chapter 9 the gaseous state - bakersfield college notes...2 4 effect of temperature and density 9 -...

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Chapter 9The Gaseous State

• The Behavior of Gases

• Factors that Affect the Properties of

Gases

• The Ideal Gas Law

• Kinetic-Molecular Theory of Gases

• Gases and Chemical Reactions

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Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

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Properties of Gas Molecules• Gases consist of particles that are

relatively far apart (as compared to liquids and solids).– This results in observed properties:

• Lower densities• Compressible by applying external pressure

• Gas particles move about rapidly.

• Gas particles have little effect on one another unless they collide.– When they collide, gas particles simply

bounce off one another.

• Gases expand to fill containers.– Take volume and shape of their containers

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Effect of Temperature and Density

• All gases expand if heated and contract if cooled.– Heat increases the kinetic energy of gas particles,

making them move faster and farther apart.

• Since the gas particles move farther apart, there are fewer particles within a given volume.– Therefore, warm gases have lower densities.

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Effect of Temperature and Density

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Pressure• Is amount of force

applied per unit

area

P = force

area

• For a gas in a

container:

P = force of gas particles

area of container9 -

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Pressure

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Figure 9.9

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Pressure

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Pressure• Pressure is

measured using a

barometer.

• Units of pressure

1 atm = 760 mm Hg

(mm Hg and torr are

the same)

1 atm = 14.7 lb/in2 (psi)

1 atm = 101, 325 Pa9 -

Figure 9.11

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Practice – Pressure Conversions

• The pressure of a gas is 8.25 x 104 Pa. What is this pressure expressed in units

of atm and torr?

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Practice Solutions – Pressure Conversions

• The pressure of a gas is 8.25 x 104 Pa. What is this pressure expressed in units of atm and torr?

• Knowing the conversion between atm and Pa is: 1 atm = 101, 325 Pa

8.25 x 104 Pa x 1 atm = 8.14 x 10-1 atm101,325 Pa

• Knowing the conversion between atm and torr is: 1 atm = 760 torr

8.14 x 10-1 atm x 760 torr = 6.19 x 102 torr

1 atm

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Factors That Affect Gases• Volume

– Measured in liters (L)

• Pressure

– Measured in atmospheres (atm)

• Temperature

– Measured in Kelvin (K)

• Amount of particles

– Measured in moles

• An ideal gas is a gas that behaves according to predicted linear relationships

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Volume vs. Pressure

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Figure 9.14

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Volume vs. Pressure• Boyle’s law

– For a given mass of gas at constant temperature, volume varies inversely with pressure.

V ∞ 1

P

P1V1 = P2V2

• As volume increases, pressure decreases.

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Figure 9.15

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Volume vs. Temperature

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Volume vs. Temperature• Charles’s law

– For a given mass of gas at constant pressure, volume is directly proportional to temperature on an absolute (kelvin) scale.

TK = T°C + 273.15

V ∞ T (in K)

V1 = V2

T1 T2

• As volume increases,

temperature increases.9 -

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Practice – Boyle’s and Charles’s Law

• What pressure is needed to compress 455 mL of oxygen gas at 2.50 atm to a volume of 282 mL?

• A sample of carbon monoxide gas occupies 150.0 mL at 25.0°C. It is then cooled at constant pressure until it occupies 100.0 mL. What is the new temperature in degrees Celsius?

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Practice Solutions – Boyle’s and Charles’s Law

• What pressure is needed to compress 455 mL of oxygen gas at 2.50 atm to a volume of 282 mL?

P1 = 2.50 atm P2 = ?

V1 = 455 mL V2 = 282 mL

Using Boyle’s Law:

P1V1 = P2V2

2. Solving for P2: P1V1 = P2

V2

P2 = (2.50 atm x 455 mL) = 4.03 atm

282 mL9 -

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Practice Solutions – Boyle’s and Charles’s Law

• A sample of carbon monoxide gas occupies 150.0 mL at 25.0°C. It is then cooled at

constant pressure until it occupies 100.0 mL. What is the new temperature in degrees

Celsius?

1. You MUST convert temperatures to Kelvin when dealing with gases – ALWAYS!

T1 = 25.0°C + 273.15 = 298.2 K

T2 = ?

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Practice Solutions – Charles’s LawV1 = 150.0 mL V2 = 100.0 mL

T1 = 298.2 K T2 = ?

