Unit 6Gases and Gas Laws

Gases in the Atmosphere

• The atmosphere of Earth is a layer of gases surrounding the planet that is retained by Earth's gravity.

• By volume, dry airis 78% nitrogen, 21% oxygen, 0.9% argon, 0.04% CO2, and small amounts of other gases.

Air Pollution

• Human activity has polluted the air with other gases:– Sulfur Oxides (SO2 & SO3) – produced from

coal burning. Contribute to acid rain.– Nitrogen Oxides (NO & NO2) – produced by burning

fossil fuels. Contribute to acid rain.– Carbon Monoxide (CO) –

emitted by motor vehicles.– Ground-level Ozone (O3) –

produced when products offossil fuel combustion reactin the presence of sunlight.

The Ozone Layer• O3 in the troposphere (ground-level ozone) is a

pollutant, but O3 in the stratosphere is a necessary part of our atmosphere.

• Stratospheric O3 protectsus by absorbing UV light.

• CFCs destroy stratospheric O3, and have been banned in the US.

• Ozone: Good up high,bad nearby.

Atmospheric Pressure

• Atmospheric pressure is the force per unit area exerted on a surface by the weight of the gases that make up the atmosphere above it.

Pressure =Force

Area

Measuring Pressure

• A common unit of pressure is millimeters of mercury (mm Hg).

• 1 mm Hg is also called 1 torr in honor of Evangelista Torricelli whoinvented the barometer (used tomeasure atmospheric pressure).

• The average atmospheric pressure at sea level at 0°C is 760 mm Hg, so one atmosphere (atm) of pressure is 760 mm Hg.

Measuring Pressure (continued)

• The pressure of a gas sample in the laboratory is often measured with a manometer.

• the difference in the liquid levels is a measure of the difference in pressure between the gas and the atmosphere.

For this sample, the gas has a larger pressure than the atmosphere.

Measuring Pressure (continued)

•Pressure can also be measured in pascals (Pa): 1 Pa = 1 N/m2.

•One pascal is verysmall, so usually kilopascals (kPa) are used instead.

•One atm is equal to 101.3 kPa.

1 atm = 760 mm Hg (Torr) = 101.3 kPa

Units of Pressure

Converting PressureSample Problem

The average atmospheric pressure in Denver, CO is 0.830 atm. Express this pressure in:

a. millimeters of mercury (mm Hg)

b. kilopascals (kPa)

0.830 atmatm

mm Hg1

760x =

0.830 atmatm

kPa1

101.3x =

631 mm Hg

84.1 kPa

Dalton’s Law of Partial Pressures

• Dalton’s law of partial pressures - the total pressure of a gas mixture is the sum of the partial pressures of the component gases.

PT = P1 + P2 + P3 …

Dalton’s Law of Partial PressuresSample Problem

A container holds a mixture of gases A, B & C. Gas A has a pressure of 0.5 atm, Gas B has a pressure of 0.7 atm, and Gas C has a pressure of 1.2 atm.

a.What is the total pressure of this system?

b. What is the total pressure in mm Hg?

PT = P1 + P2 + P3 …PT = 0.5 atm+ 0.7 atm + 1.2 atm = 2.4 atm

2.4 atmatm

mm Hg1

760x = 1800 mm Hg

The Kinetic-Molecular Theory: A Model for Gases

• Matter is composed of particles which are constantly moving.

• The average kinetic energyof a particle is proportionalto its Kelvin temperature.

• The size of a particle is negligibly small.

• Collisions are completelyelastic – energy may beexchanged, but not lost(like billiard balls.)

Ideal Gases

• The kinetic-molecular theory assumes:1) no attractions between gas molecules2) gas molecules do not take up space

• An Ideal Gas is a hypothetical gas that perfectly fits the assumptions of the kinetic-molecular theory.

• Many gases behave nearlyideally if pressure is not veryhigh and temperature is not very low.

Properties of Gases: Fluidity

• Gas particles glide easily past one another.

Because liquids and gases flow, they are both referred to as fluids.

Properties of Gases: Expansion

• Since there’s no significantattraction between gasmolecules, they keep moving around and spreading out until they fill their container.

• As a result, gases take the shape and the volume of the container they are in.

Properties of Gases: Low Density

• Gas particles are very far apart.There is a lot of unoccupied space in the structure of a gas.

