copyright © cengage learning. all rights reserved.5 | 1 contents and concepts gas laws we will...

Copyright © Cengage Learning. All rights reserved. 5 | 1

Contents and Concepts

Gas LawsWe will investigate the quantitative relationships that describe the behavior of gases.

1. Gas Pressure and Its Measurement2. Empirical Gas Laws3. The Ideal Gas Law4. Stoichiometry Problems Involving Gas Volumes5. Gas Mixtures; Law of Partial Pressures


Kinetic-Molecular Theory

This section will develop a model of gases as molecules in constant random motion.

6. Kinetic Theory of Gases

7. Molecular Speeds; Diffusion and Effusion

8. Real Gases

4

• Gases assume the volume and shape of their containers.

• Gases are the most compressible state of matter.

• Gases will mix evenly and completely when confined to the same container.

• Gases have much lower densities than liquids and solids.

Physical Characteristics of Gases

NO2 gas


Gases differ from liquids and solids:

They are compressible.

Pressure, volume, temperature, and amount are related.

Some applications

• How does a pressure cooker work?

• How is gas pressure applied in spray cans?

• How does a hot air balloon work?

• Why do we not want our tires to be full during hot summer days?

• Why do balloons deflate when left outside on a cold weather?

7

Units of Pressure

1 pascal (Pa) = 1 N/m2

1 atm = 760 mmHg = 760 torr

1 atm = 101,325 Pa

Pressure = ForceArea

(force = mass x acceleration)

8

Manometers Used to Measure Gas Pressures

closed-tube open-tube


Boyle’s LawThe volume of a sample of gas at constant temperature varies inversely with the applied pressure.

The mathematical relationship:

In equation form:

PV

1

ffii

constant

VPVP

PV

10

P a 1/V

P x V = constant

P1 x V1 = P2 x V2

Boyle’s Law

Constant temperatureConstant amount of gas

11

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

P x V = constant

• 1. The pressure on a 2.50 L of anesthetic gas changes from 105 kPa to 40.5 kPa. What will be the new volume if the temperature remains constant?

• 2. A gas with a volume of 4.00 L at a pressure of 205 kPa is allowed to expand to a volume of 12.0 L. What is the pressure in the container of the temperature remains constant?

13

Chemistry in Action:

Scuba Diving and the Gas Laws

P V

Depth (ft) Pressure (atm)

0 1

33 2

66 3

14As T increases V increases

Variation in Gas Volume with Temperature at Constant Pressure

15

Variation of Gas Volume with Temperatureat Constant Pressure

V a T

V = constant x T

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

Charles’ & Gay-Lussac’s

Law

Temperature must bein Kelvin


A balloon was immersed in liquid nitrogen (black container) and is shown immediately after being removed. It shrank because air inside contracts in volume.

As the air inside warms, the balloon expands to its orginial size.

Charles’ Law

• Volume vs. Temperature at constant pressure

• Volume is directly proportional to temperature

• Vα T @ constant P• V=kT; V/T= k• V1/T1 = V2/T2

• 1. If a sample of gas occupies 6.80 L at 3250 C, what will be its volume at 250C if the pressure does not change?

Note: convert Celsius to Kelvin scale by adding 273 to the given 0C.

• 2. Exactly 5.00 L of air at -50.00C is warmed to 100.00C. What is the new volume if the pressure remains constant?

19

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

T1 = 125 (0C) + 273.15 (K) = 398.15 K

Gay-Lussac’s Law

• Pressure and Temperature at constant Volume

• P is proportional to T @ constant volume

• PαT @ constant V• P=kT; P/T=k• P1/T1 = P2/T2

• A gas has a pressure of 6.58 kPa at 539K. What will be the pressure at 211 K if the volume does not change?

• The pressure in an automobile tire is 198 kPa at 270C. At the end of a trip on a hot sunny day, the pressure has risen to 225 kPa. What is the temperature of the air in the tire?

• Helium gas in a 2.00-L cylinder is unter 1.12 atm pressure. At 36.5 0C, that same gas sample has a pressure of 2.56 atm. What was the initial temperature of the gas?

