5-1 dr. wolf’s chm 101 chapter 5 gases and the kinetic molecular theory

Dr. Wolf’s CHM 101

5-1

Chapter 5Gases and the Kinetic

Molecular Theory


5-2

Gases and the Kinetic Molecular Theory

5.1 An Overview of the Physical States of Matter

5.2 Gas Pressure and Its Measurement

5.3 The Gas Laws and Their Experimental Foundations

5.4 Further Applications of the Ideal Gas Law

5.5 The Ideal Gas Law and Reaction Stoichiometry

5.6 The Kinetic-Molecular Theory: A Model for Gas Behavior

5.7 Real Gases: Deviations from Ideal Behavior


5-3

An Overview of the Physical States of Matter

The Distinction of Gases from Liquids and Solids

1. Gas volume changes greatly with pressure.

2. Gas volume changes greatly with temperature.

3. Gas have relatively low viscosity.

4. Most gases have relatively low densites under normal conditions.

5. Gases are miscible.


5-4

Figure 5.1The three states of matter.


5-5

Measuring Gas Pressure

Barometer - A device to measure atmospheric pressure. Pressure is defined as force divided by area. The force is the force of gravity acting on the air molecules.

Manometer - A device to measure gas pressure in a closed container.


5-6

Figure 5.3 A mercury barometer.


5-7

Figure 5.4 Two types of manometer


5-8

Table 5.2 Common Units of Pressure

Atmospheric PressureUnit Scientific Field

chemistryatmosphere(atm) 1 atm

pascal(Pa); kilopascal(kPa)

1.01325x105Pa; 101.325 kPa

SI unit; physics, chemistry

millimeters of mercury(Hg)

760 mmHg chemistry, medicine, biology

torr 760 torr chemistry

pounds per square inch (psi or lb/in2)

14.7lb/in2 engineering

bar 1.01325 bar meteorology, chemistry, physics


5-9

Sample Problem 5.1 Converting Units of Pressure

PROBLEM: A geochemist heats a limestone (CaCO3) sample and collects the CO2 released in an evacuated flask attached to a closed-end manometer. After the system comes to room temperature, h = 291.4mmHg. Calculate the CO2 pressure in torrs, atmospheres, and kilopascals.

SOLUTION:

PLAN: Construct conversion factors to find the other units of pressure.

291.4mmHg 1torr

1mmHg

= 291.4torr

291.4torr 1atm

760torr= 0.3834atm

0.3834atm 101.325kPa

1atm= 38.85kPa


5-10

Boyle’s Law - The relationship between volume and the pressure of a gas.

(Temperature is kept constant.)


5-11

Boyle’s Law V n and T are fixed1

P

PV = constant V = constant / P


5-12

Charles’s Law - The relationship between volume and the temperature of a gas.

(Pressure is kept constant.)


5-13

Boyle’s Law n and T are fixedV 1

P

Charles’s Law V T P and n are fixed

V

T= constant V = constant x T

Amonton’s Law P T V and n are fixed

P

T= constant P = constant x T

combined gas law V T

PV = constant x

T

P

PV

T= constant

When the amount of gas, n, is constant


5-14

An experiment to study the relationship between the volume and amount of a gas.

Avogadro’s Law - The volume of a gas is directly proportionate to the amount of gas. (Pressure and temperature kept constant.)

V = constant x n

V

n= constant

V n

Twice the amountgives twice the volume


5-15

If the constant pressure is 1 atm and the constant temperature is 0o C, 1 mole of any gas has a volume of 22.4 L .

This is known as the Standard Molar Volume


5-16

IDEAL GAS LAW

Boyle’s Law

= constantV

PV = V =

Charles’s Law

constant X T

Avogadro’s Law

constant X n

fixed n and T fixed n and P fixed P and T

THE IDEAL GAS LAW

PV = nRT

R = PV

nT=

1atm x 22.414L

1mol x 273.15K=

0.0821atm-L

Mol-K


5-17

Sample Problem 5.2 Applying the Volume-Pressure Relationship

PROBLEM: Boyle’s apprentice finds that the air trapped in a J tube occupies 24.8cm3 at 1.12atm. By adding mercury to the tube, he increases the pressure on the trapped air to 2.64atm. Assuming constant temperature, what is the new volume of air (inL)?