• Since gases do not have a lot of mass in a given volume, they have a very low density

• The density of a gas is about 1/1000 the density of the same substance as a liquid or solid.

Properties of Gases: Compressibility

• Because there is a lot of unoccupied space in the structure of a gas, the gas molecules can easily be squeezed closer together.

Diffusion and Effusion

• Diffusion is the gradual mixing of two or more gases due to their spontaneous, random motion.

• Effusion is the process whereby the molecules of a gas confined in a container randomly pass through a tiny opening in the container.

Rate of Diffusion

• Light molecules move faster than heavy ones. • The greater the molar mass of a gas,

the slower it will diffuse and/or effuse.

Graham’s Law of Effusion

• Graham’s Law of Effusion states that the rate of effusion is inversely proportional to the square root of the molar mass of the gas.

• The ratio of effusion rates of two different gases is given by the following equation:

Agas

B gas

B gas

Agas

MassMolar

MassMolar

rate

rate

Graham’s Law of EffusionSample Problem

Calculate the molar mass of a gas that effuses at a rate 0.462 times N2 .

Solution:

(0.462)2 = 28.0 g/molMMunknown

MMunknown = = 131 g/mol

Agas

B gas

B gas

Agas

MassMolar

MassMolar

rate

rate

28.0 g/mol(0.462)2

= 28.0 g/mol(0.213)

Gases and Pressure• Gas pressure is caused by collisions of the gas

molecules with each other and with the walls of their container.

• The greater the number of collisions, the higher the pressure will be.

Pressure – Volume Relationship

• When the volume of a gas is decreased, more collisions will occur.

• Pressure is caused by collisions.

• Therefore, pressure will increase.

Boyle’s Law

• Boyle’s Law – The volume of a fixed mass of gas varies inversely with the pressure at a constant temperature.

• P1 and V1 representinitial conditions, andP2 and V2 representanother set of conditions.

P1V1 = P2V2

Boyle’s LawSample Problem

A sample of oxygen gas has a volume of 150.0 mL when its pressure is 0.947 atm. What will the volume of the gas be at a pressure of 0.987 atm if the temperature remains constant?Solution:P1V1 = P2V2(0.947 atm)(150.0 mL) = (0.987 atm)V2

V2 =(0.947 atm)(150.0 mL)

(0.987 atm)= 144 mL

Volume – Temperature Relationship

• the pressure of gas inside and outside the balloon are the same.

• at low temperatures, the gas molecules don’t move as much – therefore the volume is small.

• at high temperatures, the gas molecules move more – causing the volume to become larger.

• Charles’s Law – The volume of a fixed mass of gas at constant pressure varies directly with the Kelvin temperature.

Charles’s Law

V1 V2=T1 T2

• V1 and T1 represent initial conditions, and V2 and T2 represent another set of conditions.

The Kelvin Temperature Scale

• Absolute zero – The theoretical lowest possible temperature where all molecular motion stops.

• The Kelvin temperature scale starts at absolute zero (-273oC.)

• This gives the followingrelationship between the two temperature scales:

K = oC + 273

Charles’s LawSample Problem

A sample of neon gas occupies a volume of 752 mL at 25°C. What volume will the gas occupy at 50°C if the pressure remains constant?Solution:V1 V2=T1 T2

752 mL V2=298 K 323 K

K = oC + 273T1 = 25 + 273 = 298T2 = 50 + 273 = 323

752 mLV2 =298 K

323 Kx = 815 mL

Pressure – Temperature Relationship

• Increasing temperature means increasing kinetic energy of the particles.

• The energy and frequency of collisions depend on the average kinetic energy of the molecules.

• Therefore, if volume is kept constant, the pressure of a gas increases with increasing temperature.

Gay-Lussac’s Law

• Gay-Lussac’s Law – The pressure of a fixed mass of gas varies directly with the Kelvin temperature.

• P1 and T1 representinitial conditions.P2 and T2 representanother set of conditions.

P1 P2=T1 T2

Gay-Lussac’s LawSample Problem

The gas in a container is at a pressure of 3.00 atm at 25°C. What would the gas pressure in the container be at 52°C?Solution:P1 P2=T1 T2

3.00 atm P2=298 K 325 K

K = oC + 273T1 = 25 + 273 = 298T2 = 52 + 273 = 325

3.00 atmP2 =298 K

325 Kx = 3.27 atm

The Combined Gas Law

• The combined gas law is written as follows:

• Each of the other simple gas laws can be obtained from the combined gas law when the proper variable is kept constant.

