Volume (V)  The volume of a gas is simply the volume of the container it is contained in.  The metric unit of volume, liter (L), is often used.  There might also be problems that use cubic meters as the unit for volume. 1L = 1 m 31L = 1 m 3

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

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

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

Boyle’s Law  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.As the pressure on a gas increased the volume of the gas will decrease.  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 changeDuring his experiments Temperature and amount of gas weren’t allowed to change

Boyle’s Mathematical Law: 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? Ex: A gas has a volume of 3.0 L at 2 atm. What will its volume be at 4 atm? 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?

Boyle’s Mathematical Law: 1)List the variables or clues given: 2)determine which law is being represented:  P 1 = 2 atm  V 1 = 3.0 L  P 2 = 4 atm  V 2 = ? 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  Jacques Charles studied the direct mathematical relationship between temp- erature and volume of a gas. As temperature increases the volume of the gas increasesAs temperature increases the volume of the gas increases  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.During his experiments pressure of the system and amount of gas were held constant.

Charles’s Mathematical Law: since V/T = k Ex: A gas has a volume of 3.0 L at 400K. What is its volume at 500K? Ex: A gas has a volume of 3.0 L at 400K. What is its volume at 500K? = V 1 V 2 T 1 T 2 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?

Charles’s Mathematical Law: 2)Rearrange the formula to solve for missing variable  T 1 = 400K  V 1 = 3.0 L  T 2 = 500K  V 2 = ? 1)List the variables or clues given: 3) Plug in the variables & calculate:

since P/T = k P 1 P 2 T 1 T 2 = Ex: A gas has a pressure of 3.0atm at 400K. What is its pressure at 500K? Ex: A gas has a pressure of 3.0atm at 400K. What is its pressure at 500K? 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? Gay-Lussac’s Mathematical Law:

2)Rearrange the formula to solve for missing variable  T 1 = 400K  P 1 = 3.0 atm  T 2 = 500K  P 2 = ? 1)List the variables or clues given: 3) Plug in the variables & calculate:

LAW RELAT- IONSHIP EQUATION CON- STANTS Boyle’s P VP VP VP V P 1 V 1 = P 2 V 2 T Charles’ V TV TV TV T V 1 T 2 = V 2 T 1 P Gay- Lussac’s P TP TP TP T P 1 T 2 = P 2 T 1 V Summary of the Named Gas-Laws:

Ex:A cylinder contain a gas of volume 30 L, at a pressure of 110 kPa and a temperature of 420 K. Find the temperature of the gas which has a volume 40 L at a pressure of 120 kPa Ex:A cylinder contain a gas of volume 30 L, at a pressure of 110 kPa and a temperature of 420 K. Find the temperature of the gas which has a volume 40 L at a pressure of 120 kPa Combined Gas Law:

2)Rearrange the formula to solve for missing variable  T 1 = 420K  V 1 = 30 L  P 1 = 110 kPa  T 2 = ?  V 2 = 40 L  P 2 = 120 kPa 1)List the variables or clues given: 3) Plug in the variables & calculate:

Ex: Find the volume from the moles gas at 200 kPa and 300K temperature Ex: Find the volume from the moles gas at 200 kPa and 300K temperature Ideal Gas Law: P = pressure V = volume n = # moles R = constant ( L ۰ atm/ K۰mol) T = temperature

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

Gas Stoichiometry:  At STP, 1 mol of gas = 22.4 liters  We can add this to our Stoichiometry graphic organizer!  We can add this to our Stoichiometry graphic organizer!

Mole #1 Atoms Particles molecules grams Mole #2 Atoms Particles molecules grams Using Avogadro’s Number 6.02 x Calculate molar mass using Periodic Table Mole Ratios from Balanced Equations liters 1 mol = 22.4 L

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 + ZnCl 2 Zn + 2 HCl  H 2 + ZnCl 2 Example 1:

6.54 g Zn 6.54 g Zn 1 Example 1:

6.54 g Zn 1 mol Zn 6.54 g Zn 1 mol Zn g Zn g Zn Example 1:

6.54 g Zn 1 mol Zn 1 mol H g Zn 1 mol Zn 1 mol H g Zn 1 mol Zn g Zn 1 mol Zn Example 1:

6.54 g Zn 1 mol Zn 1 mol H L H g Zn 1 mol Zn 1 mol H L H g Zn 1 mol Zn 1 mol H g Zn 1 mol Zn 1 mol H 2 Example 1:

6.54 g Zn 1 mol Zn 1 mol H L H g Zn 1 mol Zn 1 mol H L H g Zn 1 mol Zn 1 mol H g Zn 1 mol Zn 1 mol H 2 Example 1:

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

0.112 L Cl L Cl 2 1 Example 2:

0.112 L Cl 2 1 mol Cl L Cl 2 1 mol Cl L Cl L Cl 2 Example 2:

0.112 L Cl 2 1 mol Cl 2 2 mol Na L Cl 2 1 mol Cl 2 2 mol Na L Cl 2 1 mol Cl L Cl 2 1 mol Cl 2 Example 2:

0.112 L Cl 2 1 mol Cl 2 2 mol NaCl g NaCl L Cl 2 1 mol Cl 2 1 mol NaCl L Cl 2 1 mol Cl 2 1 mol NaCl Example 2:

0.112 L Cl 2 1 mol Cl 2 2 mol NaCl g NaCl L Cl 2 1 mol Cl 2 1 mol NaCl L Cl 2 1 mol Cl 2 1 mol NaCl Example 2: