is high
- An Inverse relationship. Like when I buy clothes

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?

P1V1 = P2V2

(before) (after)

What do you know?

P1 (before pressure) =

1.70 atm

(1.70 atm)(4.35L)=(2.40 atm)V2

V1 (before volume)=

4.35 L

7.40atm/L = (2.40atm)V2

2.4 atm

P2 (after pressure) =

V2 =3.01L

V2 = ??

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!

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)

1.5atm

P1 (before pressure) =

5.6 L

V1 (before volume)=

(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 relates volume and temperature, while keeping pressure the same

V1 = V2

T1 T2

How could we test the theory that temperature and volume are related?

Think about kinetic theory and molecules.

COLD

HOT

What’s going on with the temp?

T= High

T = Low

V = Low

V= High

Charles’ law says that as the temp increases, so does volume.

A direct relationship.

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?

Must convert to Kelvin.

0 °C + 273 = 273K

80 °C + 273 = 353K

V1 =

625 L

T1 =

0°C

T2 =

80°C

??

V2=

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

625L= V2

273K 353K

V1 = 625 L

T1 = 273K

T2 = 353K

V2 = ??L

2.29L/K= V2

353K

808L = V2

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

What’s the equation?

V1=

6.00 L

Must convert to Kelvin.

27 °C + 273 = 300K

150°C + 273 = 423K

T1=

27°C

V2=

??

T2=

150.0°C

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

6.00L = V2

300K 423K

V1=

6.00 L

T1=

300K

0.02L/K = V2

423K

V2=

??

T2=

423K

8.46L = V2

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 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.

n1 = n2

V1 V2

n1 = n2

V1 V2

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?

n1=n2=

V1 = V2=

50g

mol?

79L

40L

When doing Avogadro's law, “n” MUST be in moles!

BeforeAfter

n1=50gn2 = g?

V1 = 48LV2 = 79L

n1 = n2

V1 V2

1.6mol

When doing Avogadro's law, “n” MUST be in moles!

50g O2x 1 mol O2

32g O2

= 1.6 mol O2

1.6 mol O2

48L

= n2

79L

0.03 = n2

79L

2.6 mol = n2

This law only applies to gases held at a constant volume. Only the pressure and temperature will change.

Pi =initial pressure

Pf = final pressure

Ti = initial temperature (kelvin)

Tf = final temperature (kelvin)

Pi = Pf

TiTf

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?

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?

Pi = Pf

TiTf

Must convert to Kelvin

20°C + 273 = 293K

48°C + 273 = 321K

Pi= 235 kPa

Pf= ?

Ti= 20°C

Tf= 48°C

235

293

= Pf

321

0.80 = Pf

321

Pi= 235 kPa

Pf= ?

Ti= 293K

Tf= 321K

257.5 kPa = Pf

Charle’s Law

V1 = V2

T1 T2

They are all pretty much the same equation, just different variables!

Avogadro’s Law

V1 = V2

n1n2

Write the equations for each variable and for each law.

Gay Lussac’s Law

P1 = P2

T1 T2

Charle’s Law

V1 = V2

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.

Boyle’s Law

(P1)(V1) = (P2)(V2)

I can combine the laws!

Gay Lussac’s Law

P1 = P2

T1 T2

Combined Gas Law

(P1)(V1) = (P2)(V2)

T1 T2

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

25°C + 273 = 298K

4°C + 273 = 277K

(P1)(V1) = (P2)(V2)

T1 T2

P1= 1 atm

P2= 11 atm

V1= 40 L

V2= ??

T1= 25 °C

T2= 4 °C

P1= 1 atm

P2= 11 atm

V1= 40 L

V2= ??

T1= 298K

T2= 277K

(1)(40) = (11)(V2)

298K277K

0.13 = (11)(V2)

277K

36.01 = (11)(V2)

3.27 L = V2