GasesGases Chapter 11Chapter 11 GasesGases

Chapter 11Chapter 11

Tro, 2Tro, 2ndnd ed. ed.

WHY WE STUDY GASESOur atmosphere: thin layer surrounds us

and is critical to life on earth78% N2 & 21% O2 at sea level, plus CO2

& water vapor & noble gases (next slide)

POLLUTANTS: (see page 378) SOx & NOx cause acid rain CFC's destroy O3 uv protective layer CO2 may increase global warming (some is necessary or we would cool too much at night)

Average Composition of Dry Air

N2 78.08%

O2 20.95%

Ar 0.93%

CO2 0.033%

0.0018%Ne

He 0.0005%

Gas Volume Percent

CH4 0.0002%

Kr

Xe, H2, and N2O Trace

0.0001%

Gas Volume Percent

BEHAVIOR & PROPERTIESOF GASES:

- can be compressed greatly- can expand to fill container uniformly- have low density compared to liquids & solids- may be mixed - always homogeneous mixtures because always in motion- a confined gas exerts constant pressure on walls of its container uniformly in all directions

The Kinetic-Molecular Theory

KMT is based on the motions of gas particles.

A gas that behaves exactly as outlined by KMT is known as an ideal gas.

While no ideal gases are found in nature, real gases can approximate ideal gas behavior under certain conditions of temperature and pressure.

Principle Assumptions of the KMT (similar to pages 343-344)

1. Gases consist of tiny atomic particles.

2. The distance between particles is large compared with the size of the particles themselves.

3. Gas particles have no attraction for one another.

Principle Assumptions of the KMT

4. Gas particles move in straight lines in all directions, colliding frequently with one another and with the walls of the container.

5. No energy is lost by the collision of a gas particle with another gas particle or with the walls of the container. All collisions are perfectly elastic.

Principle Assumptions of the KMT

6. The average kinetic energy for particles is the same for all gases at the same temperature, and its value is directly proportional to the Kelvin temperature.

21KE =

2mv

Although each gas particle (atom or molecule) moves with its own velocity, the average velocity of all the particles gives an average kinetic energy to a container of gases.

GAS PROPERTIES: four quantities define state of a gas

1. Quantity in moles or mass 2. Temperature in Kelvin3. Volume in Liters4. Pressure in Atmospheres (usually)

We already know about mass/moles, temperature & volume.

Learn about pressure!

PRESSUREPressure = Force/unit area

Force = mass*accelerationForce causes something to move a

distance D in work. Gravity is a weak force, g F = m*g, units = Newtons or lbsCommon English unit of pressure is

lbs/in2 (psi)

The pressure resulting from the collisions of gas molecules with the walls of the balloon keeps the balloon inflated.

GAS PRESSUREThe pressure exerted by a gas depends on:

- the number of gas molecules present- the temperature of the gas- the volume in which the gas is

confined

V = 22.414 LT = 0.000oC

The pressure exerted by a gas is directly proportional to the number of molecules present.

Dependence of Pressure on Temperature

The pressure of a gas in a fixed volume increases with increasing temperature.

When the pressure of a gas increases, its kinetic energy increases.

The increased kinetic energy of the gas results in more frequent and energetic collisions of the molecules with the walls of the container.

Mercury Barometer

A tube of mercury is inverted and placed in a dish of mercury.

The barometer is used to measure atmospheric pressure.

The atmosphere above us exerts a pressure, called atmospheric pressure, which is measured by a barometer as shown above.

Memorize:

1 torr = 1 mm Hg

1 atm = 760 torr exactly

1 atm = 14.7 lb/in2 (psi)

1 atm = 33.9 ft water

PRESSURE CONVERSION PRACTICE

A storm is heralded by falling atmospheric pressure. The weather report says the pressure is down to 28.5 inches of Hg. Convert this to torr and atmospheres.

28.5 inches * 25.4 torr = 723.9 torr 1 inch

723.9 torr * 1 atm = 0.953 atm 760 torr

Now convert 684 torr to mm Hg, atm and psi.

Boyle’s LawAt constant temperature (T), the

volume (V) of a fixed mass of gas is inversely proportional to the Pressure (P).

