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- Theory developed to explain gas behavior.
- Theory based on properties at the molecular level.
- Kinetic molecular theory gives us a model for understanding pressure and temperature at the molecular level.
- Pressure of a gas results from the number of collisions per unit time on the walls of container.

- There is a spread of individual energies of gas molecules in any sample of gas.
- As the temperature increases, the average kinetic energy of the gas molecules increases.

- Assumptions:
- Gases consist of a large number of molecules in constant random motion.
- Volume of individual molecules negligible compared to volume of container.
- Intermolecular forces (attractive or repulsive forces between gas molecules) are negligible.
- Energy can be transferred between molecules, but total kinetic energy is constant at constant temperature.
- Average kinetic energy of molecules is proportional to temperature.

- Magnitude of pressure given by how often and how hard the molecules strike.
- Gas molecules have an average kinetic energy.
- Each molecule may have a different energy.

- As kinetic energy increases, the velocity of the gas molecules increases.
- Root mean square speed, u, is the speed of a gas molecule having average kinetic energy.
- Average kinetic energy, , is related to root mean square speed:

Do you remember how to calculate

vxy from vx and vy ?

And how about v from all three

components?

Remember these

equations!!

They’ll pop

up again in Chap. 11.

The Maxwell-Boltzmann Distribution of Velocities

Note that the mean value of velocity is zero!

The Maxwell-Boltzmann Distribution of Velocities

This is also the

form of a Gaussian

(normal) distribution,

where

ump = <u> = urms.

- Application to Gas Laws
- As volume increases at constant temperature, the average kinetic of the gas remains constant. Therefore, u is constant. However, as the volume increases the gas molecules have to travel further to hit the walls of the container. Therefore, pressure decreases.
- If temperature increases at constant volume, the average kinetic energy of the gas molecules increases. Therefore, there are more collisions with the container walls and the pressure increases.

- Molecular Effusion and Diffusion
- As kinetic energy increases, the velocity of the gas molecules increases.
- Average kinetic energy of a gas is related to its mass:
- Consider two gases at the same temperature: the lighter gas has a higher velocity than the heavier gas.
- Mathematically:

- Molecular Effusion and Diffusion
- The lower the molar mass, M, the higher the urms.

SAMPLE EXERCISE 10.14 Calculating a Root-Mean-Square Speed

Calculate the rms speed, u, of an N2 molecule at 25°C.

Solution

Analyze: We are given the identity of the gas and the temperature, the two quantities we need to calculate the rms speed.

Plan: We will calculate the rms speed using Equation 10.22.

Solve: In using Equation 10.22, we should convert each quantity to SI units so that all the units are compatible. We will also use R in units of J/mol-K (Table 10.2) in order to make the units cancel correctly.

(These units follow from the fact that 1 J = 1 kg-m2/s2 )

Comment: This corresponds to a speed of 1150 mi/hr. Because the average molecular weight of air molecules is slightly greater than that of N2, the rms speed of air molecules is a little slower than that for N2. The speed at which sound propagates through air is about 350 m/s, a value about two-thirds the average rms speed for air molecules.

- Graham’s Law of Effusion
- As kinetic energy increases, the velocity of the gas molecules increases.
- Effusion is the escape of a gas through a tiny hole.
- The rate of effusion can be quantified.

- Graham’s Law of Effusion
- Consider two gases with molar masses M1 and M2, the relative rate of effusion is given by:
- Only those molecules that hit the small hole will escape through it.
- Therefore, the higher the urms the greater the likelihood of a gas molecule hitting the hole.

SAMPLE EXERCISE 10.15 Applying Graham’s Law

An unknown gas composed of homonuclear diatomic molecules effuses at a rate that is only 0.355 times that of O2 at the same temperature. Calculate the molar mass of the unknown, and identify it.

Solution

Analyze: We are given the rate of effusion of an unknown gas relative to that of O2, and we are asked to find the molar mass and identity of the unknown. Thus, we need to connect relative rates of effusion to relative molar masses.

Plan: We can use Graham’s law of effusion, Equation 10.23, to determine the molar mass of the unknown gas.

If we let rx and represent the rate of effusion and molar mass of the unknown gas, Equation 10.23 can be written as follows:

Solve: From the information given,

Thus,

We now solve for the unknown molar mass,

PRACTICE EXERCISE

Calculate the ratio of the effusion rates of

SAMPLE EXERCISE 10.15continued

Because we are told that the unknown gas is composed of homonuclear diatomic molecules, it must be an element. The molar mass must represent twice the atomic weight of the atoms in the unknown gas. We conclude that the unknown gas is I2.