P1 P2=T1 T2

V1 V2

The Combined Gas LawSample Problem

A helium-filled balloon has a volume of 50.0 L at 25°C and 1.08 atm. What volume will it have at 0.855 atm and 10.0°C?Solution:

(1.08 atm) (0.855 atm)=

298 K 283 K

K = oC + 273T1 = 25 + 273 = 298T2 = 10 + 273 = 283

(1.08 atm)V2 =(298 K)

(283 K)= 60.0 L

P1 P2=T1 T2

V1 V2

(50.0 L) V2

(50.0 L)(0.855 atm)

Avogadro’s Law

• In 1811, Amedeo Avogadro discovered that the volume of a gas is proportional to the number of molecules (or number of moles.)

• Avogadro’s Law - equal volumes of gases at the same temperature and pressure contain equal numbers of molecules, or:

V1 V2=n1 n2

The Ideal Gas Law

• All of the gas laws you have learned so far can be combined into a single equation, the ideal gas law:

• R represents the ideal gas constant which has a value of 0.0821 (L•atm)/(mol•K).

PV = nRT

The Ideal Gas LawSample Problem

What is the pressure in atmospheres exerted by a 0.500 mol sample of nitrogen gas in a 10.0 L container at 298 K?Solution:

PV = nRTP (10.0 L) = (0.500 mol)(0.0821 L•atm/mol•K)

P =(0.500 mol) (298 K)

(10.0 L)= 1.22

atm

(298 K)

(0.0821 L•atm/mol•K)

Molar Mass of a Gas

• The ideal gas law can beused in combination withmass measurements to calculate the molar mass of an unknown gas.

• Molar mass is calculated by dividing the mass (in grams) by the amount of gas (in moles.)

Molar Massg

=mol

Molar Mass of a GasSample Problem

Calculate the molar mass of a gas with mass 0.311 g that has a volume of 0.225 L at 55°C and 886 mmHg.Solution:

PV = nRT

PVRT=

886 mmHg

T =

P =

(1.17 atm)

(328 K)

(0.225 L) =

31.8 g/mol

55 + 273 = 328 K

(0.0821 L•atm/mol•K)n =

760. mmHg1 atm

1.17 atm=

0.00978 mol

gramsmole=MM =

0.311 g0.00978 mol =

Standard Molar Volume

• Standard Temperature and Pressure (STP) is 0oC (273 K) and 1 atm.

• The Standard Molar Volume of a gas is the volume occupied by one mole of a gas at STP. It has been found to be 22.4 L.

Molar Volume Conversion Factor

• Standard Molar Volume can be used as a conversion factor to convert from the number of moles of a gas at STP to volume (L), or vice versa.

Molar Volume ConversionSample Problem

a. What quantity of gas, in moles, is contained in 5.00 L at STP?

b. What volume does 0.768 moles of a gas occupy at STP?

5.00 LL

mol22.41x = 0.223 mol

0.768 molmol

L1

22.4x = 17.2 L

The Mole Map Revisited

• As you recall, you can convert between number of particles, mass (g), and volume (L) by going through moles.

Stoichiometry Revisited• Remember that you can use mole ratios and

volume ratios (gases only) as conversion factors:

2CO(g) + O2(g) → 2CO2(g)2 molecules 1 molecule 2 molecules2 mole 1 mole 2 mol2 volumes 1 volume 2 volumes

• Example: What volume of O2 is needed to react completely with 0.626 L of CO at the same temperature & pressure conditions to form CO2?

0.626 L COL COL O2

21x = 0.313 L O2

Gas StoichiometrySample Problem

What volume (in L) of H2 at 355 K and 738 mmHg is required to synthesize 35.7 g of methanol, given:

CO(g) + 2H2(g) → CH3OH(g)Solution:First, use stoichiometry to solve for moles of H2:

Then, use the ideal gas law to find the volume of H2:

35.7 g CH3OHg CH3OH

mol CH3OH32.01 = 2.23 mol H2

66.9 L H2

mol H22mol CH3OH1

nRTP=

(0.971 atm)(355 K)(2.23mol)

=(0.0821 L•atm/mol•K)

V =