Gases

Chapter 14

Properties of Gases as 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 Theory

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 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 Boyle’s Law

Pressure (kPa) • As pressure increases, volume decreases.

Volume (L) • P * V = constant • So, P 1 V 1 = P 2 V 2

Measuring 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 Chemistry class

• Pascals (Pa) or kilopascals (kPa) • 1 Pa = the amount of force 1 kg exerts on 1 m 2 for every 1 s 2 .

1 Pa = 1 x 10 -6 atm

• 1 kPa = the amount of force 1000 kg exerts on 1 m 2 for every 1 s 2 .

1 kPa = 1 x 10 -3 atm

Boyle’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?

P 1 V 1 = P 2 V 2 P 1 = 103 kPa V P 2 1 = 20.0 L = 15.0 kPa V 2 = ?

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

Charles’ 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 of Charles’ Law

Volume (L) • As temperature increases, volume increases.

Temperature (K) • So, V/T = constant

V

1

T

1 

V

2

T

2

Measuring 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 Celsius and Kelvin

• Kelvin, K = o C + 273 So, 32 o C = 32 + 273 = 305 K • 100 o C = 100 + 273 = 373 K

Charles’ 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?

Make a

qualitative

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

V

1

T

1 

V

2

T

2 1 .

00

L

273

K



V

2 300

K

Gay 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.ht

ml

Symbolic Representation of Gay Lussac’s Law

Pressure (kPa) • As temperature increases, pressure increases.

• So, P/T = constant Temperature (K)

P

1

T

1 

P T

2 2

Gay 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?

Step #1: Qualitative Assessment of the Problem Step #2: Quantitative Assessment of the Problem

P

1

T

1 

P T

2 2 1 , 000

kPa

 293

K P

2 323

K

Combined Gas Law • Assume the number of gas molecules remains constant,

temperature

,

pressure

and

volume

related.

are

P

1

T V

1 1 

P

2

V

2

T

2

Combined Gas Law Example • A sample of gas has a volume of 283 mL at 25 o C and 0.500 atm pressure. What is the volume at 100 o C and 1.00 atm?

P

1

V

1

T

1 

P

2

V

2

T

2 V 2 = 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 Avogadro’s Law

Volume (L) • As number of particles increases, volume increases.

Number of Particles • So, V/n = constant

V

1

n

1 

V

2

n

2

Ideal Gas Law

• An

ideal gas

collisions

is defined as one in which all 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 Ideal 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 Law

• 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 Ideal 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 ( S tandard T emperature and P ressure) occupies 22.4 L.

Ideal Gas Law Example

• What volume is needed to store 0.050 moles of helium gas at 202.6 kPa and 400 K? PV = nRT P = 202.6 kPa (202.6)(V) = (0.050)(8.314)(400) n = 0.050 mol T = 400K (202.6)(V) = 166.28

V = ? L R = 8.314 J/mol*K V=166.28/202.6

V = 0.821

Another Example

• What pressure will be exerted by 20.16 g hydrogen gas in a 7.5 L cylinder at 20 o C?

What are we trying to find?

Pressure!

Do we need to change any units?

Celsius



Kelvin

20 C + 273 = 293

Mass:

Grams

mass

 moles = 20.16 g

molar mass of H 2

= 2 x 1.008 =2.016 g/mol

What do we know?

V= 7.5 L T= 20 Celsius R= 8.314 J/mol*K n= 20.16 g 20 .

16

g

 1

mole

2 .

016

g

 10

mol

NOW SOLVE!