1. Section 9.2

2.  Definition: a measure of the average kinetic energy of a substance’s particles  Generally measured in o C  The Kelvin temperature scale (K) is based on absolute temperature  absolute zero (point where gas molecules have no motion) is 0 K and/or -273 o C  Kelvin temperatures must be used in many gas law equations in which temperature enters directly into the calculations.

3.  The Celsius and Kelvin scale are related unit for unit. One degree unit on the Celsius scale is equivalent to one degree unit on the Kelvin scale. The only difference between these two scales is the zero point.

4.  so, this means that 0 K equals -273 o C and 0 o C equals 273 K. Thermometers are never marked in the Kelvin scale.  to convert temperature units, use the following equations. T K = t c + 273 or t c = T K - 273

5.  Cold Saunas are available at high end resorts. Instead of heating the sauna and causing the people inside to sweat, a cold sauna is cooled down to -110 ° C. What would the temperature be in Kelvin?  T K = t c + 273  T K = -110 + 273  T K = 163 K

6.  Helium has the lowest boiling point of any gas. It boils at 4 K. What is this in o C?  t c = T K - 273  t c = 4K - 273  t c = -269 o C

7. Section 9.2

8.  The direct relationship between the volume and temperature of a gas is known as Charles Law. According to this law,  as the temperature of a gas increases, the volume increases proportionally, provided that the pressure and the amount of gas remains the same.

9.  the graph below shows this relationship.  Charles’s Law can be written as V 1 = V 2 T 1 T 2

10.  The volume of a sample of gas is 23.2 cm 3 at 20 o C. If the pressure remains unchanged, what is its volume at 80 o C?

11. G V 1 = 23.2 cm 3 T 1 = 20 o C = 293 K T 2 = 80 o C = 353K S V 2 = (23.2 cm 3 x 353 K) 293 K = 27.95 cm 3 RV 2 = ?P Therefore, when the temperature increased by 60 K, it increased the volume from 23.2 cm 3 to 27.95 cm 3 AV 1 = V 2 T 1 T 2 V 2 = V 1 T 2 T 1

12.  Compressed oxygen is widely used in hospitals and retirement homes. To make it easier to transport, the oxygen is cooled. How many degrees Celsius would the gas be if 200 L at 23 degrees Celsius is compressed to 40 L?

13. G V 1 = 200 L T 1 = 23 o C = 296 K V 2 = 40 L S T 2 = (40 L x 296 K ) 200 L = 59. 2 K = -213. 8 o C RT 2 = ?P Therefore, when the gas is compressed for transport it is cooled to - 213. 8 o C. AV 1 = V 2 T 1 T 2 T 2 = V 2 T 1 / V 1

14.  Read pg. 429 – 434 Questions:  page 432 # 13-14  page 434 # 16-19, 21  Worksheet: Charles’ Law