gas laws: introduction at the conclusion of our time together, you should be able to:

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 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 GasesGas 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 GasesIdeal 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 Pa 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)= about 34 feet of water!

And now, we pause for this commercial message from STP

OK, so it’s really not THIS kind of STP…

STP in chemistry stands for Standard Temperature and Pressure

Standard Pressure = 1 atm (or an equivalent)

Standard Temperature = 0 deg C (273 K)

STP allows us to compare amounts of gases between

different pressures and temperatures

Let’s Review:Standard Temperature and

Pressure“STP”

P = 1 atmosphere, 760 torr T = 0°C, 273 Kelvins 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?

= 1.52 x 103 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

Gas Laws: Boyle’s and Charles’ Law

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

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

pressure or volume3. Describe Charles’ Law with a formula.4. Use Charles’ Law to determine either a

temperature or volume

Boyle’s LawP 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 Summary

Pressure is inversely proportional to volume when temperature is held constant.

1 1 2 2PV PV

Boyle’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.

Example

P1 V1 P2 V2

(0.947 atm) (0.15 L) (0.987 atm) V2

0.14 Lor 140 mL

Boyle’s Law Practice #1

P1 V1 P2 V2

(1 atm) (0.5 L) (0.5 atm) V2

1 L

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 VT T

Charles’s Original Balloon

Modern Long-Distance Balloon

Charles’ Law Summary  The volume of a gas is directly proportional to temperature, and extrapolates to zero at zero Kelvin.

(P = constant)

1 2

1 2

V VT T

Charles’ Law

Example

V1

T1

V2

T2

0.075 L

298 K

V2

323 K

0.081 L

Charles’ Law Practice #1

V1

T1

V2

T2

2.75 L

293.0 K

2.46 L

T2

262 K or -10.9 oC

Gas Laws: Gay-Lussac’s and Combined

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

1. Describe Gay-Lussac’s Law with a formula2. Use Gay-Lussac’s Law to determine either

a temperature or volume3. Combine all three laws into the Combined

Gas Law4. Use the Combined Gas Law to determine

either temperature, volume or pressure

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 PT T

Gay Lussac’s Law SummaryThe pressure and temperature of a gas are directly related, provided that the volume remains constant.

1 2

1 2

P PT T

Example

P1

T1

P2

T2

3.00 atm

298 K

P2

325 K

3.3 atm

Gay-Lussac’s Law Practice #1

P1

T1

P2

T2

1.8 atm

293 K

1.9 atm

T2

310 K or 36 oC

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

Combined Gas Law

• 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!

No, it’s not related to R2D2

1 1 2 2

1 2

PV PVT T

The Combined Gas LawThe combined gas law expresses the relationship between pressure, volume and temperature of a fixed amount of gas.

1 1 2 2

1 2

PV PVT T

Boyle’s law, Gay-Lussac’s law, and Charles’ law are all derived from this by holding a variable constant.

Combined Gas Law

If you should only need one of the other gas laws, you can cover up the item that is constant and you will get that gas law!

= P1 V1

T1

P2 V2

T2

Boyle’s Law

Charles’ Law

Gay-Lussac’s Law

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 TableP1 = 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 = P1 V1 T2 = P2 V2 T1 T1 T2

T2 = 3.20 atm x 90.0 mL x 302 K

0.800 atm x 180.0 mL

T2 = 604 K - 273 = 331 °C

= 604 K

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 LawEqual volumes of gases at the same T and P have

the same number of molecules.V = anV 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).

V = an a = proportionality constant V = volume of the gas n = number of moles of gas

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

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

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

Holds closely at P < 1 atm

Review of Kinetic Molecular

TheoryParticles of matter are ALWAYS in

motion Volume of individual particles is

zero. Collisions of particles with container

walls cause pressure exerted by gas.

Particles exert no forces on each other.

Average kinetic energy µ Kelvin temperature of a gas.

Deviations from Ideal Gas Law• Real molecules have volume.The ideal gas consumes the entire amount

of available volume. It does not account for the volume of the molecules themselves.

• There are intermolecular forces.An ideal gas assumes there are no

attractions between molecules. Attractions slow down the molecules and reduce the amount of collisions.– Otherwise a gas could not condense

to become a liquid.

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 He1 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 ExampleHow 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

32 8.0

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 DENSITY

High density

Lower density

22.4 L of ANY gas AT STP = 1 mole

Gas Density

molar massmolar volume

massDensityvolume

… so at STP…

molar mass22.4 L

Density

Density and the Ideal Gas Law

Combining the formula for density with the Ideal Gas law, substituting and rearranging algebraically:

MPDRT

M = Molar Mass P = Pressure R = Gas Constant T = Temperature in

Kelvins

Gas Stoichiometry #4

nRTV P

How many liters of oxygen gas, at 37.0°C 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 O250.0 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)