1. It’s a Gas Combined and Ideal Gas Law (Pivnert)

2. Charles’s Law: V 1 /T 1 = V 2 /T 2 This is a proportional relationship.This is a proportional relationship. As the temperature of a gas increases, the volume increases.As the temperature of a gas increases, the volume increases. Similarly, as the temperature of a gas decreases, the volume decreases.Similarly, as the temperature of a gas decreases, the volume decreases. BEFORE SUBSTITUTION INTO THE EQUATION, CHANGE TEMPERATURE FROM ˚C TO K BY ADDING 273.BEFORE SUBSTITUTION INTO THE EQUATION, CHANGE TEMPERATURE FROM ˚C TO K BY ADDING 273. To solve, isolate the unknown variable.To solve, isolate the unknown variable.

3. Gay Lussac’s Law: P 1 /T 1 = P 2 /T 2 This is a proportional relationship.This is a proportional relationship. As the temperature of a gas increases, the pressure increases.As the temperature of a gas increases, the pressure increases. Similarly, as the temperature of a gas decreases, the pressure decreases.Similarly, as the temperature of a gas decreases, the pressure decreases. BEFORE SUBSTITUTION INTO THE EQUATION, CHANGE TEMPERATURE FROM ˚C TO K BY ADDING 273.BEFORE SUBSTITUTION INTO THE EQUATION, CHANGE TEMPERATURE FROM ˚C TO K BY ADDING 273. To solve, isolate the unknown variable.To solve, isolate the unknown variable.

4. Boyle’s Law: P 1 V 1 = P 2 V 2 This is an inverse relationship.This is an inverse relationship. As pressure increases, the volume decreases.As pressure increases, the volume decreases. Similarly, as pressure decreases, the volume increases.Similarly, as pressure decreases, the volume increases. To solve, isolate the unknown variable.To solve, isolate the unknown variable.

5. The Math of Boyle’s Law P 1 V 1 = P 2 V 2P 1 V 1 = P 2 V 2 10 liters of air at 1 atm is compressed to a pressure of 4 atm. What is the volume of the compressed air.10 liters of air at 1 atm is compressed to a pressure of 4 atm. What is the volume of the compressed air. (1 atm)(10 l) = (4 atm)V 2 (1 atm)(10 l) = (4 atm)V 2 (1 atm)(10 l) = (4 atm)V 2 (4 atm) (4 atm)(1 atm)(10 l) = (4 atm)V 2 (4 atm) (4 atm)

6. The Math of Boyle’s Law (continued) P 1 V 1 = P 2 V 2P 1 V 1 = P 2 V 2 (1 atm)(10 l) = (4 atm)V 2 (4 atm) (4 atm)(1 atm)(10 l) = (4 atm)V 2 (4 atm) (4 atm) V 2 =(1 atm)(10 l) (4 atm)V 2 =(1 atm)(10 l) (4 atm) V 2 =2.5 litersV 2 =2.5 liters

7. Kinetic Molecular Theory and Boyle’s Law In a smaller volume, the number of collisions between the particles and the walls of the container is concentrated on a smaller area (think high heels), so the pressure is greater.In a smaller volume, the number of collisions between the particles and the walls of the container is concentrated on a smaller area (think high heels), so the pressure is greater. In a larger volume, the number of collisions between the particles and the walls of the container is spread out over a larger area (think snowshoes), so the pressure is less.In a larger volume, the number of collisions between the particles and the walls of the container is spread out over a larger area (think snowshoes), so the pressure is less.

8. Avogadro’s Hypothesis At constant temperature and pressure, equal volumes of gases contain equal numbers of particles.At constant temperature and pressure, equal volumes of gases contain equal numbers of particles. Or restated in mole-speak, at constant temperature and pressure, equal volumes of gases contain equal numbers of moles.Or restated in mole-speak, at constant temperature and pressure, equal volumes of gases contain equal numbers of moles.

9. Avogadro’s Hypothesis: V 1 /n 1 = V 2 /n 2 This is a proportional relationship.This is a proportional relationship. As moles of gas increase, the volume increases.As moles of gas increase, the volume increases. Similarly, as moles of gas decrease, the volume decreases.Similarly, as moles of gas decrease, the volume decreases. To solve, isolate the unknown variable.To solve, isolate the unknown variable.

10. The Math of Avogadro’s Hypothesis V 1 = V 2 n 1 n 2 Where V is Volume and n is moles.V 1 = V 2 n 1 n 2 Where V is Volume and n is moles. One mole of ozone gas (O 3 ) occupies 22.4 L. The ozone decomposes to 1.5 moles of molecular oxygen (O 2 ). What is the volume of the resulting molecular oxygen?One mole of ozone gas (O 3 ) occupies 22.4 L. The ozone decomposes to 1.5 moles of molecular oxygen (O 2 ). What is the volume of the resulting molecular oxygen? 22.4 L= V 2 ___ 1 mol 1.5 mol22.4 L= V 2 ___ 1 mol 1.5 mol

11. The Math of Avogadro’s Hypothesis (continued) 22.4 L= V 2 ___ 1 mol 1.5 mol22.4 L= V 2 ___ 1 mol 1.5 mol 22.4 L ( 1.5 mol )=V 2 1.0 mol22.4 L ( 1.5 mol )=V 2 1.0 mol V 2 = 33.6 LV 2 = 33.6 L

12. Kinetic Molecular Theory and Avogadro’s Hypothesis As the number of gas particles increases, the frequency of collisions with the walls of the container must increase. This, in turn, leads to an increase in the pressure of the gas. Flexible containers, such as a balloon, will expand until the pressure of the gas inside the balloon once again balances the pressure of the gas outside. Thus, the volume of the gas is proportional to the number of gas particles.As the number of gas particles increases, the frequency of collisions with the walls of the container must increase. This, in turn, leads to an increase in the pressure of the gas. Flexible containers, such as a balloon, will expand until the pressure of the gas inside the balloon once again balances the pressure of the gas outside. Thus, the volume of the gas is proportional to the number of gas particles.

