chapter 12 gases and their properties chapter goals understand the basis of the gas laws and know...

Chapter 12Chapter 12

Gases and Their PropertiesGases and Their Properties

Chapter GoalsChapter Goals• Understand the basis of the gas laws and Understand the basis of the gas laws and

know how to use those laws (Boyle’s law, know how to use those laws (Boyle’s law, Charles’ law, Avogadro’s hypothesis, Charles’ law, Avogadro’s hypothesis, Dalton’s law).Dalton’s law).

• Use the (model) ideal gas law.Use the (model) ideal gas law.• Apply the gas laws to stoichiometric Apply the gas laws to stoichiometric

calculations.calculations.• Understand kinetic-molecular theory as it is Understand kinetic-molecular theory as it is

applied to gases, especially the distribution applied to gases, especially the distribution of molecular speeds (energies).of molecular speeds (energies).

• Recognize why real gases do not behave Recognize why real gases do not behave like ideal gases.like ideal gases.

Gases in EarthGases in Earth

Eleven elements are gases under normal Eleven elements are gases under normal

conditions: five diatomic, Hconditions: five diatomic, H22, N, N22, O, O22, F, F22, Cl, Cl22, and , and

the six monatomic noble gases, He, Ne, Ar, Kr, the six monatomic noble gases, He, Ne, Ar, Kr, Xe, Rn.Xe, Rn.

The thickness of the atmosphere is The thickness of the atmosphere is ~ 1/250 the ~ 1/250 the diameter of the Earth. Yet this delicate layer is diameter of the Earth. Yet this delicate layer is vital for our life. It shields us from harmful vital for our life. It shields us from harmful radiation and supplies substances needed for radiation and supplies substances needed for life, such as oxygen, nitrogen, carbon dioxide, life, such as oxygen, nitrogen, carbon dioxide, and water.and water.

Composition of the Atmosphere and Composition of the Atmosphere and Some Common Properties of GasesSome Common Properties of Gases

Composition of Dry Air

GasGas % by Volume% by Volume

NN22 78.0978.09

OO22 20.9420.94

ArAr 0.930.93

COCO22 0.030.03

He, Ne, Kr, XeHe, Ne, Kr, Xe 0.0020.002

CHCH44 0.000150.00015

HH22 0.000050.00005

Comparison of Solids, Liquids, and gasesComparison of Solids, Liquids, and gases

The density of gases is much less than that The density of gases is much less than that

of solids or liquids.of solids or liquids.

Gas molecules must be very far apart

compared to liquids and solids.

Density Density (g/mL)(g/mL)

SolidSolid LiquidLiquid GasGas

HH22OO 0.9170.917 0.9980.998 0.0005880.000588

CClCCl44 1.701.70 1.591.59 0.005030.00503

General Properties of General Properties of GasesGases

• There is a lot of “free” There is a lot of “free” space in a gas.space in a gas.

• Gases can be expanded Gases can be expanded infinitely.infinitely.

• Gases occupy containers Gases occupy containers uniformly and completely.uniformly and completely.

• Gases diffuse and mix Gases diffuse and mix rapidly.rapidly.

Importance of GasesImportance of Gases

Airbags fill with NAirbags fill with N22 gas in an accident. gas in an accident.

Gas is generated by the decomposition of Gas is generated by the decomposition of

sodium azide, NaNsodium azide, NaN33..

2 NaN2 NaN33(s) (s) 2 Na(s) + 3 N 2 Na(s) + 3 N22(g)(g)

Properties of GasesProperties of Gases

Gas properties can beGas properties can be modeledmodeled using math. using math. Model depends on four Model depends on four quantities (parameters): quantities (parameters): V = volume of the gas (L)V = volume of the gas (L)T = temperature (K)T = temperature (K)n = amount (moles)n = amount (moles)P = pressureP = pressure

(atmospheres) (atmospheres)

PressurePressurePressure is force per unit area.Pressure is force per unit area.

force Fforce FPressure = Pressure = ──── P = ────── P = ── area A area A

NNSI unit: pascal, 1 Pa = SI unit: pascal, 1 Pa = ────── = 1kg = 1kgmm−1−1ss−2−2 mm22

PressurePressureAtmospheric pressure (pressure of the Atmospheric pressure (pressure of the atmosphere) is measured with aatmosphere) is measured with a barometerbarometer,, invented by invented by Torricelli Torricelli (1643).(1643).Definitions of standard pressure Definitions of standard pressure 76 cm Hg =76 cm Hg == 760 mm Hg = 760 = 760 mm Hg = 760 torrtorr = == 1 atmosphere = 1 atm= 1 atmosphere = 1 atm1 atm = 101.3 kPa = 1.013 1 atm = 101.3 kPa = 1.013 10 1055 Pa = Pa = = 14.7 psi (pounds per square inch)= 14.7 psi (pounds per square inch)1 bar = 11 bar = 1101055 Pa = 0.987 atm Pa = 0.987 atm

1 atm = 14.7 psi1 atm = 14.7 psi

HH22O density O density ~ 1 g/mL ~ 1 g/mL Hg density = 13.6 g/mL

Pressure Unit ConversionsPressure Unit ConversionsNow we can use pressure units as conversion factorsNow we can use pressure units as conversion factors

Convert 202.6 kPa to Hg mm, bars, and atm.Convert 202.6 kPa to Hg mm, bars, and atm.

760 Hg mm760 Hg mm202.6 kPa 202.6 kPa ──────── = 1520 Hg mm──────── = 1520 Hg mm 101.3 kPa101.3 kPa

101033 Pa 1 bar Pa 1 bar202.6 kPa 202.6 kPa ───── ───── ───── = 2.026───── = 2.026 bar bar 1 kPa 101 kPa 1055 Pa Pa

1 atm1 atm202.6 kPa 202.6 kPa ────── = 2.000 atm────── = 2.000 atm 101.3 kPa101.3 kPa

Example – A high-performance bicycle tire has a Example – A high-performance bicycle tire has a pressure of 132 psi. What is the pressure in mmHg?pressure of 132 psi. What is the pressure in mmHg?

since mmHg are smaller than psi, the since mmHg are smaller than psi, the answer makes senseanswer makes sense

1 atm = 14.7 psi, 1 atm = 760 mmHg1 atm = 14.7 psi, 1 atm = 760 mmHg

132 psi132 psi

mmHgmmHg

Check:Check:

Solution:Solution:

Concept Plan:Concept Plan:

Relationships:Relationships:

Given:Given:

Find:Find:

psi 14.7

atm 1

mmHg 10.826 atm 1

mmHg 760

psi 14.7

atm 1psi 132 3

atm 1

mmHg 760

psi atm mmHg

ManometersManometers• the pressure of a gas trapped in a container the pressure of a gas trapped in a container

can be measured with an instrument called a can be measured with an instrument called a manometermanometer

• manometers are U-shaped tubes, partially manometers are U-shaped tubes, partially filled with a liquid, connected to the gas filled with a liquid, connected to the gas sample on one side and open to the air on sample on one side and open to the air on the otherthe other

• a competition is established between the a competition is established between the pressure of the atmosphere and the gaspressure of the atmosphere and the gas

• the difference in the liquid levels is a the difference in the liquid levels is a measure of the difference in pressure measure of the difference in pressure between the gas and the atmospherebetween the gas and the atmosphere

ManometerManometer

for this sample, the gas has a larger pressure than the atmosphere, so

(mm) levels Hgin difference (mmHg)Pressure (mmHg)Pressure

Pressure Pressure Pressure

atmosp heregas

hatmosp heregas

IDEAL GAS LAWIDEAL GAS LAW

Brings together gas Brings together gas

properties.properties.

