Chapter 09 Slide 1

Gases: Their Properties & Behavior

Gases: Their Properties & Behavior

9

Chapter 09 Slide 2

Gas Pressure 01Gas Pressure 01

Chapter 09 Slide 3

Gas Pressure 02Gas Pressure 02

• Units of pressure: atmosphere (atm)

Pa (N/m2, 101,325 Pa = 1 atm) Torr (760 Torr = 1 atm)

bar (1.01325 bar = 1 atm)

mm Hg (760 mm Hg = 1 atm)

lb/in2 (14.696 lb/in2 = 1 atm)

in Hg (29.921 in Hg = 1 atm)

Chapter 09 Slide 4

• Pressure–Volume Law (Boyle’s Law):

Boyle’s Law 01Boyle’s Law 01

Chapter 09 Slide 5

• Pressure–Volume Law (Boyle’s Law):

Boyle’s Law 01Boyle’s Law 01

Chapter 09 Slide 6

Boyle’s Law 02Boyle’s Law 02

• Pressure–Volume Law (Boyle’s Law):

• The volume of a fixed

amount of gas maintained at

constant temperature is

inversely proportional to the

gas pressure.

Pressure1

Volume

P1 x V1 = P2 x V2

P1 x V1 = K1

P2 x V2 = K1

Chapter 09 Slide 7

A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL?

P1 x V1 = P2 x V2

P1 = 726 mmHg

V1 = 946 mL

P2 = ?

V2 = 154 mL

P2 = P1 x V1

V2

726 mmHg x 946 mL154 mL

= = 4460 mmHg

5.3

Chapter 09 Slide 8As T increases V increases

Chapter 09 Slide 9

Charles’ Law 01Charles’ Law 01

• Temperature–Volume Law (Charles’ Law):

T(K)=Temperature in KelvinT (K) = t (0C) + 273.15

Chapter 09 Slide 10

Charles’ Law 01Charles’ Law 01

• Temperature–Volume Law (Charles’ Law):

• The volume of a fixed amount of gas at constant

pressure is directly proportional to the Kelvin

temperature of the gas.

V T

V1

T1=k1

Chapter 09 Slide 11

Variation of gas volume with temperatureat constant pressure.

V T

V = constant x T

V1/T1 = V2/T2T (K) = t (0C) + 273.15

Temperature must bein Kelvin

P1<P2<P3<P4

Third Law of Thermodynamics states that it's impossible to reach absolute zero.

Chapter 09 Slide 12

A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant?

V1 = 3.20 L

T1 = 398.15 K

V2 = 1.54 L

T2 = ?

T2 = V2 x T1

V1

1.54 L x 398.15 K3.20 L

= = 192 K

V1/T1 = V2/T2

Chapter 09 Slide 13

Avogadro’s Law 01Avogadro’s Law 01

• The Volume–Amount Law (Avogadro’s Law):

Chapter 09 Slide 14

Avogadro’s Law 01Avogadro’s Law 01

• The Volume–Amount Law (Avogadro’s Law):

• At constant pressure and temperature, the volume

of a gas is directly proportional to the number of

moles of the gas present.

nV

11

1 knV

Chapter 09 Slide 15

Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of NO are obtained from one volume of ammonia at the same temperature and pressure?

4NH3 + 5O2 4NO + 6H2O

1 mole NH3 1 mole NO

At constant T and P

1 volume NH3 1 volume NO

Chapter 09 Slide 16

Ideal Gas EquationIdeal Gas Equation

Charles’ law: V T(at constant n and P)

Avogadro’s law: V n(at constant P and T)

Boyle’s law: V (at constant n and T)1P

V nT

P

V = constant x = RnT

P

nT

PR is the gas constant

PV = nRT

Chapter 09 Slide 17

The conditions 0 0C and 1 atm are called standard temperature and pressure (STP).

PV = nRT

R = PVnT

=(1 atm)(22.414L)

(1 mol)(273.15 K)

R = 0.082057 L • atm / (mol • K)

Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L.

Chapter 09 Slide 18

The Ideal Gas Law01The Ideal Gas Law01

• Ideal gases obey an equation incorporating the

laws of Charles, Boyle, and Avogadro.

• R =The gas constant

• R = 0.08206 L·atm·K–1·mol–1

TRnVP

Chapter 09 Slide 19

What is the volume (in liters) occupied by 49.8 g of HCl at STP?

