chapter 5 gases. why study gases? an understanding of real world phenomena. an understanding of how...

Chapter 5

Gases

Why study gases?

• An understanding of real world phenomena.• An understanding of how science “works.”

Copyright © Cengage Learning. All rights reserved 2

• 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.

• Gases exert pressure on their surroundings

Physical Characteristics of Gases

The Gaseous State

Ideal Gas Concept

• Ideal gas - a model of the way that particles of a gas behave at the microscopic level.

• We can measure the following of a gas:– temperature,– volume,– pressure and – quantity (mass)

We can systematically change one of the properties and see the effect on each of the others.

Measurement of Gases

• The most important gas laws involve the relationship between– number of moles (n) of gas– volume (V)– temperature (T)– pressure (P)

• Pressure - force per unit area.

• Gas pressure is a result of force exerted by the collision of particles with the walls of the container.

Pressure

• Force exerted per unit area of surface by molecules in motion.

1 atmosphere = 14.7 psi 1 atmosphere = 760 mm Hg = 760 torr 1 atmosphere = 101,325 Pascals 1 Pascal = 1 kg/m.s2 = Newton/meter2

forcePr essure = area

Manometer

• Device used for measuring the pressure of a gas in a container.


• Barometer - measures atmospheric pressure.

– Invented by Evangelista Torricelli

• A commonly used unit of pressure is the atmosphere (atm).

• 1 atm is equal to:

760 mmHg

760 torr

76 cmHg

Worked Example 5.1

Pressure Conversions: An Example

The pressure of a gas is measured as 2.5 atm. Represent this pressure in both torr and pascals.


3760 torr2.5 atm = 1.9 10 torr

1 atm

5101,325 Pa2.5 atm = 2.5 10 Pa

1 atm

Elements that exist as gases at 250C and 1 atmosphere

Liquid Nitrogen and a Balloon



• What happened to the gas in the balloon?• A decrease in temperature was followed by a

decrease in the pressure and volume of the gas in the balloon.



• This is an observation (a fact).• It does NOT explain “why,” but it does tell us “what

happened.”


• Gas laws can be deduced from observations like these.

• Mathematical relationships among the properties of a gas (Pressure, Volume, Temperature and Moles) can be discovered.


Boyle’s Law

• Boyle’s Law - volume of a gas is inversely proportional to pressure if the temperature and number of moles is held constant.

PV = k1

or

PiVi = PfVf

Boyle’s law


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A Problem to Consider• A sample of chlorine gas has a volume of 1.8 L at

1.0 atm. If the pressure increases to 4.0 atm (at constant temperature), what would be the new volume?

iiff VPVP using

)atm 0.4()L 8.1()atm 0.1(

PVP

Vf

iif

L 45.0Vf

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

A sample of helium gas occupies 12.4 L at 23°C and 0.956 atm. What volume will it occupy at 1.20 atm assuming that the temperature stays constant?

9.88 L


EXERCISE!EXERCISE!

1. A 5.0 L sample of a gas at 25oC and 3.0 atm is compressed at constant temperature to a volume of 1.0 L. What is the new pressure?

2. A 3.5 L sample of a gas at 1.0 atm is expanded at constant temperature until the pressure is 0.10 atm. What is the volume of the gas?

2

Charles’ Law

• Charles’ Law - volume of a gas varies directly with the absolute temperature (K) if pressure and number of moles of gas are constant.

2T

Vk or

f

f

i

i

T

V

T

V

Charles’s Law




A Problem to Consider• A sample of methane gas that has a volume of 3.8 L

at 5.0 oC is heated to 86.0 oC at constant pressure. Calculate its new volume.

)K278()K359)(L8.3(

TTV

fi

fiV

L 9.4Vf

i

i

f

f

TV

TV

using

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

Suppose a balloon containing 1.30 L of air at 24.7°C is placed into a beaker containing liquid nitrogen at –78.5°C. What will the volume of the sample of air become (at constant pressure)?