- Graham’s Law of Effusion
- Consider two gases with molar masses M1 and M2, the relative rate of effusion is given by:
- Only those molecules that hit the small hole will escape through it.
- Therefore, the higher the rms the more likelihood of a gas molecule hitting the hole.

- Diffusion and Mean Free Path
- Diffusion of a gas is the spread of the gas through space.
- Diffusion is faster for light gas molecules.
- Diffusion is significantly slower than rms speed (consider someone opening a perfume bottle: it takes while to detect the odor but rms speed at 25C is about 1150 mi/hr).
- Diffusion is slowed by gas molecules colliding with each other.

- Diffusion and Mean Free Path
- Average distance of a gas molecule between collisions is called mean free path.
- At sea level, mean free path is about 6 10-6 cm.

Real Gases: Deviations from Ideal Behavior

- From the ideal gas equation, we have
- For 1 mol of gas, PV/nRT = 1 for all pressures.
- In a real gas, PV/nRT varies from 1 significantly and is called Z.
- The higher the pressure the more the deviation from ideal behavior.

Real Gases: Deviations from Ideal Behavior

- From the ideal gas equation, we have
- For 1 mol of gas, PV/RT = 1 for all temperatures.
- As temperature increases, the gases behave more ideally.
- The assumptions in kinetic molecular theory show where ideal gas behavior breaks down:
- the molecules of a gas have finite volume;
- molecules of a gas do attract each other.

Real Gases: Deviations from Ideal Behavior

- As the pressure on a gas increases, the molecules are forced closer together.
- As the molecules get closer together, the volume of the container gets smaller.
- The smaller the container, the more space the gas molecules begin to occupy.
- Therefore, the higher the pressure, the less the gas resembles an ideal gas.

Real Gases: Deviations from Ideal Behavior

- As the gas molecules get closer together, the smaller the intermolecular distance.

Real Gases: Deviations from Ideal Behavior

- The smaller the distance between gas molecules, the more likely attractive forces will develop between the molecules.
- Therefore, the less the gas resembles and ideal gas.
- As temperature increases, the gas molecules move faster and further apart.
- Also, higher temperatures mean more energy available to break intermolecular forces.

Real Gases: Deviations from Ideal Behavior

- Therefore, the higher the temperature, the more ideal the gas.

Real Gases: Deviations from Ideal Behavior

- The van der Waals Equation
- We add two terms to the ideal gas equation one to correct for volume of molecules and the other to correct for intermolecular attractions
- The correction terms generate the van der Waals equation:
- where a and b are empirical constants characteristic of each gas.

Real Gases: Deviations from Ideal Behavior

- The van der Waals Equation
- General form of the van der Waals equation:

Corrects for molecular volume

Corrects for molecular attraction

SAMPLE EXERCISE 10.16 Using the van der Waals Equation

If 1.000 mol of an ideal gas were confined to 22.41 L at 0.0°C, it would exert a pressure of 1.000 atm. Use the van der Waals equation and the constants in Table 10.3 to estimate the pressure exerted by 1.000 mol of Cl2(g) in 22.41 L at 0.0°C.

Solution

Analyze: The quantity we need to solve for is pressure. Because we will use the van der Waals equation, we must identify the appropriate values for the constants that appear there.

Plan: Using Equation 10.26, we have

Solve: Substituting n = 1.000 mol, R = 0.08206 L-atm/mol-K, T = 273.2 K, V = 22.41 L, a = 6.49 L2-atm/mol2, and b = 0.0562 l/mol:

Check: We expect a pressure not far from 1.000 atm, which would be the value for an ideal gas, so our answer seems very reasonable.

SAMPLE EXERCISE 10.16continued

Comment: Notice that the first term, 1.003 atm, is the pressure corrected for molecular volume. This value is higher than the ideal value, 1.000 atm, because the volume in which the molecules are free to move is smaller than the container volume, 22.41 L. Thus, the molecules must collide more frequently with the container walls. The second factor, 0.013 atm, corrects for intermolecular forces. The intermolecular attractions between molecules reduce the pressure to 0.990 atm. We can conclude, therefore, that the intermolecular attractions are the main cause of the slight deviation of Cl2(g) from ideal behavior under the stated experimental conditions.