2. Using Charles’s Law:

V1 = V2

T1 T2

3. Solving for T2: T2 = V2T1

V1

T2 = (100.0 mL x 298.2 K) = 198.8 K

150.0 mL

T2 (in °C) = 198.8 K – 273.15 = -74.4°C

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Volume, Pressure and Temperature

• Combined gas law

– For a constant amount of gas, volume is proportional to absolute temperature

divided by pressure.

V ∞ (T/P)

P1V1 = P2V2

T1 T2

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Combining Volumes• Gay-Lussac’s law of combining volumes

– Gases combine in simple whole-number volume proportions at constant temperature

and pressure.

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Figure 9.19

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Volume vs. Amount• Avogadro’s hypothesis

– The volume occupied by a gas at a given temperature and pressure is directly proportional to the number of gas particles and thus to the moles of gas.

V ∞ n

V1 = V2

n1 n2

• As volume increases, the moles of gas increase.

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Volume vs. Amount• Avogadro’s hypothesis Cont’d

– At a given pressure and temperature, equal volumes of all gases contain equal numbers of moles (or particles).

• This hypothesis was measured at standard temperature and pressure (STP):

0°C and 1 atm

• The volume of an ideal gas at STP is called its molar volume:

1 mole of gas = 22.414 L9 -

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Practice – The Combined Gas Law and Avogadro’s Hypothesis

• A sample of hydrogen gas occupies

1.25 L at 80.0°C and 2.75 atm. What volume will it occupy at 185°C and

5.00 atm?

• If a balloon initially contains 0.35 mol of gas and occupies 8.2 L, what is the

volume of the balloon when 1.2 mol of gas is present if temperature and

pressure remain constant?9 -

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Practice Solutions – The Combined

Gas Law and Avogadro’s Law

• A sample of hydrogen gas occupies 1.25 L at 80.0°C and 2.75 atm. What volume will it occupy at 185°C and 5.00 atm?

1. First, convert the temperatures into Kelvin:

T1 = 80.0°C + 273.15 K = 353.2 K

T2 = 185°C + 273.15 K = 458 K

V1 = 1.25 L V2 = ?

T1 = 353.2 K T2 = 458 K

P1 = 2.75 atm P2 = 5.00 atm

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Practice Solutions – The Combined Gas Law

V1 = 1.25 L V2 = ?

T1 = 353.2 K T2 = 458 K

P1 = 2.75 atm P2 = 5.00 atm

2. The equation we want to use is the Combined Gas Law:

P1V1 = P2V2

T1 T2

3. Solving for V2: V2 = P1V1T2

T1P2

V2 = (2.75 atm x 1.25 L x 458 K) = 0.891 L

(353.2 K x 5.00 atm)9 -

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Practice Solutions – The Combined

Gas Law and Avogadro’s Law

• If a balloon initially contains 0.35 mol of gas and occupies 8.2 L, what is the volume of the balloon when 1.2 mol of gas is present if temperature and pressure remain constant?

n1 = 0.35 mol n2 = 1.2 molV1 = 8.2 L V2 = ?

1. Using Avogadro’s Hypothesis equation:V1 = V2

n1 n2

2. Solving for V2:V2 = V1n2 = (8.2 L x 1.2 mol) = 28 L

n1 0.35 mol9 -

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Summary – The Gas Laws• Boyle’s law

V ∞ 1 at constant T, n

P• Charles’s law

V ∞ T at constant P, n

• Avogadro’s hypothesisV ∞ n at constant T, P

• Combining them all into one proportionality:

V ∞ nTP

• To express this as an equality, we use a constant:

V = constant x nTP 9 -

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The Ideal Gas LawV = constant x nT

P• This constant is called the ideal gas constant,

R:V = RnT

P

• This gives us the Ideal Gas Law:

PV = nRT• R is calculated at STP using the molar volume

of a gas:

R = (1.000 atm x 22.414 L) = 0.08206 L atm

(1.000 mol x 273.15 K) K mol9 -

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Practice - Calculations with the Ideal Gas Law

• The volume of an oxygen cylinder used as a portable breathing supply is 2.025 L. When the cylinder is empty at 29.2°C, it has a pressure of 723 torr. How many moles of oxygen gas remain in the cylinder?

• The volume of an oxygen cylinder used as a portable breathing supply is 1.85 L. What mass of oxygen gas remains in the cylinder when it is empty if the pressure is 755 torr and the temperature is 18.1°C?

