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Page 1: Gas Laws. Pressure The force per unit area on a surface Standard Pressure = 1 atmosphere (atm.) Conversions: –1 atmospheres (atm) –760 millimeters of

Gas LawsGas Laws

Page 2: Gas Laws. Pressure The force per unit area on a surface Standard Pressure = 1 atmosphere (atm.) Conversions: –1 atmospheres (atm) –760 millimeters of

Pressure Pressure • The force per unit area on a surface

• Standard Pressure = 1 atmosphere (atm.)

• Conversions:– 1 atmospheres (atm)– 760 millimeters of Mercury (mmHg)– 760 Torricelli's (torr)– 1.01 x 105 Pascals (Pa)– 14.7 pounds per square inch (psi)

Page 3: Gas Laws. Pressure The force per unit area on a surface Standard Pressure = 1 atmosphere (atm.) Conversions: –1 atmospheres (atm) –760 millimeters of

Conversion Practice Conversion Practice • Pressure

1) 350 mmHg = __________ psi2) 345 Pa = ____________ mmHg3) 1.2 atm = __________ torr

Page 4: Gas Laws. Pressure The force per unit area on a surface Standard Pressure = 1 atmosphere (atm.) Conversions: –1 atmospheres (atm) –760 millimeters of

Temperature Temperature • Average kinetic energy of the

particles in a sample of matter (***ALL TEMPERATURES MUST BE CONVERTED TO KELVIN***)

• Standard Temp. = 0 0C or 273 K

• Conversions:o Kelvin = 0C + 273oCelsius = K – 273

Page 5: Gas Laws. Pressure The force per unit area on a surface Standard Pressure = 1 atmosphere (atm.) Conversions: –1 atmospheres (atm) –760 millimeters of

Conversion PracticeConversion Practice• Temperature

1) 26 0C = _______ K2) 70 0C = _______ K3) 456 K = _______ 0C

Page 6: Gas Laws. Pressure The force per unit area on a surface Standard Pressure = 1 atmosphere (atm.) Conversions: –1 atmospheres (atm) –760 millimeters of

Volume Volume • The amount of space occupied by an

object

• Conversions:o 1 L = 1000 mL

Page 7: Gas Laws. Pressure The force per unit area on a surface Standard Pressure = 1 atmosphere (atm.) Conversions: –1 atmospheres (atm) –760 millimeters of

Boyle’s Law Boyle’s Law • The volume of a fixed mass of gas

varies inversely with pressure at a constant temperature

• Formula: (P1)(V1) = (P2)(V2)

Page 8: Gas Laws. Pressure The force per unit area on a surface Standard Pressure = 1 atmosphere (atm.) Conversions: –1 atmospheres (atm) –760 millimeters of

Charles’ LawCharles’ Law• The volume of a fixed mass of gas at

constant pressure varies directly with the temperature

• Formula: V1 = V2or V1T2 = V2T1

T1 T2

Page 9: Gas Laws. Pressure The force per unit area on a surface Standard Pressure = 1 atmosphere (atm.) Conversions: –1 atmospheres (atm) –760 millimeters of

Boyle’s and CharlesBoyle’s and Charles

1) How hot will a 2.3 L balloon have to get to expand to a volume of 400 L? Assume that the initial temperature of the balloon is 25 0C.

2) 1.00 L of a gas at standard pressure is compressed to 473 mL. What is the new pressure of the gas?

Page 10: Gas Laws. Pressure The force per unit area on a surface Standard Pressure = 1 atmosphere (atm.) Conversions: –1 atmospheres (atm) –760 millimeters of

Boyle’s and CharlesBoyle’s and Charles

3) Helium occupies 3.8 L at -45°C. What volume will it occupy at 45°C?

4) Ammonia gas occupies a volume of 450 mL at 720 mm Hg. What volume will it occupy at standard pressure?

Page 11: Gas Laws. Pressure The force per unit area on a surface Standard Pressure = 1 atmosphere (atm.) Conversions: –1 atmospheres (atm) –760 millimeters of

Combined Gas LawCombined Gas Law• Relationship between the pressure,

volume, temperature, and number of moles of a fixed amount of gas

• Formula: P1V1 = P2V2

T1 T2

Page 12: Gas Laws. Pressure The force per unit area on a surface Standard Pressure = 1 atmosphere (atm.) Conversions: –1 atmospheres (atm) –760 millimeters of

Combined Gas Law PracticeCombined Gas Law Practice

1. If I initially have a gas at a pressure of 12 atm, a volume of 23 liters, and a temperature of -73 0C, and then I raise the pressure to 1418.2 kPa and increase the temperature to 27 0C, what is the new volume of the gas?

Page 13: Gas Laws. Pressure The force per unit area on a surface Standard Pressure = 1 atmosphere (atm.) Conversions: –1 atmospheres (atm) –760 millimeters of

2. A gas takes up a volume of 17 liters, and is placed at STP. If I raise the temperature to 77 0C and lower the volume to 9 liters, what is the new pressure of the gas?

