principles of chemistry i chem 1211 chapter 6

PRINCIPLES OF CHEMISTRY I

CHEM 1211

CHAPTER 6

DR. AUGUSTINE OFORI AGYEMANAssistant professor of chemistryDepartment of natural sciences

Clayton state university

CHAPTER 6

THE GASEOUS STATE

CHARACTERISTICS OF GASES

- Gases are composed of nonmetals

- Gases form homogeneous mixtures of each other

Air mixture of 78% N2

21% O2

0.9% Arother substances (CO2, H2, Ne, Kr, He)

Gases at Ordinary Temperature and Pressure

Noble gases (monatomic gases)He, Ne, Ar, Kr, Xe

Diatomic GasesH2, N2, O2, F2, Cl2

Other Common Gasespropane, ammonia, carbon dioxide,

hydrogen sulfide, methane, carbon monoxide, sulfur dioxide

CHARACTERISTICS OF GASES

PRESSURE

- Gases exert pressure on any surface they come in contact with

- Pressure is force applied per unit area

A

FP

A = area (m2)F = force (newton, N = kg-m/s2)P = pressure (N/m2 = pascal, Pa)

F = m x a

m = mass (kg)

a = acceleration (m/s2)

PRESSURE

THE GAS LAWS

Four Variables Define the Physical State of any GasAmount (mole)

Temperature (K)Volume (L)

Pressure (bar, Pa, mm Hg, torr, atm, psi)

1 bar = 105 Pa1 atm = 760 mmHg

= 760 torr = 1.01325 x 105 Pa

= 101.325 kPa = 14.7 psi

mm Hg: millimeters mercuryatm: atmosphere (atmospheric pressure = 1atm)

Pa: Pascalpsi: pound per square inch (Ib/in2)

Pressure Instrumentsbarometers, manometers, gauges

760, 700, 650 mm Hg- Considered to have 3 significant figures

THE GAS LAWS

BOYLE’S LAW

- The volume of a fixed amount of a gas is inversely proportional to the pressure applied to the gas

if the temperature is kept constant

PV = constant

P1V1 = P2V2

- P1 and V1 are the pressure and volume of a gas at an initial set of conditions

- P2 and V2 are the pressure and volume of the same gas under a new set of conditions

- The temperature and amount of gas remain constant

A sample of N2 gas occupies a volume of 3.0 L at 6.0 atm pressure . What is the new pressure if the gas is

allowed to expand to 4.8 L at constant temperature?

P1 = 6.0 atm V1 = 3.0 LP2 = ? V2 = 4.8 L

P1V1 = P2V2

(6.0 atm)(3.0 L) = (P2)(4.8 L)

P2 = 3.8 atm

BOYLE’S LAW

CHARLES’S LAW

- The volume of a fixed amount of gas is directly proportional to its absolute temperature

if the pressure is kept constant

- V1 and T1 are the volume and absolute temperature of a gas at an initial set of conditions

- V2 and T2 are the volume and absolute temperature of the same gas under a new set of conditions

- The pressure and amount of gas remain constant

constantT

V

2

2

1

1

T

V

T

V

A sample of Ar gas occupies a volume of 1.2 L at 125 oCand a pressure of 1.0 atm. What is the new temperature, in Celsius, if the volume of the gas is decreased to 1.0 L at the

same pressure?

V1 = 1.2 L T1 = 125 oC = 398 K

V2 = 1.0 L T2 = ?

2T

L1.0

K398

L1.2

T2 = 332 K = 59 oC

CHARLES’S LAW

AVOGADRO’S LAW

- The volume of a gas maintained at constant temperature and pressure is directly proportional to the

number of moles of the gas

n = number of moles of a gas

constantn

V

2

2

1

1

n

V

n

V

AVOGADRO’S LAW

Avogadro’s Hypothesis

- Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules

At Standard Temperature and Pressure (STP)1 mol of any gas (= 6.022 x 1023 molecules)

occupies a volume of 22.41 L

Conditions of STPTemperature = 0 oC = 273 K = 32 oF

Pressure = 1.00 atm (101.325 kPa or 100 kPa)

