Chapter 13 Outline

Gases and their properties

Gases and their properties

Standard #4The kinetic molecular theory

describes the motion of atoms and molecules and explains the

properties of gases.

* Nitrogen: 78%

* Oxygen: 21%

* Carbon Dioxide:

* Noble Gases trace amounts

* Water Vapor (the amount varies)

A. Air composed of several kinds of colorless gases

Relative Composition of air:

1%

B. Some gases have color

Chlorine

Nitrogen Dioxide is SMOG!

C. Any 1 mole of a gas occupies

22.4 L

Nature of Gases

1. Gases have Mass

2. Gases are easy to Compress

3. Gases fill their container

4.4. Gases DiffuseGases DiffuseMolecules travel from a high concentration to a low concentration

I smell something weird!

5. Gases Effuse

The escape of gas molecules through a tiny hole into an evacuated

space.

5. Gases Effuse

Gases Effuse:

6. Exerts Pressure on Wall of container

The pressure of the gas depends on the

temperature!

What do you think will happen to the gas and the amount of

pressure if this balloon is placed near a heat source/light?

Volume of balloon at

room temperature

Volume of balloon at

room temperature

Volume of balloon at 5°C

Volume of balloon at 5°C

The Kinetic Molecular Theory1. Gases consist of small particles

2. Gases are separated by large distances

3. Gases are in constant Rapid Motion

4. Collisions are elastic

No loss of energy with collisions. Energy is conserved

There is a loss of energy with collisions.

Measuring Pressure:A. Pressure:

Force

Area1. Pressure is high…area is

2. Pressure is low…area is

small

LARGE

Units for measuring PRESSURE

1atm = 760.mmHg

101.3kPa

14.7lb/in2

or psi

= =

End of day 1

Barometer

Stop here!

D. Instrument that measures:

A. Pressure:

Barometer

B. Temperature:

Thermometer

Units for measuring PRESSURE

1atm = 760.mmHg

101.3kPa

14.7lb/in2

or psi

= =

Boyle’s Law: P and VStates that the volume of a gas is

inversely proportional to the pressure at constant

temperature

Robert Boyle

25 January 1627 – 30 December 1691

Low pressure= High Volume High pressure= Low Volume

Constant temperature and Constant Number of gas particles

Boyle’s law animation

Jacues Charles

In 1808, Charles’ Law was

developed. It states that the

volume of a gas is directly related to its temperature at

constant pressure.

Nov. 1746 - Apr. 1823

Charles’ Law Visuals

John DaltonDalton's law of

partial pressures was stated by in

1801:

The total pressure of a mixture of gases is equal to the sum of the partial pressures of the individual component gases.  The partial

pressure is the pressure that each gas would exert if it alone occupied the volume of the mixture at the same temperature

Combined Gas Law Problems

• 500.0 mL of a gas was collected at 20.0 °C and 720.0 mm Hg. What is its volume at STP?

• 2.00 liters of hydrogen, originally at 25.0 °C and 750.0 mm of mercury, are heated until a volume of 20.0 liters and a pressure of 3.50 atmospheres is reached. What is the new temperature?

Ideal Gas Law

The Ideal Gas Law

PV = nRTPV = nRT

Ideal GasesIdeal GasesAn “ideal” gas exhibits certain theoretical

properties. Specifically, an ideal gas …• Obeys all of the gas laws under all conditions.• Does not condense into a liquid when cooled.• Shows perfectly straight lines when its V and T

& P and T relationships are plotted on a graph.In reality, there are no gases that fit this

definition perfectly. We assume that gases are ideal to simplify our calculations.

We have done calculations using several gas laws (Boyle’s Law, Charles’s Law, Combined Gas Law). There is one more to know…

The Ideal Gas LawThe Ideal Gas LawPV = nRT

P = Pressure (in kPa) V = Volume (in L)T = Temperature (in K) n = moles

R = 8.31 kPa • LK • mol

R is constant. If we are given three of P, V, n, or T, we can solve for the unknown value.

Recall, From Boyle’s Law:P1V1 = P2V2 or PV = constant

From combined gas law:P1V1/T1 = P2V2/T2 or PV/T = constant

Developing the ideal gas law equationDeveloping the ideal gas law equationPV/T = constant. What is the constant?At STP: T= 273K, P= 101.3 kPa, V= 22.4 L/mol

PV = constantT • mol

Mol is represented by n, constant by R:

PV = R Tn

Rearranging, we get: PV = nRT

Because V depends on mol, we can change equation to:

At STP: (101.3 kPa)(22.4 L) = (1 mol)(R)(273K)

R = 8.31 kPa • LK • mol

Note: always use kPa, L, K, and mol in ideal gas law questions (so units cancel)

Sample problemsSample problemsHow many moles of H2 is in a 3.1 L sample of H2

measured at 300 kPa and 20°C?

