GasesGasesChapter 14Chapter 14

Properties of Gasesas demonstrated by experiments

• Gases have mass

• Gases exert pressure

• Gases can expand to take up large volumes

• Gases can be compressed to fit in small volumes

Kinetic Molecular Kinetic Molecular TheoryTheory

• The gas consists of very small particles, all with non-zero mass.

• These molecules are in constant, random motion. The rapidly moving particles constantly collide with the walls of the container.

• The collisions of gas particles with the walls of the container holding them are perfectly elastic.

• The interactions among molecules are negligible. They exert no forces on one another except during collisions.

Boyle’s LawBoyle’s Law• Boyle’s law states that at constant

temperature, the absolute pressure and the volume of a gas are inversely proportional. Meaning, as pressure increases, volume decreases.

• For Example:http://www.grc.nasa.gov/WWW/K12/airplane/aboyle.html

Symbolic Representation of Symbolic Representation of Boyle’s LawBoyle’s Law

• As pressure increases, volume decreases.

• P * V = constant

• So, P1V1 = P2V2

Pressure

(kPa)

Volume (L)

Measuring pressureMeasuring pressure• Many ways to measure pressure of a gas.

– Atmospheres (atm) force per unit area exerted against a surface by the weight of air above that surface in the Earth’s atmosphere.

– mm Hg (mmHg) pressure of 1 mm of mercury in a column of mercury.

– torr equal to 1 mmHg – pascals or kilopascals

Measuring pressure in Measuring pressure in Chemistry classChemistry class

• Pascals (Pa) or kilopascals (kPa)

• 1 Pa = the amount of force 1 kg exerts on 1 m2 for every 1 s2.

1 Pa = 1 x 10-6 atm

• 1 kPa = the amount of force 1000 kg exerts on 1 m2 for every 1 s2.

1 kPa = 1 x 10-3 atm

Boyle’s Law ProblemBoyle’s Law Problem• A balloon contains 20.0 L of helium gas at

103kPa. The balloon rises to an altitude where the pressure is only 15.0 kPa. What happens to the volume of the balloon?

P1V1 = P2V2

P1= 103 kPaV1 = 20.0 LP2 = 15.0 kPaV2 = ?

(103kPa)(20.0L) = (15.0kPA)V2

Charles’ LawCharles’ Law• When pressure is held constant, the volume and

temperature of a gas are directly proportional. Meaning as temperature increases, volume increases.

• For example, http://www.wisc-online.com/objects/index_tj.asp?objID=GCH1704

• http://www.youtube.com/watch?v=ZvrJgGhnmJo

Symbolic Representation Symbolic Representation of Charles’ Lawof Charles’ Law

• As temperature increases, volume increases.

• So, V/T = constant

Temperature (K)

Volume (L)

2

2

1

1

T

V

T

V

Measuring TemperatureMeasuring Temperature• Fahrenheit

– (based on the freezing and melting point of water)

• Celsius – (based on the freezing and melting point of

water but in base 10)

• Kelvin – (based on the idea that at 0 Kelvin or absolute

zero, atoms are NOT in motion)

Converting between Converting between Celsius and KelvinCelsius and Kelvin

• Kelvin, K = oC + 273

So, 32oC = 32 + 273 = 305 K

• 100oC = 100 + 273 = 373 K

Charles’ Law PracticeCharles’ Law Practice• A balloon has a volume of 1.00 L. The

temperature of the room where the balloon is located is 273 K. If the room is heated to 300 K, what happens to the volume of the balloon?

2

2

1

1

T

V

T

V

Make a qualitative assessment of what is going to happen to the volume of the balloon after an increase in temperature.

K

V

K

L

300273

00.1 2

Gay-Lussac’s LawGay-Lussac’s Law• If the volume is constant, as the

temperature of the gas increases, the pressure increases. Temperature and pressure are directly proportional.

• For example, http://www.marymount.k12.ny.us/marynet/06stwbwrk/06gas/1amcslussac/amcsgaylussac.html

Symbolic Representation Symbolic Representation of Gay-Lussac’s Lawof Gay-Lussac’s Law

• As temperature increases, pressure increases.

• So, P/T = constant

Temperature (K)

Pressure (kPa)

2

2

1

1

T

P

T

P

Gay-Lussac’s PracticeGay-Lussac’s Practice• A sealed cylinder of gas contains nitrogen gas at

1,000 kPa and a temperature of 293K. When the cylinder is left in the sun, the temperature of the gas increases to 323K. What happens to the pressure in the cylinder?

2

2

1

1

T

P

T

P

Step #1: Qualitative Assessment of the Problem

K

P

K

kPa

323293

000,1 2

Step #2: Quantitative Assessment of the Problem

Combined Gas Law

• Assume the number of gas molecules remains constant, temperature, pressure and volume are related.

2

22

1

11

T

VP

T

VP

Combined Gas Law Example

• A sample of gas has a volume of 283 mL at 25oC and 0.500 atm pressure. What is the volume at 100oC and 1.00 atm?

2

22

1

11

T

VP

T

VP

V2 = 177 mL

Avogadro’s Law

If the temperature and pressure are constant, as the number of particles of the gas increases, the volume increases. Number of particles and volume are directly proportional.

Symbolic Representation of Symbolic Representation of Avogadro’s LawAvogadro’s Law

• As number of particles increases, volume increases.

• So, V/n = constantNumber of Particles

Volume

(L)

2

2

1

1

n

V

n

V

Ideal Gas LawIdeal Gas Law• An ideal gas is defined as one in which all

collisions between atoms or molecules are perfectly elastic and in which there are no intermolecular attractive forces.

• Think of a gas as a collection of perfectly hard spheres which collide but which otherwise do not interact with each other. Like Billiards!!

More about More about Ideal Gas LawIdeal Gas Law

• An ideal gas can be characterized by three variables: absolute pressure (P), volume (V), and absolute temperature (T).

• The relationship between them may be deduced from kinetic theory.

Exceptions to the Ideal Gas Exceptions to the Ideal Gas LawLaw

• at low temperatures (close to 0 K) the gas molecules have less kinetic energy (move around less) so they do attract each other.

• at high pressures (like many, many kPa) the gas molecules are forced closer together so that the volume of the gas molecules becomes significant compared to the volume the gas occupies.

Symbolic Representation Symbolic Representation Ideal Gas LawIdeal Gas Law

PV = nRT

• n = number of moles • R = universal gas constant = 8.3145 J/mol*K • One mole of an ideal gas at STP (Standard

Temperature and Pressure) occupies 22.4 L.

Ideal Gas Law ExampleIdeal Gas Law Example• What volume is needed to store 0.050

moles of helium gas at 202.6 kPa and 400 K?

P = 202.6 kPa n = 0.050 mol T = 400KV = ? L R = 8.314 J/mol*K

PV = nRT

(202.6)(V) = (0.050)(8.314)(400)

(202.6)(V) = 166.28

V=166.28/202.6

V = 0.821

Another ExampleAnother Example• What pressure will be exerted by 20.16 g

hydrogen gas in a 7.5 L cylinder at 20oC?

What are we trying to find?

Pressure!

What do we know?

V= 7.5 LT= 20 CelsiusR= 8.314 J/mol*Kn= 20.16 g

Do we need to change any units?

Celsius Kelvin20 C + 273 = 293

Mass:Grams molesmass = 20.16 gmolar mass of H2 = 2 x 1.008 =2.016 g/mol

molg

moleg 10

016.2

116.20

NOW SOLVE!