PV = nRT

V = nRTP

T = 0 0C = 273.15 K

P = 1 atm

n = 49.8 g x 1 mol HCl36.45 g HCl

= 1.37 mol

V =1 atm

1.37 mol x 0.0821 x 273.15 KL•atmmol•K

V = 30.6 L

Chapter 09 Slide 20

Argon is an inert gas used in lightbulbs to retard the oxidation and vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)?

PV = nRT n, V and R are constant

nRV

= PT

= constant

P1

T1

P2

T2

=

P1 = 1.20 atm

T1 = 291 K

P2 = ?

T2 = 358 K

P2 = P1 x T2

T1

= 1.20 atm x 358 K291 K

= 1.48 atm

Chapter 09 Slide 21

Calculation of Molar Mass USING Calculation of Molar Mass USING GAS DENSITYGAS DENSITY

Calculation of Molar Mass USING Calculation of Molar Mass USING GAS DENSITYGAS DENSITY

The density of air at The density of air at 15 15 ooCC and and 1.00 atm1.00 atm is is 1.23 g/L1.23 g/L. .

What is the molar mass of air?What is the molar mass of air?

Molar Mass = Mass / Number of molesMolar Mass = Mass / Number of moles

1. Calc. moles of air.1. Calc. moles of air.

V = 1.00 L, m = 1.23 g, V = 1.00 L, m = 1.23 g, P = 1.00 atmP = 1.00 atm,,

T = (273.15+15 )= 288 KT = (273.15+15 )= 288 K

n = PV/RT = 0.0423 mol,n = PV/RT = 0.0423 mol, m = 1.23 gm = 1.23 g

2.2. Calc. molar massCalc. molar mass

mass/mol = 1.23 g/0.0423 mol = 29.1 g/molmass/mol = 1.23 g/0.0423 mol = 29.1 g/mol

Chapter 09 Slide 22

Dalton’s Law of Partial Pressures

V and T are

constant

P1 P2 Ptotal = P1 + P2

Chapter 09 Slide 23

Consider a case in which two gases, A and B, are in a container of volume V.

PA = nART

V

PB = nBRT

V

nA is the number of moles of A

nB is the number of moles of B

Chapter 09 Slide 24

PT = PA + PB

nART

V

nBRTV

+ = (nA + nB) RTV

PT =

PA = XA PT

PA = nART

V

PB = XB PT

PB = nBRT

V

PA / PT = nA

nA + nB

XA = nA

nA + nB

PA / PT = XA

XB = nB

nA + nB

Pi = Xi PT

Chapter 09 Slide 25

A sample of natural gas contains 8.24 moles of CH4, 0.421 moles of C2H6, and 0.116 moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)?

Pi = Xi PT

Xpropane = 0.116

8.24 + 0.421 + 0.116

PT = 1.37 atm

= 0.0132

Ppropane = 0.0132 x 1.37 atm = 0.0181 atm

Chapter 09 Slide 26

Gas Stoichiometry 01Gas Stoichiometry 01

• In gas stoichiometry, for a constant temperature

and pressure, volume is proportional to moles.

• Assuming no change in temperature and pressure,

calculate the volume of O2 (in liters) required for the

complete combustion of 14.9 L of butane (C4H10):

2 C4H10(g) + 13 O2(g) 8 CO2(g) + 10 H2O(l)

Chapter 09 Slide 27

Gas Stoichiometry

What is the volume of CO2 produced at 370 C and 1.00 atm when 5.60 g of glucose are used up in the reaction:

C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l)

g C6H12O6 mol C6H12O6 mol CO2 V CO2

5.60 g C6H12O6

1 mol C6H12O6

180 g C6H12O6

x6 mol CO2

1 mol C6H12O6

x = 0.187 mol CO2

V = nRT

P

0.187 mol x 0.0821 x 310.15 KL•atmmol•K

1.00 atm= = 4.76 L

Chapter 09 Slide 28

Gases: P, V, T & n

Chapter 09 Slide 29

Kinetic Molecular Theory Kinetic Molecular Theory

• This theory presents physical properties of gases in

terms of the motion of individual molecules.

• Average Kinetic Energy Kelvin Temperature

• Gas molecules are points separated by a great distance

• Particle volume is negligible compared to gas volume

• Gas molecules are in rapid random motion

• Gas collisions are perfectly elastic ( No Change of Total

Kinetic Energy)

• Gas molecules experience no attraction or repulsion

Chapter 09 Slide 30

Kinetic-Molecular Theory Kinetic-Molecular Theory

Energy can be transferred between molecules during collisions, but the average kinetic energy of the molecules does not change with time, as long as the temperature of the gas remains constant.

