Gases

Chapter 13

Kinetic-Molecular Theory of Matter

• Model for gases

• Explains why gases behave the way that they do

• Based on the idea that particles of matter are always in motion.

Kinetic-Molecular Theory of Gases

• Provides a model for the ideal gas

• An “ideal gas” is an imaginary gas that perfectly fits all the assumptions of the kinetic-molecular theory.

5 Assumptions of the KMT of gases

1. Gases consist of large numbers of tiny particles that are far apart relative to their size

• Explains the lower density of gases compared with solids and liquids

• Explains why gases are easily compressed

5 Assumptions of the KMT of gases cont…

2. Collisions between gas particlesand container walls are elastic collisions

• Elastic collision- collision in which there is no net loss of kinetic energy

5 Assumptions of the KMT of gases cont…

3. Gas particles are in continuous, rapid, random motion. They posses kinetic energy.

4. There are no forces of attraction or repulsion between gas particles.

5. The average kinetic energy of gas particles is directly proportional to the Kelvin temperature of the gas.

Kinetic-Molecular Theory and the Nature of Gases

• Real gases behave like ideal gases at low pressure and high temperature

• Expansion: Gases do not have a definite shape or volume– Completely fill any container in which enclosed

Kinetic-Molecular Theory and the Nature of Gases

• Characteristics of Gases

– Fluidity: Gas particles easily glide past one another– Low Density: density of a substance in the gaseous

state = about 1/1000 the density of the same substance in the solid/liquid state

– Compressibility: gas particles can be easily compressed under pressure to increase the amount of gas stored in a container

KMT and the Nature of GasesDiffusion and Effusion

Diffusion: Spontaneous mixing of the particles of two substances caused by random motion

• Example- removing the cap from a closed container of ammonia will cause the ammonia gas to mix with the air and spread throughout the room

• Rate of diffusion depends on three properties of the gas particles involved:

– Speed– Diameter– Attractive forces present

Effusion: process by which gas particles pass through a tiny opening

Deviations of Real Gases from Ideal Behavior

• KMT applies only to ideal gases

• Real Gas: a gas that does not behave completely according to the assumptions of the kinetic-molecular theory.**No gas does!!

13.1 Pressure• Pressure and Force

– Pressure (P): defined as the force per unit area on a surface.

• P = force / area• SI Unit for force: Newton (N)

– Depends on the volume, temperature and number of molecules present

– Example: • Gas in a small container (low volume) at a very high

temperature… will the pressure be high or low?– High temperatures make the gas molecules move very quickly and

because the gas is in a small container there will be many collisions between the gas molecules and container walls creating high pressure

Pressure• Measuring Pressure

– Barometer: device used to measure atmospheric pressure

• Introduced by Torricelli in the early 1600’s• Placed a glass rod in a tub of Mercury (Hg). • When placed in the mercury the glass rod would

fill to 760 mm above the dish. • This measurement would become the standard

for atmospheric pressure -- 760 mm Hg

– Manometer: device used to measure the pressure of an enclosed gas sample

Pressure• Units of pressure

1. Millimeter of Mercury (mm Hg) - 1 mm Hg is now called 1 torr, honoring Torricelli- A.P at 0°C = 760 mm Hg or torr

2. Atmosphere of Pressure (atm) - exactly equal to 760 mm Hg

- 1 atm = 760 mm Hg - A.P. at 0°C = 1 atm

3. Pascal (Pa) - the pressure exerted by a force of one newton (1N) acting

on an area of one square meter - A.P. at 0°C = 1.01325 x 105 Pa or 101.3 KPa

PressureConversion Factors:

1.00 atm = 760. mm Hg760. torr1.01325 x 105 Pa101.3 KPa

**Hint: write these down on a separate sheet of paper entitled

CONVERSIONS, you will be permitted to use this on quizzes and tests

Standard Temperature and Pressure

• For purposes of comparison, scientists have agreed on standard conditions of…

– Pressure = 1.00 atm– Temperature = 0°C = 273K

• Abbreviated STP

Example Problems• The average atmospheric pressure in Denver is

0.830 atm. Express the pressure in mm Hg and kPa. **Hint: look at your conversion factors