0.849 L


EXERCISE!EXERCISE!

1. A 2.5 L sample of gas at 25oC is heated

to 50oC at constant pressure. Will the

volume double?

2. What would be the volume in question 1?

3. What temperature would be required to

double the volume in question 1?

2

The Empirical Gas Laws

• Gay-Lussac’s Law: The pressure exerted by a gas at constant volume is directly proportional to its absolute temperature.

P Tabs (constant moles and V)

or

i

i

f

f

TP

TP

A Problem to Consider• An aerosol can has a pressure of 1.4 atm at 25 oC. What

pressure would it attain at 1200 oC, assuming the volume remained constant?

i

i

f

f

TP

TP

using

)K298()K1473)(atm4.1(

TTP

fi

fiP

atm9.6Pf

Combined Gas Law

f

ff

i

ii

T

VP

T

VP

• This law is used when a sample of gas undergoes change involving volume, pressure, and temperature simultaneously.

1

A Problem to Consider• A sample of carbon dioxide occupies 4.5 L at 30 oC

and 650 mm Hg. What volume would it occupy at 800 mm Hg and 200 oC?

f

ff

i

iiTVP

TVP

using

)K 303)(Hg mm 800()K 473)(L 5.4)(Hg mm 650(

TPTVP

Vif

fiif

L7.5Vf

2

Calculate the temperature when a 0.50 L sample of gas at 1.0 atm and 25oC is compressed to 0.05 L of gas at 5.0 atm.

Avogadro’s Law

• Avogadro’s Law - equal volumes of an ideal gas contain the same number of moles if measured under the same conditions of temperature and pressure.

3

Vk

n or

f

f

i

i

nn

VV

1

If 2.45 mol of argon gas occupies a volume of 89.0 L, what volume will 2.10 mol of argon occupy under the same conditions of temperature and pressure?

76.3 L


EXERCISE!EXERCISE!

Assuming no change in temperature and pressure, how many moles of gas would be needed to double the volume occupied by 0.50 moles of gas?

2

Molar Volume of a Gas

• Molar Volume - the volume occupied by 1 mol of any gas.

• STP - Standard Temperature and Pressure T = 273 K (or 0oC) P = 1 atm

• At STP the molar volume of a gas is 22.4 L/mol– We will learn to calculate the volume

later.

The molar volume of a gas. Photo courtesy of James Scherer.

Figure 5.11

Gas Densities

• We know: density = mass/volume

• Let’s calculate the density of H2.

• What is the mass of 1 mol of H2?• 2.0 g

• What is the volume of 1 mol of H2?• 22.4 L

• Density = 2.0 g/22.4 L = 0.089 g/L

The Ideal Gas Law

• Combining Boyle’s Law, Charles’ Law and Avogadro’s Law gives the Ideal Gas Law:

PV=nRT

R (ideal gas constant) = 0.08206 L.Atm/mol.K

Ideal Gas Equation

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

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

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

V nT

P

V = constant x = RnT

P

nT

PR is the gas constant

PV = nRT

The conditions 0 0C and 1 atm are called standard temperature and pressure (STP).

PV = nRT

R = PVnT

=(1 atm)(22.414L)

(1 mol)(273.15 K)

R = 0.082057 L • atm / (mol • K)

Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L.

Let’s calculate the molar volume at STP using the ideal gas law:

PV = nRT

• What would be the pressure? 1 atm

• What would be the temperature? 273 K

• What would be the number of moles? 1 mol

atm 1

K 273)Kmol

atmL6mol(0.08201

P

RTV

n22.4 L

An experiment calls for 3.50 moles of chlorine, Cl2. What volume would this be if the gas volume is measured at 34 oC and 2.45 atm?