22

Avogadro’s Law

V a number of moles (n)

V = constant x n

V1 / n1 = V2 / n2

Constant temperatureConstant pressure

Avogadro’s Principle

• Volume is proportional to number of moles• V α n where n is the number of moles• More moles occupy greater volume

24

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

Combined Gas Law

• P α T VP=k T VPV = k T

• At 0.000C and 1.00 atm pressure a sample of gas occupies 30.0 mL. if the temperature is increased 30.00C and the gas sample is transferred to 20.0 mL container, what is the gas pressure?


Standard Temperature and Pressure (STP)

The reference condition for gases, chosen by convention to be exactly 0°C and 1 atm pressure.

The molar volume, Vm, of a gas at STP is 22.4 L/mol.

The volume of the yellow box is 22.4 L. To its left is a basketball.

STP

• At STP,• T=00C or 273 K• P= 1atm

• At Standard temperature and pressure, molar volume is 22.4 L

• Meaning, 1 mole of any gas occupies 22.4 liters at STP

Ideal Gas Law

• Ideal gas behaves as if there is no IMF present among the molecules of gas.

• PV = R nTR = 1atm (22.4L) 1 mole ( 273)R= .0821 atm-L/mol-K

29

Ideal Gas Equation

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

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

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

V a nT

P

V = constant x = RnT

P

nT

PR is the gas constant

PV = nRT

30

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 = 30.7 L

Sample problems for Ideal gas law

• If the pressure exerted by the gas at 250C in a volume of 0.044 L is 3.81 atm, how many moles of gas are present?

• Determine the celsius temperature of 2.49 moles of gas contained in a 1.00-L vessek at a pressure of 143 kPa.

32

Density (d) Calculations

d = mV =

PMRT

m is the mass of the gas in g

M is the molar mass of the gas

Molar Mass (M ) of a Gaseous Substance

dRTP

M = d is the density of the gas in g/L

33

A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and 27.0 0C. What is the molar mass of the gas?

dRTP

M = d = mV

4.65 g2.10 L

= = 2.21 g

L

M =2.21

g

L

1 atm

x 0.0821 x 300.15 KL•atmmol•K

M = 54.5 g/mol

Density of gas from ideal gas equation

• D=M/V• V=M/D• PV=nRT• P(Mass/D)= nRT• Since n= mass/MW• Then,• P(Mass/D)=(Mass/MW)RT• Therefore: P(MW)=DRT• D=P(MW) RT

• What is the density of a gas at STP that has a molar mass of 44.0 g/mol?

D=1atm(44.0g/mol) .0821 (273K)D= 1.96 g/L


Stoichiometry and Gas VolumesUse the ideal gas law to find moles from a given volume, pressure, and temperature, and vice versa.

36

Gas Stoichiometry

What is the volume of CO2 produced at 37 0C 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


Gas MixturesDalton found that in a mixture of unreactive gases, each gas acts as if it were the only gas in the mixture as far as pressure is concerned.

38

Dalton’s Law of Partial Pressures

V and T are constant

P1P2 Ptotal = P1 + P2

39

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

PT = PA + PB XA = nA

nA + nB

XB = nB

nA + nB

PA = XA PT PB = XB PT

Pi = Xi PT mole fraction (Xi ) = ni

nT

40

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

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5 | 41

A 100.0-mL sample of air exhaled from the lungs is analyzed and found to contain 0.0830 g N2, 0.0194 g O2, 0.00640 g CO2, and 0.00441 g water vapor at 35°C. What is the partial pressure of each component and the total pressure of the sample?