Chapter 09 Slide 31

• Average Kinetic Energy (KE) is given by:

KE 1

2mu2

u2 u2N

Kinetic Molecular TheoryKinetic Molecular Theory

U: The Root–Mean–Square Speed

Chapter 09 Slide 32

• Average Kinetic Energy (KE) is given by:

KE 1

2mu2

u2 u2N

Kinetic Molecular Theory 03Kinetic Molecular Theory 03

U: The Root–Mean–Square Speed

(1000 miles/hr)

rms speed at 25 °C

Chapter 09 Slide 33

Kinetic Molecular Theory 05Kinetic Molecular Theory 05

• Maxwell speed distribution curves.

Chapter 09 Slide 34

Kinetic Molecular Theory 04Kinetic Molecular Theory 04

• The Root–Mean–Square Speed: is a measure of the average molecular speed.

• R=8.314 J/K.mol 1J = 1Kg.m2/s2

• Calculate the root–mean–square speeds of helium atoms and nitrogen

molecules in miles/hr at 25°C.

Taking square root of both sides gives the equation M

RTurms

3MRT

u32

He : miles/hr

Diffusion of GasesDiffusion of Gases

Diffusion: The mixing of different gases by molecular motion with frequent molecular collisions.

Chapter 09 Slide 36

Graham’s Law 02Graham’s Law 02

• Effusion is when gas

molecules escape ,

through a tiny hole into

a vacuum.

Chapter 09 Slide 37

Graham’s Law 03Graham’s Law 03

• Graham’s Law: Rate of effusion is proportional to its rms speed, urms.

• For two gases at same temperature and pressure:

Rate urms 3RT

M

Rate1

Rate2

M2

M1

M2

M1

Chapter 09 Slide 38

A Problem to ConsiderA Problem to Consider

• How much faster would H2 gas effuse through an opening than methane, CH4?

)M(H

)M(CH

CH of Rate

H of Rate

2

4

4

2

8.2g/mol 2.0g/mol 16.0

CH of RateH of Rate

4

2

So hydrogen effuses 2.8 times faster than CH4

Chapter 09 Slide 39

Behavior of Real Gases01Behavior of Real Gases01

• Deviations result from assumptions about ideal gases.

1. Molecules in gaseous state do not exert any

force, either attractive or repulsive, on one

another.

2. Volume of the molecules is negligibly small

compared with that of the container.

Chapter 09 Slide 40

Behavior of Real Gases02Behavior of Real Gases02

• At higher pressures, particles are much closer

together and attractive forces become more

important than at lower pressures.

As a result, the pressure of real gases will be smaller than the ideal value

Chapter 09 Slide 41

Behavior of Real Gases03Behavior of Real Gases03

• The volume taken up by gas particles is actually less important at lower pressures than at higher pressure. As a result, the volume at high pressure will be greater than the ideal value.

Chapter 09 Slide 42

Behavior of Real Gases05Behavior of Real Gases05

• Corrections for non-ideality require van der Waals equation.

nRTbnVV

naP –

2

2

IntermolecularAttractions

ExcludedVolume

Chapter 09 Slide 43

A Problem to ConsiderA Problem to Consider

• If sulfur dioxide were an “ideal” gas, the pressure at 0 oC exerted by 1.000 mol occupying 22.41 L would be 1.000 atm. Use the van der Waals equation to estimate the “real” pressure.

Use the following values for SO2

a = 6.865 L2.atm/mol2

b = 0.05679 L/mol

Chapter 09 Slide 44

What is the Real pressure of one mole of a gas, with a Volume of 22.4 L at STP?What is the Real pressure of one mole of a gas, with a Volume of 22.4 L at STP?

2

2

V

an -

nb-VnRT

P

R= 0.0821 L. atm/mol. K

T = 273.2 K

V = 22.41 L

a = 6.865 L2.atm/mol2

b = 0.05679 L/mol

nRTbnVV

naP –

2

2

•First, let’s rearrange the van der Waals equation to solve for pressure.

Chapter 09 Slide 45

A Problem to ConsiderA Problem to Consider

• The “real” pressure exerted by 1.00 mol of SO2 at STP is slightly less than the “ideal” pressure.

2

2

V

an -

nb-VnRT

P

L/mol) 79mol)(0.056 (1.000 - L 22.41

)K2.273)( 06mol)(0.082 (1.000 P Kmol

atmL

2mol

atmL2

L) 41.22(

) (6.865mol) (1.000-

2

2

atm 0.989 P