GASESGASES

Chapter 10

The AtmosphereThe Atmosphere

The atmosphere is a gaseous solution of nitrogen, N2, and oxygen, O2.

The atmosphere both supports life and acts as a waste receptacle for the exhaust gases that accompany many industrial processes leading to many types of pollution, including smog and acid rain.

A GasA Gas

- Uniformly fills any container.

- Mixes completely with any other gas

- Exerts pressure on its surroundings.

- Is easily compressed.

05_47

h = 760mm Hgfor standardatmosphere

Vacuum

Simple barometer invented by Evangelista Torricelli.A barometer is used to measure atmospheric pressure.

Atmospheric PressureAtmospheric Pressure

Atmospheric pressure results from the mass of the air being pulled toward the center of the earth by gravity--the weight of the air.

At sea level, atmospheric pressure is 760mm.

Atmospheric pressure varies with altitude--at Breckenridge, CO (elevation 9600 ft), the pressure is 520 mm.

Figure 12.1: The pressure exerted by the gases in the Figure 12.1: The pressure exerted by the gases in the atmosphere can demonstrated by boiling water in a canatmosphere can demonstrated by boiling water in a can

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

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h h

Atmosphericpressure (Patm)

Atmosphericpressure (Patm)

Gaspressure (Pgas)less thanatmospheric pressure

Gaspressure (Pgas)greater thanatmospheric pressure

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

(a) (b)

A simple manometer s a device for measuringthe pressure of a gas in a container.

Pressure Unit ConversionsPressure Unit ConversionsThe pressure of a tire is measured to be 28

psi. What would the pressure be in atmospheres, torr, and pascals.

(28 psi)(1.000 atm/14.69 psi) = 1.9 atm

(28 psi)(1.000 atm/14.69 psi)(760.0 torr/1.000atm) = 1.4 x 103 torr

(28 psi)(1.000 atm/14.69 psi)(101,325 Pa/1.000 atm) = 1.9 x 105 Pa

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Pext

Pext

Volume is decreased

Volume of a gas decreases as pressure increases at constant temperature

05_50

P (

in H

g)

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(in3 )

BOYLE’S LAW DATA

P vs V V vs 1/P

Figure 12.6: Illustration of Boyle’s lawFigure 12.6: Illustration of Boyle’s law

Boyle’s LawBoyle’s Law**

Robert Boyle -- Irish scientist (1627-1691)

(Pressure)( Volume) = Constant (T = constant)

P1V1 = P2V2 (T = constant)

V 1/P (T = constant)

(*Holds precisely only at very low pressures.)

The temperature and amount of gas must remain unchanged.

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

ideal gas.ideal gas.

Boyle’s Law CalculationsBoyle’s Law CalculationsA 1.5-L sample of gaseous CCl2F2 has a pressure of 56

torr. If the pressure is changed to 150 torr, will the volume of the gas increase or decrease? What will the new volume be?

Decrease

P1 = 56 torr

P2 = 150 torr

V1 = 1.5 L

V2 = ?

V1P1 = V2P2

V2 = V1P1/P2

V2 = (1.5 L)(56 torr)/(150 torr)V2 = 0.56 L

Boyle’s Law CalculationsBoyle’s Law CalculationsIn an automobile engine the initial cylinder

volume is 0.725 L. After the piston moves up, the volume is 0.075 L. The mixture is 1.00 atm, what is the final pressure?

P1 = 1.00 atm

P2 = ?

V1 = 0.725 L

V2 = 0.075 L

V1P1 = V2P2

P2 = V1P1/V2

P2 = (0.725 L)(1.00 atm)/(0.075 L)P2 = 9.7 atm

Is this answer reasonable?

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

Plot of V vs. T(oC) for several gases

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Pext

Pext

Energy (heat) added

Volume of a gas increases as heat is addedwhen pressure is held constant.

Charles’s LawCharles’s Law

Jacques Charles -- French physicist (1746-1823)

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

V = bT (P = constant)

y = mx + b

b = a proportionality constant

Charles’s LawCharles’s Law

VT

VT

P1

1

2

2 ( constant)

Temperature must be in Kelvin!!

Charles’s Law CalculationsCharles’s Law CalculationsA 2.0-L sample of air is collected at 298 K and

then cooled to 278 K. The pressure is held constant. Will the volume increase or decrease? What will the new volume be?

V1 = 2.0 L

V2 = ?