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Practice Solutions - Calculations with the Ideal Gas Law

• The volume of an oxygen cylinder used as a portable breathing supply is 2.025 L. When the cylinder is empty at 29.2°C, it has a pressure of 723 torr. How many moles of oxygen gas remain in the cylinder?

1. When using the Ideal Gas Law, you want to make sure the units of each variable (V, P, n, and T) match those of R:

R = 0.08206 L atm/(mol K)

T = 29.2°C + 273.15 = 302.4 K V = 2.025 L

P = 723 torr x 1 atm = 0.951 atm n = ?

760 torr9 -

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R = 0.08206 L atm/(mol K)

T = 29.2°C + 273.15 = 302.4 K V = 2.025 L

P = 723 torr x 1 atm = 0.951 atm n = ?

760 torr

2. Using the Ideal Gas Law:

PV = nRT

3. Solving for n:

n = PV = (0.951 atm x 2.025 L) = 0.0776 mol

RT ((0.08206 L atm) x 302.4 K)

mol K9 -

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• The volume of an oxygen cylinder used as a portable breathing supply is 1.85 L. What mass of oxygen gas remains in the cylinder when it is empty if the pressure is 755 torr and the

temperature is 18.1°C?

1. When using the Ideal Gas Law, you want to make sure the units of each variable (V, P, n, and T) match those of R:

R = 0.08206 L atm/(mol K) MM = 32.0 g/mol

T = 18.1°C + 273.15 = 291.3 K V = 1.85 L

P = 755 torr x 1 atm = 0.993 atm mass = ?

760 torr9 -

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R = 0.08206 L atm/(mol K) MM = 32.0 g/mol

T = 18.1°C + 273.15 = 291.3 K V = 1.85 L

P = 755 torr x 1 atm = 0.993 atm mass = ?

760 torr


PV = nRT

3. Solving for n (n is the only variable we don’t have and we will use it to get m):

n = PV = (0.993 atm x 1.85 L) = 0.0769 mol

RT ((0.08206 L atm) x 291.3 K)

mol K9 -

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4. Now we need to convert from moles (n) to mass (m):

MM = 32.0 g/mol n = 0.0769 mol

n = mass

MM

5. Solving for mass: mass = n x MM

mass = 0.0769 mol x 32.0 g = 2.46 g

1 mol9 -

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Density of a GasDensity (d) = mass (m)

Volume (V)

• We can solve for density in the Ideal Gas Law using substitution:

PV = nRT

n = m/MM

PV = mRT

MM

d = m = P(MM)

V RT

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Dalton’s Law of Partial Pressures

• Gases in a mixture behave independently and exert the same pressure they would exert if they were in a container alone.

Ptotal = PA + PB + PC + …

Ptotal = Pdry air + Pwater9 -

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Table 9.2 Vapor Pressure of Water at

Various Temperatures

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Practice – Partial Pressures

• Suppose 2.25 L of H2 gas is collected over water at 18.0°C and 722.8 torr. How many moles of H2 are produced in this reaction?

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Practice Solutions – Partial Pressures

• Suppose 2.25 L of H2 gas is collected over water at 18.0°C and 722.8 torr. How

many moles of H2 are produced in this reaction?

V = 2.25 L n = ?

T = 18.0°C + 273.15 = 291.2 K

P = 722.8 torr x 1 atm = 0.951 atm

760 torr

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Practice Solutions – Partial Pressures

V = 2.25 L R = 0.08206 L atm/(mol K)

T = 18.0°C + 273.15 = 291.2 K n = ?

P = 722.8 torr x 1 atm = 0.951 atm

760 torr


PV = nRT

2. Solving for n: n = PV

RT

n = (0.951 atm x 2.25 L) = 0.0895 mol

(0.08206 L atm x 291.2 K)

mol K 9 -

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Kinetic-Molecular Theory of Gases• A model that explains experimental

observations about gases under normal temperature and pressure conditions that we encounter in our environment

• Has 5 postulates:

1. Gases are composed of small and widely separated particles (molecules or atoms).– Low volumes of particles

– Low densities of gas

– High compressibility

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Kinetic-Molecular Theory of Gases• Postulates continued:

2. Particles of a gas behave independently of one another.– Independent movement, unless two

particles collide.

– No forces of attraction or repulsion operate between and among gas particles.