Page 14: Gas Laws. Pressure The force per unit area on a surface Standard Pressure = 1 atmosphere (atm.) Conversions: –1 atmospheres (atm) –760 millimeters of

Avogadro’s LawAvogadro’s Law• Equal volumes of gases at the same

temperature and pressure have the same number of molecules

• Formula: V1 = V2

n1 n2

Page 15: Gas Laws. Pressure The force per unit area on a surface Standard Pressure = 1 atmosphere (atm.) Conversions: –1 atmospheres (atm) –760 millimeters of

Avogadro PracticeAvogadro Practice

• A sample of N2 has 1.70 moles and occupies 3.80 L at 25.0˚C.

– What volume will it occupy with 2.60 moles?

– How many moles are in a sample that occupies 1.45 L?

Page 16: Gas Laws. Pressure The force per unit area on a surface Standard Pressure = 1 atmosphere (atm.) Conversions: –1 atmospheres (atm) –760 millimeters of

Ideal Gas LawIdeal Gas Law• An imaginary gas that perfectly fits

all the assumptions of the kinetic molecular theory

Page 17: Gas Laws. Pressure The force per unit area on a surface Standard Pressure = 1 atmosphere (atm.) Conversions: –1 atmospheres (atm) –760 millimeters of

Kinetic Molecular TheoryKinetic Molecular Theory• A theory based on the idea that gases are always

in motion• Assumptions:

1. Gas particles are in constant motion2. The average kinetic energy of gas particles

depends on the temperature of the gas3. Gases consist of large numbers of tiny particles

spread far apart4. Collisions between gas particles are elastic

collisions5. There are no forces of attraction or repulsion

between gas particles

Page 18: Gas Laws. Pressure The force per unit area on a surface Standard Pressure = 1 atmosphere (atm.) Conversions: –1 atmospheres (atm) –760 millimeters of

Ideal Gas Law Cont’dIdeal Gas Law Cont’d• Formula:

PV = nRT• Parts of the Equation:

– P = Pressure must be in unit that matches R

– V = Volume must be in Liters– n = number of moles must be in Moles– R = ideal gas constant

– T = Temperature must be in Kelvin

Page 19: Gas Laws. Pressure The force per unit area on a surface Standard Pressure = 1 atmosphere (atm.) Conversions: –1 atmospheres (atm) –760 millimeters of

Ideal Gas Law PracticeIdeal Gas Law Practice1) If I have an unknown quantity of gas

held at a temperature of 922 0C in a container with a volume of 25 mL and a pressure of 560 atm, how many moles of gas do I have?

Page 20: Gas Laws. Pressure The force per unit area on a surface Standard Pressure = 1 atmosphere (atm.) Conversions: –1 atmospheres (atm) –760 millimeters of

Ideal Gas Law PracticeIdeal Gas Law Practice2) At what temperature would 2.10

moles of N2 gas have a pressure of 950 mm Hg and in a 25.0 L?

Page 21: Gas Laws. Pressure The force per unit area on a surface Standard Pressure = 1 atmosphere (atm.) Conversions: –1 atmospheres (atm) –760 millimeters of

Starter (The faster you do this the Starter (The faster you do this the better your day will become!!!)better your day will become!!!)

1) Get out your homework and KEEP it on your desk!

2) Get out your lab notebook.

3) Have a calculator and something to write with.

Page 22: Gas Laws. Pressure The force per unit area on a surface Standard Pressure = 1 atmosphere (atm.) Conversions: –1 atmospheres (atm) –760 millimeters of

(MW)PV = gRT(MW)PV = gRT

• If you had 14 liters of propane (C3H8) under 3.7 atmospheres of pressure, being stored at 50 degrees Celsius, how many grams of propane would you have?

Page 23: Gas Laws. Pressure The force per unit area on a surface Standard Pressure = 1 atmosphere (atm.) Conversions: –1 atmospheres (atm) –760 millimeters of

(MW)PV = gRT(MW)PV = gRT• If you had 7.2 grams of Hydrogen

being stored at -25 degrees Celsius, under 2.4 atmospheres of pressure, what would the volume of the container be?

Page 24: Gas Laws. Pressure The force per unit area on a surface Standard Pressure = 1 atmosphere (atm.) Conversions: –1 atmospheres (atm) –760 millimeters of

(MW)PV = gRT(MW)PV = gRT• A 540 ml Frito bag contains 40 grams

of an inert gas. If its molecular weight is 2.02 g/mol and the temperature is 48oC, what is the pressure in atm?