Can be derived from Can be derived from

experiment and theory.experiment and theory.

P V = n R P V = n R TT

Boyle’s LawBoyle’s LawIf n and T are constant, then If n and T are constant, then

PV = k (constant)PV = k (constant)

This means, for example, This means, for example,

that P goes up as V goes that P goes up as V goes

down.down.

A bicycle pump is a good A bicycle pump is a good

example of Boyle’s law. example of Boyle’s law.

As the volume of the air As the volume of the air

trapped in the pump is trapped in the pump is

reduced, its pressure goes reduced, its pressure goes

up, and air is forced into the up, and air is forced into the

tire.tire.

Robert Boyle Robert Boyle

(1627-1691).(1627-1691).

Boyle’s Law: Boyle’s Law: The Volume-Pressure RelationshipThe Volume-Pressure RelationshipPV = k (PV = k (hyperbolahyperbola) V) V 11or V= k or V= k ──── or V or V 1/P 1/P P 1/PP 1/P

PP11VV11 = k = k1 1 for one sample of a gas.for one sample of a gas.

PP22VV22 = k = k2 2 for a second sample of a gas.for a second sample of a gas.

kk11 = k = k2 2 for for the same sample of a gas (same the same sample of a gas (same

number of moles at the same T.)number of moles at the same T.)Thus we can write Boyle’s Law Thus we can write Boyle’s Law

mathematically as mathematically as PP11VV11 = P = P22VV22

for constant n and Tfor constant n and T

Boyle’s Law: The Volume-Pressure RelationshipBoyle’s Law: The Volume-Pressure Relationship

Example: At 25Example: At 25ooC a sample of He has a C a sample of He has a

volume of 4.00 x 10volume of 4.00 x 1022 mL under a pressure of mL under a pressure of

7.60 x 107.60 x 1022 torr. What volume would it occupy torr. What volume would it occupy

under a pressure of 2.00 atm at the same T?under a pressure of 2.00 atm at the same T?

FirstlyFirstly, we need to convert atm to torr:, we need to convert atm to torr:

760 torr760 torr2.00 atm2.00 atm───── =───── = 1 atm1 atm

= 1520 torr= 1520 torr

Then, using Boyle’s lawThen, using Boyle’s law

mL 1000.2

torr1520

mL 400 torr760

P

V PV

V PV P

2

2

112

2211

Boyle’s Law: The Volume-Pressure RelationshipBoyle’s Law: The Volume-Pressure Relationship

• Notice that in Boyle’s law we can use any Notice that in Boyle’s law we can use any

pressure or volume units as long as we pressure or volume units as long as we

consistently use the same units for both consistently use the same units for both

PP11 and P and P22 or V or V11 and V and V22..

• Use your intuition to help you decide if Use your intuition to help you decide if the volume will go up or down as the the volume will go up or down as the pressure is changed and vice versa.pressure is changed and vice versa.

Charles’s LawCharles’s LawIf n and P are If n and P are

constant, thenconstant, then

V = kTV = kT

V and T are directly V and T are directly

related (proportional.)related (proportional.)Jacques Charles Jacques Charles (1746-1823). (1746-1823). Isolated boron Isolated boron and studied and studied gases. gases. Balloonist.Balloonist.

Charles’ Law: Charles’ Law: The Volume-Temperature Relationship; The Volume-Temperature Relationship; The Absolute Temperature ScaleThe Absolute Temperature Scale

0

5

10

15

20

25

30

35

0 50 100 150 200 250 300 350 400

Volume (L) vs.

Temperature (K)

Gases liquefy before reaching 0 K

absolute zero = −273.15 0C

This part is extrapolated

Charles’ Law: Charles’ Law: The Volume-Temperature Relationship; The Volume-Temperature Relationship; The Absolute Temperature ScaleThe Absolute Temperature Scale

Charles’s law states that the volume of a gas is directly proportional to the absolute temperature at constant pressure.

Gas laws must use the Kelvin scale to be correct.

Relationship between Kelvin and centigrade. K = °C + 273.15

Charles’ Law: Charles’ Law: The Volume-Temperature RelationshipThe Volume-Temperature Relationship

Mathematical form of Charles’ law.Mathematical form of Charles’ law. VV

V V T or V = k T or T or V = k T or ──── = k = k n & P constn & P const TT

VV11 V V22── ── = k and ── = k = k and ── = k TT11 T T22

VV11 V V22 ── ── = ── = ── in the most useful formin the most useful form TT11 T T22

Charles’ Law: Charles’ Law: Example: A sample of hydrogen, HExample: A sample of hydrogen, H22, ,

occupies 1.00 x 10occupies 1.00 x 1022 mL at 25.0 mL at 25.0ooC and 1.00 atm. C and 1.00 atm.

What volume would it occupy at 50.0What volume would it occupy at 50.0ooC C

under the same pressure?under the same pressure?

TT11 = 25 + 273 = 298 T = 25 + 273 = 298 T22 = 50 + 273 = 323 = 50 + 273 = 323

VV11 V V2 2 V V11 T T22

── ── = ── V= ── V22 = ──── = ──── TT11 T T22 T T11

1.00 x 101.00 x 1022 mL mL 323 K323 KVV22 = = ───────────── = 108 mL───────────── = 108 mL 298 K298 K

Standard Temperature and PressureStandard Temperature and PressureStandard temperature and pressure is given the Standard temperature and pressure is given the symbol symbol STPSTP..

It is a reference point for some gas calculations. It is a reference point for some gas calculations.

