Gas Laws: IntroductionAt the conclusion of our time together,

you should be able to:

1. List 5 properties of gases2. Identify the various parts of the kinetic

molecular theory3. Define pressure4. Convert pressure into 3 different units5. Define temperature6. Convert a temperature to Kelvin

Importance of Gases

• Airbags fill with N2 gas in an accident.

• Gas is generated by the decomposition of sodium azide, NaN3.

• 2 NaN3 ---> 2 Na + 3 N2

THREE STATES OF MATTERTHREE STATES OF MATTER

General Properties of Gases

• There is a lot of “free” space in a gas.

• Gases can be expanded infinitely.• Gases fill containers uniformly and

completely.• Gases diffuse and mix rapidly.

To Review

Gases expand to fill their containers

Gases are fluid – they flow Gases have low density

1/1000 the density of the equivalent liquid or solid

Gases are compressible Gases effuse and diffuse

Properties of Gases

Gas properties can be modeled using math. This model depends on —

• V = volume of the gas (L)• T = temperature (K)

– ALL temperatures in the entire unit MUST be in Kelvin!!! No Exceptions!

• n = amount (moles)• P = pressure

(atmospheres)

Ideal Gases

Ideal gases are imaginary gases that perfectly fit all of the assumptions of the kinetic molecular theory. Gases consist of tiny particles that

are far apart relative to their size. Collisions between gas particles and between particles and the walls of the container are elastic collisions

No kinetic energy is lost in elastic collisions

Ideal Gases (continued)

Gas particles are in constant, rapid motion. They therefore possess kinetic energy, the energy of motion

There are no forces of attraction between gas particles

The average kinetic energy of gas particles

depends on temperature, not on the identity of the particle.

Pressure

Is caused by the collisions of molecules with the walls of a container

Is equal to force/unit area SI units = Newton/meter2 = 1 Pascal (Pa) 1 atmosphere = 101.325 kPa (kilopascal)1 atmosphere = 1 atm = 760 mm Hg = 760

torr1 atm = 29.92 in Hg = 14.7 psi = 0.987 bar

= 10 m column of water.

Measuring Pressure

The first device for measuring atmospheric pressure was developed by Evangelista Torricelli during the 17th century.The device was called a “barometer”

  Baro = weight   Meter = measure

An Early Barometer

The normal pressure due to the atmosphere at sea level can support a column of mercury that is 760 mm high.

PressureColumn height measures

Pressure of atmosphere• 1 standard atmosphere (atm)

*= 760 mm Hg (or torr) *= 29.92 inches Hg *= 14.7 pounds/in2 (psi)= 101.325 kPa (SI

unit is PASCAL)

Let’s Review:Standard Temperature and

Pressure“STP”

Allows us to compare amounts of different gases by converting them all to STP (the same temperature and pressure conditions) P = 1 atmosphere, 760 torr T = 0°C, 273 Kelvin The molar volume of an ideal gas is

22.42 liters at STP

Pressure Conversions

A. What is 475 mm Hg expressed in atm?

B. The pressure of a tire is measured as 29.4 psi. What is this pressure in mm Hg?

= 1520 mm Hg

= 0.625 atm475 mm Hg x

29.4 psi x

1 atm760 mm Hg

760 mm Hg 14.7 psi

101.325 kPa 14.7 psi

Pressure Conversions – Your Turn

A. What is 2.00 atm expressed in torr?

B. The pressure of a tire is measured as 32.0 psi. What is this pressure in kPa?

= 1520 torr2.00 atm x 760 torr 1 atm

= 221 kPa32.0 psi x

Converting Celsius to Kelvin

Gas law problems involving temperature require that the temperature be in KELVINS!

Kelvins = C + 273

°C = Kelvins - 273

Don’t use temp. when determining sig. figs. for your answer

Charles’ Law

If n and P are constant, then V α T

V and T are directly proportional.

• If one temperature goes up, the volume goes up!

Jacques Charles (1746-1823). Isolated

boron and studied gases. Balloonist.

1 2

1 2

V V

T T

Charles’ Law

Gay-Lussac’s Law

If n and V are constant, then P α T

P and T are directly proportional.

• If one temperature goes up, the pressure goes up!

Joseph Louis Gay-Lussac (1778-1850)

1 2

1 2

P P

T T

Boyle’s Law

P α 1/VThis means Pressure and

Volume are INVERSELY PROPORTIONAL if moles and temperature are constant (do not change). For example, P goes up as V goes down.

Robert Boyle (1627-1691). Son of Earl of Cork, Ireland.

1 1 2 2PV PV

Boyle’s Law ExampleBoyle’s Law Example

A bicycle pump is a good example of Boyle’s law.

As the volume of the air trapped in the pump is reduced, its pressure goes up, and air is forced into the tire.

Now let’s put all 3 laws together into one big

combined gas law…….