PLAN: SOLUTION:

V1 in cm3

V1 in mL

V1 in L

V2 in L

unit conversion

gas law calculation

P1 = 1.12atm P2 = 2.64atm

V1 = 24.8cm3 V2 = unknown

n and T are constant

24.8cm3 1mL

1cm3

L

103mL= 0.0248L = V1

P1V1

n1T1

P2V2

n2T2

=P1V1 = P2V2

P1V1

P2

V2 = = 0.0248L1.12atm

2.46atm= 0.0105L

1cm3=1mL

103mL=1L

xP1/P2= R =


5-18

Sample Problem 5.3 Applying the Temperature-Pressure Relationship

PROBLEM: A 1-L steel tank is fitted with a safety valve that opens if the internal pressure exceeds 1.00x103 torr. It is filled with helium at 230C and 0.991atm and placed in boiling water at exactly 1000C. Will the safety valve open?

PLAN: SOLUTION:

P1(atm) T1 and T2(0C)

P1(torr) T1 and T2(K)

P1 = 0.991atm P2 = unknown

T1 = 230C T2 = 100oC

P2(torr)

1atm=760torr

x T2/T1

K=0C+273.15

P1V1

n1T1

P2V2

n2T2

=P1

T1

P2

T2

=

0.991atm1atm

760 torr = 753 torr

P2 = P1 T2

T1 = 753 torr

373K

296K= 949 torr


5-19

Sample Problem 5.4 Applying the Volume-Amount Relationship

PROBLEM: A scale model of a blimp rises when it is filled with helium to a volume of 55dm3. When 1.10mol of He is added to the blimp, the volume is 26.2dm3. How many more grams of He must be added to make it rise? Assume constant T and P.

PLAN:

SOLUTION:

We are given initial n1 and V1 as well as the final V2. We have to find n2 and convert it from moles to grams.

n1(mol) of He

n2(mol) of He

mol to be added

g to be added

x V2/V1

x M

subtract n1

n1 = 1.10mol n2 = unknown

V1 = 26.2dm3 V2 = 55.0dm3

P and T are constant

P1V1

n1T1

P2V2

n2T2

=

V1

n1

V2

n2

= n2 = n1 V2

V1

n2 = 1.10mol55.0dm3

26.2dm3= 2.31mol

4.003g He

mol He= 4.84g Hen2 - n1 = 2.31 -1.10 = 1.21 mol He


5-20

Sample Problem 5.5 Solving for an Unknown Gas Variable at Fixed Conditions

PROBLEM: A steel tank has a volume of 438L and is filled with 0.885kg of O2. Calculate the pressure of O2 at 210C.

PLAN:

SOLUTION:

V, T and mass, which can be converted to moles (n), are given. We use the ideal gas law to find P.

V = 438L T = 210C (convert to K)

n = 0.885kg (convert to mol) P = unknown

210C + 273.15 = 294K0.885kg103g

kg

mol O2

32.00g O2

= 27.7mol O2

P = nRT

V=

27.7mol 294Katm*L

mol*K0.0821x x

438L= 1.53atm


5-21

Sample Problem 5.6 Calculating Gas Density

PROBLEM: Calculate the density (in g/L) of carbon dioxide and the number of molecules per liter (a) at STP (00C and 1 atm) and (b) at ordinary room conditions (20.0C and 1.00atm).

PLAN:

SOLUTION:

Density is mass/unit volume; substitute for volume in the ideal gas equation. Since the identity of the gas is known, we can find the molar mass. Convert mass/L to molecules/L with Avogardro’s number.

d = mass/ V PV = nRT

n / V = P / RTd =

RTM x P

d =44.01g/mol x 1atm

atm*L

mol*K0.0821 x 273K

= 1.96g/L

1.96g

L

mol CO2

44.01g CO2

6.022x1023molecules

mol= 2.68x1022molecules CO2/L

(a)

Mass = n Mmult. both sides by M


5-22

Sample Problem 5.6 Calculating Gas Density

continued

(b) = 1.83g/Ld =44.01g/mol x 1atm

x 293Katm*L

mol*K0.0821

1.83g

L

mol CO2

44.01g CO2

6.022x1023molecules

mol= 2.50x1022molecules CO2/L


5-23

Calculating the Molar Mass, M, of a GasSince PV = nRT

Then n = PV / RT

And n = mass / M

So m / M = PV / RT

And M = mRT / PV

Or M = d RT / P


5-24

Sample Problem 5.7 Finding the Molar Mass of a Volatile Liquid

PROBLEM: An organic chemist isolates from a petroleum sample a colorless liquid with the properties of cyclohexane (C6H12). She uses the Dumas method and obtains the following data to determine its molar mass:

PLAN:

SOLUTION:

Use unit conversions, mass of gas and density-M relationship.