Standard P Standard P 1.00000 atm or 101.3 kPa 1.00000 atm or 101.3 kPa

Standard T Standard T 273.15 K or 0.00 273.15 K or 0.00 ooCC

Avogadro’s HypothesisAvogadro’s HypothesisAvogadro’s Hypothesis states that at the same T and P,Avogadro’s Hypothesis states that at the same T and P,equal volumesequal volumes of two gases contain the of two gases contain the same number same number of molecules (of molecules (or molesor moles)) of gas. of gas. V = k nV = k n n = moles n = moles k = const. k = const. On those basis, the following equation meansOn those basis, the following equation means

2 H2 H22(g) + O(g) + O22(g) (g) 2 H 2 H22O(g)O(g)

2 moles H2 moles H22 reacts with 1 mole of O reacts with 1 mole of O22 to produce 2 moles H to produce 2 moles H22OO

ratios 2 : 1 : 2ratios 2 : 1 : 2

2 L of H2 L of H22 reacts with 1 L of O reacts with 1 L of O22 to produce 2 L of H to produce 2 L of H22OO

oror

8 volumes + 4 volumes 8 volumes + 4 volumes 8 volumes (H 8 volumes (H22O) 8 = 2x4O) 8 = 2x4

Here, volume means any unit (mL or L).Here, volume means any unit (mL or L).

Gases in this experiment are measured at same T & P.Gases in this experiment are measured at same T & P.

Avogadro’s Hypothesis and the Standard Molar Avogadro’s Hypothesis and the Standard Molar VolumeVolumeAvogadro’s Hypothesis states that at the same Avogadro’s Hypothesis states that at the same temperature and pressure, equal volumes of two temperature and pressure, equal volumes of two gases contain the same number of molecules (or gases contain the same number of molecules (or moles) of gas.moles) of gas.If we set the temperature and pressure for any gas If we set the temperature and pressure for any gas to be STP, then one mole of that gas has a volume to be STP, then one mole of that gas has a volume called the called the standard molar volumestandard molar volume..* The standard molar volume is 22.4 L at STP.* The standard molar volume is 22.4 L at STP.

This is another way to measure moles.This is another way to measure moles.For For gasesgases, the volume is proportional to the , the volume is proportional to the number of moles. number of moles. V = k nV = k n

11.2 L of a gas at STP = 0.500 mole11.2 L of a gas at STP = 0.500 mole44.8 L = ? moles44.8 L = ? moles

Avogadro’s Hypothesis and the Standard Molar Avogadro’s Hypothesis and the Standard Molar VolumeVolumeExample: One mole of a gas occupies 36.5 L Example: One mole of a gas occupies 36.5 L and its density is 1.36 g/L at a given temperature and its density is 1.36 g/L at a given temperature and pressure. (a) What is its molar mass? and pressure. (a) What is its molar mass? (b) What is its density at STP?(b) What is its density at STP?

? . ..

gmol

Lmol

gL

g / mol 36 5 13649 6

g/L 21.2L 4.22

mol 1

mol

g 6.49

L

g ?

STP

The Combined Gas Law EquationThe Combined Gas Law EquationBoyle’s and Charles’ Laws combined into Boyle’s and Charles’ Laws combined into

one statement that is called the combined gas one statement that is called the combined gas

law equation.law equation.

Useful when the V, T, and P of a gas are Useful when the V, T, and P of a gas are

changing, changing, but n is constantbut n is constant..

2

22

1

11

2

2

1

12211

T

V P

T

V P k

T

V P

:is law gas combined The :gas of samplegiven aFor

T

V

T

V VPVP

Law Charles' Law sBoyle'

The Combined Gas Law EquationThe Combined Gas Law EquationExample: A sample of nitrogen gas, NExample: A sample of nitrogen gas, N22, ,

occupies 7.50 x 10occupies 7.50 x 1022 mL at 75.0 mL at 75.000C under a C under a

pressure of 8.10 x 10pressure of 8.10 x 1022 torr. What volume torr. What volume

would it occupy at STP?would it occupy at STP?

mL 627

K 348 torr760

K 273mL 750 torr810

T P

T V P=Vfor Solve

torr760=P torr 810=P

K 273=TK 348=T

?=V mL 750=V

12

2112

21

21

21

The Combined Gas Law EquationThe Combined Gas Law EquationExample: A sample of methane, CHExample: A sample of methane, CH44, occupies 2.60 , occupies 2.60

x 10x 1022 mL at 32 mL at 32ooC under a pressure of 0.500 atm. At C under a pressure of 0.500 atm. At

what temperature would it occupy 5.00 x 10what temperature would it occupy 5.00 x 1022 mL mL

under a pressure of 1.20 x 10under a pressure of 1.20 x 1033 torr? torr?

C1580 K 1852=

mL 260 torr380

mL 500 torr1200K 305

V P

V P T= T

? = T K 305 = T

torr 380 =

torr1200 = P atm 0.500 = P

mL 500 = V mL 260 = V

o

11

2212

21

21

21

Summary of Gas Laws:Summary of Gas Laws:The Ideal Gas LawThe Ideal Gas Law

• Boyle’s Law Boyle’s Law −− V V 1/P (at constant T & n) 1/P (at constant T & n)• Charles’ Law – V Charles’ Law – V T (at constant P & n) T (at constant P & n)• Avogadro’s Law – V Avogadro’s Law – V n (at constant T & P) n (at constant T & P)Combine these three laws into one statementCombine these three laws into one statement

V V nT/P nT/PConvert the proportionality into an equality.Convert the proportionality into an equality.

n R Tn R TV = V = ───── ───── PP

This provides the This provides the Ideal Gas LawIdeal Gas Law PV = nRT PV = nRTRR is a proportionality constant called the universal is a proportionality constant called the universal gas constant.gas constant.

Summary of Gas Laws:Summary of Gas Laws:The Ideal Gas LawThe Ideal Gas Law

We must determine the value of We must determine the value of RR..

Recognize that for one mole of a gas at Recognize that for one mole of a gas at

1.00 atm, and 273 K (STP), the volume is 1.00 atm, and 273 K (STP), the volume is

22.4 L.22.4 L.

Use these values in the ideal gas law. Use these values in the ideal gas law.

R = PV

nT

1.00 atm L

1.00 mol K

L atm

mol K

22 4

273

0 0821

.

.

Summary of Gas Laws: The Ideal Gas LawSummary of Gas Laws: The Ideal Gas Law

RR has other values if the units are changed. has other values if the units are changed.• R = 8.314 J/mol KR = 8.314 J/mol K

Use this value in thermodynamics.Use this value in thermodynamics.

• R = 8.314 kg mR = 8.314 kg m22/s/s22 K mol K molUse this later in this chapter for gas velocities.Use this later in this chapter for gas velocities.

• R = 8.314 dmR = 8.314 dm33 kPa/K mol kPa/K molThis is R in all metric units.This is R in all metric units.

• R = 1.987 cal/K molR = 1.987 cal/K molThis is the value of R in calories rather than J.This is the value of R in calories rather than J.

