Chemistry Notes – Gas Laws

  1. Gas Properties
  1. Gas is the state of matter where the particles are widely separated and have little interaction. All substances can be gases under the right conditions.
  2. The volume of a gas is determined by temperature and pressure. These three characteristics are related in mathematical equations called gas laws.
  3. Pressure is the collisions that occur between gas molecules and other matter. The more collisions that occur, the greater the pressure. Standard pressure is the average pressure at sea level. 1.0 atmospheres, 760 mm/Hg, or 101 kPa.
  4. Gases have no definite shape or volume. Gases can be forced into smaller volumes, a characteristic that is called compressibility.
  5. Temperature is the kinetic energy of a substance. Kinetic energy is the energy of motion. The greater the motion, the higher the temperature. Standard temperature is 273 K
  6. Kinetic-Molecular Theory

1)Gases are composed of particles.

2)Gas particles are widely separated.

3)Gas particles are in constant, rapid, random motion.

4)Collisions between gas particles are elastic (they do not lose energy on contact).

5)The motion of gas particles is dependent only on the temperature.

6)Gas particles exert no forces on each other.

  1. Temperature must be expressed in Kelvin in gas laws. The Kelvin scale starts at absolute zero (the coldest temperature possible) so there are no negative temperatures. K = oC + 273.
  1. Boyles Law
  1. Developed by Sir Robert Boyle, an English chemist and physicist.
  2. At constant temperature, the pressure and volume of a gas are inversely proportional to each other. P1V1 = P2V2
  3. Example – 12 liters of gas is at 1.3 atm. of pressure. At constant temperature, what will be the volume of the gas at 1.8 atm.?

P1V1 = P2V2 P1V1= V21.3 atm. x 12 liters = 8.7 liters

P2 1.8 atm.

  1. Charles Law
  1. Developed by Jacques Charles, a French physicist.
  2. At constant pressure, the volume and temperature of a gas is directly proportional. V1= V2

T1 T2

  1. Example – 35 ml. of a gas is at 21 oC. At constant pressure, what is the volume of the gas at 32 oC.?

V1 = V2V2 = V1 x T2 V2 = 35 ml. x 305 K = 36 ml.

T1 T2 T1 294 K

  1. Gay-Lussac’s Law
  1. Developed by Joseph Gay-Lussac, a French chemist.
  2. At constant volume, the pressure and temperature of a gas are directly proportional. P1 = P2

T1 T2

  1. Gay-Lussac’s Law is the principle behind the pressure cooker used to sterilize equipment and cook food faster.
  2. Example – A gas has a pressure of 104 kPa at 33 o C. If the volume is constant, what is the pressure of the gas at 55 o C.?

P1 = P2 P2 = P1 x T2P2 = 104 kPa x 328K = 111 kPa

T1 T2 T1 306 K

  1. Combined Gas Law
  1. Combines Boyles, Charles, and Gay-Lussac’s Law
  2. Allows you to solve for changes in volume, temperature, and pressure.
  3. P1 V1 = P2 V2

T1T2

  1. Example – 3.7 liters of gas is at 750 mm/Hg and 15 oC. What is the volume of the gas at STP?

P1 V1 = P2 V2P1 V1 T2 = V2 750 mm/Hg x 3.7 liters x 273 K = 3.5 lit

T1T2 T1 P2 288 K x 760 mm/Hg

* Read in Prentice Hall Chemistry pages 413 through 425.

  1. Avagadro’s Principle
  1. At equal temperatures and pressures, equal volumes of gases contain the same number of molecules.
  2. One mole of any gas at STP occupies a volume of 22.4 liters.
  1. Dalton’s Law of Partial Pressures
  1. The total pressure in a container is the sum of the partial pressures of gases in the container. The total air pressure in the room is the sum of the pressures of nitrogen, oxygen, and carbon dioxide.
  2. P1 + P2 + P3 …. = Ptotal
  3. Example – A gas is collected over water at 15 oC in a 410 ml. bottle. The total pressure in the bottle is 98 kPa. What volume would the dry gas take up at standard pressure and 15 oC.?

Pdry gas = P bottle - P water Vapor pressure of water at 15 oC = 1.7 kPa

Pdry gas = 98 kPa – 1.7 kPa = 96.3 kPa

P1 V1 = P2 V2 P1 V1 = V2 410 ml. x 96.3 kPa = 390 ml.

P2101 kPa

  1. Ideal Gas Law
  1. Relates the number of moles, pressure, temperature, and volume of a gas.
  2. PV = nRT

R is the ideal gas law constant. 8.31 liters – kPa

mole – K

.0821 liters – atm.

mole – K

  1. If you know three of the variables you can find the fourth since R is a constant.
  2. Example – How many grams of hydrogen gas occupies a volume of 1.6 liters at 10.0 oC and 81 kPa?

First you need to solve for the number of moles present. Then multiply the number of moles by the mass of hydrogen gas (2.0 grams/mole).

PV = nRT n = PV

RT

n = 81 kPa x 1.6 liters = .055 mole x 2.0 grams = .11 grams

8.31lit-atm x 283 K mole

mole-K

* Read in Prentice Hall Chemistry pages 426 through 434.