Combined Gas Law (for a fixed amount, moles, of gas)

• The good news is that you don’t have to remember all three gas laws! Since they are all related to each other, we can combine them into a single equation. BE SURE YOU KNOW THIS EQUATION! (if a variable is not given in the problem, just leave it out)

1 1 2 2

1 2

PV PV

T T

Combined Gas Law Problem

A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29°C. What is the new temperature(°C) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm?

Set up Data Table

P1 = 0.800 atm V1 = 180 mL T1 = 302 K

P2 = 3.20 atm V2 = 90 mL T2 = ??

Calculation

• P1 = 0.800 atm V1 = 180 mL T1 = 302 K

• P2 = 3.20 atm V2 = 90 mL T2 = ??

P1 V1 P2 V2 = T1 T2

Solving for T2 = 604 K

T2 = 604 K - 273 = 331 °C

Gas Laws: Avogadro’s and Ideal

At the conclusion of our time together, you should be able to:

1. Describe Avogadro’s Law with a formula.2. Use Avogadro’s Law to determine either

moles or volume3. Describe the Ideal Gas Law with a formula.4. Use the Ideal Gas Law to determine either

moles, pressure, temperature or volume5. Explain the Kinetic Molecular Theory

Avogadro’s Law

Equal volumes of gases at the same T and P have the same number of molecules.

V and n are directly related.

twice as many molecules

Avogadro’s Law Summary

For a gas at constant temperature and pressure, the volume is directly proportional to the number of moles of gas (at low pressures).

V1

n1

V2

n2

Standard Molar Volume

Equal volumes of all gases at the same temperature and pressure contain the same number of molecules.

- Amedeo Avogadro

Avogadro’s Law Practice #1

V1

n1

V2

n2

4.00 L

0.21 mol

7.12 L

n2

0.37 mol total0.16 mol added

Ultimate Comined Gas Law(include Aavogadro’s law in the combined gas law)

1 1 2 2

1 1 1 2

PV PV

T Tn n

The relationship between T, P, n, and V for a gas is a constant which we call the ideal gas constant, R (=0.08206 L atm/ mole K). So we can can set one side as equal to R (the gas constant) and it is now called the ideal gas law.

IDEAL GAS LAW

Brings together gas properties.

Can be derived from experiment and theory.

BE SURE YOU KNOW THIS EQUATION!

P V = n R T

Ideal Gas Law

PV = nRT P = pressure in atm (you may need to

convert)

V = volume in liters n = moles R = proportionality constant

= 0.08206 L atm/ mol·K T = temperature in Kelvin

R is a constant, called the Ideal Gas ConstantInstead of learning a different value for R for all the

possible unit combinations, we can just memorize one value and convert the units to match R.

R = 0.08206

L • atm mol • K

Using PV = nRT

How much N2 is required to fill a small room with a volume of 960 cubic feet (27,000 L) to 745 mm Hg at 25 oC?

Solution1. Get all data into proper units

V = 27,000 L T = 25 oC + 273 = 298 K P = 745 mm Hg (1 atm/760 mm Hg)

= 0.98 atmAnd we always know R, 0.08206 L atm / mol K

How much N2 is required to fill a small room with a volume of 960 cubic feet (27,000 L) to P = 745 mm Hg at 25 oC?

Solution2. Now plug in those values and solve for the

unknown. PV = nRT

RT RT

Using PV = nRT

n = (0.98 atm)(2.7 x 10 4 L)

(0.0821 L • atm/K • mol)(298 K)

n = 1.1 x 103 mol (or about 30 kg of gas)

Ideal Gas Law Problems #1

Using PV = nRT

(5.6 atm) (12 L) (0.08206 atm*L/mol*K) (T)

200 K

(4 mol)

Gas Laws: Dalton’s LawAt the conclusion of our time together,

you should be able to:

1. Explain Dalton’s Law 2. Use Dalton’s Law to solve a problem

Dalton’s Law

John Dalton1766-1844

Dalton’s Law of Partial Pressures

For a mixture of gases in a container,

PTotal = P1 + P2 + P3 + . . .

This is particularly useful in calculating the pressure of gases collected over water.

Dalton’s Law of Partial Pressures

What is the total pressure in the flask?Ptotal in gas mixture = PA + PB + ...

Therefore, Ptotal = PH2O + PO2

= 0.48 atm

Dalton’s Law: total P is sum of PARTIAL pressures.

2 H2O2 (l) ---> 2 H2O (g) + O2 (g)

0.32 atm 0.16 atm

Gases in the Air

The % of gases in air Partial pressure (STP)

78.08% N2 593.4 mm Hg

20.95% O2 159.2 mm Hg

0.94% Ar 7.1 mm Hg

0.03% CO2 0.2 mm Hg

PAIR = PN + PO + PAr + PCO = 760 mm Hg 2 2 2

Total Pressure = 760 mm Hg

Collecting a gas “over water”

• Gases, since they mix with other gases readily, must be collected in an environment where mixing can not occur. The easiest way to do this is under water because water displaces the air. So when a gas is collected “over water”, that means the container is filled with water and the gas is bubbled through the water into the container. Thus, the pressure inside the container is from the gas AND the water vapor. This is where Dalton’s Law of Partial Pressures becomes useful.