Using PV = nRTUsing PV = nRTWhat volume would 50.0 g of ethane, CWhat volume would 50.0 g of ethane, C22HH66, occupy , occupy

at 140at 140 o oC under a pressure of 1.82 x 10C under a pressure of 1.82 x 1033 torr? torr?T = 140 + 273 = 413 K 1 atmT = 140 + 273 = 413 K 1 atm P = 1820 torr P = 1820 torr ──────────── = 2.39 atm = 2.39 atm 1 mol 760 torr1 mol 760 torrn = 50.0 g n = 50.0 g ────────── = 1.67 mol = 1.67 mol 30.0 g30.0 g

PV = n RTPV = n RT

L 6.23atm 39.2

K 413K mol

atm L 0821.0mol 67.1

P

T Rn = V

Using PV = nRTUsing PV = nRT

Calculate the pressure exerted by 50.0 g of ethane, Calculate the pressure exerted by 50.0 g of ethane,

CC22HH66, in a 25.0 L container at 25.0, in a 25.0 L container at 25.0ooC. C.

MW = 30.07 g/molMW = 30.07 g/mol

atm 63.1PL 25.0

K 298K mol

atm L0.0821mol 1.67

P

V

T Rn = P

K 298 = T and mol 1.67 =n

A flying balloon contains 1.2x10A flying balloon contains 1.2x1077 L of He at a L of He at a pressure of 737 mm Hg and 25 °C. What mass of He pressure of 737 mm Hg and 25 °C. What mass of He does the balloon contain? A.W. He = 4.00g/moldoes the balloon contain? A.W. He = 4.00g/mol

1 atm1 atm737 mm Hg 737 mm Hg ─────── = 0.970 atm ─────── = 0.970 atm 760 mm Hg760 mm Hg

mmP V = n R T P V = ── R TP V = n R T P V = ── R T MM

PVM 0.970 atm PVM 0.970 atm 1.2x10 1.2x107 7 L L 4.00g/mol 4.00g/mol m = ──── = ───────────────────────m = ──── = ─────────────────────── R T 0.0821 atm L/K mol R T 0.0821 atm L/K mol 298 K 298 K

m = 1.9 m = 1.9 10 1066 g g

Determination of Molecular Weights and Determination of Molecular Weights and Molecular Formulas of Gaseous SubstancesMolecular Formulas of Gaseous SubstancesExample: A gaseous compound is 80.0% carbon Example: A gaseous compound is 80.0% carbon

and 20.0% hydrogen by mass. At STP, 546 mL of theand 20.0% hydrogen by mass. At STP, 546 mL of the

gas has a mass of 0.732 g. What is its molecular gas has a mass of 0.732 g. What is its molecular

formula?formula?

100 g of compound contains 80 g of C and 20 g of H.100 g of compound contains 80 g of C and 20 g of H.

15 = mass with CH is formula empirical the36.67

19.8

ratio.number holesmallest w theDetermine

H mol 8.19H g 1.01

H mol 1H g 20.0 = atoms H mol ?

C mol 67.6C g 12.0

C mol 1C g 80.0 = atoms C mol ?

3

Determination of Molecular Weights and Determination of Molecular Weights and Molecular Formulas of Gaseous SubstancesMolecular Formulas of Gaseous SubstancesExample: At STP, 546 mL of the gas has a mass of Example: At STP, 546 mL of the gas has a mass of 0.732 g. 0.732 g. Now, we can calculate the molar mass (M) Now, we can calculate the molar mass (M) of the compound:of the compound: m(g) m m R Tm(g) m m R TPV = PV = nnRT, RT, n =n = ───, PV = ── RT, M = ────────, PV = ── RT, M = ───── M M P VM M P V

0.732 g 0.732 g 0.0821 L 0.0821 Latm/molatm/molK K 273 K g 273 K gM = ─────────────────────── = 30.0─── M = ─────────────────────── = 30.0─── 1.00 atm 1.00 atm 0.546 L mol 0.546 L mol

M 30.0M 30.0 ─── ─── = ─── = 2 Then, 2 = ─── = 2 Then, 2 CH CH33 = = CC22HH66, the formula, the formula..W EF 15.0W EF 15.0

The Density of Gases and Molar MassThe Density of Gases and Molar MassAs previously said, density of a gas is given in As previously said, density of a gas is given in g/L.g/L.

Calculate the density of methane, CHCalculate the density of methane, CH44, at 37 °C and , at 37 °C and

a pressure of 1.50 atm.a pressure of 1.50 atm. m(g) m R Tm(g) m R TPV = nRT, n = PV = nRT, n = ─── , PV = ──────── , PV = ───── M MM M

m P Mm P Md = ──= ─── M(CHd = ──= ─── M(CH44) = 12.01 + 4) = 12.01 + 41.008 = 16.04 g/mol1.008 = 16.04 g/mol V R T V R T T = 37 + 273 = 310. KT = 37 + 273 = 310. K

1.50 atm 1.50 atm 16.04 g/mol g16.04 g/mol gd = ────────────────── = 0.945 ─── d = ────────────────── = 0.945 ─── 0.0821 L0.0821 Latm/molatm/molK K 310. K L 310. K L

He

The Density of Gases and Molar MassThe Density of Gases and Molar MassThe density of a gas is 0.391 g/L at 70.5 torr The density of a gas is 0.391 g/L at 70.5 torr

and 22.3 °C. Calculate its molar mass (M).and 22.3 °C. Calculate its molar mass (M).

T = 22.3 + 273.15 = 295.4 KT = 22.3 + 273.15 = 295.4 K

1 atm1 atmP = 70.5 torr P = 70.5 torr ─────── = 0.0928 atm ─────── = 0.0928 atm 760 torr760 torr

P M d R TP M d R Td = ─── Then, M = ──── d = ─── Then, M = ──── R T PR T P

0.391 g/L0.391 g/L0.0821 L0.0821 Latm/molatm/molK K 295.4K 295.4K gg

M = ──────────────────────── = 102 ─── M = ──────────────────────── = 102 ─── 0.0928 atm mol0.0928 atm mol

Mass-Volume Relationships Mass-Volume Relationships in Reactions Involving Gasesin Reactions Involving Gases

• In this section we are looking at reaction stoichiometry, like in Chapter 4, just including gases in the calculations.

2 mol KClO2 mol KClO33 yields 2 mol KCl and 3 mol O yields 2 mol KCl and 3 mol O22

2(122.6g) yields 2 (74.6g) and 3 (32.0g)2(122.6g) yields 2 (74.6g) and 3 (32.0g)

Those 3 moles of OThose 3 moles of O22 can also be thought of as: can also be thought of as:

3(22.4L) or 67.2 L at STP3(22.4L) or 67.2 L at STP

g)(2(s)&MnO

(s)3 O 3 + KCl 2KClO 2 2

What volume of oxygen measured at STP, can What volume of oxygen measured at STP, can be produced by the thermal decomposition of be produced by the thermal decomposition of 120.0 g of KClO120.0 g of KClO33??

