The Chemistry of GasesThe Chemistry of Gases

A GasA Gas

- Uniformly fills any container.

- Mixes completely with any other gas

- Exerts pressure on its surroundings.

PressurePressure

- is equal to force/unit area

- SI units = Newton/meter2 = 1 Pascal (Pa)

- 1 standard atmosphere = 101,325 Pa

- 1 standard atmosphere = 1 atm =

760 mm Hg = 760 torr

05_47

h = 760mm Hgfor standardatmosphere

Vacuum

Effect of Atmospheric Effect of Atmospheric PressurePressure

05_48

h h

Atmosphericpressure (Patm)

Atmosphericpressure (Patm)

Gaspressure (Pgas)less thanatmospheric pressure

Gaspressure (Pgas)greater thanatmospheric pressure

(Pgas) = (Patm) - h (Pgas) = (Patm) + h

(a) (b)

Standard Pressure UnitsStandard Pressure Units

1 atmosphere =

760 mm Hg = 760 torr

1.013 x 105 Pa = 101.3 kPa

14.7 psi (pounds per square inch)

Pressure ConversionsPressure Conversions

The pressure of a gas is measured as 49 torr. Represent this as atmospheres and as Pascals.

49 torr| 1 atm = 6.4 x 10-2 atm

| 760 torr

49 torr | 1.013 x 105 Pa = 6.5 x 103 Pa

| 760 torr

Boyle’s LawBoyle’s Law**

Pressure Volume = Constant (T = constant)

P1V1 = P2V2 (T = constant)

V 1/P (T = constant)

(*Holds precisely only at very low pressures.)

A gas that A gas that strictly obeys strictly obeys Boyle’s Law is called an Boyle’s Law is called an

ideal gasideal gas..

05_1541

Pext

Pext

Volume is decreased

05_50P

(in

Hg)

0

V(in3)

20 40 60

50

100

PP2

V

2V

(a)

00

1/P (in Hg)

0.01 0.02 0.03

20

40slope = k

(b)

V(in

3 )

1.53 L of sulfur dioxide gas at 5600 Pa is changed to 15000 Pa at constant temperature. What is its new volume?

P1V1 =P2V2 5600 x 1.53 = 15000 x V2

V2 = 0.57 L

1.0 L of hydrogen gas at 345 torr is connected

to an empty 2.5 L flask and allowed to combine. What is its new pressure?

345 x 1.0 = P2 x 3.5 P2 = 98.6 torr

Charles’s LawCharles’s Law

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

V = bT (P = constant)

b = a proportionality constant

05_1543

Pext

Pext

Energy (heat) added

Balloon with Liquid NBalloon with Liquid N22

05_53

V(L

)

-300

T(ºC)

-200 -100 0 100 200

1

2

3

6

4

5

300

-273.2 ºC

N2O

H2

H2O

CH4

He

Charles’s LawCharles’s Law

VT

VT

P1

1

2

2 ( constant)

Always Change to KELVIN!Always Change to KELVIN!

K = C + 273

A sample of gas at 15o C and 1 atm has a volume of 2.58 L. What is its new volume at 38o C and 1 atm pressure?

Pressure remains constant at 1 atm

T1 = 15 + 273 = 288 K

T2 = 38 + 273 = 311 K

V1 = V2 or V1T2 = V2T1

T1 T2

2.58 x 311 = V2 x 288 V2 = 2.79 L

Gay-Lussac’s LawGay-Lussac’s Law

Pressure is directly proportional to the Kelvin Temperature

P = k T

or P1 = P2

T1 T2

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Pext

Pext

Temperature is increased

Standard Temperature Standard Temperature and Pressureand Pressure

“STP”

P = 1 atmosphere

T = C

The molar volume of an ideal gas is 22.42 liters at STP

Combined Gas LawCombined Gas Law

When all three change use:

P1 V1 = P2 V2

T1 T2

or P1V1 T2 = P2V2 T1

A cylinder contains 255 mL of CO2 at 27o C and 812 torr. What volume would this occupy at STP?

P1= 812 torr T1= 27o C + 273= 300 K

V1= 255 mL P2= 1 atm= 760 torr

T2 = 0o C = 273 K

P1V1T2=P2V2T1 (812)(255)(273)=V2(760)(300)

V2= 247.9 mL

Avogadro’s LawAvogadro’s Law

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

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Increase volume to return to originalpressure

Gas cylinder

Moles of gasincreases

Pext Pext

Pext

Ideal Gas LawIdeal Gas Law

- An equation of state for a gas.

- “state” is the condition of the gas at a given time.

