Gases

Dr. Chin Chu

River Dell Regional High School

Particles in a Solid, Liquid and Gas

Random Motion of Gas Particles

II. Gas Pressure A. Pressure is force per unit area

1. result of particle collisions

2. measured by a barometer

3. influenced by temperature, gas

volume, and the number of gas

particles

a. as the number of particle

collisions increases the pressure

increases

I. Kinetic TheoryA. Assumptions

1. gas particles do not attract each

other

2. gas particles are very small

3. particles are very far apart

4. constant, random motion

5. elastic collisions

6. kinetic energy varies with temperature

B. Properties of Gases 1. low density (grams/liter) 2. can expand and can be compressed 3. can diffuse and effuse a. rate related to molar mass b. diffusion is the movement of particles from an area of greater concentration to an area of lesser concentration c. effusion is the movement of gas particles through a small opening

B. Units of Pressure

1. pascal - 1Pa = 1N(newton)/m2

(meter squared)

2. psi (pounds/square inch)

3. mm Hg or torr

4. atm (atmospheres)

* 1 atm = 760 mm Hg = 760 torr = 101.3 KPa = 14.7 psi (pounds per square inch)

Measuring Atmospheric Pressure

Aneroid Baramoter Mercury Barometer

II. The Gas LawsA.Boyle’s Law (P1V1 = P2V2 )inverse relationship

1.As the volume of a gas increases the pressure decreases (temperature remains constant)

2. Example A sample of gas in a balloon is

compressed from 7.00 L to 3.50 L. The pressure

at 7.00L is 125 KPa. What will the pressure be at

2.50L if the temperature remains constant?

P1 = 125 KPa P2 = X V1 = 7.00L V2 = 3.50L

(125)(7.00) = (X) (3.50)

X = 250.KPa

Boyle’s Law

As volume increases the pressure decreases when temperature remains constant

Boyles Law

B. Charles’ Law V1 = V2 must use kelvin

T1 T2 temperature

1. As the temperature of a gas increases the volume increases (direct relationship)

2. Example A gas sample at 20.0 C occupies

a volume of 3.00 L. If the temperature is raised to

50.0 C, what will the volume be if the pressure

remains constant?

V1 = 3.00L V2 = X T1 = 293K T2 = 323K

3.00 = X 293X = (3)(323) X = (3)(323)

293 323 293

X = 3.31 L

Charles’ Law – Temperature increases – volume increases

Charles’ Law

C. Gay Lussac’s Law P1 = P2

T1 T2

1. as the temperature increases the pressure

increases when the volume remains constant

2. Example The pressure of a gas in a tank is

4.00 atm at 200.0C. If the temperature rises

rises to 280.0C, what will be the pressure of the

gas in the tank?

P1 = 4.00 atm P2 = X T1 = 473K T2 = 553K

4.00 = X 473X = (4)(553) X = (4)(553)

473 553 473

X = 4.68 atm

D. Combined Gas Law P1 V1 = P2 V2

T1 T2

1. Combines Boyle’s, Charle’s and Gay Lussac’s

2. Example A gas at 70.0KPa and 10.0C fills a

flexible container with an initial volume of 4.00L

If the temperature is raised to 60C and the pressure

is raised to 80.0 KPa, what is the new volume?

P1 = 70.0 KPa P2 = 80.0 KPa V1 = 4.00L V2 = X T1 = 283K T2 = 333K

(70.0)(4.00) = (80.0)(X)

283 333

X = (333)(70.0)(4.00)

(283)(80.0) X = 4.12L

E. Dalton’s Law of Partial Pressures

Ptotal = P1 + P2 + P3 + .....Pn

The total pressure of a mixture of gases is equal to

the sum of the pressures of all the gases in the

mixture

1. Example Find the total pressure for a mixture that contains four gases with partial pressures of 5.00 kPa, 4.56 kPa, 3.02 kPa and 1.20kPa.

Dalton’s Law Partial Pressures

Dalton’s Law of Partial Pressures

2. Suppose two gases in a container have a total

pressure of 1.20 atm. What is the pressure of gas

B if the partial pressure of gas A is 0.75 atm?

3. What is the partial pressure of hydrogen gas in

a mixture of hydrogen and helium if the total

pressure is 600.0mmHg and the partial pressure of

helium is 439 mmHg?

III. Avogadro’s PrincipleA. Equal volumes of gases at the same

temperature and pressure have the same number of particles

B. Molar Volume (22.4 L at STP)

1. volume of one mole of gas particles at

STP(standard temperature and pressure)

0C and 1.00 atm (760mm Hg)

* 1 mole of any gas at STP = 22.4 L

2. conversion factors 1 mol 22.4 L

22.4 L 1 mol

Avogadro’s Principle

Equal volumes of gases at the same temperature and pressure contain the same number of particles

C. Sample Problems

1. Calculate the volume occupied by .250 mol of

oxygen gas at STP.

2. Calculate the number of moles of methane gas

in a 11.2 L flask at STP.

3. Calculate the volume of 88.0 g of CO2 at STP.

4. How many grams of He are found in a 5.60L balloon at STP?

5. Calculate the density of H2 at STP.

D = molar mass

molar volume

6. Calculate the molar mass of a gas that has a

density of 3.2 g/L.

IV. Ideal vs Real Gases

A. Ideal compared to Real Gases

1. ideal gas

a) particles do not have volume

b) there are no intermolecular

attractions

c) all particle collisions are elastic

d) obey all kinetic theory

assumptions

2. real gases behave like ideal gases except when

a) pressure is very high

b) temperatures are low

c) molecules are very large

d) spaces between particles is small

(small volume)

B. Ideal Gas Law - PV = nRT

1. pressure ( atm,mm Hg, KPa)

2. volume (liters)

3. temperature (kelvin)

4. number of moles (n)

5. R = constant (L) (pressure unit*)

(mol) (K)

• unit for pressure determines whichconstant must be used in the Ideal Gas

Law PV = nRT

a) R = 62.4 (pressure is mm Hg) b) R = .0821 (pressure is atm) c) R = 8.314 (pressure is KPa)

Ideal Gas Law PV = nRT

What Principles and/or Laws are Ilustrated?

C. Application Problems (PV = nRT)

1. How many moles of O2 are in a 2.00L container

at 2.00 atm pressure and 200K?

2. Calculate the volume occupied by 2.00 mol of N2 at 300K and .800 atm pressure.

3. What is the pressure in mm Hg of .200 moles of

gas in a 5.00 L container at 27C?

4. Calculate the number of grams of oxygen in a 4.00 L sample of gas at 1.00 atm and 27 C.