FW (KClO FW (KClO33) = 122.6 g/mol) = 122.6 g/mol

g)(2(s)&MnO

(s)3 O 3 + KCl 2KClO 2 2

2STP2STP

2

2STP

3

2

3

332STP

O L 9.32O L ?

O mol 1

O L 4.22

KClO mol 2

O mol 3

KClO g 122.6

KClO mol 1 KClO g 120.0O L ?

31. Iron reacts with HCl(aq) to produce iron (II) 31. Iron reacts with HCl(aq) to produce iron (II) chloride and hydrogen gas. The gas from the chloride and hydrogen gas. The gas from the reaction of 2.2 g Fe with excess acid is collected in reaction of 2.2 g Fe with excess acid is collected in a 10.0 L flask at 25 °C. What is the Ha 10.0 L flask at 25 °C. What is the H22 pressure? pressure?Fe(s) + 2 HCl(aq) Fe(s) + 2 HCl(aq) FeCl FeCl22(aq) + H(aq) + H22(g)(g)

1 mol Fe 1 mol H1 mol Fe 1 mol H22 2.2 g Fe 2.2 g Fe ────── ────── ────── = 0.039 mol H────── = 0.039 mol H22

55.85 g Fe55.85 g Fe 1 mol Fe 1 mol Fe n R Tn R TP V = n R T P = ──── P V = n R T P = ──── VV

0.039 mol0.039 mol 0.0821 atm L/K mol 0.0821 atm L/K mol 298 K 298 KP = ──────────────────────── = 0.095 atm P = ──────────────────────── = 0.095 atm 10.0 L10.0 L

Gases Laws and Chemical ReactionsGases Laws and Chemical Reactions

2 H2 H22OO22(liq) (liq) 2 H 2 H22O(g) + OO(g) + O22(g)(g)

Decompose 1.1 g of HDecompose 1.1 g of H22OO22 in a flask with volume of 2.50 in a flask with volume of 2.50

L. What is the pressure of OL. What is the pressure of O22 at 127 at 127 ooC? Of HC? Of H22O?O?

M(HM(H22OO22) = 34.0 g/mol) = 34.0 g/mol

Firstly, calculate moles of HFirstly, calculate moles of H22OO22 and O and O22::

1 mol H1 mol H22OO22 1 mol O 1 mol O221.1 g 1.1 g HH22OO22──────── ──────── ─────── = ─────── = 0.016 mol O0.016 mol O22 34.0 g H34.0 g H22OO22 2 mol H 2 mol H22OO22

Bombardier beetle Bombardier beetle uses decomposition of uses decomposition of hydrogen peroxide to hydrogen peroxide to defend itself.defend itself.

Gases Laws and Chemical ReactionsGases Laws and Chemical Reactions


Now, with the moles of ONow, with the moles of O22 we calculate P of O we calculate P of O22

n R T 0.016 mol n R T 0.016 mol 0.0821 0.0821 400 K 400 K P=P=──── = ───────────────── = ──── = ───────────────── = 0.21 atm O0.21 atm O22

V 2.50 L V 2.50 L

We can calculate moles and P of HWe can calculate moles and P of H22O or just, O or just,

because the coefficient of Hbecause the coefficient of H22O is 2 and OO is 2 and O22’s is one,’s is one,

moles and P of Hmoles and P of H22O are the double of OO are the double of O22’s.’s.

HH22O, n = 0.032 mol P = 0.42 atmO, n = 0.032 mol P = 0.42 atm

Dalton’s Law of Partial Dalton’s Law of Partial PressuresPressuresFor the reactionFor the reaction


final pressures are: 0.42 atm 0.21 atmfinal pressures are: 0.42 atm 0.21 atm

They are said to be They are said to be partial pressures of Hpartial pressures of H22O and OO and O22..What is the total pressure in the flask?What is the total pressure in the flask?

PPtotal total in gas mixture = Pin gas mixture = PAA + P + PBB + … Therefore, + … Therefore,

PPtotaltotal = P(H = P(H22O) + P(OO) + P(O22) = 0.42 + 0.21 = 0.63 atm) = 0.42 + 0.21 = 0.63 atm

Example of Example of Dalton’s LawDalton’s Law, P, Ptotaltotal = = ΣΣ(partial pressures) (partial pressures)

holds at holds at constant volume and temperatureconstant volume and temperature. . Each Each gas occupies the whole volume regardless of othergas occupies the whole volume regardless of other..

Dalton’s Law of Partial PressuresDalton’s Law of Partial Pressures

John DaltonJohn Dalton1766-18441766-1844


Vapor Pressure of a liquid is the pressure exerted Vapor Pressure of a liquid is the pressure exerted

by a substance’s vapor over the substance’s by a substance’s vapor over the substance’s

liquid at equilibrium.liquid at equilibrium.


Zn + HCl(aq) Zn + HCl(aq) ZnCl ZnCl22(aq) + (aq) + HH22(g)(g)

PPatmatm = PH = PH2 2 + P + P HH22OO


Example: A sample of hydrogen was collected Example: A sample of hydrogen was collected

by displacement of water at 25.0 by displacement of water at 25.0 ooC. The C. The

atmospheric (atmospheric (barometricbarometric) pressure was 748 torr. ) pressure was 748 torr.

What pressure would the dry hydrogen exert inWhat pressure would the dry hydrogen exert in

the same container?the same container?

PPtotaltotal = P(H = P(H22) + P(H) + P(H22O) O) P(H P(H22) = P) = Ptotaltotal – P(H – P(H22O)O)

P(HP(H22O) = 24 torr (O) = 24 torr (taken from tablestaken from tables) at 25.0 °C) at 25.0 °C

P(HP(H22) = (748 – 24) torr = 724 torr) = (748 – 24) torr = 724 torr

Dalton’s Law of Partial PressuresDalton’s Law of Partial PressuresA sample of oxygen was collected by displacement A sample of oxygen was collected by displacement

of water. The oxygen occupied 742 mL at 27.0 of water. The oxygen occupied 742 mL at 27.0 ooC. C.

The The barometricbarometric pressure was 753 torr. What volume pressure was 753 torr. What volume

would the dry oxygen occupy at STP?would the dry oxygen occupy at STP?

P(HP(H22O) = 27 torr (O) = 27 torr (taken from tablestaken from tables) at 27.0 °C) at 27.0 °C

Use combined gas law as all parameters changed.Use combined gas law as all parameters changed.

V 742 mL V ?

