Tải bản đầy đủ - 0 (trang)
9 Avogadro’s Law: The Relation between Volume and Molar Amount

9 Avogadro’s Law: The Relation between Volume and Molar Amount

Tải bản đầy đủ - 0trang

www.downloadslide.net

SECTION 8.9



Avogadro’s Law: the relation between Volume and molar Amount



Because the particles in a gas are so tiny compared to the empty space surrounding

them, there is no interaction among gas particles as proposed by the kinetic–molecular

theory. As a result, the chemical identity of the particles does not matter and the value

of the constant k in the equation V>n = k is the same for all gases. It is therefore possible to compare the molar amounts of any two gases simply by comparing their volumes

at the same temperature and pressure.

Notice that the values of temperature and pressure do not matter; it is only necessary that T and P be the same for both gases. To simplify comparisons of gas samples,

however, it is convenient to define a set of conditions called standard temperature

and pressure (STP), which specifies a temperature of 0 °C (273 K) and a pressure of 1

atm (760 mmHg).

At STP, 1 mol of any gas 16.02 * 1023 particles2 has a volume of 22.4 L, a quantity

called the standard molar volume (Figure 8.18).



Standard temperature and pressure

(STP) 0 °C (273.15 K); 1 atm

(760 mmHg)

Standard molar volume Volume

of one mole of any ideal gas at STP,

22.4 L>mol.







O2



He



F2



Ar



1.00 mol

32.0 g

22.4 L



1.00 mol

4.00 g

22.4 L



1.00 mol

38.0 g

22.4 L



1.00 mol

39.9 g

22.4 L



Figure 8.18



Avogadro’s law.

Each of these 22.4 L bulbs contains

1.00 mol of gas at 0 °C and 1 atm pressure. Note that the volume occupied by

1 mol of gas is the same even though

the mass (in grams) of 1 mol of each

gas is different.



Worked Example 8.8 Using Avogadro’s Law: Finding moles in a Given Volume at stP

Use the standard molar volume of a gas at STP (22.4 L) to find how many moles of air at STP are in a room

measuring 4.11 m wide by 5.36 m long by 2.58 m high.

AnAlySiS We first find the volume of the room and then use standard molar volume as a conversion factor to



find the number of moles.



Solution

StEP 1: Identify known information.

We are given the room dimensions.



Length = 5.36 m

Width = 4.11 m

Height = 2.58 m



volume of the room is the product of

its three dimensions. Once we have the

volume 1in m3 2, we can convert to liters

and use the molar volume at STP as a

conversion factor to obtain moles of air.



Volume = 14.11 m2 15.36 m2 12.58 m2 = 56.8 m3

1000 L

4

= 56.8 m3 *

3 = 5.68 * 10 L

1m

1 mol

1 mol = 22.4 L S

22.4 L



StEP 4: Solve. Use the room volume and



5.68 * 104 L *



StEP 2: Identify answer and units.

StEP 3: Identify the equation. The



the molar volume at STP to set up an equation, making sure unwanted units cancel.



Moles of air = ?? mol



1 mol

= 2.54 * 103 mol

22.4 L



ProBlEM 8.15

How many moles of methane gas, CH4, are in a 100 m3 storage tank at STP? How

many grams of methane is this? How many grams of carbon dioxide gas could the

same tank hold?



269



www.downloadslide.net

270



CHAPTER 8



Gases, Liquids, and solids



8.10 The Ideal Gas Law

Learning Objective:

• Use the ideal gas law to calculate the pressure, temperature, volume, or number of

moles of an ideal gas.

The relationships among the four variables P, V, T, and n for gases can be combined

into a single expression called the ideal gas law. If you know the values of any three of

the four quantities, you can calculate the value of the fourth.

ideal gas law



Gas constant (R) The constant R in

the ideal gas law, PV = nRT.



PV

= R 1A constant value2

nT

or PV = nRT



The constant R in the ideal gas law (instead of the usual k) is called the gas constant.