THE IDEAL GAS LAW

PV = nRT

Considering all three gas laws

V α 1/P V α T V α n

P

nTαV

P

nTRV

R is the ideal gas constant

= 0.08206 L-atm/mol.K

= 8.314 J/mol-K

= 8.314 m3-Pa/mol-K

= 1.987 cal/mol-K

= 62.36 L-torr/mol-K

THE IDEAL GAS LAW

RELATING THE GAS LAWS

PV = nRT

If n is constant

constantT

PV

2

22

1

11

T

VP

T

VP

A 1.00-L container is filled with 0.500 mole of CO gas at35.0 oC. Calculate the pressure, in atmospheres, exerted

by the gas in the container

PV = nRT

P = ? V = 1.00 L n = 0.500 molT = 35.0 oC = 308 K R = 0.08206 atm.L/mol.K

(P)(1.00 L) = (0.500 mol)(0.08206 atm.L/mol.K)(308 K)

P = 12.6 atm


A balloon filled with helium initially has a volume of 1.00 x 106 L at 25 oC and a pressure of 752 mm Hg.Determine the volume of the balloon after a certain

time when it encounters a temperature of -33 oC and a pressure of 75.0 mm Hg

P1 = 752 mm Hg V1 = 1.00 x 106 L T1 = 25 oC = 298 K P2 = 75.0 mm Hg V2 = ? T2 = -33 oC = 240 K

L10x8.08K)Hg)(298mm(75.0

K)L)(24010xHg)(1.00mm(752

TP

TVPV 6

6

12

2112


GAS DENSITIES AND MOLAR MASS

- Density (d) = mass/volume (m/V)

- Rearrange the ideal gas equation and multiply both sides by molar mass (M) to give

RT

MP

V

Mn

M x n = mass (m) and m/V = d

RT

MPd

Calculate the density of carbon dioxide gas at 0.45 atm and 252 K

K)K)(252-L/mol-atm(0.08206

atm) 5g/mol)(0.4 (44.01d

Density (d) = 0.96 g/L


Calculate the average molar mass of dry air if its density is 1.17 g/L at 21 oC and 740.0 torr

Molar mass = 29.0 g/mol

P

dRTM

torr)(740.0

K) K)(294-torr/mol-L g/L)(62.36 (1.17M


REACTION STOICHIOMETRY

Consider the following reaction

4Al(s) + 3O2(g) → 2Al2O3(s)

Calculate the mass of aluminum that would react completely with 2.00 L of pure oxygen gas at STP

1 mol O2 at STP = 22.4 L


22

222 Omol0.0893

)O L (22.4

)O mol )(1O L (2.00O moles

Almol0.119)O mol (3

Al) mol )(4O mol (0.0893Al moles

2

2

Alg21.3Al) mol (1

Al) g Al)(26.98 mol (0.119Al mass

- At the same conditionsvolumes of gases combine in the same proportions as

the coefficients of the chemical equation

Example3H2(g) + N2(g) → 2NH3(g)

This implies that

- 3 mol H2 reacts with 1 mol N2 to produce 2 mol NH3

- 3 L H2 reacts with 1 L N2 to produce 2 L NH3


DALTON’S LAW OF PARTIAL PRESSURES

- The total pressure exerted by a mixture of gases is the sum of the partial pressures of the individual gases present

- The partial pressure is the pressure that a gas in a mixture of gases would exert if it were present alone under the same

conditions

- Ptotal is the total pressure of a gaseous mixture

- P1, P2, P3,…. are the partial pressures of the individual gases

- ntotal is the total number of moles of a gaseous mixture

- n1, n2, n3,….are the number of moles of the individual gases

V

RTn

V

RT......nnn......PPPP total321321total


The total pressure by a mixture of He, Ne, and Ar gases is3.50 atm. Find the partial pressure of Ar if the partial pressures

of He and Ne are 0.50 atm and 0.75 atm, respectively

Ptotal = 3.50 atmP1 = 0.50 atmP2 = 0.75 atm

P3 = ?