T K T K

P = 726 torr P torr

V mL273 K300 K

726 torr torr

645 mL @ STP

1 2

1 2

1 2

2

300 273

753 27 760

742760

Total P, Partial Pressures, and Mole Total P, Partial Pressures, and Mole FractionFraction

For a mixture of several gases (A, B, C, …)For a mixture of several gases (A, B, C, …)

nnAA R T n R T nBB R T n R T nCC R T R TPPTT = P = PAA + P + PBB + P + PCC + … = + … = ──── + ──── + ──── + …──── + ──── + ──── + … V V VV V V (n(nAA+ n+ nBB + n + nCC)R T n)R T n ii RT RTPPTT = = ─────────── P─────────── Pii = ──── i = A, B, C, … = ──── i = A, B, C, … V VV V

There is a simple relation between every PThere is a simple relation between every P ii and P and PTT

Total P, Partial Pressures, and Mole Total P, Partial Pressures, and Mole FractionFraction nnAA R T R T ──── ──── PPAA V n V nAA n nAA

XXAA= = ── = ───────────── = ────────── = ──── = ───────────── = ────────── = ── PPTT (n (nAA+ n+ nBB + n + nC C + + )R T n)R T nAA+ n+ nBB + n + nCC + + n nTT

────────────────────────── VVXXAA is the is the mole fraction of Amole fraction of A, ratio of moles of A to the , ratio of moles of A to the

total, or ratio of partial pressure to total pressuretotal, or ratio of partial pressure to total pressure

nnA A n nBB n nCC XXAA + X + XBB + X + XCC + + = = ─── + ─── + ─── + ─── + ─── + ─── + nnTT n nTT n nTT

nnA A + n+ nBB + n + nCC + + PPii They They addadd

XXAA + X + XBB + X + XCC + + = = ─────────── = ── ─────────── = ── = 1= 1 up to up to oneone

nnTT P PTT

A 5.25 L sample of argon was collected A 5.25 L sample of argon was collected over water at 30 °C and at a pressure of over water at 30 °C and at a pressure of 830.0 torr. What were the mole 830.0 torr. What were the mole fractions of Ar and water?fractions of Ar and water?Vapor pressure of HVapor pressure of H22O is 31.8 torr at 30 O is 31.8 torr at 30 °C.°C. PPTT = P = PArAr + P + PWW P PArAr = P = PTT −− P PWW W = water W = water

PPArAr = (830.0 – 31.8) = 798.2 torr = (830.0 – 31.8) = 798.2 torr

PPArAr 798.2 torr 798.2 torr XXArAr = = ─── = ─────── = 0.9617 ─── = ─────── = 0.9617 (no units)(no units) PPTT 830.0 torr 830.0 torr XXArAr + X + XWW = 1 = 1 X XWW = 1 – X = 1 – XAr Ar = 1 – 0.9617 = 0.0383= 1 – 0.9617 = 0.0383

We didn’t use V and T. We didn’t need themWe didn’t use V and T. We didn’t need them..


If 1.00 x 10If 1.00 x 1022 mL of hydrogen, measured at 25.0 mL of hydrogen, measured at 25.0 ooC C

and 3.00 atm pressure, and 1.00 x 10and 3.00 atm pressure, and 1.00 x 1022 mL of oxygen, mL of oxygen,

measured at 25.0 measured at 25.0 ooC and 2.00 atm pressure, were C and 2.00 atm pressure, were

forced into one of the containers at 25.0 forced into one of the containers at 25.0 ooC, what C, what

would be the pressure of the mixture of gases?would be the pressure of the mixture of gases?

V and T of the two gases are the same: the Ps won’t changeV and T of the two gases are the same: the Ps won’t change

What is the mol fraction of each gas in the mixture?What is the mol fraction of each gas in the mixture?

P(HP(H22) 3.00 atm) 3.00 atmX(HX(H22)=)=──── = ────── = 0.600 X(O──── = ────── = 0.600 X(O22) = 1 – 0.600 = 0.400) = 1 – 0.600 = 0.400 PPTT 5.00 atm 5.00 atm

P P P

3.00 atm + 2.00 atm

= 5.00 atm

Total H O2 2


10.0 g of argon and 20.0 g of Ne are placed in a closed 10.0 g of argon and 20.0 g of Ne are placed in a closed

container at 25.0 container at 25.0 ooC and the pressure of the mixture is 25.5 C and the pressure of the mixture is 25.5

atm. What is the partial pressure of Ne in the mixture?atm. What is the partial pressure of Ne in the mixture?

We need to calculate the moles of the two gases and, then, We need to calculate the moles of the two gases and, then,

the mole fraction and partial pressure of Ne.the mole fraction and partial pressure of Ne.

Moles:Moles:

Ne, 20.0/20.179 = 0.991 mole Ar, 10.0/39.948 = 0.250 moleNe, 20.0/20.179 = 0.991 mole Ar, 10.0/39.948 = 0.250 mole

nnNeNe 0.991 mole 0.991 moleXXNeNe = = ─── = ──────── = 0.799 ─── = ──────── = 0.799 nnT T 1.241 mole1.241 mole

PPNeNe XXNeNe= = ─── ─── PPNeNe == XXNeNe × P × PTT == 0.799 × 25.5 atm = 20.4 atm 0.799 × 25.5 atm = 20.4 atm PPTT

KINETIC MOLECULAR THEORY KINETIC MOLECULAR THEORY (KMT)(KMT)

Theory used to explain gas laws. KMT Theory used to explain gas laws. KMT assumptions areassumptions are

• Gases consist of molecules in Gases consist of molecules in constant, random motion.constant, random motion.

• P arises from collisions with container P arises from collisions with container walls.walls.

• No attractive or repulsive forces No attractive or repulsive forces between molecules. Collisions elastic.between molecules. Collisions elastic.

• Volume of molecules is Volume of molecules is negligiblenegligible, , compared to the V of container.compared to the V of container.

Kinetic Molecular TheoryKinetic Molecular TheoryBecause we assume molecules are in motion, Because we assume molecules are in motion,

they have a kinetic energy. they have a kinetic energy. Average for 1 moleAverage for 1 mole

m: mass of a single moleculem: mass of a single molecule m um u2 2 M uM u22

KE = KE = ──── u = speed ──── u = speed KE = ────KE = ──── 2 2 22

At the same T, all gases At the same T, all gases have the samehave the same average KE. average KE. 3 3

KE = KE = ── RT (for 1 mol of gas)── RT (for 1 mol of gas) 22

KE and T are directly proportionalKE and T are directly proportional

At the same T, all gases At the same T, all gases have the samehave the same average KE. average KE. 3 3

KE = KE = ── RT (for 1 mol of gas)── RT (for 1 mol of gas) 22

KE and T are directly proportionalKE and T are directly proportional

As T goes up for a gas, KE also As T goes up for a gas, KE also increases — and so does speed.increases — and so does speed.As T goes up for a gas, KE also As T goes up for a gas, KE also increases — and so does speed.increases — and so does speed.