V = cb * 1/P, or V*P =cb

P1V1 = cb = P2V2

P1V1 = P2V2

Memorize this!!!

Graph of pressure versus volume. This shows the inverse PV relationship of an ideal gas.

The effect of pressure on the volume of a gas.

Boyle’s Law ProblemAn 8.00 L sample of N2 is at a pressure of 500.0 torr.

What must be the pressure to change the volume to 3.00 L? (T is constant).

First ask yourself if P will increase or decrease:V decreased, therefore P should increase.

Second, rearrange Boyle’s Law: P2 = P1V1/V2

Third, plug in the data given: P2 = 500.0 torr * 8.00 L/ 3.00 L

= 1333 torr (or 1.33 x 103 torr)

Charles’ LawAt constant pressure the volume of a

fixed mass of gas is directly proportional to the absolute temperature.

V = cc * T, or V/T = cc

V1/T1 = cc = V2/T2

V1/T1 = V2/T2

Memorize this!

Volume-temperature relationship of methane (CH4).

Absolute Zero on the Kelvin Scale

-273oC (more precisely –273.15oC) is the zero point on the Kelvin scale. It is the temperature at which an ideal gas would have zero volume.

Effect of temperature on the volume of a gas. Pressure is constant at 1 atm. When temperature increases at constant pressure, the volume of the gas increases.

Charles’ Law ProblemA 255 mL sample of nitrogen at 75oC is confined at

a pressure of 3.0 atmospheres. If the pressure remains constant, what will be the volume of the nitrogen if its temperature is raised to 250.oC?

First, convert temperature to Kelvin ALWAYS!T1 = 75oC = 348 K T2 = 250.oC = 523 K

Second, rearrange Charles’ Law V2 = V1T2/T1

Third, plug in data: V2 = 255 mL*523K/348K

= 383 mL

Gay-Lussac’s Law (added to chp)

The pressure of a fixed mass of gas, at constant volume, is directly proportional to the Kelvin temperature.

P = cg*T, or P/T = cc

P1/T1 = cc = P2/T2

P1/T1 = P2/T2

Memorize this!

Gay-Lussac’s Law ProblemAt a temperature of 40.oC an oxygen container is at

a pressure of 2.15 atmospheres. If the temperature of the container is raised to 100.oC what will be the pressure of the oxygen?

First convert to Kelvin:

T1 = 40.oC = 313 K T2 = 100.oC = 373 K

Second, write and solve the equation for the unknown: P2 = P1T2/T1 = 2.15 atm*373K/313K

= 2.56 atm

Standard Temperature and Pressure

273.15 K or 0.00oC

Exactly 1 atm or 760 torr

or 760 mm Hg or 14.7 psi or…

Combined Gas LawA combination of Boyle’s, Gay-

Lussac’s and Charles’ Law.P1V1 = P2V2

T1 T2

Used when pressure and temperature change at the same time while moles/mass is constant.

Solve the equation for any one of the 6 variables

Combined Gas Law ProblemA sample of hydrogen occupies 465 ml at STP. If

the pressure is increased to 950 torr and the temperature is decreased to –15oC, what would be the new volume?

First convert to Kelvin: T1 = 273 K, T2 = 258 K

Second, rearrange for V2 = P1V1T2/P2T1

Third, plug in data:V2 = 760 torr * 465 mL * 258 K = 352 mL

950 torr * 273 K

Dalton’s Law ofPartial Pressures

Each gas in a mixture exerts a pressure that is independent of the other gases present.

The total pressure of a mixture of gases is the sum of the partial pressures exerted by each of the gases in the mixture.

Ptotal = Pa + Pb + Pc + Pd + ….

Dalton’s Law ProblemA container contains He at a pressure of 0.50

atm, Ne at a pressure of 0.60 atm, and Ar at a pressure of 1.30 atm. What is the total pressure in the container?

Ptotal = PHe + PNe+ PAr

Ptotal = 0.5 atm + 0.6 atm + 1.30 atm = 2.40 atm

(Yes, it’s that easy.)

The pressure in the collection container is equal to the atmospheric pressure.

The pressure of the gas collected plus the pressure of water vapor at the collection temperature is equal to the atmospheric pressure.