Its value depends on the units chosen for pressure, with the two most common values

being

For P in atmospheres:



R = 0.0821



L # atm

mol # K



L # mmHg

mol # K

In using the ideal gas law, it is important to choose the value of R having pressure

units that are consistent with the problem and, if necessary, to convert volume into liters

and temperature into kelvins.

Table 8.3 summarizes the various gas laws, and Worked Examples 8.10 and 8.11

show how to use the ideal gas law.

For P in millimeters Hg:



R = 62.4



table 8.3  A Summary of the Gas Laws

Gas Law

V1 >T1 = V2 >T2

P1V1 = P2V2



Boyle’s law

Charles’s law



P1 >T1 = P2 >T2



Avogadro’s law



P1V1 >T1 = P2V2 >T2



Ideal gas law



PV = nRT



Gay-Lussac’s law

Combined gas law



V1 >n1 = V2 >n2



Variables



Constant



P, V



n, T



V, T



n, P



P, T



n, V



P, V, T



n



V, n



P, T



P, V, T, n



R



Worked Example 8.9 Using the ideal Gas Law: Finding moles

How many moles of air are in the lungs of an average person with a total lung capacity of 3.8 L? Assume that

the person is at 1.0 atm pressure and has a normal body temperature of 37 °C.

AnAlySiS This is an ideal gas law problem because it asks for a value of n when P, V, and T are known:



n = PV>RT. The volume is given in the correct unit of liters, but temperature must be converted to kelvins.



Solution

StEP 1: Identify known information. We

know three of the four variables in the ideal

gas law.

StEP 2: Identify answer and units.



P = 1.0 atm

V = 3.8 L

T = 37 °C = 310 K

Moles of air, n = ?? mol



www.downloadslide.net

SECTION 8.10



StEP 3: Identify the equation. Knowing



PV = nRT 1 n =



three of the four variables in the ideal gas

law, we can rearrange and solve for the

unknown variable, n. Note: Because

pressure is given in atmospheres, we use

the value that is expressed in atm:

L # atm

R = 0.0821

mol # K

StEP 4: Solve. Substitute the known



n =



information and the appropriate value

of R into the ideal gas law equation and

solve for n.



PV

=

RT



PV

RT



11.0 atm2 13.8 L2



a 0.0821



the ideal Gas Law



L # atm

b 1310 K2

mol # K



= 0.15 mol



Worked Example 8.10 Using the ideal Gas Law: Finding Pressure



Methane gas is sold in steel cylinders with a volume of 43.8 L containing 5.54 kg. What is the pressure in atmospheres inside the cylinder at a temperature of 293.15 K (20.0 °C)? The molar mass of methane 1CH4 2 is

16.0 g>mol.



AnAlySiS This is an ideal gas law problem because it asks for a value of P when V, T, and n are given.

Although not provided directly, enough information is given so that we can calculate the value of n

1n = g>molar mass2.



Solution

StEP 1: Identify known information.

We know two of the four variables in

the ideal gas law—V and T—and can

calculate the third, n, from the information provided.

StEP 2: Identify answer and units.

StEP 3: Identify equation. First,



calculate the number of moles, n, of

methane in the cylinder by using molar mass (16.0 g>mol) as a conversion

factor. Then use the ideal gas law to

calculate the pressure.

StEP 4: Solve. Substitute the known



information and the appropriate value

of R into the ideal gas law equation

and solve for P.



V = 43.8 L

T = 37 °C = 310 K



Pressure, P = ?? atm



n = 15.54 kg methane2 a



1000 g

1 mol

ba

b = 346 mol methane

1 kg

16.0 g

nRT

PV = nRT 1 P =

V



P =



nRT

=

V



1346 mol2 a 0.0821



L # atm

b 1293 K2

mol # K

= 190 atm

43.8 L



ProBlEM 8.16

An aerosol spray can of deodorant with a volume of 350 mL contains 3.2 g of propane

gas 1C3H8 2 as propellant. What is the pressure (in Pa) in the can at 20 °C?



