1. Gas Laws and Heat Measurement Formulas (Use with the Gas Laws/Heat and Its Measurement Worksheets

2. Boyle’s Law A balloon has a volume of 8.0 L at a pressure of 90 kPa. What will be the new volume if the pressure drops to 40.0 kPa? Boyle’s Formula = P 1 V 1 = P 2 V 2 P 1 = 90.0 kPAV 1 = 8.0 P 2 = 40.0 kPAV 2 = ? So… P 1 V 1 (90.0 kPa) (8.0 L) V 2 = P 2 = 40.0 kPa = 18 L

3. Charles’ Law Charles’ law states that the volume of a gas varies directly with Kelvin temperature. You add 273 to Celsius ( o C + 273) and find the proportions: V 1 V 2 T 1 = T 2 (Pressure must remain constant) A 3.0 L balloon at 10.0 o C was placed in a container of ice water (2.0 o C). What is the resulting volume? V 1 = 3.0 L T 1 = 283 K (10.0 o C + 273) V 2 = ? T 2 = 275 K (2.0 o C + 273)

4. Charles’ Law 3.0 L = ? L 283 K 275K (3.0 x 275) 825 283 = 283 = 2.92 L Notice that, when the temperature was lowered, the volume became lower.

5. Heat and Its Measurement Heat energy (Q) can be measured in units of calories (Cal) or joules (J). When there is a temperature change (Δ), you find the heat energy by mass x ΔT (T F – T I ) x the specific heat capacity of the substance.

6. Heat and Its Measurement Example: How many joules of heat are given off when 8.0 g of water heats up from 20 o C to 80 o C? The specific heat of water is 4.18 j/g o C. Mass = 8.0 ΔT = 60 specific heat = 4.18 Q = Mass x ΔT x specific heat Q = 8.0 x 60 x 4.18 = 2006.4 J

7. Heat and Its Measurement Example: How many joules of heat are given off when 8.0 g of water cools from 80 o C to 20 o C? The specific heat of water is 4.18 j/g o C. Mass = 8.0 ΔT = - 60 specific heat = 4.18 Q = Mass x ΔT x specific heat Q = 8.0 x 60 x 4.18 = - 2006.4 J Because the water is cooling, it is an exothermic change (losing heat)

8. Heat and Its Measurement If you are measuring heat during a phase change, then use the formula Q = mass x heat of fusion (freezing) or vaporization Ex: How many calories are given off when 60g of water at 0 o C freezes? The heat of fusion is 79.72 cal/g. Q = 60g x 79.72 = 4783.2 Cal