5 | 42

mL10

L1mL100.0

K308Kmol

atmL0.08206

Ng28.01

Nmol1Ng0.0830

3

2

22

N2P

mL10

L1mL100.0

K308Kmol

atmL0.08206

Og32.00

Omol1Og0.0194

3

2

22

O2P

mL10

L1mL100.0

K308Kmol

atmL0.08206

COg44.01

COmol1COg0.00640

3

2

22

CO2P

mL10

L1mL100.0

K308Kmol

atmL0.08206

OHg18.01

OHmol1OHg0.00441

3

2

22

OH2P

atm0.749

atm0.153

atm0.0368

atm0.0619


5 | 43

atm0.7492N P

atm0.1532O P

atm0.03682CO P

atm0.0619OH2P

OHCOON 2222PPPPP

P = 1.00 atm


5 | 44

The partial pressure of air in the alveoli (the air sacs in the lungs) is as follows: nitrogen, 570.0 mmHg; oxygen, 103.0 mmHg; carbon dioxide, 40.0 mmHg; and water vapor, 47.0 mmHg. What is the mole fraction of each component of the alveolar air?

mmHg40.02CO P

mmHg570.02N P

mmHg47.0OH2P

mmHg103.0 2OP


5 | 45

OHCOON 2222PPPPP

570.0 mmHg103.0 mmHg

40.0 mmHg47.0 mmHg

P = 760.0 mmHg


5 | 46

Mole fraction of N2

Mole fraction of H2OMole fraction of CO2

Mole fraction of O2

mmHg760.0

mmHg47.0

mmHg760.0

mmHg40.0

mmHg760.0

mmHg103.0

mmHg760.0

mmHg570.0

Mole fraction N2 = 0.7500

Mole fraction O2 = 0.1355

Mole fraction CO2 = 0.0526

Mole fraction O2 = 0.0618

47

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

PT = PO + PH O2 2

Collecting a Gas over Water


Collecting Gas Over WaterGases are often collected over water. The result is a mixture of the gas and water vapor.

The total pressure is equal to the sum of the gas pressure and the vapor pressure of water.

The partial pressure of water depends only on temperature and is known (Table 5.6).

The pressure of the gas can then be found using Dalton’s law of partial pressures.

49

Vapor of Water and Temperature

?


You prepare nitrogen gas by heating ammonium nitrite:

NH4NO2(s) N2(g) + 2H2O(l)

If you collected the nitrogen over water at 23°C and 727 mmHg, how many liters of gas would you obtain from 5.68 g NH4NO2?

Molar mass NH4NO2

= 64.06 g/mol

P = 727 mmHgPvapor = 21.1 mmHgPgas = 706 mmHgT = 23°C = 296 K


P = 727 mmHgPvapor = 21.1 mmHgPgas = 706 mmHgT = 23°C = 296 K P

nRTV

Molar mass NH4NO2

= 64.06 g/mol

4 2 24 2

4 2 4 2

1mol NH NO 1mol N5.68 g NH NO

64.04 g NH NO 1mol NH NO

= 0.8887 mol N2 gas


P = 727 mmHgPvapor = 21.1 mmHgPgas = 706 mmHgT = 23°C = 296 Kn = 0.0887 mol

P

nRTV

mmHg760

atm1mmHg706

K)(296Kmol

atmL0.08206mol0.0887

V

= 2.32 L of N2

(3 significant figures)


2K 2

1mvE

Kinetic-Molecular Theory (Kinetic Theory)A theory, developed by physicists, that is based on the assumption that a gas consists of molecules in constant random motion.

Kinetic energy is related to the mass and velocity:

m = massv = velocity


Postulates of the Kinetic Theory1. Gases are composed of molecules whose

sizes are negligible.2. Molecules move randomly in straight lines in all

directions and at various speeds.3. The forces of attraction or repulsion between

two molecules (intermolecular forces) in a gas are very weak or negligible, except when the molecules collide.

4. When molecules collide with each other, the collisions are elastic.

5. The average kinetic energy of a molecule is proportional to the absolute temperature.


An elastic collision is one in which no kinetic energy is lost. The collision on the left causes the ball on the right to swing the same height as the ball on the left had initially, with essentially no loss of kinetic energy.

57

Kinetic theory of gases and …

• Compressibility of Gases

• Boyle’s Law

P a collision rate with wallCollision rate a number densityNumber density a 1/VP a 1/V

• Charles’ LawP a collision rate with wallCollision rate a average kinetic energy of gas moleculesAverage kinetic energy a TP a T

58

Kinetic theory of gases and …

• Avogadro’s Law

P a collision rate with wallCollision rate a number densityNumber density a nP a n

• Dalton’s Law of Partial Pressures

Molecules do not attract or repel one another

P exerted by one type of molecule is unaffected by the presence of another gas

Ptotal = SPi


Molecular SpeedsAccording to kinetic theory, molecular speeds vary over a wide range of values. The distribution depends on temperature, so it increases as the temperature increases.