Table of Vapor Pressures for Water

Solve This!

A student collects some hydrogen gas over water at 20 degrees C and 768 torr. What is the pressure of the H2 gas?

768 torr – 17.5 torr = 750.5 torr

Dalton’s Law of Partial Pressures

Also, for a mixture of gases in a container, because P, V and n are directly proportional if the other gas law variables are kept constant:

nTotal = n1 + n2 + n3 + . . .

VTotal = V1 + V2 + V3 + . . .

This is useful in solving problems with differing numbers of moles or volumes of the gases that are mixed together.

Gas Laws: GasStoichiometry

At the conclusion of our time together, you should be able to:

1. Use the Ideal Gas Law to solve a gas stoichiometry problem.

Gases and Stoichiometry

2 H2O2 (l) ---> 2 H2O (g) + O2 (g)

Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L. What is the volume of O2 at STP?

Bombardier beetle uses decomposition of hydrogen peroxide to defend itself.

Gases and Stoichiometry

2 H2O2 (l) ---> 2 H2O (g) + O2 (g)

Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L. What is the volume of O2 at STP?

Solution1.1 g H2O2 1 mol H2O2 1 mol O2 22.4 L O2

34 g H2O2 2 mol H2O2 1 mol O2

= 0.36 L O2 at STP

Gas Stoichiometry: Practice!

How many grams of He are present in 8.0 L of gas at STP?

= 1.4 g He

8.0 L He x 1 mol He

22.4 L He

x 4.00 g He

1 mol He

Gas Stoichiometry Trick

If reactants and products are at the same conditions of temperature and pressure, then mole ratios of gases are also volume ratios.

3 H2(g) + N2(g) 2NH3(g)

3 moles H2 + 1 mole N2 2 moles NH3 67.2 liters H2 + 22.4 liter N2 44.8 liters NH3

Gas Stoichiometry Trick Example

How many liters of ammonia can be produced when 12 liters of hydrogen react with an excess of nitrogen in a closed container at constant temperature?

3 H2(g) + N2(g) 2NH3(g)

12 L H2

L H2

= L NH3 L NH3

3

28.0

What if the problem is NOT at STP?

• 1. You will need to use PV = nRT

Gas Stoichiometry Example on HO

How many liters of oxygen gas, at 1.00 atm and 25 oC, can be collected from the complete decomposition of 10.5 grams of potassium chlorate?

2 KClO3(s) 2 KCl(s) + 3 O2(g)

10.5 g KClO3 1 mol KClO3

122.55 g KClO3

3 mol O2

2 mol KClO3

0.13 mol O2

Gas Stoichiometry Example on HO

How many liters of oxygen gas, at 1.00 atm and 25 oC, can be collected from the complete decomposition of 10.5 grams of potassium chlorate?

2 KClO3(s) 2 KCl(s) + 3 O2(g)

3.2 L O2

(1.0 atm)(V) (0.08206 atm*L/mol*K) (298 K)(0.13 mol)

Gas Laws: Dalton, Density and Gas Stoichiometry

At the conclusion of our time together, you should be able to:

1. Explain Dalton’s Law and use it to solve a problem.

2. Use the Ideal Gas Law to solve a gas density problem.

3. Use the Ideal Gas Law to solve a gas stoichiometry problem.

Gas Stoichiometry #4

nRT

VP

How many liters of oxygen gas, at 37.0C and 0.930 atmospheres, can be collected from the complete decomposition of 50.0 grams of potassium chlorate?

2 KClO3(s) 2 KCl(s) + 3 O2(g)

= “n” mol O2

50.0 g KClO3 1 mol KClO3

122.55 g KClO3

3 mol O2

2 mol KClO3 = 0.612 mol O2

L atm(0.612mol)(0.0821 )(310K)

mol K0.930atm

= 16.7 L

Try this one!

How many L of O2 are needed to react 28.0 g NH3 at 24°C and 0.950 atm?

4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)

Gas Laws: Finding “R”At the conclusion of our time together, you

should be able to:

1. Use the Ideal Gas Law to find the value of “R”.

See the problem on p. 50

Partial Pressure of CO2

Carbon Dioxide gas over water at 17.0 degrees C and 97.932 kPa. What is the pressure of the CO2 gas?

97.932 kPa – 1.90 kPa = 96.032 kPa = 0.948 atm

1.90 kPa

Determine the Moles of CO2 Used:

Mass of canister before-mass of canister after

17.099 g – 16.524 g= 0.575 g0.575 g CO2 1 mol CO2

44.01 g CO2

0.0131 mol CO2

Using Ideal Gas Law solve for “R”

(0.948 atm) (0.347 L)

(R) (290 K)

0.0868 atm*L/mol*K

(0.0131 mol)

% Error

0.08206 atm*L/mol*K (standard)

0.0868 atm*L/mol*K (experimental results)

- 0.0048 atm*L/mol*K 0.08206 atm*L/mol*K x 100

- 5.85 % error

- 0.00474 atm*L/mol*K (experimental error)