ProBlEM 8.17

A helium gas cylinder of the sort used to fill balloons has a volume of 0.180 m3 and a

pressure of 150 * 105 Pa (150 atm) at 298 K (25 °C). How many moles of helium are

in the tank? How many grams?



271



www.downloadslide.net

272



CHAPTER 8



Gases, Liquids, and solids



KEy ConCEPt ProBlEM 8.18

Show the approximate level of the movable piston in drawings (a) and (b) after the indicated changes have been made to the initial gas sample (assume a constant pressure

of 1 atm).

1.0 atm



(initial)

T = 300 K

n = 0.300 mol



(a)

T = 450 K

n = 0.200 mol



(b)

T = 200 K

n = 0.400 mol



8.11 Partial Pressure and Dalton’s Law

Learning Objective:

• Use Dalton’s law to calculate the partial pressure or the number of moles of a gas in a

mixture.



Partial pressure The contribution of

a given gas in a mixture to the total

pressure.



According to the kinetic–molecular theory, each particle in a gas acts independently of

all others because there are no attractive forces between them and they are so far apart.

To any individual particle, the chemical identity of its neighbors is irrelevant. Thus,

mixtures of gases behave the same as pure gases and obey the same laws.

Dry air, for example, is a mixture of about 21% oxygen, 78% nitrogen, and 1% argon

by volume, which means that 21% of atmospheric air pressure is caused by O2 molecules, 78% by N2 molecules, and 1% by Ar atoms. The contribution of each gas in a

mixture to the total pressure of the mixture is called the partial pressure of that gas.

According to Dalton’s law, the total pressure exerted by a gas mixture 1Ptotal 2 is the

sum of the partial pressures of the components in the mixture.

Dalton’s law: Ptotal = Pgas 1 + Pgas 2 + Pgas 3

In dry air at a total air pressure of 101,325 Pa, the partial pressure caused by the

contribution of O2 is 0.21 * 101,325 Pa = 21,278 Pa, the partial pressure of N2 is

0.78 * 760 mmHg = 79,034 Pa, and that of argon is 1013 Pa. The partial pressure

exerted by each gas in a mixture is the same pressure that the gas would exert if it were

alone. Put another way, the pressure exerted by each gas depends on the frequency of

collisions of its molecules with the walls of the container. However, this frequency does

not change when other gases are present because the different molecules have no influence on one another.

To represent the partial pressure of a specific gas, we add the formula of the gas as

a subscript to P, the symbol for pressure. You might see the partial pressure of oxygen

represented as PO2, for instance. Moist air inside the lungs at 37 °C and atmospheric

pressure has the following average composition at sea level. Note that Ptotal is equal to

atmospheric pressure, 101,325 Pa.

Ptotal = PN2

+

PO2

+

PCO2

+

PH2O

= 76,394 Pa + 13,332 Pa + 5333 Pa + 6266 Pa

= 101,325 Pa

The composition of air does not change appreciably with altitude, but the total pressure decreases rapidly. The partial pressure of oxygen in air therefore decreases with

increasing altitude, and it is this change that leads to difficulty in breathing at high

elevations.



www.downloadslide.net

SECTION 8.12



Liquids



273



Worked Example 8.11 Using Dalton’s Law: Finding Partial Pressures

Humid air on a warm summer day is approximately 20% oxygen, 75% nitrogen, 4% water vapor, and

1% argon. What is the partial pressure of each component if the atmospheric pressure is 1.0 * 105 Pa?

AnAlySiS According to Dalton’s law, the partial pressure of any gas in a mixture is equal to the percent



concentration of the gas times the total gas pressure (750 mmHg). In this case,



Solution

Oxygen partial pressure 1PO2 2:

Nitrogen partial pressure 1PN2 2:

Water vapor partial pressure 1PH2O 2:

Argon partial pressure 1PAr 2:

Total pressure = 1.0 * 105 Pa



Ptotal = PO2 + PN2 + PH2O + PAr

0.20

0.75

0.04

0.01



*

*

*

*



105 Pa

105 Pa

105 Pa

105 Pa



Note that the sum of the partial pressures must equal the total pressure (within rounding error).