Analyze: First we need to determine the molecular formula of a compound from elemental analysis data and data on the properties of the gaseous substance. Thus, we have two separate calculations to do.

(a) Plan: We can use the percentage composition of the compound to calculate its empirical formula.• (Section 3.5) Then we can determine the molecular formula by comparing the mass of the empirical formula with the molar mass. • (Section 3.5)

Solve: To determine the empirical formula, we assume that we have a 100-g sample of the compound and then calculate the number of moles of each element in the sample:

Because the ratio of the moles of the two elements is essentially 1:1, the empirical formula is CN.

To determine the molar mass of the compound, we use Equation 10.11.

SAMPLE INTEGRATIVE EXERCISE Putting Concepts Together

Cyanogen, a highly toxic gas, is composed of 46.2% C and 53.8% N by mass. At 25°C and 751 torr, 1.05 g of cyanogen occupies 0.500 L. (a) What is the molecular formula of cyanogen? (b) Predict its molecular structure. (c) Predict the polarity of the compound.

(b) Plan: To determine the molecular structure of the molecule, we must first determine its Lewis structure.

• (Section 8.5) We can then use the VSEPR model to predict the structure. • (Section 9.2)

Solve: The molecule has 2(4) + 2(5) = 18 valence-shell electrons. By trial and error, we seek a Lewis structure with 18 valence electrons in which each atom has an octet and in which the formal charges are as low as possible. The following structure meets these criteria:

(This structure has zero formal charges on each atom.)

SAMPLE INTEGRATIVE EXERCISEcontinued

The molar mass associated with the empirical formula, CN, is 12.0 + 14.0 = 26.0 g/mol. Dividing the molar mass of the compound by that of its empirical formula gives (52.0 g/mol)/(26.0 g/mol) = 2.00. Thus, the molecule has twice as many atoms of each element as the empirical formula, giving the molecular formula C2N2

The Lewis structure shows that each atom has two electron domains. (Each nitrogen has a nonbonding pair of electrons and a triple bond, whereas each carbon has a triple bond and a single bond.) Thus the electron-domain geometry around each atom is linear, causing the overall molecule to be linear.

(c) Plan: To determine the polarity of the molecule, we must examine the polarity of the individual bonds and the overall geometry of the molecule.

Solve: Because the molecule is linear, we expect the two dipoles created by the polarity in the carbon–nitrogen bond to cancel each other, leaving the molecule with no dipole moment.

Chapter 11 -- Intermolecular Forces, Liquids, and Solids

In many ways, this chapter is simply a

continuation of our earlier discussion of

‘real’ gases.

This plot includes a realistic one for Volume as a function of Temperature!

What determines whether a substance exists

as a gas, liquid, or solid?

Two primary factors are involved:

Kinetic Energy of the particles.

Strength of attractions between

the particles.

What are the important Intermolecular Forces

i.e, forces between molecules ?

Note that earlier chapters concentrated on Intramolecular Forces, those within the molecule.

Important ones:

ion-ion similar to atomic systems

ion-dipole (review definition of dipoles)

dipole-dipole

dipole-induced dipole

London Dispersion Forces:

(induced dipole-induced dipole)

related to polarizability

Hydrogen Bonding

van der Waals

forces

How do you know the relative strengths

of each?

Virtually impossible experimentally!!!

Most important though:

Establish which are present.

London Dispersion Forces: Always

All others depend on defining property

such as existing dipole for d-d.

It has been possible to calculate the

relative strengths in a few cases.

Relative Energies of Various Interactions

d-d d-id disp

Ar 0 0 50

N2 0 0 58

C6H6 0 0 1086

C3H8 0.0008 0.09 528

HCl 22 6 106

CH2Cl2 106 33 570

SO2 114 20 205

H2O 190 11 38

HCN 1277 46 111

Ion-Dipole Interactions

- A fourth type of force, ion-dipole interactions are an important force in solutions of ions.
- The strength of these forces are what make it possible for ionic substances to dissolve in polar solvents.

Let’s take a closer look at these interactions:

Ion-dipole interaction

This is the simple one.

But we also have to consider other shapes.

Review hybridization and molecular shapes.

Recall the discussion of sp, sp2, and sp3

hybridization?

polarizability 1.3 4.6 6.7 10.2 2.6

molecular wt. 37 71 160 254 16

Molecular Weight predicts the trends in the boiling points of atoms or molecules without dipole moments because polarizability tends to increase with increasing mass.

But polarizability also depends on shape,

as well as MW.