A Problem to Consider

PnRT V since

atm 2.45K) )(307 1mol)(0.082 (3.50 Kmol

atmL

V then

L 36.0 V then

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 HCl

36.45 g HCl= 1.37 mol

V =1 atm

1.37 mol x 0.0821 x 273.15 KL•atmmol•K

V = 30.6 L

Worked Example 5.3

Worked Example 5.4

An automobile tire at 23°C with an internal volume of 25.0 L is filled with air to a total pressure of 3.18 atm. Determine the number of moles of air in the tire.

3.27 mol


EXERCISE!EXERCISE!

What is the pressure in a 304.0 L tank that contains 5.670 kg of helium at 25°C?

114 atm


EXERCISE!EXERCISE!

At what temperature (in °C) does 121 mL of CO2 at 27°C and 1.05 atm occupy a volume of 293 mL at a pressure of 1.40 atm?

696°C


EXERCISE!EXERCISE!

A sample of oxygen gas has a volume of 2.50 L at STP. How many grams of O2 are present?

3.57 g


EXERCISE!EXERCISE!

1. What is the volume of gas occupied by 5.0 g CH4 at 25oC and 1 atm?

2. What is the mass of N2 required to occupy 3.0 L at 100oC and 700 mmHg?

Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)?

PV = nRT n, V and R are constant

nRV

= PT

= constant

P1

T1

P2

T2

=

P1 = 1.20 atmT1 = 291 K

P2 = ?T2 = 358 K

P2 = P1 x T2

T1

= 1.20 atm x 358 K291 K

= 1.48 atm

Worked Example 5.5

Worked Example 5.6

Worked Example 5.7

Molecular Weight Determination

• The relationship between moles and mass.

mass molecular massmoles

or

mMmn

Molecular Weight Determination

• If we substitute this in the ideal gas equation, we obtain

RT)(PVmM

mIf we solve this equation for the molecular mass, we obtain

PVmRT Mm


• A 15.5 gram sample of an unknown gas occupied a volume of 5.75 L at 25 oC and a pressure of 1.08 atm. Calculate its molecular mass.

PVmRT M Since m

L) atm)(5.75 (1.08

K) )(298g)(0.0821 (15.5 M then Kmol

atmL

m

g/mol 61.1 Mm

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 =54.6 g/mol

Density Determination

• If we look again at our derivation of the molecular mass equation,

RT)(PVmM

mwe can solve for m/V, which represents density.

RTPM

D Vm m

A Problem to Consider• Calculate the density of ozone, O3 (Mm = 48.0g/mol), at

50 oC and 1.75 atm of pressure.

RTPM

D Since m

K) )(323(0.0821g/mol) atm)(48.0 (1.75

D thenKmol

atmL

g/L 17.3 D

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

Worked Example 5.8

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

Stoichiometry Problems Involving Gas Volumes

Suppose you heat 0.0100 mol of potassium chlorate, KClO3, in a test tube. How many liters of oxygen can you produce at 298 K and 1.02 atm?

)g(O 3 KCl(s) 2 (s)KClO 2 23

Consider the following reaction, which is often used to generate small quantities of oxygen.


3

23 KClO mol 2

O mol 3 KClO mol 0100.0

2O mol 5001.0

• First we must determine the number of moles of oxygen produced by the reaction.


PnRT V

atm 02.1K) )(298 0821.0)(O mol (0.0150 Kmol

atmL2V

• Now we can use the ideal gas equation to calculate the volume of oxygen under the conditions given.

L 0.360 V

Dalton’s Law of Partial Pressures 1

• Dalton’s Law - a mixture of gases exerts a pressure that is the sum of the pressures that each gas would exert if it were present alone under the same conditions.

Pt=p1+p2+p3+...

• For example, the total pressure of our atmosphere is equal to the sum of the pressures of N2 and O2.

22 ONair ppP

Consider the following apparatus containing helium in both sides at 45°C. Initially the valve is closed.– After the valve is opened, what is the

pressure of the helium gas?


3.00 atm3.00 L

2.00 atm9.00 L

EXERCISE!EXERCISE!