Volume of flask = 213mL

Mass of flask + gas = 78.416g

T = 100.00C

Mass of flask = 77.834g

P = 754 torr

Is the calculated molar mass consistent with the liquid being cyclohexane?

m = (78.416 - 77.834)g = 0.582g

M = m RT

P V

0.582gatm*L

mol*K0.0821 373Kx

0.213L x 0.992atm= 84.4g/mol

M of C6H12 is 84.16g/mol and the calculated value is within experimental error.

x

=


5-25

Dalton’s Law of Partial Pressures

Ptotal = P1 + P2 + P3 + ...

P1= 1 x Ptotal where 1 is the mole fraction

1 = n1

n1 + n2 + n3 +...=

n1

ntotal

Partial Pressure of a Gas in a Mixture of Gases

Gases mix homogeneously.

Each gas in a mixture behaves as if it is the only gas present, e.g. its pressure is calculated from PV = nRT with n equal to the number of moles of that particular gas......called the Partial Pressure.

The gas pressure in a container is the sum of the partial pressures of all of the gases present.


5-26

Often in gas experiments the gas is collected “over water.” So the gases in the container includes water vapor as a gas. The water vapor’s partial pressure contributes to the total pressure in the container.

Collecting Gas over Water


5-27

Sample Problem 5.8 Applying Dalton’s Law of Partial Pressures

PROBLEM: In a study of O2 uptake by muscle at high altitude, a physiologist prepares an atmosphere consisting of 79 mol% N2, 17 mol% 16O2, and 4.0 mol% 18O2. (The isotope 18O will be measured to determine the O2 uptake.) The pressure of the mixture is 0.75atm to simulate high altitude. Calculate the mole fraction and partial pressure of 18O2 in the mixture.

PLAN:

SOLUTION:

Find the and P from Ptotal and mol% 18O2. 18O218O2

mol% 18O2

18O2

partial pressure P18O2

divide by 100

multiply by Ptotal

18O2

=4.0mol% 18O2

100= 0.040

= 0.030atmP = x Ptotal = 0.040 x 0.75atm 18O2

18O2


5-28

Sample Problem 5.9 Calculating the Amount of Gas Collected Over Water

PLAN:

SOLUTION:

The difference in pressures will give us the P for the C2H2. The ideal gas law will allow us to find n. Converting n to grams requires the molar mass, M.

PROBLEM: Acetylene (C2H2), an important fuel in welding, is produced in the laboratory when calcium carbide (CaC2) reaction with water:

CaC2(s) + 2H2O(l) C2H2(g) + Ca(OH)2(aq)

For a sample of acetylene that is collected over water, the total gas pressure (adjusted to barometric pressure) is 738torr and the volume is 523mL. At the temperature of the gas (230C), the vapor pressure of water is 21torr. How many grams of acetylene are collected?

Ptotal PC2H2

nC2H2

gC2H2

PH2O n =

PV

RT

x M

PC2H2

= (738-21)torr = 717torr

717torratm

760torr

= 0.943atm


5-29

Sample Problem 5.9 Calculating the Amount of Gas Collected Over Water

continued

0.943atm 0.523Lxn

C2H2

=atm*L

mol*K0.0821 x 296K

= 0.203mol

0.203mol26.04g C2H2

mol C2H2

= 0.529 g C2H2


5-30

P,V,T

of gas A

amount (mol)

of gas A

amount (mol)

of gas B

P,V,T

of gas B

ideal gas law

ideal gas law

molar ratio from balanced equation

Summary of the stoichiometric relationships among the amount (mol,n) of gaseous reactant or

product and the gas variables pressure (P), volume (V), and temperature (T).


5-31

Sample Problem 5.10 Using Gas Variables to Find Amount of Reactants and Products

PROBLEM: A laboratory-scale method for reducing a metal oxide is to heat it with H2. The pure metal and H2O are products. What volume of H2 at 765torr and 2250C is needed to form 35.5g of Cu from copper (II) oxide?