ProBlEM 8.19

Assuming a total pressure of 9.5 * 105 Pa, what is the partial pressure of each component

in the mixture of 98% helium and 2.0% oxygen breathed by deep-sea divers? How does

the partial pressure of oxygen in diving gas compare with its partial pressure in normal air?

ProBlEM 8.20

Determine the percent composition of air in the lungs from the following composition

in partial pressures: PN2 = 0.76 * 105 Pa, PO2 = 0.13 * 105 Pa, PCO2 = 0.050 * 105 Pa,

and PH2O = 0.060 * 105 Pa; all at 37 °C and 1.0 * 105 Pa pressure.

ProBlEM 8.21

The atmospheric pressure on the top of Mt. Everest, an altitude of 8850 m, is only

0.35 * 105 Pa. What is the partial pressure of oxygen in the lungs at this altitude

(assuming that the percent O2 is the same as in dry air)?



KEy ConCEPt ProBlEM 8.22

Using the image in the margin, assume that you have a mixture of He (blue spheres)

and Xe (green spheres) at 300 K. The total pressure of the mixture is 105 Pa. What are

the partial pressures of each of the gases?



8.12 Liquids

Learning Objective:

• identify how the vapor pressure, boiling point, and surface tension of liquids are related

to temperature and intermolecular forces.

Molecules are in constant motion in the liquid state, just as they are in gases. If a molecule

happens to be near the surface of a liquid, and if it has enough energy, it can break free of

the liquid and escape into the gas state, called vapor. In an open container, the now gaseous molecule will wander away from the liquid, and the process will continue until all the

molecules escape from the container (Figure 8.19a). This, of course, is what happens during evaporation. We are all familiar with puddles of water evaporating after a rainstorm.

If the liquid is in a closed container, the situation is different because the gaseous

molecules cannot escape. Thus, the random motion of the molecules occasionally

brings them back into the liquid. After the concentration of molecules in the gas state

has increased sufficiently, the number of molecules reentering the liquid becomes equal



Vapor The gas molecules are in equilibrium with a liquid.



www.downloadslide.net

274



CHAPTER 8



Gases, Liquids, and solids



Vapor pressure The partial pressure

of vapor molecules in equilibrium with

a liquid.



to the number escaping from the liquid (Figure 8.19b). At this point, a dynamic equilibrium exists, exactly as in a chemical reaction at equilibrium. Evaporation and condensation take place at the same rate, and the concentration of vapor in the container is

constant as long as the temperature does not change.

Once molecules have escaped from the liquid into the gas state, they are subject to

all the gas laws previously discussed. In a closed container at equilibrium, for example,

the vapor molecules will make their own contribution to the total pressure of gases

above the liquid according to Dalton’s law (Section 8.11). We call this contribution the

vapor pressure of the liquid.

Equilibrium

vapor

pressure







Figure 8.19



The transfer of molecules between

liquid and gas states.

(a) Molecules escape from an open container and drift away until the liquid

has entirely evaporated. (b) Molecules

in a closed container cannot escape.

Instead, they reach an equilibrium in

which the rates of molecules leaving

the liquid and returning to the liquid

are equal, and the concentration of

molecules in the gas state is constant.



Normal boiling point The boiling

point at a pressure of exactly 1 atm.



(a)



Mercury-filled

manometer



(b)



Vapor pressure depends on both temperature and the chemical identity of a liquid. As the temperature rises, molecules become more energetic and more likely to

escape into the gas state. Thus, vapor pressure rises with increasing temperature until

ultimately it becomes equal to the pressure of the atmosphere. At this point, bubbles

of vapor form under the surface and force their way to the top, giving rise to the violent action observed during a vigorous boil. At an atmospheric pressure of exactly

760 mmHg, boiling occurs at the normal boiling point.