3. Each particle in a gas is in rapid, straight-line motion, until it collides with another molecule or with its container.– When collisions occur, the collisions are

perfectly elastic.

– Energy transferred from one particle to another with no net loss of energy.

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Kinetic-Molecular Theory of Gases

• Postulates continued:

4. The pressure of a gas arises from the sum of the collisions of the particles

with the walls of the container.

– The smaller the container, the more collisions between the gas particles and the walls of the container, resulting in

higher pressures.

– Predicts pressure should be proportional

to the number of gas particles.

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Kinetic-Molecular Theory of Gases• Postulates continued:

5. The average kinetic energy of gas particles depends on the absolute temperature.– Relationship between kinetic energy (KE)

and velocity (v) of the gas particles:

KEav = ½m(vav)2

– The average velocity for gas particles is greater at higher temperatures.

– Thus, if the particles move faster, they hit the walls of the container more often, resulting in higher pressure.

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• 5th Postulate continued:

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Figure 9.22

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Figure 9.23

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Diffusion• The movement of gas particles from regions of

high concentration to regions of low concentration.

KEav = ½m(vav)2

• Gases have different average velocities even though they all have the same average kinetic energy.

• Light molecules or atoms move faster than heavy molecules or atoms.

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Effusion

• The passage of a

gas through a

small opening.

• Smaller particles

effuse faster than

larger ones.

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Figure 9.25

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Gases and Chemical Reactions• Product volume from reactant

volume

– We convert from volume to moles or

moles to volume using the Ideal Gas Law as our conversion factor.

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VolumeA

Mole A Mole BVolume

B

PV=nRT PV=nRTMole

ratio

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Gases and Chemical Reactions

• Moles and mass from volume

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VolumeA

Mole A Mole BMass

B

PV=nRT Molar

Mass

Mole

ratio

MassA

Mole A Mole BMass

B

Molar

Mass

Molar

Mass

Mole

ratio

Modifying the pattern to include volume:

Remember the stoichiometry pattern we used before:

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Practice – Gas Reactions• A sample of hydrogen gas has a volume of 7.49

L at a pressure of 22.0 atm and a temperature of 32.0°C. What volume of gaseous water is produced by the following reaction at 125.0°C and 0.975 atm, if all the hydrogen gas reacts with iron(III) oxide?

Fe2O3(s) + 3 H2(g) �� 2 Fe(s) + 3 H2O(l)

• How many moles of H2SO4 form if 2.38 L of SO3

gas, measured at 65.0°C and 1.05 atm, react with sufficient water? How many grams? The balanced equation is as follows:

SO3(g) + H2O(l) �� H2SO4(l)

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Practice Solutions – Gas Reactions

• A sample of hydrogen gas has a volume of 7.49 L at a pressure of 22.0 atm and a temperature of

32.0°C. What volume of gaseous water is produced by the following reaction at 125.0°C and 0.975 atm, if all the hydrogen gas reacts

with iron(III) oxide?

Fe2O3(s) + 3 H2(g) �� 2 Fe(s) + 3 H2O(l)

Vhydrogen = 7.49 L Vwater = ?

P1 = 22.0 atm P2 = 0.975 atm

T1 = 32.0°C + 273.15 = 305.2 K

T2 = 125.0°C + 273.15 = 398.2 K

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Fe2O3(s) + 3 H2(g) �� 2 Fe(s) + 3 H2O(l)


P1 = 22.0 atm P2 = 0.975 atm

T1 = 305.2 K T2 = 398.2 K

1. First, we need to find moles of hydrogen gas. The way to do that is to use the Ideal Gas Law.

PV = nRT

nhydrogen = PV = (22.0 atm x 7.49 L) = 6.58 mol H2

RT (0.08206 L atm x 305.2 K)

mol K

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Fe2O3(s) + 3 H2(g) �� 2 Fe(s) + 3 H2O(l)


P1 = 22.0 atm P2 = 0.975 atm

T1 = 305.2 K T2 = 398.2 K

2. Then, we need to find moles of water from moles of hydrogen gas. We use a mole ratio from the balanced chemical equation to do this.