Ptotal = P1 + P2 + P3

P3 = Ptotal - (P1 + P2) = 3.50 atm - (0.50 atm + 0.75 atm) = 2.25 atm


For gas 1 in a mixture of gases with n1 molesMole fraction (x1) = n1/ntotal

total1totaltotal

11 PxP

n

nP

O2 is 21% of airMole fraction of O2 (x1) = 0.21

Total atmospheric pressure = 1 atmPartial pressure of O2 = P1 = (0.21)(1 atm) = 0.21 atm

= (0.21)(760 mm Hg) = 160 mm Hg


EXPERIMENT TO COLLECT GAS OVER WATER

- Magnesium reacts with HCl to produce hydrogen gas according to the following equation

Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

- The H2 gas is collected in a gas collection tube initially filled with water and inverted in a water pan

- The volume of H2 gas is measured by raising or lowering the tube such that the water levels in the tube and pan are the same

For this condition

The pressure inside the tube is equal to the atmospheric pressure outside

Ptotal = Pgas + Pwater

Ptotal = total atmospheric pressure

Pgas = pressure of gas in tube

Pwater = vapor pressure of water at experimental temperature

EXPERIMENT TO COLLECT GAS OVER WATER

- Gases consist of large numbers of small molecules that are in continuous random motion

- Attractive and repulsive forces between gas molecules are negligible

- The combined volume of all the molecules of a gas is negligible relative to the total volume in which the

gas is contained

KINETIC MOLECULAR THEORY OF GASES

- Collisions between molecules are perfectly elastic (the average kinetic energy of the colliding molecules does

not change at constant temperature)

- The average kinetic energy of the molecules is proportional to the absolute temperature (molecules of all gases have the

same average kinetic energy at a given temperature)

KINETIC MOLECULAR THEORY OF GASES

AVERAGE KINETIC ENERGY

2mu2

1ε

ε = average kinetic energy

m = mass of an individual molecule

u = root-mean-square (rms) velocity at a given temperature

M

3RTu

AVERAGE KINETIC ENERGY

Root-Mean-Square Velocity

- Square-root of the average of the squares

- The lower the molar mass of a gas the higher the u

Diffusion - The spread of one substance throughout space or

throughout a another substance

Effusion - Escape of gas molecules through a tiny hole

into an evacuated space

GRAHAMS LAW OF EFFUSION


- r1 and r2 are the effusion rates of gases 1 and 2 respectively- M1 and M2 are the molar masses of gases 1 and 2 respectively

- The rate of effusion of a gas is inversely proportional to the square root of its molar mass

1

2

2

1

M

M

r

r

- The rate of effusion of a gas is directly proportional to the rms velocity of the molecules

1

2

2

1

2

1

2

1

M

M

3RT/M

3RT/M

u

u

r

r


- The time it takes for a gas to effuse is inversely proportional to the rate of effusion

1

2

1

2

2

1

M

M

t

t

r

r

It took 4.5 minutes for 1.0 L helium to effuse through a porousbarrier. How long will it take for 1.0 L Cl2 gas to effuse under

identical conditions?

g/mol 4.00

g/mol 70.90

minutes 4.5

t 2

t2 = 19 minutes


REAL GASES

Mean Free Path

- The average distance traveled by a molecule between collisions

- Real Gases deviate from ideal gas behavior at high pressures

- Deviation is very small at low pressures (usually below 10 atm)

- Deviation increases with decreasing temperature

- Deviation is significant near the temperature at which a given gas changes to the liquid state

- Molecules of real gases have finite volumes and attract one another

- van der Waals equation

nRTnb)(VV

anP

2

2

REAL GASES

a = constant (measure of how strongly gas molecules attract each other

b = constant (measure of the small but finite volume occupied by gas molecules)

n2a/V2 accounts for the attractive forces and adjusts pressure upwards

nb accounts for the finite volume occupied by molecules

REAL GASES

Calculate the pressure exerted by 0.2500 mol N2 gas in a 5.000 L container at 22.5 oC using both the ideal gas law

and the van der Waals equation. Compare the results.

PV = nRT

P = nRT/V

= (0.2500 mol)(0.08206 L-atm/mol-K)(295.5 K)/(5.000 L)

= 1.212 atm

REAL GASES

nRTnb)(VV

anP

2

2

atm1.211P

)5.295)(08206.0)(2500.0(

)]0391.0)(2500.0([(5.000)(5.000)

(1.39)(0.2500)P

2

2

REAL GASES

principles of chemistry i chem 1211 chapter 6

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