The vapor pressure and boiling point of a liquid will also depend on the intermolecular

forces (discussed in Section 8.2) at work between liquid molecules. Ether molecules, for

example, can engage in dipole–dipole interactions, which are weaker than the hydrogen

bonds formed between water molecules. As a result, ether exhibits a higher vapor pressure

at a given temperature and a lower boiling point than water, as seen in Figure 8.20.



Vapor pressure (Pa)



135 3 103



Normal

boiling points

1 atm pressure



104 3 103

100 3 103

78 3



Ether

bp 307.6 K

(34.6 °C)



103



Ethanol

bp 351.3 K

(78.3 °C)



Water

bp 373 K

(100 °C)



52 3 103

26 3 103

0







Because bromine is colored, it is

possible to see its gaseous reddish

vapor above the liquid.





0



20



40

60

Temperature (°C)



80



100



Figure 8.20



A plot of the change of vapor pressure with temperature for ethyl ether, ethanol, and water.

At a liquid’s boiling point, its vapor pressure is equal to atmospheric pressure. Commonly reported

boiling points are those at 100 * 103 Pa.



www.downloadslide.net

SECTION 8.13



If atmospheric pressure is higher or lower than normal, the boiling point of a liquid

changes accordingly. At high altitudes, for example, atmospheric pressure is lower than

at sea level, and boiling points are also lower. On top of Mt. Everest (8850 m), atmospheric pressure is about 0.33 * 105 and the boiling temperature of water is only 344 K

(71 °C). If the atmospheric pressure is higher than normal, the boiling point is also

higher. This principle is used in strong vessels known as autoclaves, in which water at

high pressure is heated to the temperatures needed for sterilizing medical and dental

instruments (443 K/170 °C).

Many familiar properties of liquids can be explained by the intermolecular forces

just discussed. We all know, for instance, that some liquids, such as water or gasoline, flow easily when poured, whereas others, such as motor oil or maple syrup, flow

sluggishly.

The measure of a liquid’s resistance to flow is called its viscosity. Not surprisingly, viscosity is related to the ease with which individual molecules move around in

the liquid and thus to the intermolecular forces present. Substances such as gasoline,

which have small, nonpolar molecules, experience only weak intermolecular forces

and have relatively low viscosities, whereas more polar substances such as glycerin

3C3H5 1OH2 3 4 experience stronger intermolecular forces and so have higher viscosities.

Another familiar property of liquids is surface tension, the resistance of a liquid to

spreading out and increasing its surface area. Water beading up on a newly waxed car

and the ability of a water strider to walk on water are both due to surface tension.

The difference between the intermolecular forces experienced by molecules at the

surface of the liquid and those experienced by molecules in the interior causes surface

tension. Molecules in the interior of a liquid are surrounded and experience maximum

intermolecular forces, whereas molecules at the surface have fewer neighbors and feel

weaker forces. Surface molecules are therefore less stable, and the liquid acts to minimize their number by minimizing the surface area (Figure 8.21).



solids



275



A bench-top autoclave, used to sterilize medical and dental instruments.





Surface tension allows a water strider

to walk on water without penetrating the

surface.





Recall from Section 1.11 that specific heat is the amount of heat

required to raise the temperature of 1g

of a substance by 1 K/°C.





Figure 8.21



Surface tension.

Surface tension is caused by the different forces experienced by molecules

in the interior of a liquid and those on

the surface. Molecules on the surface

are less stable because they feel fewer

attractive forces, so the liquid acts to

minimize their number by minimizing

surface area.



8.13 Solids

Learning Objective:

• Distinguish between the different types of solids and explain their physical properties.