2total atm gas H OP = P = P + P

Collecting a Gas Sample Over Water

Oxygen collected over Oxygen collected over water.water.

Dalton’s Law Problem (again)

A sample of O2 was collected in a bottle over water at a temperature of 25oC when the atmospheric pressure was 760.0 torr. The vapor pressure of water at 25oC is 23.8 torr. What is the pressure of the O2 gas?

Pt = 760.0 torr = PO2 + PH2O

PO2 = 760.0 torr – 23.8 torr = 736.2 torr

Avogadro’s LawEqual volumes of different gases at the

same temperature and pressure contain the same number of molecules. (Section 11.7)

V is directly proportional to n, where n is moles (if at same T & P)

V1/n1 = Ca = V2/n2

V1/n1 = V2/n2

Memorize this!

Avogadro’s Law ProblemIf 0.500 moles of CO2 occupies 11.2

Liters, what volume will 0.670 moles of CO2 occupy at the same T & P?

V2 = V1n2/n1

= 11.2 L * 0.670 mol/0.500 mol= 15.0 L

IDEAL GAS LAW

PV = nRTR = Ideal Gas Constant

= 0.082057 L-atm/mol-K

MEMORIZE THIS R VALUE AND UNITS!

Ideal Gas Law ProblemA balloon filled with 5.00 moles of helium gas is at a

temperature of 25oC. The atmospheric pressure is 750. torr. What is the balloon’s volume?

Convert T to K = 25oC + 273 = 298K

Convert pressure to atmospheres:

750. torr (1 atm/760 torr) = 0.987 atmRearrange: V = nRT/P = 5.00 mol(0.082057 L-atm/mol-K)298K

0.987 atm= 124 L

Another Ideal Gas Law ProblemA big balloon of H2 has a volume of 3.20 x 104 L, T is

20.0°C, P = 750.0 torr. How many moles of gas are in the balloon?

n = PV RT

= (750.0 torr/760 torr/atm) * 3.20 x 104 L 0.082057 L.atm/mol.K * 293.15 K= 1.31 x 103 moles of H2 gas

PRACTICE WITH IDEAL GAS LAW:

1. What pressure will 1.00 mol H2 exert in a 250.0 mL container at 27oC?

(98.5 atm)2. What pressure will 1.00 mol H2 exert in a

250.0 mL container at 327oC?(197 atm)

3. What volume will 2.00 mol H2 exert if P is 98.5 atm and T is 27oC?

(0.500 L)

SUPER-COMBINEDGAS LAW

Rearrange Ideal Gas Law toR = PV/nTAt first set of conditions R = P1V1/n1T1

At second set R = P2V2/n2T2

Set them equal to each other:P1V1 = P2V2

n1T1 n2T2

This can be used to find any of the four simple gas laws.

GROUP WORK:Derive Boyle’s Law, Charles’ Law, Gay-

Lussac’s Law, and Avogadro’s Law from the super-combined gas law.

Mole-Mass-Volume Relationships

Volume of one mole of any gas at STP = 22.414 L.

22.414 L at STP is known as the molar volume of any gas.

22.414L

Standard Molar Volume Problem

The density of neon at STP is 0.900 g/L. What is the molar mass of neon?

(0.900g/L)(22.414L/mol) = 20.2 g/mol

Density of Gases

md =

vliters

grams

Density of Gases

md =

vdepends

on T and P

Gas Density ProblemThe molar mass of SO2 is 64.07 g/mol.

Determine the density of SO2 at STP.

D = (64.07 g/mol)(1mol/22.414 L)

= 2.858 g/L at STP

VARIATIONS OF THE IDEAL GAS LAW ARE USEFUL!

Relationship to molar mass: moles = mass/molar mass

n = m PV = mRT M MRearrange to calculate molar mass or even

density (D = m/V):M = mRT = DRT/P m = PM =

Density PV V RT

Molar Mass & Density Problem

What is molar mass of a gas if 0.681 grams occupies 442 mL (think Density) at 49oC and 0.629 atm?

D = 0.681 g/0.442 L = 1.5407 g/LM = DRT/P=1.5407g/L(0.082057L.atm/mol.K)322 K

(0.629 atm)= 64.7 g/mol

Molar Mass & Density Problem (again)

An unknown gas A has a density of 1.429 g/L at 0.00oC and 1.00 atm. Find its molar mass.