– atm → mm Hg: Use times sign fraction bar!• Start with given: 0.830 atm x 760 mm Hg

1 atm

= 631 mm Hg

– atm → kPa: Use times sign fraction bar!• Start with given: 0.830 atm x 101.325 kPa

1 atm

= 84.1 kPa

You Try!!• Complete the following problems on your outline or on a separate sheet of paper

using your pressure conversion factors:

1. Convert a pressure of 748 torr to mmHg748 mmHg

2. Convert a pressure of 500. torr to atm and kPa0.658 atm 66.6 kPa

3. Convert a pressure of 1.87 atm to torr1420 torr

4. Convert a pressure of 1.75 atm to kPa and mmHg177 kPa 1330 mmHg

13.2 Boyles Law

• Robert Boyle discovered that gas pressure and volume are related mathematically

• Gas laws deal with: volume, temperature, pressure and amount of a gas

Boyles Law• Boyles Law: Pressure-Volume Relationship

• Boyles Law: the volume of a fixed mass of a gas is inversely related to the pressure at constant temperature

• Mathematically: P1V1 = P2V2

– P1V1 : initial conditions

– P2V2 : new conditions

• Used: To compare the changing conditions of a gas

Boyles Law

Use KMT to understand why this occurs:

Using Boyles Law• Example 1: Suppose that 1.0 L of gas is initially at 1.0

atm of pressure (V1 = 1.0 L P1 = 1.0 atm). The gas is allowed to expand fivefold at constant temperature to 5.0 L (V2 = 5.0 L), calculate the new pressure.

P1V1 = P2V2 → (1.0 atm x 1.0 L) = P2 x 5.0L

1.0 atm x 1.0 L = P2

5.0 L

P2 = 0.20 atm

You Try!!• Complete the following problems on your outline or

on a separate sheet of paper using Boyle’s Law:1. A sample of oxygen gas has a volume of 150 mL when its pressure is 0.947

atm. What will the volume of the gas be at a pressure of 0.987 atm if the temperature remains constant?

140 mL2. A gas has a pressure of 1.26 atm and occupies a volume of 7.40 L. If the gas is

compressed to a volume of 2.93 L, what will its pressure be, assumingconstant temperature?

3.18 atm3. A sample of gas collected in a 350 cm3 container exerts a pressure of 103

kPa. What would be the volume of this gas at 150 kPa of pressure? (Assume that the temperature remains constant.)

240 cm3

Charles Law

• Charles Law: Volume-Temperature Relationship

• Discovered: At constant pressure, when the temperature of gas molecules increase, the volume of the gas increases as well

Charles Law

• Charles Law: the volume of a fixed mass of a gas at constant pressure varies directly with the Kelvin temperature– Kelvin (K) = 273.15 + °C– Kelvin scale gets rid of (-) temperatures

• Mathematically: V1 / T1 = V2 / T2 – V1 and T1 = initial conditions

– V2 and T2 = new conditions

Charles Law• Example 1: A sample of neon gas occupies a volume of 752 mL at 25°C. What volume will the gas occupy at 50°C if the pressure remains constant?

V1 / T1 = V2 / T2 __________________________________________________________________________________________________________________________________________________________

First- Identify each variable: given and unknown V1 = 752 mL V2 = ?

T1 = 25°C + 273K = 298K T2 = 50°C + 273K = 323K__________________________________________________________________________________________________________________________________________________________

752 mL / 298K = V2 / 323K

(752 mL / 298K) 323K = V2

V2 = 815 mL Ne

You Try!!• Complete the following problems on your outline or on a separate

sheet of paper using Charles Law:

1. A helium filled balloon has a volume of 2.75 L at 20.0°C. The volume ofthe balloon decreases to 2.46 L after it is placed outside on a cold day. What is the outside temperature in K? In °C?

262K -11°C

2. A gas is collected and found to fill 2.85 L at 25.0°C. What will be itsvolume at standard temperature? (**look at your notes on STP)

2.61L

3. 4.40 L of a gas is collected at 50.0°C. What will be its volume uponcooling to 25.0°C?