Page 25: Gas Laws. Pressure The force per unit area on a surface Standard Pressure = 1 atmosphere (atm.) Conversions: –1 atmospheres (atm) –760 millimeters of

(MW)P = DRT(MW)P = DRT

• A group of scientist discovered a new gas. The gas is being stored in a container that is at 412 K, and 3.5 atmospheres of pressure. The gas they learn has a density of 2.8 g/L. What is the molecular weight of the gas?

Page 26: Gas Laws. Pressure The force per unit area on a surface Standard Pressure = 1 atmosphere (atm.) Conversions: –1 atmospheres (atm) –760 millimeters of

(MW)P = DRT(MW)P = DRT• Radon has a density of 8.75 g/L when

it is at 54oC. What pressure is on the gas in p.s.i.?

Page 27: Gas Laws. Pressure The force per unit area on a surface Standard Pressure = 1 atmosphere (atm.) Conversions: –1 atmospheres (atm) –760 millimeters of

(MW)P = DRT(MW)P = DRT• Suppose that you found a new

molecule known as the Cannonium. Cannonium has a molecular weight of 253.56 g/mol, and you have it at 14.2 oC. If it is under 1250 torr of pressure, what is the density of Cannonium?

Page 28: Gas Laws. Pressure The force per unit area on a surface Standard Pressure = 1 atmosphere (atm.) Conversions: –1 atmospheres (atm) –760 millimeters of

Dalton’s Law (Partial Dalton’s Law (Partial Pressure)Pressure)

• The total pressure of a mixture of gases is equal to the sum of all the partial pressures

• Formula: PT = P1 + P2 + P3 + ...

Page 29: Gas Laws. Pressure The force per unit area on a surface Standard Pressure = 1 atmosphere (atm.) Conversions: –1 atmospheres (atm) –760 millimeters of

DaltonDalton• A 10.0 L flask contains 0.200 moles of

methane, 0.300 moles of hydrogen, and 0.400 moles of nitrogen at 25 C. What is the partial pressure of each component of the mixture? What is the total pressure in atmospheres inside the flask?

Page 30: Gas Laws. Pressure The force per unit area on a surface Standard Pressure = 1 atmosphere (atm.) Conversions: –1 atmospheres (atm) –760 millimeters of

DaltonDalton• The stopcock connecting a 1.0 L bulb

containing oxygen gas at a pressure of 540 torr and a 3.0 L bulb containing helium gas at a pressure of 776 torr is opened, and the gases are allowed to mix. Assuming that the temperature remains constant, what is the final pressure in the system?

Page 31: Gas Laws. Pressure The force per unit area on a surface Standard Pressure = 1 atmosphere (atm.) Conversions: –1 atmospheres (atm) –760 millimeters of

DaltonDalton• A 450 ml container that contains 2.8 grams

of Oxygen is connected to a 300 ml container that contains 95 grams of Hydrogen. They are both at 25C before and after the valve is opened. What is the partial pressure of each (in atm), what is the total pressure (in atm)?

Page 32: Gas Laws. Pressure The force per unit area on a surface Standard Pressure = 1 atmosphere (atm.) Conversions: –1 atmospheres (atm) –760 millimeters of

GAS LAW STOICHIOMETRYGAS LAW STOICHIOMETRY

Page 33: Gas Laws. Pressure The force per unit area on a surface Standard Pressure = 1 atmosphere (atm.) Conversions: –1 atmospheres (atm) –760 millimeters of

Gas Law StoichiometryGas Law Stoichiometry• After this combustion reaction 12.2 L

of carbon dioxide is left. How many grams of water were produced if the reaction took place at 1.3 atm and 600 degrees celsius.

C3H8 + O2 CO2 + H2O

Page 34: Gas Laws. Pressure The force per unit area on a surface Standard Pressure = 1 atmosphere (atm.) Conversions: –1 atmospheres (atm) –760 millimeters of

Gas Law StoichiometryGas Law Stoichiometry

• How many liters of F2 at 30o C and 2.8 atm of pressure are needed to react with 8 grams of Hydrogen in the following reaction?

H2 + F2 HF

Page 35: Gas Laws. Pressure The force per unit area on a surface Standard Pressure = 1 atmosphere (atm.) Conversions: –1 atmospheres (atm) –760 millimeters of

Gas Law StoichiometryGas Law Stoichiometry• Aluminum bonds with Oxygen to

make Aluminum Oxide (Al2O3). If you formed 86.5 grams of Aluminum Oxide at 68 degrees Celsius and 845 torr, what would the volume of the Oxygen be?

Al + O2 Al2O3

Page 36: Gas Laws. Pressure The force per unit area on a surface Standard Pressure = 1 atmosphere (atm.) Conversions: –1 atmospheres (atm) –760 millimeters of

• Ni(CO)4 can be made by the reaction of finely divided nickel with gaseous CO. If you have CO in a 1.50 L flask at a pressure of 418 mmHg at 25oC, what is the maximum number of grams of Ni(CO)4 that can be made?

Ni + CO Ni(CO)4