Chapter 10

Gases

The Pressure-Volume Relationship Boyle′s Law

British chemist Robert Boyle (1627–1691) first investigated the relationship between the pressure of a gas and its volume, using a J-shaped tube like that shown in FIGURE 10.6. In the tube on the left, a quantity of gas is trapped above a column of mercury. Boyle then changed the pressure on the gas by adding mercury to the tube. He found that the volume of the gas decreased as the pressure increased. For example, doubling the pressure caused the gas volume to decrease to half its original value.

Fig 10.7

Boyle’s law, which summarizes these observations, states that the volume of a fixed

quantity of gas maintained at constant temperature is inversely proportional to the

pressure. When two measurements are inversely proportional, one gets smaller as the

other gets larger. Boyle’s law can be expressed mathematically as

t

cons

V

or

P

t

cons

V

tan

1

tan

=

´

=

[10.2]

The Temperature-Volume Relationship

Charles′s Law

As _ FIGURE 10.8 illustrates, the volume of an inflated balloon

increases when the temperature of the gas inside the

balloon increases and decreases when the temperature of the gas decreases.

Jacques Charles

The relationship between gas volume increases as temperature increases and decreases as temperature decreases. Some typical volume–temperature data are shown in FIGURE 10.9. Notice that the extrapolated (dashed) line passes through -273 °C. (the absolute temperature)

The Quantity-Volume relationship Avogadro′s Law:

Avogadro’s hypothesis: Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. For example, 22.4 L of

any gas at 0 (C and 1 atm contain 6.02 × 1023 gas molecules (that is, 1 mol).

Avogadro’s law : The volume of a gas maintained

at constant temperature and pressure is directly proportional to the number of moles

of the gas. That is,

V = constant × n [10.4]

where n is number of moles. Thus, for instance, doubling the number of moles of gas

causes the volume to double if T and P remain constant.

The Ideal Gas Equation

We can combine these relationships into a general

PV = nRT the ideal-gas equation

The term R in the ideal-gas equation is the gas constant. The value and units of R

depend on the units of P, V, n, and T.

R values

8.314 J mol-1K-1

0.08206 L. atm mol-1K-1

Cal mol-1K-1

8.314 m3 Pa mol-1K-1

62.36 torr mol-1K-1

The volume of the gas is 22.4 L at STP (standard conditions)

1.0 mol of an ideal gas at 1.000 atm and 0 (C (273.15 K).

Ex 10.5

The gas pressure in an aerosol can is 1.5 atm at 25 °C Assuming that the gas obeys the ideal-gas equation,

what is the pressure when the can is heated to 450 °

Solution

Ideal gas equation : PV = nRT

P1 = 1.5 atm

T1 = 298 K

T2 = 723 K

V1 = V2

P2 = ?

atm

K

K

atm

P

6

.

3

298

723

5

.

1

2

=

÷

÷

ø

ö

ç

ç

è

æ

=

PRACTICE EXERCISE 10.5

The pressure in a natural-gas tank is maintained at 2.20 atm. On a day when the temperature is –15 °C, the

volume of gas in the tank is 3.25 × 103 m3. What is the volume of the same quantity of gas on a day when

the temperature is 31 °C?

Answer: 3.83 × 103 m3

Ex 10.6

An inflated balloon has a volume of 6.0 L at sea level (1.0 atm) and is allowed to ascend until

the pressure is 0.45 atm. During ascent, the temperature of the gas falls from 22 °C to -21 °C.

Calculate the volume of the balloon at its final altitude.

From ideal gas equation :

L

K

K

atm

atm

V

T

T

P

P

V

V

T

V

P

T

V

P

11

252

295

45

.

0

0

.

1

0

.

6

2

2

1

2

1

1

2

2

2

2

1

1

1

=

´

´

=

´

´

=

=

PRACTICE EXERCISE 10.6

A 0.50-mol sample of oxygen gas is confined at and 1.0 atm in a cylinder with a movable piston. The piston compresses the gas so that the final volume is half the initial volume and the final pressure is 2.2 atm. What is the final temperature of the gas in degrees Celsius?