6.58 mol H2 x 3 mol H2O = 6.58 mol H2O

3 mol H2

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Fe2O3(s) + 3 H2(g) �� 2 Fe(s) + 3 H2O(l)


P1 = 22.0 atm P2 = 0.975 atm

T1 = 305.2 K T2 = 398.2 K

3. Finally, we need to find the volume of water. The way to do that is to use the Ideal Gas Law.

PV = nRTVwater = nwaterRT = (6.58 mol H2O x 0.08206 (L atm/mol K) x 398.2 K)

P (0.975 atm)

Vwater = 221 L9 -

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Practice Solutions – Gas Reactions• How many moles of H2SO4 form if 2.38 L of SO3 gas,

measured at 65.0°C and 1.05 atm, react with sufficient water? How many grams? The balanced equation is as follows:

SO3(g) + H2O(l) �� H2SO4(l)

VSO3= 2.38 L nH2SO4

= ?

T = 65.0°C + 273.15 = 338.2 K

P = 1.05 atm mH2SO4= ?

1. First, we need to find moles of SO3.

PV = nRT

nSO3= PV = (1.05 atm x 2.38 L) = 0.0900 mol SO3

RT (0.08206 L atm x 338.2 K)

mol K

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Practice Solutions – Gas ReactionsSO3(g) + H2O(l) �� H2SO4(l)

VSO3= 2.38 L nH2SO4

= ?

T = 65.0°C + 273.15 = 338.2 K

P = 1.05 atm mH2SO4= ?

2. Next, we will find moles of H2SO4. We use a mole ratio to convert from SO3 to H2SO4.

nH2SO4= 0.0900 mol SO3 x 1 mole H2SO4 = 0.0900 mol H2SO4

1 mole SO3

3. Finally, we need to find the mass of H2SO4. The conversion factor between moles and mass is Molar Mass.

mH2SO4= 0.0900 mol H2SO4 x 98.09 g H2SO4 = 8.83 g H2SO4

1 mol H2SO4

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Solving Simple Algebraic Equations Math Toolbox 9.1

• We can manipulate equations in any way that does not destroy the equality.

• Operations that maintain equalities are:– adding the same number to both sides of

the equation.

– subtracting the same number from both sides of the equation.

– multiply or dividing both sides of the equation by the same number.

– raising both sides of the equation to the same power.

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Solving Simple Algebraic Equations Math Toolbox 9.1

• For example, let’s say we need to solve for volume (V) in the Ideal Gas Law.

PV = nRT• When we solve for a variable (this time V), we

want it alone by itself on one side of the equality. To rearrange the Ideal Gas Law, we need to divide both sides by P:

PV = nRTP P

• On the left side of the equation, P is both in the numerator and the denominator; it therefore cancels out.

V = nRTP

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Graphing Math Toolbox 9.2

• Steps for graphing (by hand):

1. Examine and

organize the data to be plotted.

• Placing the data in a table may be

easiest to transfer to the graph.

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Graphing Math Toolbox 9.2• Steps for graphing (by hand):

2. Identify and label the axes.

– The dependent variable is the variable measured experimentally. Plot it on the vertical axes (the y axis).

– The independent variable is the variable controlled and varied by the experimenter. Plot it on the horizontal axis (the x axis).

– Clearly label each axis with the name of the variable and the units. Put the units either in parenthesis after the name of the variable or separated from the variable name by a comma: volume (L) or volume, L.

9 -Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display

63


3. Decide on the scales and limits of values to be plotted along each axis.

– The graph should nearly fill the graph paper.

– Adjust the scale so the data can be entered easily on the graph and read off the graph with about the same precision as they were measured.

– You don’t need to begin each axis scale with zero if it is not convenient.

– Increments on the x axis can differ from those on the y axis.

4. After selecting the scales for each axis, clearly mark the value of each major division along each axis.

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5. Place dots on the graph at the intersection of x and y values to represent a data point.

– Place a circle 1-2 mm in diameter around each data point to emphasize its location.

– Use squares or triangles to represent different sets of data.

6. Draw a smooth line through the data set.

– Do NOT connect the dots with separate straight-line segments.

– Not all data points may actually fall on the line.

– Label the completed graph with an appropriately identifying title in a clear space at the top.

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7005.17

4503.30

6004.37

5003.71

4002.95

3502.59

3252.37

3002.20

Temperature (K)

Volume (L)

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Gas volume versus absolute temperature

2.00

3.00

4.00

5.00

6.00

200 300 400 500 600 700 800

Temperature (K)

Vo

lum

e (

L)

chapter 9 the gaseous state - bakersfield college notes...2 4 effect of temperature and density 9 -...

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