A brief look around us reveals that most substances are solids rather than liquids or

gases. It is also obvious that there are many different kinds of solids. Some, such as iron

and aluminum, are hard and metallic; others, such as sugar and table salt, are crystalline and easily broken; and still others, such as rubber and many plastics, are soft and

amorphous.

The most fundamental distinction between solids is that some are crystalline and

some are amorphous. A crystalline solid is one whose particles—whether atoms, ions,

or molecules—have an ordered arrangement extending over a long range. This order on

the atomic level is also seen on the visible level because crystalline solids usually have

flat faces and distinct angles.



Crystalline solid A solid whose

atoms, molecules, or ions are rigidly

held in an ordered arrangement.



www.downloadslide.net

276



CHAPTER 8



Gases, Liquids, and solids



Crystalline solids, such as pyrite (left) and fluorite (right), have flat faces and distinct angles.

The octahedral shape of pyrite and the cubic shape of fluorite reflect similarly ordered

arrangements of particles at the atomic level.









Figure 8.22



Diamond.

Diamond is a covalent network solid—

one very large molecule of carbon

atoms linked by covalent bonds.



Amorphous solid A solid whose

particles do not have an orderly

arrangement.



Crystalline solids can be further categorized as ionic, molecular, covalent network, or metallic. Ionic solids are those like sodium chloride, whose constituent particles are ions. A crystal of sodium chloride is composed of alternating Na+ and Cl ions ordered in a regular three-dimensional arrangement held together by ionic bonds

(see Figure 3.4). Molecular solids are those like sucrose or ice, whose constituent particles are molecules held together by the intermolecular forces discussed in Section 8.2.

Covalent network solids are those like diamond (Figure 8.22) or quartz 1SiO2 2, whose

atoms are linked together by covalent bonds into a giant three-dimensional array. In

effect, a covalent network solid is one very large molecule.

Metallic solids, such as silver or iron, can be viewed as vast three-dimensional arrays of metal cations immersed in a sea of electrons that are free to move about. This

continuous electron sea acts both as a glue to hold the cations together and as a mobile

carrier of charge to conduct electricity. Furthermore, the fact that bonding attractions

extend uniformly in all directions explains why metals are malleable rather than brittle.

When a metal crystal receives a sharp blow, no spatially oriented bonds are broken; instead, the electron sea simply adjusts to the new distribution of cations.

An amorphous solid, by contrast with a crystalline solid, is one whose constituent

particles are randomly arranged and have no ordered long-range structure. Amorphous

solids often result when liquids cool before they can achieve internal order or when

their molecules are large and tangled together, as happens in many polymers. Glass is

an amorphous solid, as are tar, the gemstone opal, and some hard candies. Amorphous

solids differ from crystalline solids by softening over a wide temperature range rather

than having sharp melting points and by shattering to give pieces with curved rather

than planar faces. Table 8.4 gives a summary of the different types of solids and their

characteristics.



table 8.4  Types of Solids

NaCl, KI, Ca3 1PO4 2 2



Substance



Smallest Unit



Interparticle Forces



Properties



Ionic solid



Ions



Attraction between positive and

negative ions



Brittle and hard; high melting point;

crystalline



Examples



Molecular solid



Molecules



Intermolecular forces



Soft; low to moderate melting point;

crystalline



Covalent

network



Atoms



Covalent bonds



Very hard; very high melting point;

crystalline



Metal or alloy



Metal atoms



Metallic bonding (attraction

between metal ions and

surrounding mobile electrons)



Lustrous; soft (Na) to hard (Ti); high

melting point; crystalline



Elements (Fe, Cu, Sn, c),

bronze (CuSn alloy), amalgams

(Hg+ other metals)



Amorphous

solid



Atoms, ions, or

molecules (including

polymer molecules)



Any of the above



Noncrystalline; no sharp melting

point; able to flow (may be very slow);

curved edges when shattered



Glasses, tar, some plastics



Ice, wax, frozen CO2, all solid

organic compounds

Diamond, quartz 1SiO2 2,

tungsten carbide (WC)



www.downloadslide.net

SECTION 8.14



Most substances are more dense as solids than as liquids because molecules are

more closely packed in the solid than in the liquid state. Water, however, is unique.