Further Applications of the Ideal Gas Equation

Gas densities and molar mass

The density of the gas is given by the expression:

RT

PM

V

nM

d

=

=

(10.10)

From the equation it is seen that the density of a gas depends on its pressure, molar mass,

and temperature. The higher the molar mass and pressure, the denser the gas. The higher the temperature, the less dense the gas.

Also, for equal molar masses of two gases (at the same pressure but different temperatures) the hotter gas is less dense than the cooler one, so the hotter gas rises.

Example 10.7

What is the density of carbon tetrachloride vapor at 714 torr and 25 ºC.

Solution

L

g

K

mol

g

atm

RT

PM

V

nM

d

/

42

.

4

398

0821

.

0

/

8

.

153

935

.

0

=

´

´

=

=

=

Equation 10.10 can be rearranged to solve for the molar mass of a gas:

P

dRT

M

=

[10.11]

Thus, we can use the experimentally measured density of a gas to determine the molar

mass of the gas molecules.

example 10.8

A large evacuated flask initially has a mass of 134.567 g. When the flask is filled with a gas of unknown molar mass to a pressure of 735 torr at 31 °C, its mass is 137.456 g. When the flask is evacuated again and then filled with water at 31 °C, its mass is 1067.9 g. (The density of water at this temperature is 0.997 g/mL.) Assuming the ideal-gas equation applies, calculate the molar mass of the gas.

Solution

Mass of the gas : 137.456 g - 134.567 g = 2.889 g

Mass of water = 1067.9 g - 134.567 g = 933.3 g

Volume of the flask = Volume of water = m/density = 933.3/0.997 = 936 mL = 0.936 L

Density of the gas = 2.889 g>0.936 L = 3.09 g/L

Molar mass of the gas is :

mol

g

atm

P

dRT

M

/

7

.

79

09671

.

0

08206

.

0

09

.

3

=

´

=

=

Volumes of gases in chemical reactions

From the balanced chemical equation it is possible to calculate the volumes of gases consumed

or produced in reactions. The coefficients in a balanced chemical equation tell us the relative amounts (in moles) of reactants and products in a reaction.

Example 10.9

Automobile air bags are inflated by nitrogen gas generated by the rapid decomposition of sodium azide, NaN3:

2 NaN3 (s) → Na (s) + 3 N2 (g)

If an air bag has a volume of 36 L and is to be filled with nitrogen

gas at 1.15 atm and 26 (C, how many grams of NaN3 must be decomposed?

n =PV/RT =(1.15 atm)(36 L) /(0.08206 L-atm/mol-K)(299 K) = 1.69 mol N2

3

3

3

2

3

2

3

2

73

/

0

.

65

12

.

1

NaN

,

NaN

mol

1.12

.

N

mol

3

NaN

mol

2

)

N

mol

(1.69

NaN

moles

69

.

1

299

08206

.

0

36

15

.

1

NaN

g

mol

g

of

g

So

N

mole

RT

PV

n

=

´

=

=

=

´

=

=

´

´

=

=

Gas mixtures and partial pressures

Dalton’s law of partial pressures: The total pressure of a mixture of gases equals the sum of the pressures that each would exert if it were present alone..

Pt = P1 + P2 + P3 +......

Partial pressures and mole fractions

1

1

1

1

/

/

X

n

n

V

RT

n

V

RT

n

P

P

t

t

t

=

=

=

Where X1 is the mole fraction of gas1\

P1 = X1× Pt

Example 10.11

Collecting gases over water

Decomposition of KClO3

2 KClO3(s)،2 KCl(s) + 3 O2(g)

The oxygen gas is collected in a bottle that is initially filled with water and inverted in a

water pan.