4.06L

Gay-Lussac’s Law

• Pressure-Temperature Relationship

• Gay Lussac’s Law: the pressure of a fixed mass of gas at constant volume varies directly with the Kelvin temperature

• Mathematically: P1 / T1 = P2 / T2

Gay-Lussac’s Law• Example 1: The gas in an aerosol can is at a pressure of 3.00 atm at 25°C.

Directions on the can warn the user not to keep the can in a place where the temperature exceeds 52°C. What would the gas pressure on the can be at 52°C?

P1 / T1 = P2 / T2 ___________________________________________________

First- Identify each variable: given and unknown P1 = 3.00 atm P2 = ? T1 = 25°C + 273K = 298K T2 = 52°C + 273K = 325K

____________________________________________________ 3.00 atm / 298K = P2 / 325K

(3.00atm / 298K) 325K = P2

P2 = 3.27 atm

You Try!!• Complete the following problems on your outline or on a separate

sheet of paper using Gay-Lussac’s Law:

1. A sample of helium gas has a pressure of 1.20 atm at 22.0°C. At what Celsius temperature will the helium reach a pressure of 2.00 atm?

219°C

2. Determine the pressure change when a constant volume of gas at 1.00 atm is heated from 20.0 °C to 30.0 °C.

1.03 atm… pressure change = 1.03 – 1.00 = 0.03 atm

3. A gas has a pressure of 699.0 mm Hg at 40.0 °C. What is the temperature at standard pressure?

340.K

Avogadro’s Law

• Volume-Mole Relationship

• Avogadro’s Law: for a gas at constant temperature and pressure, the volume is directly proportional to the number of moles of gas.

• Mathematically: V1 / n1 = V2 / n2

Avogadro’s Law• Example 1: Suppose we have a 12.2L sample containing 0.50 mol of

O2 gas at a pressure of 1atm and a temperature of 25°C. If all of this O2 is converted to ozone, O3, at the same temperature and pressure, what will be the volume of the ozone be if 0.33 mol is formed?

• Equation: 3O2 (g) 2O3 (g)

V1 / n1 = V2 / n2 ___________________________________________________

First- Identify each variable: given and unknown V1 = 12.2 L V2 = ? n1 = 0.50 mol n2 = 0.33 mol ___________________________________________________

Avogadro’s Law

• Example 1 cont.Determine n2:

0.50mol O2 x (2 mol O3 / 3 mol O2) = 0.33 mol O3

Plug into equation:12.2 L / 0.50 mol = V2 / 0.33 mol

Solve for V2: V2 = 8.1 L

You Try!

• Consider two samples of N2 (g). Sample 1 contains 1.5 mol N2 and has a volume of 36.7 L at 25°C and 1 atm. Sample 2 has a volume of 16.5 L at 25°C and 1 atm. Calculate the number of moles of N2 in sample 2.

The Combined Gas Law

• The combined gas law: P1V1 / T1 = P2V2 / T2

– From this equation, any value can be calculated as long as the other five values are known

• Preferred Units:– P = atm– V = L or mL– T = K

• MEMORIZE THIS EQUATION!!!

The Combined Gas Law• Example 1: A helium filled balloon has a volume of 50.0 L at 25°C

and 1.08 atm. What volume will this balloon have at 0.855 atm and 10°C? P1V1 / T1 = P2V2 / T2

__________________________________________________________________________________________________________________________________________________________________________________

First- Identify each variable: given and unknown P1 = 1.08 atm P2 = 0.855 atm T1 = 25°C + 273K = 298K T2 = 10°C + 273K = 283K V1 = 50.0 L V2 = ?

_________________________________________________________________________________________

(1.08 atm x 50.0 L) / 298 K = (0.855 atm x V2) / 283 K

[(1.08 atm x 50.0 L) / 298 K] x 283 K = 0.855 atm x V2

{ [(1.08 atm x 50.0 L) / 298 K] x 283 K } / 0.855 atm = V2

P2 = 60.0 L He

You Try!!• Complete the following problems on your outline or on a separate

sheet of paper using the Combined Gas Law:

1. A 350. mL air sample collected at 35.0°C has a pressure of 550. torr. What pressure will the air exert if it is allowed to expand to 425 mL at 57.0°C?

485 torr

2. A gas has a volume of 800. mL at -23.00 °C and 300.0 torr. What would the volume of the gas be at 227.0 °C and 600.0 torr of pressure?

