The gas laws

Chapter 5 of AP chemistry

Gasses

We are surrounded by gasses. We would die if we were released into

space where there is no gas pressure Because much of early chemistry involved

observations of gas behavior, there are many different units of pressure

Early measurements of pressure

Pressure was first measured with a meter long tube filled with mercury.

This was the first barometer. I’ll explain how it works.

Units of pressure

The standard unit of pressure here in the States is the atmosphere (atm)

1 atm = 760 mm Hg

760 torr

101,325 Pa (Pascal)

14.7 PSI (pound per square inch)

1.01325 bar

Boyle’s law

Boyle found that the pressure and the volume of a gas are inversely proportional.

This makes perfect sense. If you crush a plastic bottle with the top on, what happens to the pressure in the bottle?

Let’s look at the graph.

The equation

P*V=K This law states that the product of pressure

and volume will be a constant value This leads to a more useful equation

P1V1=K=P2V2

OR

P1V1=P2V2

Charles’s law

Jacques Charles determined that the temperature and the volume of a gas had a directly proportional relationship.

Again this makes sense. When you heat up a gas the particles move faster and they expand their container (if they can).

We’ll look at a graph.

Absolute Zero

Charles did his work in the late 1700’s, well before modern refrigeration.

He did note that all his separate gasses shared a common x-intercept on the graph

That x-intercept represented the temperature at which all substances would occupy zero volume, absolute zero.

Absolute Zero

-273.15 °C

Kelvins

Whenever temperature is present in gas law calculations it should be expressed in Kelvins.

°C + 273= K IMPORTANT: When your temperature

increases from 25 °C to 50 °C it DID NOT double!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

25 °C to 50 °C

25 °C = 273 + 25 = 298 K 50 °C = 273 + 50 = 323 K 298 K to 323 K is not a doubling of

temperature. NEVER EVER use °C in gas law problems

unless you’re looking up vapor pressure or something else on a chart.

The equation

The useful equation from Charles’s law is:

This equation can be combined with Boyle’s law to form the combined gas law

2

2

1

1

2

2

1

1

T

V

T

V

Or

T

VK

T

V

The combined gas law

So far we have two equations:

KPV andK T

V

These two can be combined to form the combined gas law

2

22

1

11

T

VP

T

VP ORK

T

PV

Example problems

1. A helium-filled balloon has a volume of 50.0 L at 25 °C and 1.08 atm. What volume will it have at 0.885 atm and 10.0 °C?

2. A hydrogen-filled zeppelin used on bombing missions in WW I had a volume of 4550 m3 at 25 °C and 1.00 atm. What volume will it have if it flies into a low pressure zone with a pressure of 0.95 atm and 20.0 °C?

Example problems

1. A 500 mL water bottle filled with air at 25 °C is at a pressure of 0.877 atm. What pressure would result if the volume is reduced to 300 mL and the temperature is raised to 37 °C?

2. A balloon filled with nitrogen gas is placed in a freezer at 0 °C, and 1 atm. The balloon has a volume of 255 mL in the freezer. Once it is removed from the freezer it is placed in warm water where it warms to the temperature of the water and it expands to a volume of 400 mL, and as a result, the pressure on the balloon is also increased to 2.5 atm? What is the final temperature of the balloon? What’s wrong with this problem?

Avogadro’s law

Avagadro found that for a gas at constant temperature and pressure, the volume is directly proportional to the number of moles of gas.

This is written as

2

2

1

1 VV ORK

V OR KV

nnnn

Where V is volume and n is the number of moles, and K is a constant.

Ramifications

It may seem strange but all gasses occupy the same volume if they are under the same pressure and temperature.

UF6 has a molar mass of over 300 g/mol yet a mole of it occupies the same volume as a mole of H2 which has a molar mass of just over 2 g/mol. We’ll get into how later.

Example

Suppose we have a 12.2-L sample containing 0.50 mol oxygen gas (O2) at a pressure of 1 atm and a temperature of 25 °C. If all this O2 were converted to ozone (O3) at the same temperature and pressure, what would be the volume of the ozone?

Solution

1) Write the equation

3(g)2(g) O23O

2) Determine moles of O3 produced

3O mol 33.02

5.0

3 x

x

3) Solve Avogadro’s law

L 8.1V33.0

V

5.0

12.2VV 2

2

2

2

1

1 nn

Given: V1= 12.2 L, 0.5 mol O2

Want: V2

Need: Mol O3

Will use: 2

2

1

1 VV

nn

Example problem

If 1 mole of water vapor occupies 30.0 liters at a given pressure and temperature, what volume would be occupied by a mole of water if it were decomposed into hydrogen and oxygen gas?

What are you Given? What do you want? What do you need? What will you use?

The ideal gas law

PV=nRT… Do you remember me??? We’ll look at the laws that we’ve studied so far:

Kn

V :Law sAvogadro'

KT

V :Law sCharle'

KPV :Law sBoyle'

nRTPV

orK nT

PV

:law gas Ideal

R is the gas constant

PV=nRT This is the catch all gas law. This law

incorporates all the other gas laws into one nice neat equation.

This law should be applies when mass, or number of moles of gas is mentioned.

As always, there are easy problems and there are hard problems

PS We’re always going to put pressure in atm from now on.

Easy problems

1. A 5.00 mole sample of argon is put in a steel tank with a volume of 100. L. Once the gas fills the container the temperature is found to be 25 ºC. If the container is opened, would air from the outside world be sucked in or pushed out?

2. How many moles would be sucked in or pushed out?

Easy problems Givens:

n = 5.00 mole V = 100. L T = 25 ºC R = 0.08206 P = ???

If the P < 1 atm, are is sucked in, if P > then argon is pushed out.

