1

Chapter 5

Gases

2

Properties of GasesOccupy the entire volume of their container

CompressibleFlow readily and mix easily

Have low densities, low molecular weight

3

Atmospheric PressureAtmospheric pressure

Pressure from earth’s atmosphere at sea levelEqual to 1atmosphere Unit is atm

BarometerUsed to measure atmospheric pressureHeight in mm HgAt sea level height is exactly 760mm

Other units: 1atm is equal to…..760 mmHg 760 torr 1.01325 kPa29.921 in. Hg

4

Measuring Experimental Gas PressureA manometer measures pressure of gases

Below 1 atm Above 1atmClosed tube manometer Closed tube manometerPgas = Height = mm Hg Pgas = 760mm Hg + height

5

Gas Laws

6

For a fixed amount of gas at constant temperature, volume decreases as pressure increases

Rearranging equation PV = constant so P1V1 = P2V2

pressure: volumeCuts space between molecules

volume: pressureMolecule-wall collisions increase

Boyle’s Law: Pressure/Volume Relationship

1

221 V

VPP 1

221 P

VPV

7

For a fixed amount of gas at constant pressure, volume increases as temperature increases

Rearranging equation V/T= constant so V1/T1 = V2/T2

Temp. in kelvins (T), not Celsius (t)T (K) = t (0ºC) + 273.15

As temperature increases: Molecules move faster They hit the wall harderVolume increases to hold pressure

Charles’ Law: Volume/Temp. Relationship

2

121 T

TVV 2

121 V

VTT

8

At fixed temperature and pressure, the volume of a gas depends on the # of moles of gas present

Rearranging equation V/n= constant so V1/n1 = V2/n2

More molecules need more space to maintain the same pressure and temperature

Avogadro’s law: Moles/Volume Relationship

2

121 n

nVV 2

121 V

Vnn

9

A 4.50-L cylinder contains He(g) at an unknown pressure. It is now connected to a 92.5-L evacuated cylinder. When the

connecting valve between the two cylinders is opened, the pressure falls to 1.40 atm.

What was the pressure in the 4.50-L cylinder?Initial Conditions (P1V1) Final Conditions (P2V2)

V1 =4.50L V2 =4.50L+92.5L=97.0LP1 = ? atm P2 = 1.40 atm

atmL

xLxatmVVPP 2.30

50.41

10.97

140.1

1

221

10

Decreasing the temperature of 10.00 L of H2 (g) from 25ºC to -77ºC decreases its volume to what value?

Find your data:T1 = 25ºC + 273.15 = 298K (convert to K) T2 = -77ºC + 273.15 = 196K (convert to K)V1 is given in problem = 10.00LV2 = ?L

From Charles’s Law:2

2

1

1

TV

TV

LK

KLxTTVV 58.6

29819600.10

1

212

11

The Combined Gas Law and the Ideal Gas ConstantCombines all three laws by means of varying the volume

Boyles Law constant = PVCharles’s Law constant = V / TAvogadro’s Law constant = V / n

STP: Standard Temperature and Pressure = 1atm 273KP= 1atm

V= 22.4L

Ideal Gas Constant: R= .

Acts as 1 side of combined gas law

22

22

11

11

TnVP

TnVP

RmolK

LatmKmolxLatmx

TnVP

0821.0

27314.221

11

11

12

Using the Ideal Gas LawIdeal Gas Law PV = nRT

Must convert units to match R

Identify the noble gas if 23.6 g exerts a pressure of 1293 torr at 37ºC in a 17.5-L container?

Convert T and P: T = 37 ºC = 310.15 K P= x=1.70atm

Solve for moles

Find molar mass MM= ..

= 20.2g/mol

Identify noble gas Neon: MM20.17g/mol

molKLatm

nTPVR 0821.0

molK

xLatm

molKxLxatmn 17.115.3101

0821.015.17

170.1

13

The Law of Combining VolumesAt the same T and P, the volume of a gas is directly related to

the moles of gas and the number of molecules of a gas.

The ratio is the same as that in the chemical equation

14Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many liters of NO are obtained from 1.0 liter of

ammonia at the same temperature and pressure?

Write the balanced chemical equation4NH3(g) + 5O2(g) 4NO (g)+ 6H2O (g)

Write the balanced chemical equationMole ratio: 4 NH3 4NO Volume ratio: 4L of NH3 4Lof NO

Solve for 1 liter of NH3Mole ratio: 1NH3 1NO Volume ratio: 1L of NH3 1Lof NO produced

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Gather your DataV1 = 16.2L V2 = 22.4Ln1 = ? n2 = 1 mole

Equation needed:

Convert between moles and mass using molar massMolar Mass of SF6 = 146.05 g/mol

What is the Mass of 16.2 L of SF6 at STP?

molesL

xLxmolVVnn 723.0

4.221

12.16

100.1

2

121

gmol

gxmolesg 10605.1461

723.0?