SOLUTION:

PLAN: Since this problem requires stoichiometry and the gas laws, we have to write a balanced equation, use the moles of Cu to calculate mols and then volume of H2 gas.

mass (g) of Cu

mol of Cu

mol of H2

L of H2

divide by M

molar ratio

use known P and T to find V

CuO(s) + H2(g) Cu(s) + H2O(g)

35.5g Cumol Cu

63.55g Cu

1mol H2

1 mol Cu= 0.559mol H2

0.559mol H2 x 498Katm*L

mol*K0.0821 x

1.01atm

= 22.6L


5-32

Sample Problem 5.11 Using the Ideal Gas Law in a Limiting-Reactant Problem

PROBLEM: The alkali metals [Group 1A(1)] react with the halogens [Group 7A(17)] to form ionic metal halides. What mass of potassium chloride forms when 5.25L of chlorine gas at 0.950atm and 293K reacts with 17.0g of potassium?

SOLUTION:

PLAN: After writing the balanced equation, we use the ideal gas law to find the number of moles of reactants, the limiting reactant and moles of product.

2K(s) + Cl2(g) 2KCl(s) V = 5.25L

T = 293K n = unknown

P = 0.950atm

n = PV

RT Cl2 x 5.25L

= 0.950atm

atm*L

mol*K0.0821 x 293K

= 0.207mol

17.0g39.10g K

mol K= 0.435mol K

0.207mol Cl22mol KCl

1mol Cl2

0.435mol K2mol KCl

2mol KCl2 is the limiting reactant.

0.414mol KCl74.55g KCl

mol KCl= 30.9 g KCl

= 0.414mol KCl formed

= 0.435mol KCl formed


5-33

Postulates of the Kinetic-Molecular Theory

Because the volume of an individual gas particle is so small compared to the volume of its container, the gas particles are considered to have mass, but no volume.

Gas particles are in constant, random, straight-line motion except when they collide with each other or with the container walls.

Collisions are elastic therefore the total kinetic energy(Kk) of the particles is constant.

Postulate 1: Particle Volume

Postulate 2: Particle Motion

Postulate 3: Particle Collisions


5-34

A molecular description of Boyle’s Law

Boyle’s Law V n and T are fixed1

P


5-35

A molecular description of Charles’s Law

n and P are fixedCharles’s Law V T


5-36

A molecular description of Dalton’s law of partial pressures.

Dalton’s Law of Partial Pressures

Ptotal = P1 + P2 + P3 + ...


5-37

A molecular description of Avogadro’s Law

Avogadro’s Law V n P and T are fixed


5-38

Kinetic-Molecular Theory

Gas particles are in motion and have a molecular speed, .

But they are moving at various speeds, some very slow, some very fast, but most near the average speed of all of the particles,

(avg) .

Since kinetic energy is defined as ½ mass x (speed)2, we can define the average kinetic energy, Ek(avg) = ½m

(avg)

Since energy is a function of temperature, the average energy and, hence, average molecular speed will increase with temperature.


5-39

The relationship between average kinetic energy and temperature is given as, Ek(avg) = 3/2 (R/NA) x T

(where R is the gas constant in energy units, 8.314 J/mol-KNA is Avogadro’s number, and T temperature in K.)

An increase in temperature results in an increase in average molecular kinetic energy.

To have the same average kinetic energy, heavier atoms must have smaller speeds.

The root-mean-square speed, (rms) , is the speed where a molecule has

the average kinetic energy. The relationship between (rms) and molar

mass is:

(rms) = (3RT/M ) ½

So the speed (or rate of movement) is: rate (M ) ½


5-40

Relationship between molar mass and molecular speed.


5-41

Graham’s Law of Effusion

Graham’s Law of Effusion

The rate of effusion of a gas is inversely related to the square root of its molar mass.

Effusion is the process by which a gas in a closed container moves through a pin-hole into an evacuated space. This rate is proportional to the speed of a molecule so.....

Rate of effusion (M ) ½

So doing two identical effusion experiments measuring the rates of two gases, one known, one unknown, allows the molecular mass of the unknown to be determined.

=rate A

rate B( MB / MA ) ½


5-42

Sample Problem 5.12 Applying Graham’s Law of Effusion

PROBLEM: Calculate the ratio of the effusion rates of helium and methane (CH4).

SOLUTION:

PLAN: The effusion rate is inversely proportional to the square root of the molar mass for each gas. Find the molar mass of both gases and find the inverse square root of their masses.

M of CH4 = 16.04g/mol M of He = 4.003g/mol

CH4

Herate

rate= 2.002( 16.04/ 4.003 ) ½=


5-43

End of Chapter 5

5-1 dr. wolf’s chm 101 chapter 5 gases and the kinetic molecular theory

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