Plug and chug Givens:

n = 5.00 mole V = 100. L T = 25 ºC298K R = 0.08206 P = ???

atmP

P

nRTPV

22.1.)100(

)298)(08206.0)(00.5(

)298)(08206.0)(00.5(.)100(

When the container argon will be pushed out of the container so as to equalize the pressure.

A little harder problem

How much argon will be pushed out?Let’s think about this. What will happen to the

pressure in the tank once the tank is opened to the outside world?

It will equalize with the pressure outside the tank.

How will it accomplish this? It will release some of the gas (moles)

Set up

P1= 1.22 atm

V1= 100. L

T1 = 298K R = 0.08206 n1 = 5.00

P2= 1.00 atm

V2= 100. L

T2 = 298K R = 0.08206 n2 = ???

A lot of stuff cancels.22

22

11

11

RTn

VP

RTn

VP

Plug and chug

molatm

molatmn

n

atm

mol

atm

10.422.1

)00.5)(00.1(

00.1

00.5

22.1

2

2

You started with 5.00 mol and ended with 4.10 mol.

You she 0.90 mol of argon. You can convert that to grams if you’re asked to

Alternatively

You could also simply apply the ideal gas law to the new container

Givens:P=1.00atmV=100. LT=298 KR=0.08206n=???

moln

n

n

nRTPV

09.4

29808206.0

.10000.1

29808206.0.10000.1

Gas Stoichiometry

What can we now calculate using the ideal gas law?

Now we can do stoichiometry

Sample problem Two chambers are mixed. The first occupies a volume

of 25.0 L and contains ethane gas (C2H6) under 2.00 atm of pressure and a temperature of 20.0 ºC.

The second chamber is three times as large as the first and contains oxygen at STP (standard temperature and pressure, 0 ºC, and 1 atm [know this]).

The gases are mixed, a spark is lit, the whole thing explodes in a massive combustion reaction, but the container is undamaged. The container is cooled to 0 ºC. What is the pressure in the container?

Write and balance the equation, find moles of each reactant, determine the limiting reagent and calculate the moles of CO2 produced. You have T, V, and n. Solve for P

A note on chemical reactions

Decomposition of carbonatesThe carbonate ion (CO3

2-) forms ionic bonds with many metals.

When these metals are heated the carbonate ion breaks down into the metal oxide and carbon dioxied gas.

Here’s the example of copper (II) carbonate

2(g)(s)or heat

3(s) COCuOCuCO

We’ll work a problem

A carbonate has an unknown metal cation. You find that upon thermal decomposition of 57.3705g of the carbonate 10.25 L of carbon dioxide can be collected at STP.

What mass of metal oxide would be left behind?

2(g)(s)or heat

3(s) COMOMCO

2(g)(s)or heat

3(s) COMOMCO

First convert L of CO2 to moles

Next find the mass of CO2 and subtract it from the original mass of the carbonate. Done!

2CO mol 0.4575x10.25L

x

22.4L

1mol

grams 20.13mol 1

44.01g*CO mol 0.4575 2(g)

MO of 37.24g20.13g57.3705g

Let’s make it harder

As you will soon see in lab, a metal oxide can be liberated from its oxygen by heat in an atmosphere of hydrogen or methane gas.

If upon liberation of the oxygen in a hydrogen atmosphere 8.244g of water are produced. If I tell you that the oxygen and the metal exist in a 1:1 ratio, what metal do you have?

First you must find the mass of oxygen lost. For that you need moles of oxygen.

OHMHMO 2heat

2

oxide fromoxygen g32.7mol 1

16.00g*O mol 0.4575

OH mol 4575.018.02g

mol 1*8.244g 2

You now know the mass of the oxide and the mass of oxygen. So you can solve for the mass of the metal.

37.24g(mass of MO) -7.32g (mass of oxygen)

Mass of metal in oxide =29.92g

The end at last

If you know the mass of the metal and you know that the metal and oxygen exist in a 1:1 ratio in the oxide you have all you need.

OH mol 4575.018.02g

mol 1*8.244g

OHMHMO

2

22

We know that the metal and water will be produced in equal molar quantities. Thus we know the mass and the mole of metal. So we solve for the molar mass. (Zinc) 65.40MM

metal mol 0.4575

metal g 29.92MM

mol

gmassMolar

Molar mass of a gas

If you know the density of a gas you can find its molar mass.

Recall D=mass/V

Also recall PV=nRt

Forget not MM=mass/n See pg 205 for more

This is a useful trick

A gas is found to have a density of 2.68 g/L at a temperature of 27 ºC under 1.50 atm of pressure. What’s its molar mass?

Dalton’s law of partial pressures

This law states that the total pressure is equal to the sum of all its parts. This may seem obvious but, it is very useful.

The equation is below:

321T PPPP

More Dalton

We’ll look at a former student’s work. He did a good job.

Those are basic problems

The problems get harder. Here you go. Aluminum will react with hydrochloric acid

to produce aluminum chloride and hydrogen gas. If the reaction takes place at a temperature is 20.0 ºC and a pressure is 752 torr, what mass of aluminum would be required to produce 20.00 L of gas if collected by water dispacement.

Vapor pressure of water at 20.0 ºC is 17.54 torr.

Step 1: Stoichiometry

23 H3AlCl2HCl62Al

2H mol 780.23

1.853

2

Al mol 853.126.99g

mol 1*Al 50.00g

xx

Step 2: Find partial pressure of H2

PT=PH2O+PH2

Look up PH2O at 20.0 it’s 17.54 torr

750.0 torr-17.54 torr= 732.46 torr

PV=nRT

We now have:P=732.46 torr 0.9638 atmn=2.780 mol H2

R=0.08206T= 20.0 ºC

Now we solve for v