16Determine the volume in liters of NO(g) that will be produced from 5.25 L of 02 (g) and excess NH3(g) at constant pressure and temperature.

4 NH3 (g) + 5 O2 (g) 4 NO (g) + 6 H2O (l)Using Avogadro’s Law

Using the Law of Combining Volumes

NOO

NOONO L

LLx

LL 20.4

54

125.5

?2

2

NONO

NONO Lmol

molxLnnVV 2.4

1187.04.22

2

121

NOO

NOO

O

OO molmolmolxmol

LLxmol

VxVnn 187.0

5423.0

4.2225.51

2

2

2

22

2

121

17A 16.4-g sample of a gas at 25.0 lb/in2 and 25.0ºC is confined in a 17.5 L container. This gas is moved to a 28.5 L container

at 100.0ºC and 14.0 lb/in2. How much gas was removed or added?

Get equation:

Use gas masses, not moles n1 = 16.4g n2 = ? Convert temp. to Kelvin T1 = 298K T2 = 373KPressure units cancel: P1 = 25.0 lb/in2 P2 = 14.0 lb/in2

Volume as written: V1 = 17.5L V2 = 28.5L

16.4g of gas at start - 11.9g after moving it = 4.5g removed

gLLx

inlbinlbx

KKxgn 9.11

5.175.28

/0.25/0.14

373298

14.16

2

2

22

22

11

11

TnVP

TnVP

112

22112 VPT

VPTnn

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Stoichiometry and Gas LawsWhat volume in liters of CO (g) at 250ºC and 0.904 atm is

produced from 453.6g Fe2O3 (s) in the following reaction:Fe2O3 (s) + 3C (s) 2Fe (s) +3 CO (g) Determine # moles of CO (g) produced to get n

Use the ideal gas law to determine the volumeV = nRT n = 8.52 mol R = 0.0821 L atm / mol K

P T = 523.15 K P = 0.904 atm

Latm

xKxmolK

LatmxmolV 405904.

1115.5230821.0

152.8

nmolmolmolx

gmol

xg

mol COOFe

CO

OFe

OFeOFeCO 52.8

13

6.1591

16.453

?3232

3232

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Dalton’s Law of Partial Pressures

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Mixtures of Gases: Dalton's Law of Partial Pressures

The Total Pressure is the sum of the partial pressures.Ptotal = P1 + P2 + P3 + …

The Partial Pressure is the pressure exerted by each individual gas in the container.

P1 = (n1RT)/ V; P2 = (n2RT)/ V; and so on

The Mole Fraction is the fraction of all the molecules in a mixture that are of a given type.

P1/Ptotal = n1/ntotal = x1P1 = x1

. Ptotal

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Mixtures of GasesDalton's Law of Partial Pressures

22A 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

Solving for Xi, the mole fraction:

Solving for Pi:

Pi = Xi PT= 0.0132 x 1.37atm = 0.0181atm

0132.0116.0421.024.8

116.0

83624

83

83

HCHCCH

HCHC molmolmol

molX

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Collection of Gases Over WaterAn gas is collected into a container of water

The water saturated gas rises and displaces liquid waterThe vapor pressure is the partial pressure of the water vapor

It must be deducted from the measured pressure of the gasPtotal Patm = Pgas + Pwater(g) Pgas= Patm - Pwater(g)

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The Kinetic Molecular Theory

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Kinetic Molecular TheoryThe molecules are in constant random motion

Completely fill the container Vast spaces between moleculesMolecules considered “volumeless”Move easily when force is applied

Elastic CollisionsNo gain or loss of energy

Molecules move faster as the temperature is raisedTemperature increases the average kinetic energy

Pressure is created by molecules hitting the wallsAssume no interaction between molecules

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Molar Mass & Temperature Effects

Small molecules move faster than big ones

Higher temperaturesMolecules move faster

MolarMassRTuu rms

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Diffusion and EffusionDiffusion is the process by which one gas

mixes with another as a result of molecular movement

NH4Cl collects closer to HClHCl molecules larger than NH3 molecules

Effusion is the process in which a gas under pressure escapes from its container through a small hole.

NH317g/mol

HCl36g/mol

NH4Cl

1

2

1

2

2

1

MolarMassMolarMass

tt

rr

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Deviation from Ideal Behavior

29

The Van de Waal’s EquationAssumptions of the kinetic-molecular theory not exactAt high pressure/density gases act more like liquids/solids

1. Attractive forces between molecules increase2. Molecular volume no longer negligible

Vanderwaal’s Equation: (P + an2/V2)(V-nb) = nRT

Molecular attraction correction coefficient (a)Actual P is higher than the measured P

Pactual= P + an2/V2

Volume correction coefficient (b)Actual V is lower than the container V

Vactual= V-nb