Ptotal = Pgas + PH2O

See example 10.12

The Kinetic Molecular Theory of gases

Statements

1. Gases consist of large numbers of molecules that are in continuous, random motion.

(The word molecule is used here to designate the smallest particle of any gas even though some gases, such as the noble gases, consist of individual atoms. All we learn

about gas behavior from the kinetic-molecular theory applies equally to atomic gases.)

2. The combined volume of all the molecules of the gas is negligible relative to the total volume in which the gas is contained.

3. Attractive and repulsive forces between gas molecules are negligible.

4. Energy can be transferred between molecules during collisions but, as long as temperature remains constant, the average kinetic energy of the molecules does not

change with time.

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

At any given temperature the molecules of all gases have the same average kinetic energy.

Distribution of Molecular Speed

FIGURE 10.17 (a), shows the distribution of molecular speeds for nitrogen

gas at 0 ºC and 100 ºC,we see that a larger fraction of the 100 °C molecules moves at the

higher speeds. This means that the 100 °C sample has the higher average kinetic energy.

The rms speed is important because the average kinetic energy of the gas molecules

in a sample is equal to 1/2 m(urms)2 . Because of the mass does not change with temperature, the increase in the average kinetic energy as the temperature increases

implies that the rms speed of the molecules (as well as their average speed)

increases as temperature increases.

Application of Kinetic-Molecular Theory to Gas Laws

Empirical observations of gas properties

1. An increase in volume at constant temperature causes pressure to decrease.

2. A temperature increase at constant volume causes pressure to increase.

Molecular effusion and diffusion التسرب و الإنتشار الجزيئي ( للغازات)

See example 10.14

Effusion, describes the escape of gas molecules through a tiny hole. (_ FIGURE

10.19).

Diffusion,describes the spread إنتشار of one substance throughout a space or throughout a second substance. For example, the molecules of a perfume diffuse throughout a room.

Graham΄ s Law of Fussion

If the rates of effusion of the two gases are r1 and r2 and their molar masses are and ,

Graham’s law states that

1

2

2

1

M

M

r

r

=

[10.24]

Graham΄s relation indicates that the lighter gas has the higher effusion rate.

M

RT

U

rms

2

=

[ 10.22]

From equation 10.22: The rate of effusion is directly proportional to the rms speed of the molecules. Because R and T are constant, we have, from Equation 10.22

1

2

2

1

2

1

2

1

/

3

/

3

M

M

M

RT

M

RT

U

U

r

r

rms

rms

=

=

=

REAL GASES: DEVIATIONS RFOM IDEAL BEHAVIOR

At high pressures (generally above 10 atm) the deviation from ideal behavior is large and different for each gas. Real gases, in other words, do not behave ideally at high pressure.

At lower pressures (usually below 10 atm), however, the deviation from ideal behavior is small, and we can use the ideal-gas equation without generating serious error.

Van der waals equation

Van der Waals recognized that the ideal-gas equation could be corrected to account

for the effects of intermolecular attractive forces and for molecular volumes. He introduced

two constants for these corrections: a, a measure of how strongly the gas molecules attract one another, and b, a measure of the finite volume occupied by the molecules. His description of gas behavior is known as the van der Waals equation:

The term nb accounts for the small but finite volume occupied by the gas molecules (Figure 10.24). The van der Waals equation subtracts nb to adjust the volume downward to give the volume that would be available to the molecules in the ideal case. The constants a and b, called van der Waals constants, are experimentally determined, positive quantities that differ from one gas to another.

a and b generally increase with increasing molecular mass. Larger, more massive molecules have larger volumes and tend to have greater intermolecular attractive forces.

(

)

]

27

.

10

[

2

2

nRT

nb

V

v

a

n

P

=

-

÷

÷

ø

ö

ç

ç

è

æ

+

See Example 10.16

Important Problems : 10.15, 17, 18, 21, 25, 30, 36, 38, 41, 42, 49, 53, 60-67, 72, 83, 85

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