1. It’s a Gas Part 3: Charles’s and Gay-Lussac’s Law

2. Summary of Kinetic Molecular Theory (KMT) Postulates Gas particles are in constant random motion.Gas particles are in constant random motion. Gas particles occupy no volume.Gas particles occupy no volume. Collisions between gas particles are perfectly elastic: there is no loss of kinetic energy.Collisions between gas particles are perfectly elastic: there is no loss of kinetic energy. There are neither attractive nor repulsive forces between gas particles.There are neither attractive nor repulsive forces between gas particles. The higher the absolute temperature, the higher the average kinetic energy of the gas.The higher the absolute temperature, the higher the average kinetic energy of the gas.

3. 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.

4. 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.

5. Charles’s Law At constant pressure, the volume of the gas increases or decreases by the same factor as its absolute temperature increases or decreases.At constant pressure, the volume of the gas increases or decreases by the same factor as its absolute temperature increases or decreases. What is absolute zero? What is absolute zero? 0 K on the Kelvin scale or -273˚C on the Celsius scale.0 K on the Kelvin scale or -273˚C on the Celsius scale.

6. Charles’s Law Representative Data Temperature (˚C) Volume (mL) 1065.2 2067.4 3069.2 4071.6 5073.1 6075.7 7078.0 8080.2

7. The volume goes to zero at -273˚C. A new scale was created called the Kelvin scale with units called Kelvins.

8. The Math of Charles’s Law V 1 = V 2 T 1 T 2 Where V is Volume and T is Absolute Temperature (K).V 1 = V 2 T 1 T 2 Where V is Volume and T is Absolute Temperature (K). To convert a Celsius temperature value to a temperature in the Kelvin scale, use the following: T (K) = t (˚C) + 273 where T is the Kelvin temperature and t is the Celsius temperature.To convert a Celsius temperature value to a temperature in the Kelvin scale, use the following: T (K) = t (˚C) + 273 where T is the Kelvin temperature and t is the Celsius temperature. One must make this conversion, if the temperature value is not already Kelvin.One must make this conversion, if the temperature value is not already Kelvin.

9. The Math of Charles’s Law (continued) V 1 = V 2 T 1 T 2 Where V is Volume and T is Absolute Temperature (K).V 1 = V 2 T 1 T 2 Where V is Volume and T is Absolute Temperature (K). A sample of gas occupies a volume of 100 mL at a temperature of - 73˚C. What volume would the gas occupy at -123 ˚C?A sample of gas occupies a volume of 100 mL at a temperature of - 73˚C. What volume would the gas occupy at -123 ˚C? First, convert ˚C to K: T 1 = - 73˚ + 273 = 200 K T 2 = - 123˚ + 273 = 150 KFirst, convert ˚C to K: T 1 = - 73˚ + 273 = 200 K T 2 = - 123˚ + 273 = 150 K

10. The Math of Charles’s Law (continued) V 1 = V 2 T 1 T 2V 1 = V 2 T 1 T 2 T 1 = 200 K and T 2 = 150 KT 1 = 200 K and T 2 = 150 K 100 mL=V 2 200 K 150 K100 mL=V 2 200 K 150 K (100 mL)(150 K) = V 2 (150 K) (200 K) (150 K)(100 mL)(150 K) = V 2 (150 K) (200 K) (150 K) V 2 =(100 mL)(150 K) = 75 mL (200 K)V 2 =(100 mL)(150 K) = 75 mL (200 K)

11. Kinetic Molecular Theory and Charles’s Law For a flexible container or a freely moving piston, at a given average kinetic energy for the gas particles, the volume of the gas will be determined by the number and energy of the collisions between the particles and the walls of the container.For a flexible container or a freely moving piston, at a given average kinetic energy for the gas particles, the volume of the gas will be determined by the number and energy of the collisions between the particles and the walls of the container. When the average kinetic energy of the gas is increased by raising the temperature of the gas, the frequency and energy of the collisions will increase and the volume of the gas will expand until the frequent and high energy collisions are spread out over a larger area and a balance is achieved.When the average kinetic energy of the gas is increased by raising the temperature of the gas, the frequency and energy of the collisions will increase and the volume of the gas will expand until the frequent and high energy collisions are spread out over a larger area and a balance is achieved.

12. Gay-Lussac’s Law At constant volume, the pressure of the gas increases or decreases by the same factor as its absolute temperature increases or decreases.At constant volume, the pressure of the gas increases or decreases by the same factor as its absolute temperature increases or decreases. The form of the equation is the same as that for Charles’s law, except pressure replaces volume.The form of the equation is the same as that for Charles’s law, except pressure replaces volume. Again, it is necessary to convert ˚C to K.Again, it is necessary to convert ˚C to K.

13. The Math of Gay-Lussac’s Law P 1 = P 2 T 1 T 2 Where P is Pressure and T is Absolute Temperature (K).P 1 = P 2 T 1 T 2 Where P is Pressure and T is Absolute Temperature (K). A sample of gas occupies a fixed volume. Its pressure is 1 atm at a temperature of 27 ˚C. What is the pressure of the gas at 127 ˚C?A sample of gas occupies a fixed volume. Its pressure is 1 atm at a temperature of 27 ˚C. What is the pressure of the gas at 127 ˚C? First, convert ˚C to K: T 1 = 27˚ + 273 = 300 K T 2 = 127˚ + 273 = 400 KFirst, convert ˚C to K: T 1 = 27˚ + 273 = 300 K T 2 = 127˚ + 273 = 400 K

14. The Math of Gay-Lussac’s Law (continued) P 1 = P 2 T 1 T 2P 1 = P 2 T 1 T 2 T 1 = 300 K and T 2 = 400 KT 1 = 300 K and T 2 = 400 K 1 atm=P 2 300 K 400 K1 atm=P 2 300 K 400 K (1 atm)(400 K) = P 2 (400 K) (300 K) (400 K)(1 atm)(400 K) = P 2 (400 K) (300 K) (400 K) P 2 =(1 atm)(400 K) = 1.33 atm (300 K)P 2 =(1 atm)(400 K) = 1.33 atm (300 K)

15. Kinetic Molecular Theory and \ Gay-Lussac’s Law For a rigid container, at a given average kinetic energy for the gas particles, the pressure of the gas will be determined by the number and energy of the collisions between the particles and the walls of the container.For a rigid container, at a given average kinetic energy for the gas particles, the pressure of the gas will be determined by the number and energy of the collisions between the particles and the walls of the container. When the average kinetic energy of the gas is increased by raising the temperature of the gas, the frequency and energy of the collisions will increase and the pressure of the gas will increase as more force is applied to the walls of the container.When the average kinetic energy of the gas is increased by raising the temperature of the gas, the frequency and energy of the collisions will increase and the pressure of the gas will increase as more force is applied to the walls of the container.

16. Summary Charles’s Law: V 1 = V 2 T 1 T 2Charles’s Law: V 1 = V 2 T 1 T 2 Gay-Lussac’s Law: P 1 = P 2 T 1 T 2Gay-Lussac’s Law: P 1 = P 2 T 1 T 2 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. Isolate the unknown variable and solve.Isolate the unknown variable and solve.