Liquid water has a maximum density of 1.000 g> mL at 277.13 K (3.98 °C) but then

becomes less dense as it cools. When it freezes, its density decreases still further to

0.917 g>mL.

As water freezes, each molecule is locked into position by hydrogen bonding to four

other water molecules (Figure 8.23). The resulting structure has more open space than

liquid water, accounting for its lower density. As a result, ice floats on liquid water, and

lakes and rivers freeze from the top down. If the reverse were true, fish would be killed

in winter as they became trapped in ice at the bottom.



Changes of state Calculations



277



Water has other unique properties, including a high heat of vaporization and the highest specific heat

of any liquid. The high ∆Hvap, is of

particular importance in regulating

body temperature by dissipating heat

generated by metabolic processes

(Chapters 21 and 22).







Figure 8.23



Ice.

Ice consists of individual H2O

molecules held rigidly together in an

ordered manner by hydrogen bonds. The

open, cage-like crystal structure shows

why ice is less dense than liquid water.



8.14 Changes of State Calculations

Learning Objective:

• Calculate the total amount of heat associated with changes in physical states of a

substance.

What happens when a solid is heated? As more and more energy is added, molecules

begin to stretch, bend, and vibrate more vigorously, and atoms or ions wiggle about

with more energy. Finally, if enough energy is added and the motions become vigorous

enough, particles start to break free from one another and the substance starts to melt.

Addition of more heat continues the melting process until all particles have broken free

and are in the liquid phase. The quantity of heat required to completely melt a substance

once it reaches its melting point is called its heat of fusion. After melting is complete,

further addition of heat causes the temperature of the liquid to rise.

The change of a liquid into a vapor proceeds in the same way as the change of a

solid into a liquid. When you first put a pan of water on the stove, all the added heat

goes into raising the temperature of the water. Once the water reaches its boiling point,

further absorbed heat goes into freeing molecules from their neighbors as they escape

into the gas state. The quantity of heat needed to completely vaporize a liquid once it

reaches its boiling point is called its heat of vaporization. A liquid with a low heat of

vaporization, like rubbing alcohol (isopropanol), evaporates rapidly and is said to be

volatile. If you spill a volatile liquid on your skin, you will feel a cooling effect as it

evaporates because it is absorbing heat from your body.



Heat of fusion The quantity of heat

required to completely melt 1 g of a

substance once it has reached its melting point.



Heat of vaporization The quantity of

heat needed to completely vaporize 1 g

of a liquid once it has reached its boiling point.



www.downloadslide.net

278



CHAPTER 8



Gases, Liquids, and solids



It is important to know the difference between heat that is added or removed to

change the temperature of a substance and heat that is added or removed to change the

phase of a substance. Remember that temperature is a measure of the kinetic energy

in a substance (see Section 7.1). When a substance is above or below its phase-change

temperature (i.e., melting point or boiling point), adding or removing heat will simply

change the kinetic energy and, hence, the temperature of the substance. The amount of

heat needed to produce a given temperature change was presented previously (Section

1.11) but is worth presenting again here.

Heat 1J2 = Mass 1g2 * Temperature change 1°C2 * Specific heat a



J

b

g * °C



In contrast, when a substance is at its phase-change temperature, heat that is added

is being used to overcome the intermolecular forces holding particles in that phase.

The temperature remains constant until all particles have been converted to the next

phase. The energy needed to complete the phase change depends only on the amount

of the substance and the heat of fusion (for melting) or the heat of vaporization (for

boiling).

J

Heat 1J2 = Mass 1g2 * Heat of fusion a b

g



J

Heat 1J2 = Mass 1g2 * Heat of vaporization a b

g



If the intermolecular forces are strong then large amounts of heat must be added to

overcome these forces and the heats of fusion and vaporization will be large. Table 8.5

gives a list of heats of fusion and heats of vaporization for some common substances.