PV = nRT(300 kPa)(3.1 L) = n (8.31 kPa•L/K•mol)(293 K)

(8.31 kPa•L/K•mol)(293 K)(300 kPa)(3.1 L)

= n = 0.38 mol

How many grams of O2 are in a 315 mL container that has a pressure of 12 atm at 25°C?

P = 300 kPa, V = 3.1 L, T = 293 K

PV = nRT

(8.31 kPa•L/K•mol)(298 K)(1215.9 kPa)(0.315 L) = n = 0.1547 mol

P= 1215.9 kPa, V= 0.315 L, T= 298 K

0.1547 mol x 32 g/mol = 4.95 g

Ideal Gas Law QuestionsIdeal Gas Law Questions1. How many moles of CO2(g) is in a 5.6 L sample

of CO2 measured at STP?

2. a) Calculate the volume of 4.50 mol of SO2(g) measured at STP. b) What volume would this occupy at 25°C and 150 kPa? (solve this 2 ways)

3. How many grams of Cl2(g) can be stored in a 10.0 L container at 1000 kPa and 30°C?

4. At 150°C and 100 kPa, 1.00 L of a compound has a mass of 2.506 g. Calculate its molar mass.

5. 98 mL of an unknown gas weighs 0.087 g at SATP. Calculate the molar mass of the gas. Can you determine the identity of this unknown gas?

P=101.325 kPa, V=5.6 L, T=273 K PV = nRT

(101.3 kPa)(5.6 L) = n (8.31 kPa•L/K•mol)(273 K)

1. Moles of CO2 is in a 5.6 L at STP?

(8.31 kPa•L/K•mol)(273 K)(101.325 kPa)(5.6 L)

= n = 0.25 mol

2. a) Volume of 4.50 mol of SO2 at STP.

P= 101.3 kPa, n= 4.50 mol, T= 273 K PV=nRT

(101.3 kPa)(V)=(4.5 mol)(8.31 kPa•L/K•mol)(273 K)

(101.3 kPa)

(4.50 mol)(8.31 kPa•L/K•mol)(273 K)V = = 100.8 L

2. b) Volume at 25°C and 150 kPa (two ways)?Given: P = 150 kPa, n = 4.50 mol, T = 298 K

(150 kPa)

(4.50 mol)(8.31 kPa•L/K•mol)(298 K)V = = 74.3 L

From a): P = 101.3 kPa, V = 100.8 L, T = 273 KNow P = 150 kPa, V = ?, T = 298 K

P1V1

T1

=P2V2

T2

(101.3 kPa)(100 L)(273 K)

=(150 kPa)(V2)

(298 K)(101.3 kPa)(100.8 L)(298 K)

(273 K)(150 kPa)=(V2) = 74.3 L

3. How many grams of Cl2(g) can be stored in a 10.0 L container at 1000 kPa and 30°C?

PV = nRT

(8.31 kPa•L/K•mol)(303 K)(1000 kPa)(10.0 L) = n = 3.97 mol

P= 1000 kPa, V= 10.0 L, T= 303 K

3.97 mol x 70.9 g/mol = 282 g

4. At 150°C and 100 kPa, 1.00 L of a compound has a mass of 2.506 g. Calculate molar mass.

PV = nRT

(8.31 kPa•L/K•mol)(423 K)(100 kPa)(1.00 L) = n = 0.02845 mol

P= 100 kPa, V= 1.00 L, T= 423 K

g/mol = 2.506 g / 0.02845 mol = 88.1 g/mol

5. 98 mL of an unknown gas weighs 0.081 g at SATP. Calculate the molar mass.

PV = nRT

(8.31 kPa•L/K•mol)(298 K)(100 kPa)(0.098 L) = n = 0.00396 mol

P= 100 kPa, V= 0.098 L, T= 298 K

g/mol = 0.081 g / 0.00396 mol = 20.47 g/mol

It’s probably neon (neon has a molar mass of 20.18 g/mol)

Determining the molar mass of butaneDetermining the molar mass of butaneUsing a butane lighter, balance, and graduated

cylinder determine the molar mass of butane.• Determine the mass of butane used by

weighing the lighter before and after use.• The biggest source of error is the mass of H2O

remaining on the lighter. As a precaution, dunk the lighter & dry well before measuring initial mass. After use, dry well before taking final mass. (Be careful not to lose mass when drying).

• When you collect the gas, ensure no gas escapes & that the volume is 90 – 100 mL.

• Place used butane directly into fume hood.• Submit values for mass, volume, & g/mol.