Water provides our best example

of Hydrogen Bonding.

These boiling points demonstrate the enormous

contribution of hydrogen bonding.

unusual in the relative densities of the liquid and solid phases.

The crystal structure suggests a reason for the unusual

high density of ice.

11.3 Some Properties of Liquids

Viscosity—the resistance to flow of a liquid, such as

oil, water, gasoline, molasses, (glass !!!)

Surface Tension – tendency to minimize the surface area

compare water, mercury

Cohesive forces—bind similar molecules together

Adhesive forces – bind a substance to a surface

Capillary action results when these two are not equal

Soap reduces the surface tension, permitting one

material to ‘wet’ another more easily

Viscosity

- Resistance of a liquid to flow is called viscosity.
- It is related to the ease with which molecules can move past each other.
- Viscosity increases with stronger intermolecular forces and decreases with higher temperature.

The SI unit is kg/m-s. Many tables still use the older unit of viscosity, the poise, which is 1 g/cm-s, with typical values listed as cP = 0.01 P.

Surface Tension

Surface tension results from the net inward force experienced by the molecules on the surface of a liquid.

- Surface molecules are only attracted inwards towards the bulk molecules.
- Therefore, surface molecules are packed more closely than bulk molecules.
- Surface tension is the amount of energy required to increase the surface area of a liquid, in J/m2.
- Cohesive forcesbind molecules to each other.
- Adhesive forcesbind molecules to a surface.

- Meniscusis the shape of the liquid surface.
- If adhesive forces are greater than cohesive forces, the liquid surface is attracted to its container more than the bulk molecules. Therefore, the meniscus is U-shaped (e.g. water in glass).
- If cohesive forces are greater than adhesive forces, the meniscus is curved downwards.
- Capillary Action: When a narrow glass tube is placed in water, the meniscus pulls the water up the tube.
- Remember that surface molecules are only attracted inwards towards the bulk molecules.

FUSION

- Sublimation: solid gas.
- Vaporization: liquid gas.
- Melting or fusion: solid liquid.
- Deposition: gas solid.
- Condensation: gas liquid.
- Freezing: liquid solid.

SAMPLE EXERCISE 11.4 Calculating H for Temperature and Phase Changes

Calculate the enthalpy change upon converting 1.00 mol of ice at –25°C to water vapor (steam) at 125°C under a constant pressure of 1 atm. The specific heats of ice, water, and steam are 2.09 J/g-K, 4.18 J/g-K and 1.84 J/g-K, respectively. For H2O, Hfus = 6.01 kJ/mol and Hvap = 40.67 kJ/mol.

Solution

Analyze: Our goal is to calculate the total heat required to convert 1 mol of ice at –25°C to steam at 125°C.

Plan: We can calculate the enthalpy change for each segment and then sum them to get the total enthalpy change (Hess’s law, Section 5.6).

Solve: For segment AB in Figure 11.19, we are adding enough heat to ice to increase its temperature by 25°C. A temperature change of 25°C is the same as a temperature change of 25 K, so we can use the specific heat of ice to calculate the enthalpy change during this process:

For segment BC in Figure 11.19, in which we convert ice to water at 0°C, we can use the molar enthalpy of fusion directly:

The enthalpy changes for segments CD, DE, and EF can be calculated in similar fashion:

The total enthalpy change is the sum of the changes of the individual steps:

Check: The components of the total energy change are reasonable in comparison with the lengths of the horizontal segments of the lines in Figure 11.19. Notice that the largest component is the heat of vaporization.

- Explaining Vapor Pressure on the Molecular Level, Volatility,
- Vapor Pressure, and Temperature;
- Vapor Pressure and Boiling Point
- 11.6 Phase Diagrams
- The Phase Diagrams of H2O and CO2
- 11.7 Structures of Solids
- Unit Cells
- The Crystal Structure of Sodium Chloride
- Close Packing of Spheres
- 11.8 Bonding in Solids
- Molecular Solids
- Covalent-Network Solids
- Ionic Solids
- Metallic Solids
- Chapter 13 Properties of Solutions
- 13.1 The Solution Process
- Energy Changes and Solution Formation
- Solution Formation, Spontaneity, and Disorder
- Solution Formation and Chemical Reactions

With the beaker covered, equilibrium is

soon established.

Using the Clausius-Clapeyron Equation:

Tabulate P in atm and T in K; Calculate lnP and 1/T

Plot lnP vs 1/T

The slope is = - ΔHvap /R

The slope!

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