Gas Laws Volume V v The volume of

Gas Laws

Volume (V) v The volume of a gas is simply the volume of the container it is contained in. v The metric unit of volume, liter (L), is often used. v There might also be problems that use cubic meters as the unit for volume. • 1 L = 1 m 3

Temperature (T) v The temperature of a gas is generally measured with a thermometer in Celsius. v All calculations involving gases should be made after converting the Celsius to Kelvin temperature. Kelvin = C° + 273 Celsius = K - 273

Pressure (P) v The pressure of a gas is the force exerted on the wall of the container, in which a gas is trapped. v There are several units for pressure depending on the instrument used to measure it including: 1) atmospheres (atm) 2) Millimeters of Mercury (mm. Hg) 3) Kilopascal (k. Pa) 4) Torr

Standard Temperature and Pressure (STP) v v T = 0˚C 273 K P = 1 atm

Boyle’s Law v Robert Boyle was among the first to note the inverse relationship between pressure and volume of a gas. • As the pressure on a gas increased the volume of the gas will decrease. v He measured the volume of air at different pressures, and observed a pattern of behavior. • During his experiments Temperature and amount of gas weren’t allowed to change 1621 -1691

Boyle’s Mathematical Law: If we have a given amount of a gas at a starting pressure and volume, what would happen to the pressure if we changed the volume? Or to the volume if we changed the pressure? since PV equals a constant P 1 V 1 = P 2 V 2 Ex: A gas has a volume of 3. 0 L at 2 atm. What will its volume be at 4 atm?

Boyle’s Mathematical Law: 1) List the variables or clues given: Ø P 1 = 2 atm Ø V 1 = 3. 0 L Ø P 2 = 4 atm Ø V 2 = ? 2) determine which law is being represented: P 1 V 1 = V 2 P 2 3) Plug in the variables & calculate: (2 atm)(3. 0 L) = (4 atm) (V 2)

Charles’s Law v Jacques Charles studied the direct mathematical relationship between temperature and volume of a gas. • As temperature increases the volume of the gas increases v Charles measured the volume of air at different temperatures, and recorded the results. • During his experiments pressure of the system and amount of gas were held constant. 1746 -1823

Charles’s Mathematical Law: If we have a given amount of a gas at a starting volume and temperature, what would happen to the volume if we changed the temperature? Or to the temperature if we changed the volume? since V/T = k V 1 T 1 = V 2 T 2 V 1 T 2 = V 2 T 1 Ex: A gas has a volume of 3. 0 L at 400 K. What is its volume at 500 K?

Charles’s Mathematical Law: 1) List the variables or clues given: Ø T 2 = 500 K Ø V 2 = ? 2) Rearrange the formula to solve for missing variable Ø T 1 = 400 K Ø V 1 = 3. 0 L 3) Plug in the variables & calculate:

Gay-Lussac’s Mathematical Law: If we have a given amount of a gas at a starting temperature and pressure, what would happen to the pressure if we changed the temperature? Or to the temp. if we changed the pressure? since P/T = k P 1 P 2 = T 1 T 2 P 1 T 2 = P 2 T 1 Ex: A gas has a pressure of 3. 0 atm at 400 K. What is its pressure at 500 K?

Gay-Lussac’s Mathematical Law: 1) List the variables or clues given: Ø T 1 = 400 K Ø P 1 = 3. 0 atm Ø T 2 = 500 K Ø P 2 = ? 2) Rearrange the formula to solve for missing variable 3) Plug in the variables & calculate:

Summary of the Named Gas-Laws: LAW RELATIONSHIP EQUATION CONSTANTS Boyle’s P V P 1 V 1 = P 2 V 2 T Charles’ V T V 1 T 2 = V 2 T 1 P Gay. Lussac’s P T P 1 T 2 = P 2 T 1 V

Combined Gas Law: P 1 V 1 T 2 = P 2 V 2 T 1 Ex: A cylinder contain a gas of volume 30 L, at a pressure of 110 k. Pa and a temperature of 420 K. Find the temperature of the gas which has a volume 40 L at a pressure of 120 k. Pa

Combined Gas Law: 1) List the variables or clues given: Ø T 1 = 420 K Ø V 1 = 30 L Ø P 1 = 110 k. Pa Ø T 2 = ? Ø V 2 = 40 L Ø P 2 = 120 k. Pa 2) Rearrange the formula to solve for missing variable 3) Plug in the variables & calculate:

Ideal Gas Law: PV = n. RT P = pressure V = volume n = # moles R = constant (0. 0821 L ۰ atm/ K۰ mol) T = temperature Ex: Find the volume from the 0. 250 moles gas at 200 k. Pa and 300 K temperature

Ideal Gas Law: 1) List the variables or clues given: Ø Ø Ø P = 200 k. Pa V=? n = 0. 250 mol R= 0. 0821 T = 300 K 2) Rearrange the formula to solve for missing variable 3) Plug in the variables & calculate:

Gas Stoichiometry: v At STP, 1 mol of gas = 22. 4 liters v We can add this to our Stoichiometry graphic organizer!

Atoms Particles molecules Using Avogadro’s Number 6. 02 x 10 23 1 mol = 22. 4 L liters Mole #1 Atoms Particles molecules Using Avogadro’s Number 6. 02 x 10 23 Mole #2 1 mol = 22. 4 L liters Mole Ratios from Balanced Equations Calculate molar mass using Periodic Table grams

Example 1: What volume of hydrogen at STP can be produced when 6. 54 g of Zn reacts with hydrochloric acid, HCl? Zn + 2 HCl H 2 + Zn. Cl 2

Example 1: 6. 54 g Zn 1

Example 1: 6. 54 g Zn 1 mol Zn 1 65. 4 g Zn

Example 1: 6. 54 g Zn 1 mol H 2 1 65. 4 g Zn 1 mol Zn

Example 1: 6. 54 g Zn 1 mol H 2 22. 4 L H 2 1 65. 4 g Zn 1 mol H 2

Example 1: 6. 54 g Zn 1 mol H 2 22. 4 L H 2 1 65. 4 g Zn 1 mol H 2

Example 2: How many grams of Na. Cl can be produced by the reaction of 0. 112 liters of chlorine gas at STP with excess sodium? Cl 2 + 2 Na. Cl

Example 2: 0. 112 L Cl 2 1

Example 2: 0. 112 L Cl 2 1 mol Cl 2 1 22. 4 L Cl 2

Example 2: 0. 112 L Cl 2 1 mol Cl 2 2 mol Na 1 22. 4 L Cl 2 1 mol Cl 2

Example 2: 0. 112 L Cl 2 1 mol Cl 2 2 mol Na. Cl 55. 85 g Na. Cl 1 22. 4 L Cl 2 1 mol Na. Cl

Example 2: 0. 112 L Cl 2 1 mol Cl 2 2 mol Na. Cl 55. 85 g Na. Cl 1 22. 4 L Cl 2 1 mol Na. Cl