T1 = 298 K

T2 = 278 K

V1/V2 = T1/T2

V2 = V1 T2/T1

V2 = (2.0 L)(278 K)/(298 K)V2 = 1.9 L

Decrease

Charles’s Law CalculationsCharles’s Law CalculationsConsider a gas with a a volume of 0.675 L at 35 oC and

1 atm pressure. What is the temperature (in Co) of the gas when its volume is 0.535 L at 1 atm pressure?

V1 = 0.675 L

V2 = 0.535 L

T1 = 35 oC + 273 = 308 K

T2 = ?

V1/V2 = T1/T2

T2 = T1 V2/V1

T2 = (308 K)(0.535 L)/(0.675 L)T2 = 244 K -273 T2 = - 29 oC

05_1544

Increase volume to return to originalpressure

Gas cylinder

Moles of gasincreases

Pext Pext

Pext

At constant temperature and pressure, increasing the moles of a gas increases its volume.

Figure 12.11: The total pressure of a mixture of gases Figure 12.11: The total pressure of a mixture of gases depends on the number of moles of gas particles depends on the number of moles of gas particles

(atoms or molecules) present(atoms or molecules) present

Avogadro’s LawAvogadro’s LawFor a gas at constant temperature and pressure, the volume is directly proportional to the number of moles of gas (at low pressures).

V = an

y = mx + b

a = proportionality constantV = volume of the gasn = number of moles of gas

AVOGADRO’S LAWAVOGADRO’S LAW

Temperature and pressure must remain constant!!

2

1

2

1

n

n

V

V

AVOGADRO’S LAWAVOGADRO’S LAW

A 12.2 L sample containing 0.50 mol of oxygen gas, O2, at a pressure of 1.00 atm and a temperature of 25 oC is converted to ozone, O3, at the same temperature and pressure, what will be the volume of the ozone? 3 O2(g) ---> 2 O3(g)

(0.50 mol O2)(2 mol O3/3 mol O2) = 0.33 mol O3

V1 = 12.2 L

V2 = ?

n1 = 0.50 mol

n2 = 0.33 mol

V1/V2 = n1/n2

V2 = V1 n2/n1

V2 = (12.2 L)(0.33 mol)/(0.50 mol)V2 = 8.1 L

COMBINED GAS LAWCOMBINED GAS LAW

V1/ V2 = P2 T1/ P1 T2

V2 = V1P1T2/P2T1

V2 = (309 K)(0.454 atm)(3.48 L)

(258 K)(0.616 atm)

V2 = 3.07 L

What will be the new volume of a gas under the following conditions?

V1 = 3.48 LV2 = ?P1 = 0.454 atmP2 = 0.616 atmT1 = - 15 oC + 273 = 258 KT2 = 36 oC + 273T2= 309 K

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Pext

Pext

Temperature is increased

Pressure exerted by a gas increases as temperature increases provided volume remains constant.

If the volume of a gas is held constant, then V1 / V2 = 1.Therefore:

P1 / P2 = T1 / T2

Pressure and Kelvin temperature are directlyproportional.

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 GASIDEAL GAS

1. Molecules are infinitely far apart.

2. Zero attractive forces exist between the molecules.

3. Molecules are infinitely small--zero molecular volume.

What is an example of an ideal gas?

REAL GASREAL GAS

1. Molecules are relatively far apart compared to their size.

2. Very small attractive forces exist between molecules.

3. The volume of the molecule is small compared to the distance between molecules.

What is an example of a real gas?

Real vs. Ideal GasesReal vs. Ideal Gases

When p 1 atm and T O oC, real gases approximate ideal gases.

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 Law CalculationsIdeal Gas Law CalculationsA sample of hydrogen gas, H2, has a volume of 8.56 L at a

temperature of O oC and a pressure of 1.5 atm. Calculate the number of moles of hydrogen present.

p = 1.5 atm

V = 8.56 L

R = 0.08206 Latm/molK

n = ?

T = O oC + 273

T = 273K

pV = nRTn = pV/RTn = (1.5 atm)(8.56L) (0.08206 Latm/molK)(273K)n = 0.57 mol

Ideal Gas Law CalculationsIdeal Gas Law CalculationsA 1.5 mol sample of radon gas has a volume of 21.0 L at

33 oC. What is the pressure of the gas?

p = ?

V = 21.0 L

n = 1.5 mol

T = 33 oC + 273

T = 306 K

R = 0.08206 Latm/molK

pV = nRTp = nRT/Vp = (1.5mol)(0.08206Latm/molK)(306K)

(21.0L)p = 1.8 atm

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

For a mixture of gases in a container, the total pressure exerted is the sum of the partial pressures of the gases present.