PV = nRT

Ideal Gas LawIdeal Gas Law

PV = nRT

R = proportionality constant

= 0.08206 L atm mol

P = pressure in atm

V = volume in liters

n = moles

T = temperature in Kelvins

Holds closely at P < 1 atm

Ideal Gas LawsIdeal Gas Laws

Can be used to calculate molar masses by gas density

density= mass = molar mass

volume molar volume

At STP, molar volume is 22.4 L

Suppose you have a sample of ammonia gas with a volume of 856 mL at a pressure of 932 torr and 28o C. What is the mass of the ammonia present?

P= 932 torr / 760 torr = 1.23 atm

V = 0.856 L T = 28 + 273= 301 K

n = PV = (1.23)(.856) = 0.0426 moles

RT (.0821)(301)

mass = moles x MM= 0.0426 x 17= 0.724g

Dalton’s Law of Dalton’s Law of Partial PressuresPartial Pressures

For a mixture of gases in a container,

PTotal = P1 + P2 + P3 + . . .

Used primarily when collecting

gases over water.

Mixtures of helium and oxygen are used for deep scuba diving. If 46 L oxygen and 12 L He at 1.0 atm are put into a 5.0 L tank at 25oC, calculate the partial pressures of each gas and the total pressure in the tank.

n = PV = (1)(46) and (1)(12)

RT (.0821)(298) (.0821)(298)

nO2= 1.9 mol nHe = 0.49 mol

In the tank, V= 5.0 L

PO2 =nRT / V = (1.9)(.0821)(298) / 5.0 = 9.3 atm

Phe = (0.49)(.0821)(298)/ 5.0 = 2.4 atm

Ptotal = 9.3 + 2.4 = 11.7 atm

HomeworkHomework

#1 p223ff 8, 12, 15, 17, 18, 30

Stoichiometry and GasesStoichiometry and GasesA sample of solid potassium chlorate decomposes by the equation 2 KClO3 (s) -----> 2 KCl (s) + 3 O2 (g)

If 650 mL of oxygen at 22oC and 754 torr is collected over water, how much KClO3 decomposed?

PH2O = 21 torr (from table of vapor pressure)

PO2 = 754 - 21 = 733 torr / 760 = 0.964 atm

nO2 = PV /RT = (.964)(.650) / (.0821)(295) = 0.0259

0.0259 mol O2 | 2 mol KClO3 | 122.6 g KClO3 = 2.11 g

| 3 mol O2 | 1 mol KClO3 KClO3

HomeworkHomework

#2 p226ff 32, 35, 38, 39, 41, 42, 44

Kinetic Molecular TheoryKinetic Molecular Theory

1. Volume of individual particles is zero.

2. Collisions of particles with container walls cause pressure exerted by gas.

3. Particles exert no forces on each other.

4. Average kinetic energy Kelvin temperature of a gas.

The Meaning of The Meaning of TemperatureTemperature

Kelvin temperature is an index of the random motions of gas particles (higher T means greater motion.)

(KE)32avg RT

05_58

Rel

ativ

e nu

mbe

r of

O2

mol

ecu

les

with

giv

en

velo

city

0

Molecular velocity (m/s)

4 x 102 8 x102

05_59

Rel

ativ

e nu

mbe

r of

N2

mol

ecu

les

with

giv

en

velo

city

0Velocity (m/s)1000 2000 3000

273 K

1273 K

2273 K

Effusion: describes the passage of gas into an evacuated chamber.

Diffusion: describes the mixing of gases. The rate of diffusion is the rate of gas mixing.

05_60

Gas

Vacuum

Pinhole

Diffusion RatesDiffusion Rates

Rate of effusion for gas 1Rate of effusion for gas 2

2

1

MM

Distance traveled by gas 1Distance traveled by gas 2

2

1

MM

Effusion:Effusion:

Diffusion:Diffusion:

Find the ratio of the effusion rates of UF6 versus H2.

Rate H2 = MMUF6 = 352 =13.2

Rate UF6 MMH2 2

HomeworkHomework

#3 p228ff 53, 54, 68, 70, 72, 76

#4 p229ff 77, 89, 91, 92

Super XCR #97

Real GasesReal Gases

Must correct ideal gas behavior when at high pressure (smaller volume) and low temperature (attractive forces become important).

05_62

00

P(atm)

200 400 600 800

1.0

2.0

PV

nRTCO2

H2

CH4N2

Ideal

gas

1000

Real GasesReal Gases

[ ]P a V nb nRTobs2( / ) n V

corrected pressurecorrected pressure corrected volumecorrected volume

PPidealideal VVidealideal