Kinetic Molecular TheoryKinetic Molecular Theory

At the same T, all gases have the At the same T, all gases have the same average KE.same average KE.

As T goes up, KE also increases As T goes up, KE also increases — and so does speed.— and so does speed.

Kinetic Molecular TheoryKinetic Molecular Theory

where u is the speed and M is the where u is the speed and M is the molar mass.molar mass.

• speed INCREASES with Tspeed INCREASES with T

• speed DECREASES with Mspeed DECREASES with M

Maxwell’s equationMaxwell’s equation

root mean square speed

2u M3RTuurmsrms = =

Kinetic Molecular TheoryKinetic Molecular TheoryExample: What is the Example: What is the

root mean square root mean square

speed of Nspeed of N22 molecules molecules

at room T, 25.0at room T, 25.0ooC?C?

kg mkg m22

R = 8.314 R = 8.314 ────────────── ss22 K mol K mol

What is the root mean What is the root mean

square velocity of He square velocity of He

atoms at room T, atoms at room T,

25.025.0ooC?C?

u

3 8.314kg m

sec K mol K

kg / mol

m / s = 1159 mi / hr

rms

2

2

298

0 028

515

.

u

3 8.314kg m

sec K mol K

kg / mol

m / s = 3067 mi / hr

rms

2

2

298

0 004

1363

.

M in kg

DistributioDistribution of Gas n of Gas Molecule Molecule SpeedsSpeeds

DistributioDistribution of Gas n of Gas Molecule Molecule SpeedsSpeeds

•Boltzmann plotsBoltzmann plots•Named for Ludwig Named for Ludwig

Boltzmann, Scot Physicist Boltzmann, Scot Physicist

who also developed the who also developed the

theory of electromagnetic theory of electromagnetic

radiation (1864.)radiation (1864.)

Velocity of Gas MoleculesVelocity of Gas MoleculesMolecules of a given gas have aMolecules of a given gas have a

rangerange of speeds.of speeds.

Velocity of Gas MoleculesVelocity of Gas Molecules

Average velocity decreases with Average velocity decreases with

increasing mass, at the same Temperatureincreasing mass, at the same Temperature

3 RTurms = √ ────

M

GAS DIFFUSION AND EFFUSIONGAS DIFFUSION AND EFFUSION

DIFFUSIONDIFFUSION is the gradual mixing is the gradual mixing of molecules of different gases.of molecules of different gases.

Air BrAir Br22

GAS EFFUSIONGAS EFFUSIONEFFUSIONEFFUSION is the movement of is the movement of molecules through a small hole molecules through a small hole

((pinholepinhole) into an empty container.) into an empty container.


Molecules effuse thru holes in Molecules effuse thru holes in

a rubber balloon (a rubber balloon (pinholespinholes), for ), for example, example,

at a rate (= moles/time) that isat a rate (= moles/time) that is

• proportional to Tproportional to T

• inversely proportional to M.inversely proportional to M.

Therefore, He effuses more Therefore, He effuses more

rapidly than Orapidly than O22 at same T. at same T.

HeHe


Graham’s law Graham’s law

governs effusion governs effusion

and diffusion of gas and diffusion of gas

molecules.molecules.

Thomas Graham, 1805-1869. Thomas Graham, 1805-1869. Professor in Glasgow and London.Professor in Glasgow and London.

Rate of effusion is Rate of effusion is inversely proportional inversely proportional to its molar mass.to its molar mass.

Rate of effusion is Rate of effusion is inversely proportional inversely proportional to its molar mass.to its molar mass.

M of AM of B

Rate for B

Rate for A


• Calculate the ratio of the rate of effusion of He to Calculate the ratio of the rate of effusion of He to that of sulfur dioxide, SOthat of sulfur dioxide, SO22, at the same , at the same

temperature and pressure.temperature and pressure.

R

R

M

M

g / mol4.0 g / mol

R R

He

SO

SO

He

He SO

2

2

2

641

16 4 4

.

GAS DIFFUSION AND EFFUSIONGAS DIFFUSION AND EFFUSIONA sample of hydrogen, HA sample of hydrogen, H22, was found to effuse , was found to effuse

through a pinhole 5.2 times as rapidly as the same through a pinhole 5.2 times as rapidly as the same

volume of unknown gas (at the same temperature volume of unknown gas (at the same temperature

and pressure). What is the molecular weight of the and pressure). What is the molecular weight of the

unknown gas?unknown gas?

g/mol 54 = g/mol) 0.2(27M

g/mol 0.2

M27

g/mol 0.2

M2.5

M

M

R

R

unk

unk

unk

H

unk

unk

H

2

2

GAS DIFFUSION AND EFFUSIONGAS DIFFUSION AND EFFUSIONExamples of relationships:Examples of relationships:

Gas A effuses (or diffuses) 5.2 times faster than gas BGas A effuses (or diffuses) 5.2 times faster than gas B rate A Mrate A MBB

───── ───── = 5.2 = SQRT( ────)= 5.2 = SQRT( ────) rate Brate B M MAA

Gas A effuses at half of the speed of gas BGas A effuses at half of the speed of gas B

rate A 1rate A 1 ───── ───── = ── = 0.5= ── = 0.5 rate B 2 rate B 2

Gas A takes a time the triple it takes to gas B to Gas A takes a time the triple it takes to gas B to

diffuse: the more time it takes the slower the gas is.diffuse: the more time it takes the slower the gas is.

rate A 1rate A 1 ───── ───── = ── = ── rate B 3rate B 3

Gas DiffusionGas Diffusionrelation of mass to rate of relation of mass to rate of

diffusiondiffusion

Gas DiffusionGas Diffusionrelation of mass to rate of relation of mass to rate of

diffusiondiffusion• HCl and NHHCl and NH33 diffuse diffuse

from opposite ends of from opposite ends of tube. tube.

• Gases meet to form Gases meet to form NHNH44ClCl

• HCl heavier than NHHCl heavier than NH33

• Therefore, NHTherefore, NH44Cl Cl

forms closer to HCl forms closer to HCl end of tube.end of tube.

• HCl and NHHCl and NH33 diffuse diffuse

from opposite ends of from opposite ends of tube. tube.

• Gases meet to form Gases meet to form NHNH44ClCl

• HCl heavier than NHHCl heavier than NH33

• Therefore, NHTherefore, NH44Cl Cl

forms closer to HCl forms closer to HCl end of tube.end of tube.Active Figure 12.18Active Figure 12.18

Chapter 12 # 102Chapter 12 # 102

Each four tires of a car is filled with a different gas. Each four tires of a car is filled with a different gas.