Butane, for example, has a small heat of vaporization since the predominant intermolecular forces in butane (dispersion) are relatively weak. Water, on the other hand, has

a particularly high heat of vaporization because of its unusually strong hydrogen bonding interactions. Thus, water evaporates more slowly than many other liquids, takes

a long time to boil away, and absorbs more heat in the process. A so-called heating

curve, which indicates the temperature and state changes as heat is added, is shown in

Figure 8.24.

table 8.5  Melting Points, Boiling Points, Heats of Fusion, and Heats of Vaporization of Some

Common Substances

Substance



Melting

Point (K)



Boiling

Point (K)



Ammonia



195.3



239.6



Butane



134.6



272.5



Ether



157



307.6



Ethanol



155.7



351.5



Isopropanol



183.5



Sodium



370.8



Water



273



355.4



Heat of Fusion

(J>g)

351



Heat of Vaporization

(J>g)

1370



80.3



387



98.3



358



109

89.5



837

665



1156



113



4250



373



333



2260



www.downloadslide.net

SECTION 8.14



Changes of state Calculations





Temperature (K)



323



Liquid



273



Melting point



∆H (fus)



223

173



Vapor



∆H (vap)



Figure 8.24



A heating curve for water, showing

the temperature and state changes

that occur when heat is added.

The horizontal lines at 273 K (0 °C)

and 373 K (100 °C) represent the heat

of fusion and heat of vaporization,

respectively. The sloped lines represent

temperature changes resulting from

absorbed heat relative to the specific

heat of the substance in a given phase.



423

373 Boiling point



Solid

0



2



4



6



8



10



12



14



Heat added (kJ/mol)



Worked Example 8.12 heat of Fusion: Calculating total heat of melting

Naphthalene, an organic substance often used in mothballs, has a heat of fusion of 149 J>g and a molar mass

of 128.0 g>mol. How much heat in kilojoules is required to melt 0.300 mol of naphthalene?

AnAlySiS The heat of fusion tells how much heat is required to melt 1 g. To find the amount of heat needed to



melt 0.300 mol, we need a mole-to-mass conversion.

BAllPArK EStiMAtE Naphthalene has a molar mass of 128.0 g>mol, so 0.300 mol has a mass of about one-



third this amount, or about 40 g. Approximately 150 J is required to melt 1 g, so we need about

40 times this amount of heat or 150 * 40 = 6000 J = 6.0 kJ.



Solution

Heat of fusion = 149 J>g

StEP 1: Identify known information.

We know heat of fusion (cal>g) and the Moles of naphthalene = 0.300 mol

number of moles of naphthalene.

StEP 2: Identify answer and units.

StEP 3: Identify conversion factors.



First, convert moles of naphthalene to

grams using the molar mass (128 g>mol)

as a conversion factor. Then use the heat

of fusion as a conversion factor to

calculate the total heat necessary to

melt the mass of naphthalene.

StEP 4: Solve. Multiplying the mass of



naphthalene by the heat of fusion then

gives the answer.



Heat = ?? cal or J



10.300 mol naphthalene2 a



Heat of fusion = 149 J>g



(38.4 g naphthalene2 a



279



128.0 g

b = 38.4 g naphthalene

1 mol



149 J

b = 5720 J = 5.72 kJ

1 g naphthalene



BAllPArK CHECK The calculated result agrees with our estimate (6.0 kJ)



ProBlEM 8.23

How much heat in kilojoules is required to (a) melt and (b) boil 1.50 mol of

isopropanol (rubbing alcohol; molar mass = 60.0 g>mol)? The heat of fusion and heat

of vaporization of isopropanol are given in Table 8.5.



Tài liệu bạn tìm kiếm đã sẵn sàng tải về

9 Avogadro’s Law: The Relation between Volume and Molar Amount

Tải bản đầy đủ ngay(0 tr)

×