PTotal = P1 + P2 + P3 + . . .

Dalton’s Law of Partial Dalton’s Law of Partial Pressures CalculationsPressures Calculations

A mixture of nitrogen gas at a pressure of 1.25 atm, oxygen at 2.55 atm, and carbon dioxide at .33 atm would have what total pressure?

PTotal = P1 + P2 + P3

PTotal = 1.25 atm + 2.55 atm + .33 atm

Ptotal = 4.13 atm

Water Vapor PressureWater Vapor Pressure2KClO3(s) ----> 2KCl(s) + 3O2(g)

When a sample of potassium chlorate is decomposed and the oxygen produced collected by water displacement, the oxygen has a volume of 0.650 L at a temperature of 22 oC. The combined pressure of the oxygen and water vapor is 754 torr (water vapor pressure at 22 oC is 21 torr). How many moles of oxygen are produced?

Pox = Ptotal - PHOH

Pox = 754 torr - 21 torr

pox = 733 torr

Water Vapor PressureWater Vapor PressureContinuedContinued

p = (733 torr)(1 atm/760 torr)

p = 0.964 atm

V = 0.650 L

n = ?

T = 22 oC + 273

T = 295 K

R = 0.08206 Latm/molK

pV = nRTn = pV/RTn = (0.964 atm)(0.650 L) (0.08206 Latm/molK)(295K)n = 2.59 x 10-2 mol

Kinetic Molecular TheoryKinetic Molecular Theory1. 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.

5. Gases consist of tiny particles (atoms or molecules.

05_58

Rela

tive

nu

mbe

r o

f O

2 m

ole

cule

sw

ith g

ive

n v

elo

city

0

Molecular velocity (m/s)

4 x 102 8 x102

Plot of relative number of oxygen molecules with a given velocity at STP (Boltzmann Distribution).

05_59

Rel

ativ

e nu

mbe

r of

N2

mol

ecul

esw

ith g

iven

vel

ocity

0Velocity (m/s)1000 2000 3000

273 K

1273 K

2273 K

Plot of relative number of nitrogen molecules witha given velocity at three different temperatures.

Kinetic Molecular TheoryKinetic Molecular Theory

As the temperature of a gas increases, the velocity of the molecules increases.

Molecules at a higher temperature have greater KE and hit the walls of the container harder and more often exerting more pressure.

Kinetic Molecular TheoryKinetic Molecular Theory

As the temperature of a gas increases, the velocity of the molecules increases.

Molecules at a higher temperature have greater KE and hit the walls of the container harder and more often exerting more pressure. If the container is not rigid, the container and volume of the gas will expand.

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

Molar VolumeMolar Volume

pV = nRT

V = nRT/p

V = (1.00 mol)(0.08206 Latm/molK)(273K)

(1.00 atm)

V = 22.4 L

Figure 12.12: The production of oxygen by thermal Figure 12.12: The production of oxygen by thermal decomposition of KClOdecomposition of KClO33

GAS STOICHIOMETRYGAS STOICHIOMETRYNot at STPNot at STP

Calculate the volume of oxygen gas produced at 1.00 atm and 25 oC by the complete decomposition of 10.5 g of potassium chlorate.

2KClO3(s) ----> 2KCl(s) + 3O2(g)

(10.5g KClO3)(1mol/122.6 g)(3 mol O2/2mol KClO3) = 1.28 x 10-1 mol O2

Gas StoichiometryGas StoichiometryNot at STP (Continued)Not at STP (Continued)

p = 1.00 atm

V = ?n = 1.28 x 10-1 mol

R = 0.08206 Latm/molK

T = 25 oC + 273 = 298 K

pV = nRTV = nRT/pV = (1.28 x 10-1mol)(0.08206Latm/molK)(298K)

(1.00 atm)V = 3.13 L O2

Gases at STPGases at STP

A sample of nitrogen gas has a volume of 1.75 L at STP. How many moles of N2 are present?

(1.75L N2)(1.000 mol/22.4 L) = 7.81 x 10-2 mol N2

Gas StoichiometryGas Stoichiometryat STPat STP

Quicklime, CaO, is produced by heating calcium carbonate, CaCO3. Calculate the volume of CO2 produced at STP from the decomposition of 152 g of CaCO3. CaCO3(s) ---> CaO(s) + CO2(g)

(152g CaCO3)(1 mol/100.1g)(1mol CO2/1mol CaCO3) (22.4L/1mol) = 34.1L CO2

Note: This method only works when the gas is at STP!!!!!

Chemistry is a gas!!Chemistry is a gas!!