Each tire has the same V, and each is filled to the same Each tire has the same V, and each is filled to the same

P, 3.00 atm, and 25.0 °C. One contains 116.0 g air, P, 3.00 atm, and 25.0 °C. One contains 116.0 g air,

another 80.7 g Ne, the third 16.0 g He, and the forth has another 80.7 g Ne, the third 16.0 g He, and the forth has

160.0 g of an unknown gas.160.0 g of an unknown gas.

a) # of molecules?a) # of molecules?

All tires contain the same number of molecules, All tires contain the same number of molecules, because they have the same volume, T, and P.because they have the same volume, T, and P.

PVPV n = n = ──────── # of molecules are moles (n) x 6.02 x 10 # of molecules are moles (n) x 6.02 x 102323

R TR T

Chapter 12 # 102Chapter 12 # 102b) Molar mass of unknown gas?b) Molar mass of unknown gas? Due to fact that all gases have the same number of Due to fact that all gases have the same number of

molecules, the ratio of their masses corresponds to the ratio molecules, the ratio of their masses corresponds to the ratio of the molar masses, because n(moles) is the same. The of the molar masses, because n(moles) is the same. The difference in the masses is due to their molar masses:difference in the masses is due to their molar masses:

A.W. He = 4.0 g/molA.W. He = 4.0 g/molmolar mass unknown 160.0 gmolar mass unknown 160.0 g──────────────────────────── = = ───── ───── = 10.0 times heavier x 4.0 = 40.0 = 10.0 times heavier x 4.0 = 40.0

molar mass He 16.0 molar mass He 16.0

c) Highest average speed? c) Highest average speed? According to the KE = 3/2 RT, all gases have the same According to the KE = 3/2 RT, all gases have the same

kinetic energy at the same temperature, but the average kinetic energy at the same temperature, but the average speed depends on their molar masses. According tospeed depends on their molar masses. According to

3RT3RT u = SQRT (u = SQRT (────────) ) The gas with the lowest MThe gas with the lowest M rms M rms M will have the highest speed, i.e., He. will have the highest speed, i.e., He.

Using KMT to Understand Gas Using KMT to Understand Gas LawsLaws

Recall that KMT assumptions areRecall that KMT assumptions are

• Gases consist of molecules in Gases consist of molecules in constant, random motion.constant, random motion.

• P arises from collisions with P arises from collisions with container walls.container walls.

• No attractive or repulsive forces No attractive or repulsive forces between molecules. Collisions elastic.between molecules. Collisions elastic.

• Volume of molecules is negligible, Volume of molecules is negligible, compared to the V of container.compared to the V of container.

Avogadro’s Hypothesis Avogadro’s Hypothesis and Kinetic Molecular and Kinetic Molecular

TheoryTheory

Avogadro’s Hypothesis Avogadro’s Hypothesis and Kinetic Molecular and Kinetic Molecular

TheoryTheory

P proportional to n —P proportional to n —when V and T are constantwhen V and T are constant

n R Tn R TP = ——— P = ——— VV

Gas Pressure, Gas Pressure, Temperature, and Kinetic Temperature, and Kinetic

Molecular TheoryMolecular Theory

Gas Pressure, Gas Pressure, Temperature, and Kinetic Temperature, and Kinetic

Molecular TheoryMolecular Theory

P proportional to T —P proportional to T —when n and V are constantwhen n and V are constant

n R Tn R TP = ——— P = ——— VV

Boyle’s Law (look @ notes)Boyle’s Law (look @ notes) and Kinetic Molecular and Kinetic Molecular

TheoryTheory

Boyle’s Law (look @ notes)Boyle’s Law (look @ notes) and Kinetic Molecular and Kinetic Molecular

TheoryTheory

P proportional to 1/V — P proportional to 1/V — when n and T are constantwhen n and T are constant

n R Tn R TP = ——— P = ——— VV

Deviations from Ideal Gas Deviations from Ideal Gas LawLaw

• Real molecules Real molecules havehave volume.volume.

• There areThere are intermolecular intermolecular attractive forces.attractive forces.

–Otherwise a gas Otherwise a gas could not could not become a liquid.become a liquid.


Account for volume of Account for volume of

molecules and intermolecular molecules and intermolecular

forces withforces with VAN DER WAALS’s VAN DER WAALS’s

EQUATION.EQUATION.Measured V = V(ideal)Measured P

intermol. forcesvol. correction

J. van der Waals, J. van der Waals, 1837-1923, 1837-1923, Professor of Professor of Physics, Physics, Amsterdam. Amsterdam. Nobel Prize 1910.Nobel Prize 1910.

nRTV - nbV2

n2aP + ----- )(

Deviations from Ideal Gas LawDeviations from Ideal Gas LawDeviations from Ideal Gas LawDeviations from Ideal Gas Law

ClCl22 gas has gas has aa = 6.49,= 6.49, bb = 0.0562= 0.0562

For 8.0 mol ClFor 8.0 mol Cl22 in a 4.0 L tank at 27 in a 4.0 L tank at 27 ooC.C.

nRT 8.0 nRT 8.0 0.0821 0.0821 300 300P(ideal) =P(ideal) =─── = ─────────────── = ──────────── = 49.3 atm = 49.3 atm V 4.0 V 4.0 P (van der Waals) = 29.5 atm P (van der Waals) = 29.5 atm

Measured V = V(ideal)Measured P

intermol. forces

vol. correction

nRTV - nbV2

n2aP + -----


• Calculate the pressure exerted by 84.0 g of ammonia, Calculate the pressure exerted by 84.0 g of ammonia, NHNH33, in a 5.00 L container at 200. , in a 5.00 L container at 200. ooC using the ideal gas C using the ideal gas

law.law.

atm 4.38PL 5.00

K 473K mol

atm L0821.0mol 94.4

V

nRT = P

mol 94.4g 17.0

mol 1NH g 84.0 =n 3

Deviations from Ideal Gas LawDeviations from Ideal Gas Law• Solve of ammonia using the van der Waal’s equation.Solve of ammonia using the van der Waal’s equation.

• P=? 84.0 g of NHP=? 84.0 g of NH33, in a 5.00 L container at 200. , in a 5.00 L container at 200. ooC. C.

2

2

2

2

2

2

V

an

nb-V

nRT=P

nRTnb-VV

an + P

molL0.0371=b

mol

atm L 4.17 = a mol 4.94 =n

ideal from difference 7.6% a is which atm 7.35P

)atm 1.4atm 8.39(atm 07.4L 817.4

atm L 8.191P

L 00.5

17.4mol 94.4

)0371.0)(mol 94.4(L 00.5

K4730821.0 mol 94.4P 2

molatm L2

molL

K molatm L

2

2

chapter 12 gases and their properties chapter goals understand the basis of the gas laws and know...

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