MA = DRT/P= 1.429 g/L* 0.082057 L.atm/mol.K * 273.15 K

1.00 atm = 32.03g/mol

Or:MA = (1.429 g/L)*(22.414 L/mol) = 32.03 g/mol

Molar Mass & Density Problem (again)

What is density of CO2 at STP and at 25.00oC? (Two ways to solve this.)

a. D = 44.01 g/mol = 1.96 g/L 22.414 L/mol

b. D = PM = 1.00 atm*44.01 g/mol RT 0.082057 * 298.15 K

= __________g/L

CHEMICAL REACTIONS STOICHIOMETRY OF GASES:

Stoichiometry problems that you will love to do!

At STP one mole of any gas occupies 22.4 L This means that equal volumes of gases at STP

have the same number of moles!!!! 11.2 L of N2 = 0.500 moles = 11.2 L of O2 =

0.500 moles = 11.2 L of He, etc.The volumes will all change equally with

changes in T and P, if all are at same T & P.We can relate volumes of gases just the same

as relating moles of gases, using mol/mol ratio as a vol/vol ratio.

CHEMICAL REACTIONS STOICHIOMETRY OF GASES:

Example using vol/vol ratio from mol/mol ratio: N2(g) + 3 H2(g) 2 NH3(g)

Given 355 L of hydrogen, how many L of nitrogen required? How many L of ammonia produced?

355 L H2 * 1 vol N2 = 118 L

3 vol H2

355 L H2 * 2 vol NH3 = 237 L

3 vol H2

CHEMICAL REACTIONS STOICHIOMETRY OF GASES:

If 12.0 g Zn react with excess sulfuric acid, how many moles of gas will you have at STP?

Zn(s) + H2SO4(aq) ZnSO4(aq) + H2(g)

12.0 g/(65.38 g/mol) = 0.1835 mol Zn 0.1835 mol Zn * 1 mol H2 = 0.1835 mol H2

1 mol Zn0.1835 mol H2 * 22.414 L/mol = 4.11 L

The same problem but at other conditions like 25oC and 1.00 atm, use the Ideal Gas Law for step 4.

V = nRT/P = 0.1835 mol * 0.082057* 298 K = 4.49 L1.00 atm

Gas Stoichiometry Map

Primary conversions involved in stoichiometry.

Gas Stoichiometry ProblemWhat volume of hydrogen, collected at 30.0oC and

700. torr, will be formed by reacting 50.0 g of aluminum with hydrochloric acid?

2 Al(s) + 6 HCl(aq) 2AlCl3(aq) + 3 H2(g)

50.0gAl(1mol/26.98g)(3molH2/2molAl)=2.78 molH2

Rearrange Ideal Gas Law to V = nRT/P, convert T to Kelvin and P to atm.V = 2.78 mol(0.082057)303.15 K/0.9211 atm = 75.1 L of H2

Ideal GasAn ideal gas obeys the gas laws.

The volume the molecules of an ideal gas occupy is negligible compared to the volume of the gas. This is true at all temperatures and pressures.

The intermolecular attractions between the molecules of an ideal gas are negligible at all temperatures and pressures.

Real GasesDeviations from the gas laws occur at high

pressures and low temperatures.At high pressures the volumes of the real

gas molecules are not negligible compared to the volume of the gas

At low temperatures the kinetic energy of the gas molecules cannot completely overcome the intermolecular attractive forces between the molecules.

Diffusion

The ability of two or more gases to mix spontaneously until they form a uniform mixture.

Stopcock closed No diffusion occurs

Stopcock open Diffusion occurs

Graham’s Law of EffusionEffusion: a process by which gas molecules pass through a very small orifice from a container at higher pressure to one at lower pressure.The rates of effusion of two gases at the same temperature and pressure are inversely proportional to the square roots of their densities, or molar masses.

rate of effusion of gas Arate of effusion of gas B

dB=

dAmolar mass B

= molar mass A

2

effusion rate COeffusion rate CO

2molar mass CO=

molar mass CO

44.0 g= 1.25

28.0 g

What is the ratio of the rate of effusion of CO to CO2?