Root-Mean Square (rms) Molecular Speed, u

A type of average molecular speed, equal to the speed of a molecule that has the average molecular kinetic energy

mM

RTu

3

60

The distribution of speedsfor nitrogen gas molecules

at three different temperatures

The distribution of speedsof three different gases

at the same temperature

urms = 3RTM


When using the equation

R = 8.3145 J/(mol · K).

T must be in Kelvins

Mm must be in kg/mol

?


What is the rms speed of carbon dioxide molecules in a container at 23°C?

mM

RTu

3rms

T = 23°C = 296 KCO2 molar mass =

0.04401 kg/mol


mol

kg0.04401

K296Kmol

s

mkg

8.314532

2

rmsu

25

rms 2

m1.68 10

su

2rms

m4.10 10

su

2

2

s

mkgJ

Recall

64

Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties.

NH3

17 g/molHCl

36 g/mol

NH4Cl

r1

r2

M2

M1=

molecular path


Diffusion

The process whereby a gas spreads out through another gas to occupy the space uniformly.

Below NH3 diffuses through air. The indicator paper tracks its progress.


Effusion

The process by which a gas flows through a small hole in a container. A pinprick in a balloon is one example of effusion.


Graham’s Law of EffusionAt constant temperature and pressure, the rate of effusion of gas molecules through a particular hole is inversely proportional to the square root of the molecular mass of the gas.

mM

1moleculesofeffusionofrate

?


Both hydrogen and helium have been used as the buoyant gas in blimps. If a small leak were to occur, which gas would effuse more rapidly and by what factor?

Hydrogen will diffuse more quickly by a factor of 1.4.

2.016

4.002

4.002

12.016

1

HeRate

HRate 2

69

Deviations from Ideal Behavior

1 mole of ideal gas

PV = nRT

n = PVRT

= 1.0

Repulsive Forces

Attractive Forces


Real GasesAt high pressure the relationship between pressure and volume does not follow Boyle’s law. This is illustrated on the graph below.


At high pressure, some of the assumptions of the kinetic theory no longer hold true:

1. At high pressure, the volume of the gas molecule (Postulate 1) is not negligible.

2. At high pressure, the intermolecular forces (Postulate 3) are not negligible.

72

Effect of intermolecular forces on the pressure exerted by a gas.


Van der Waals EquationAn equation that is similar to the ideal gas law, but which includes two constants, a and b, to account for deviations from ideal behavior.

The term V becomes (V – nb).

The term P becomes (P + n2a/V2).

Values for a and b are found in Table 5.7

nRTnbVV

anP

2

2

74

Van der Waals equationnonideal gas

P + (V – nb) = nRTan2

V2( )}

correctedpressure

}

correctedvolume


Use the van der Waals equation to calculate the pressure exerted by 2.00 mol CO2 that has a volume of 10.0 L at 25°C. Compare this with value with the pressure obtained from the ideal gas law.

n = 2.00 molV = 10.0 LT = 25°C = 298 K

For CO2:a = 3.658 L2 atm/mol2

b = 0.04286 L/mol


n = 2.00 molV = 10.0 LT = 25°C = 298 K

L10.0

K)(298Kmol

atmL0.08206mol2.00

P

= 4.89 atm(3 significant figures)

V

nRTP

Ideal gas law:


n = 2.00 molV = 10.0 LT = 25°C = 298 K

For CO2:a = 3.658 L2 atm/mol2

b = 0.04286 L/mol

Pactual = 4.79 atm(3 significant figures)

2

2

V

an

nbV

nRTP

22

22

L10.0

mol

atmL3.658mol2.00

mol

L0.04286mol2.00L10.0

K298Kmol

atmL0.08206mol2.00

P

atm0.146atm4.933 P

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