13. Combined Gas Law P 1 V 1 = P 2 V 2 andV 1 = V 2 and P 1 = P 2 T 1 T 2 T 1 T 2P 1 V 1 = P 2 V 2 andV 1 = V 2 and P 1 = P 2 T 1 T 2 T 1 T 2 How would one combine these into one law?How would one combine these into one law? P 1 V 1 = P 2 V 2 T 1 T 2P 1 V 1 = P 2 V 2 T 1 T 2 Combined Gas Law: What temperature scale?Combined Gas Law: What temperature scale? The same one as for Charles’s and Gay- Lussac’s Law: Add 273 to the temperature in Celsius to get temperature in Kelvin (K).The same one as for Charles’s and Gay- Lussac’s Law: Add 273 to the temperature in Celsius to get temperature in Kelvin (K).

14. The Math of the Combined Gas Law A balloon is filled with gas at a pressure of 102.3 kPa and at a temperature of 45.5 ˚C. Its volume under these conditions is 12.5 L. The balloon is taken into a decompression chamber where the volume is measured as 2.50 L. If the temperature is 36.0 ˚C, what is the pressure of the chamber?A balloon is filled with gas at a pressure of 102.3 kPa and at a temperature of 45.5 ˚C. Its volume under these conditions is 12.5 L. The balloon is taken into a decompression chamber where the volume is measured as 2.50 L. If the temperature is 36.0 ˚C, what is the pressure of the chamber? 45.5˚C + 273 = 318.5 K and 36˚C + 273 = 309 K45.5˚C + 273 = 318.5 K and 36˚C + 273 = 309 K 102.3 kPa·12.5 L=P 2 ·2.5 L 318.5 K 309 K102.3 kPa·12.5 L=P 2 ·2.5 L 318.5 K 309 K 4.015 kPa·L·K -1 = 0.0080906L·K -1 · P 24.015 kPa·L·K -1 = 0.0080906L·K -1 · P 2 P 2 =496 kPaP 2 =496 kPa

15. Ideal Gas Law: Pivnert One can go one step further and combine the Combined Gas Law with Avogadro’s Hypothesis to create the Ideal Gas Law: PV = nRTOne can go one step further and combine the Combined Gas Law with Avogadro’s Hypothesis to create the Ideal Gas Law: PV = nRT This law is applied to one condition for a gas. All of the other laws use two conditions, V 1, P 1, T 1 going to V 2, P 2, T 2, etc.This law is applied to one condition for a gas. All of the other laws use two conditions, V 1, P 1, T 1 going to V 2, P 2, T 2, etc. What temperature scale?What temperature scale? The same one as before: Add 273 to the temperature in Celsius to get temperature in Kelvin (K).The same one as before: Add 273 to the temperature in Celsius to get temperature in Kelvin (K).

16. Ideal Gas Law: PV=nRT R is called the universal gas constant and equals eitherR is called the universal gas constant and equals either 0.08206 L·atmor8.314 L·kPa mol·K mol·K0.08206 L·atmor8.314 L·kPa mol·K mol·K Which R to use depends on the problem.Which R to use depends on the problem. If pressure is in units of atmospheres or an answer in atmospheres is required, one uses the 0.08206 one.If pressure is in units of atmospheres or an answer in atmospheres is required, one uses the 0.08206 one. If pressure is in units of kPa, one uses the 8.314 one.If pressure is in units of kPa, one uses the 8.314 one.

17. The Math of Pivnert If one has 2.4 moles of gas held at a temperature of 97 °C and in a container with a volume of 45 liters, what is the pressure of the gas in kPa.If one has 2.4 moles of gas held at a temperature of 97 °C and in a container with a volume of 45 liters, what is the pressure of the gas in kPa. 97 + 273 = 370 K97 + 273 = 370 K PV = nRTPV = nRT P(45 L) = (2.4 mol)(8.314 L·kPa)(370 K) mol·KP(45 L) = (2.4 mol)(8.314 L·kPa)(370 K) mol·K P(45 L) = 7382.8 L·kPa 45 L 45 LP(45 L) = 7382.8 L·kPa 45 L 45 L P = 164 kPa P = 164 kPa

18. Using the other R value What is the volume of one mole of a gas at STP?What is the volume of one mole of a gas at STP? STP stands for Standard Temperature and Pressure and is 273 K and 1 atmosphere.STP stands for Standard Temperature and Pressure and is 273 K and 1 atmosphere. Thus, one must use the 0.08205 value of RThus, one must use the 0.08205 value of R (1 atm)V = (1 mol)(0.08206 L·atm)(273 K) mol·K(1 atm)V = (1 mol)(0.08206 L·atm)(273 K) mol·K V = 22.4 LV = 22.4 L

19. Summary Combined Gas Law: P 1 V 1 = P 2 V 2 T 1 T 2Combined Gas Law: P 1 V 1 = P 2 V 2 T 1 T 2 Ideal Gas Law: PV = nRTIdeal Gas Law: PV = nRT Convert Celsius temperatures to Kelvin.Convert Celsius temperatures to Kelvin. Put the right numbers in the right places.Put the right numbers in the right places. Use the right R value for the problem.Use the right R value for the problem. Isolate the unknown variable and solve.Isolate the unknown variable and solve.