27.4 L of oxygen gas at 25.0°C and 1.30 atm, and 8.50 L of helium gas at 25.0°C and 2.00 atm were pumped into a tank with a volume of 5.81 L at 25°C.• Calculate the new partial pressure of oxygen.

6.13 atm• Calculate the new partial pressure of helium.

2.93 atm• Calculate the new total pressure of both gases.

9.06 atm


EXERCISE!EXERCISE!

• The partial pressure of a component gas, “A”, is then defined as

Partial Pressures of Gas Mixtures

totAA P P Applying this concept to the ideal gas equation, we find that each gas can be treated independently.

RTn VP AA

• The composition of a gas mixture is often described in terms of its mole fraction.

Partial Pressures of Gas Mixtures

tot

A

tot

AA P

Pnn

Aof fraction Mole

The mole fraction, of a component gas is the fraction of moles of that component in the total moles of gas mixture.

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

A Problem to Consider• Given a mixture of gases in the atmosphere

at 760 torr, what is the partial pressure of N2 ( = 0 .7808) at 25 oC?

torr) (760 (0.7808) P then2N

torr 593 P2N

totNN P P since22

• So far we have considered “what happens,” but not “why.”

• In science, “what” always comes before “why.”


Kinetic Molecular Theory of Gases 3

1. Gases are made up of small atoms or molecules that are in constant and random motion.

2. The distance of separation is very large compared to the size of the atoms or molecules.

– the gas is mostly empty space.

Provides an explanation of the behavior of gases that we have studied in this chapter. Summary follows:

3. All gas particles behave independently. – No attractive or repulsive forces exist

between them.

4. Gas particles collide with each other and with the walls of the container without losing energy. – The energy is transferred from one atom or

molecule to another.

5. The average kinetic energy of the atoms or molecules is proportional to absolute temperature. – K.E. = 1/2mv2 so as temperature goes up, the

speed of the particles goes up.

How does the Kinetic Molecular Theory of Gases explain the following statements?

• Gases are easily compressible.

• Gases will expand to fill any available volume.

• Gases have low density.

4

Kinetic Molecular Theory




You are holding two balloons of the same volume. One contains helium, and one contains hydrogen. Complete each of the following statements with “different” or “the same” and be prepared to justify your answer.


He

H2

CONCEPT CHECK!CONCEPT CHECK!

•The pressures of the gas in the two balloons

are __________.


He

H2


•The temperatures of the gas in the two balloons

are __________.


He

H2


The numbers of moles of the gas in the two balloons are __________.


He

H2


•The densities of the gas in the two balloons are __________.


He

H2


Molecular View of Boyle’s Law




Molecular View of Charles’s Law




Molecular View of the Ideal Gas Law




Which of the following best represents the mass ratio of Ne:Ar in the balloons?

1:11:22:11:33:1


Ne

Ar

VNe = 2VAr



•You have a sample of nitrogen gas (N2) in a container fitted with a piston that maintains a pressure of 6.00 atm. Initially, the gas is at 45°C in a volume of 6.00 L.

•You then cool the gas sample.


Which best explains the final result that occurs once the gas sample has cooled?

a) The pressure of the gas increases.b) The volume of the gas increases.c) The pressure of the gas decreases.d) The volume of the gas decreases.e) Both volume and pressure change.



The gas sample is then cooled to a temperature of 15°C. Solve for the new condition. (Hint: A moveable piston keeps the pressure constant overall, so what condition will change?)

5.43 L



Root Mean Square Velocity

R = 8.3145 J/K·mol (J = joule = kg·m2/s2)

T = temperature of gas (in K)M = mass of a mole of gas in kg

• Final units are in m/s.Copyright © Cengage Learning. All rights reserved 98

3 = RTurms M

• Diffusion – the mixing of gases.• Effusion – describes the passage of a gas

through a tiny orifice into an evacuated chamber.

• Rate of effusion measures the speed at which the gas is transferred into the chamber.


Diffusion




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

NH3

17 g/molHCl36 g/mol

NH4Cl

r1

r2

M2

M1=

Effusion




Graham’s Law of Effusion

• M1 and M2 represent the molar masses of the gases.


1

2

2 gas for effusion of Rate

1 gas for effusion of Rate

M

M

• An ideal gas is a hypothetical concept. No gas exactly follows the ideal gas law.

• We must correct for non-ideal gas behavior when:– Pressure of the gas is high.– Temperature is low.

• Under these conditions: – Concentration of gas particles is high.– Attractive forces become important.


Gases behave most ideally at low pressure and high temperatures.

Ideal Gases Vs. Real Gases

• In reality there is no such thing as an ideal gas. Instead this is a useful model to explain gas behavior.

• Non-polar gases behave more ideally than polar gases because attractive forces are present in polar gases.

Deviations from Ideal Behavior

1 mole of ideal gas

PV = nRT

n = PVRT

= 1.0

Repulsive Forces

Attractive Forces

Plots of PV/nRT Versus P for Nitrogen Gas at Three Temperatures


Real Gases (van der Waals Equation)


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

corrected pressurecorrected pressure

PPidealideal

corrected volumecorrected volume

VVidealideal

• For a real gas, the actual observed pressure is lower than the pressure expected for an ideal gas due to the intermolecular attractions that occur in real gases.


Values of the van der Waals Constants for Some Gases

• The value of a reflects how much of a correction must be made to adjust the observed pressure up to the expected ideal pressure.

• A low value for a reflects weak intermolecular forces among the gas molecules.



• If sulfur dioxide were an “ideal” gas, the pressure at 0 oC exerted by 1.000 mol occupying 22.41 L would be 1.000 atm. Use the van der Waals equation to estimate the “real” pressure.

The following values for SO2

a = 6.865 L2.atm/mol2

b = 0.05679 L/mol


• First, let’s rearrange the van der Waals equation to solve for pressure.

2

2

V

an -

nb-VnRT

P

R= 0.0821 L. atm/mol. KT = 273.2 KV = 22.41 L

a = 6.865 L2.atm/mol2

b = 0.05679 L/mol


• The “real” pressure exerted by 1.00 mol of SO2 at STP is slightly less than the “ideal” pressure.

2

2

V

an -

nb-VnRT

P

L/mol) 79mol)(0.056 (1.000 - L 22.41

)K2.273)( 06mol)(0.082 (1.000 P Kmol

atmL

2mol

atmL2

L) 41.22(

) (6.865mol) (1.000-

2

2

atm 0.989 P

Air Pollution

• Two main sources:– Transportation – Production of electricity

• Combustion of petroleum produces CO, CO2, NO, and NO2, along with unburned molecules from petroleum.


Nitrogen Oxides (Due to Cars and Trucks)

• At high temperatures, N2 and O2 react to form NO, which oxidizes to NO2.

• The NO2 breaks up into nitric oxide and free oxygen atoms.

• Oxygen atoms combine with O2 to form ozone (O3).


Radiantenergy

2NO ( ) NO( ) O( ) g g g

2 3O( ) O ( ) O ( ) g g g

Concentration for Some Smog Components vs. Time of Day


Sulfur Oxides (Due to Burning Coal for Electricity)

• Sulfur produces SO2 when burned.

• SO2 oxidizes into SO3, which combines with water droplets in the air to form sulfuric acid.


2 2S(in coal) O ( ) SO ( ) g g

2 2 32SO ( ) O ( ) 2SO ( ) g g g

3 2 2 4SO ( ) H O( ) H SO ( ) g l aq

Sulfur Oxides (Due to Burning Coal for Electricity)

• Sulfuric acid is very corrosive and produces acid rain.

• Use of a scrubber removes SO2 from the exhaust gas when burning coal.


A Schematic Diagram of a Scrubber


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