800. mL

3. A gas sample occupies 3.25 liters at 24.5 °C and 1825 mm Hg. Determine the temperature at which the gas will occupy 4250 mL at 1.50 atm.

243 K

The Ideal Gas Law

• Combination of Boyle’s, Charles, and Gay-Lussac’s Laws

• Uses universal gas constant: R– Always has the value 0.08206 L • atm / mol • K

• Written: PV = nRT– P: pressure– V: volume– n: number of moles– R: constant– T: temperature

Ideal Gas Law Calculations

• A sample of hydrogen gas has a volume of 8.56 L at a temperature of 0°C and a pressure of 1.5 atm. Calculate the number of moles of H2 present in the gas sample.

• Use: PV = nRT– P = 1.5 atm n = ?– V = 8.56 L R = 0.08206 L • atm / mol • K– T = 0°C + 273 = 273 K

• Rearrange equation and solven = PV / RTn = 0.57 mol

YOU TRY!

• DO: Self-Check Exercise 13.5 on page 403 in book

• DO: Example 13.9 on page 403 in book

• DO: Self-Check Exercise 13.5 on page 403 in book

Dalton’s Law of Partial Pressures

• Dalton’s Law of Partial Pressures: the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases

• Partial pressure: the pressure of each gas in a mixture

• Daltons Law expressed: PT = P1 + P2 + P3 + P4 … – PT = total pressure – P1 + P2 + P3 + P4… = the partial pressures of component

gases 1, 2, 3, 4, and so on…

• Calculating the total pressure of a gas and water vapor inside a collection bottle is done using Dalton’s Law:

– The collection bottle is raised until the water levels inside and outside of the bottle are the same

– At this point, the total pressure inside the bottle = atmospheric pressure (Patm)

– According to Dalton’s Law:• Patm = Pgas + PH2O

– The pressure of water at the temperature it was collected would have to be found using your table

Dalton’s Law of Partial Pressures

Water Displacement• Gases produced in the laboratory are often collected over

water• The gas produced by the reaction displaces the water, which is

more dense• A gas collected by water displacement is not pure but is always

mixed with water vapor• Water vapor, like other gases, exerts pressure called water-

vapor pressure• The collection bottle is raised so that the water level inside and

outside of the container are equal and at this point the total pressure inside the collection bottle must equal Patm

• If we know the atmospheric pressure and the water temperature, we can figure out the pressure of the gas inside the

container.

Dalton’s Law of Partial Pressures• Example 1: Oxygen gas from the decomposition of KClO3 was collected by

water displacement. The barometric pressure and the temperature during the experiment were 731.0 torr and 20.0°C. What was the partial pressure of the oxygen collected?

Patm = Pgas + PH2O

First identify each variable:PT = Patm = 731.0 torr

PH2O = 17.5 torr (vapor pressure of water at 20°C)

Pgas = PO2 = ?

Plug into equation: Patm = PO2 + PH2O → PO2 = Patm - PH2O

PO2 = 731.0 torr – 17.5 torr = 713.5 torr

• Example 2: Some hydrogen gas is collected over water at 20.0°C. The levels of water inside and outside the gas-collection bottle are the same. The partial pressure of hydrogen is 742.5 torr. What is the barometric pressure at the time the gas is collected?

Patm = Pgas + PH2O

First identify each variable:PT = Patm = ?

PH2O = 17.5 torr (partial pressure of water at 20.0°C

Pgas = PH2 = 742.5 torr

Plug into equation: Patm = PH2 + PH2O

PH2O = 742.5 torr + 17.5 torr = 760.0 torr

Dalton’s Law of Partial Pressures

Dalton’s Law of Partial Pressures• Example 3: Helium gas is collected over water at 30°C. What is the

partial pressure of the helium, given that the barometric pressure is 750.0 mmHg?

Patm = Pgas + PH2O

First identify each variable:PT = Patm = 750.0 mmHg

PH2O = 31.8 mmHg (partial pressure of water at 30.0°C)

Pgas = PHe = ?

Plug into equation: Patm = PHe + PH2O → PHe = Patm - PH2O

PHe = 750.0 mmHg – 31.8 torr = 718.2 mmHg