V1 = V2 T1 T2 Charles’ Law T and V have a direct, linear relationship Temp. (⁰C) Volume (cm3) 30 80 -10 70 -50 60 -90 50 -130 40 -170 -210 20 -250 10 Charles’ Law volume (cm3) Temperature (⁰C) -273⁰C T and V have a direct, linear relationship WARMUP Graph the data How are temperature and volume related? At what temperature would volume be zero? V1 = V2 T1 T2

More Gas Laws Charles Gay-Lussac Avagadro

V1 = V2 T1 T2 Answer: 2.3 cm3 (V1) = 2.5cm3 291.5K 311.5K 1. Some students think that teachers are full of hot air. If Ms Jonson takes a deep breath of air at 18.5°C and it heats to 38.5°C and the volume expands to 2.5 cm3, what was the starting volume of air? Assume constant pressure. V1 = V2 T1 T2 (V1) = 2.5cm3 291.5K 311.5K P1 = cons. P2 = cons. V1 = ? V2 = 2.5 cm3 T1 = 18.5°C T2 = 38.5°C 291.5 K 311.5 K Answer: 2.3 cm3

2. A sample of nitrogen occupies 250 mL at 25°C 2. A sample of nitrogen occupies 250 mL at 25°C. What volume will it occupy at 95°C? V1 T1 T2 Initial After P1 = P2 = V1 = V2 = T1 = T2 = ----- ----- 250 mL ? 95°C 368 K 25°C 298 K V2 = 310 mL

? T1 V1 T2 V2 Initial After P1 = P2 = V1 = V2 = T1 = T2 = T2 = 885 K 3. Oxygen gas is at a temperature of 40.0°C when it occupies a volume of 2.30 L. To what temperature (in °C) should it be raised to occupy a volume of 6.50 L? V1 T2 V2 Initial After P1 = P2 = V1 = V2 = T1 = T2 = ----- ----- 2.30 L 6.50 L 40.0°C 313 K ? T2 = 885 K -273 wait…need Celsius T2 = 612°C

Gay-Lussac’s Law Pressure How are pressure and temperature related? P and T have a direct, linear relationship Temperature

T2 = 309.3 K or 36°C P1 T1 P2 T2 1.8 atm 1.9 atm -------- -------- 1. Before a trip from New York to Boston, the pressure of an automobile tire is 1.8 atm at 20.°C. At the end of the trip, the pressure gauge reads 1.9 atm. What is the new Celsius temperature of the air inside the tire? (Assume the tires had a constant volume throughout the whole trip). T1 P2 T2 Initial After P1 = P2 = V1 = V2 = T1 = T2 = 1.8 atm 1.9 atm -------- -------- 20°C = 293 K ? T2 = 309.3 K or 36°C

? P2 P1 T1 T2 Initial After P1 = P2 = V1 = V2 = T1 = T2 = 2. Determine the pressure (in atm) when a constant volume of gas at standard pressure is heated from 20.0°C to 30.0°C. P1 T1 T2 Initial After P1 = P2 = V1 = V2 = T1 = T2 = 1.00 atm ? ----- ----- 20°C 293 K 30°C 303 K P2 = 1.03 atm

Rising Water Demo Procedure: Half fill the dish with tap water, add some food coloring, and mix. Light the candle. Place the 250ml beaker over the candle and press it to the bottom of the dish. Observe and explain! Why does the candle go out? Lack of oxygen. Why does the water level rise? Difference in air pressure causes suction. The cause of the pressure differential is interesting  think about the combustion reaction… The balanced reaction for the combustion of candle wax is given as: 2C20H42(s) + 61O2(g)  40CO2(g) + 42H2O(l) There is a net loss of 21 moles of gas as O2 is used and CO2 is made. Also, CO2 is more water soluble than O2 (stays dissolved in the liquid rather than being present as gas). Avogadro’s Law: as moles of gas decreases, volume decreases.

Avogadro’s Law 1 L of Helium 1 L of Oxygen V1 = V2 n1 n2 equal numbers of molecules (moles) of gas at the same temperature (with same KE and speed) 1 L of Helium 1 L of Oxygen should exert the same pressure,therefore should have the same volume V1 = V2 n1 n2 1 mole of ANY gas at STP will occupy 22.4 L molar volume = 22.4 L/mole

? V1 = V2 n1 n2 n2 = n1V2 V1 Initial After P1 = P2 = V1 = V2 = n1 = 1. 4.55 moles hydrogen gas occupy 75.0 mL at STP. Under the same STP conditions, how many moles of gas would be present in a 1680 ml sample? Initial After P1 = P2 = V1 = V2 = n1 = n2 = T1 = T2 = V1 = V2 n1 n2 ----- ----- 75.0 mL 1680 ml n2 = n1V2 V1 4.55 moles ? n2 = (4.55 moles)(1680ml) (75.0ml) ----- ----- n2 = 102 moles

n1 V1 2. A 5.6 mole sample of air is inside a 2.0L soda bottle. What volume would 0.34 moles of air occupy under the same conditions? V2 n2 V1 = V2 n1 n2 Initial After P1 = P2 = V1 = V2 = n1 = n2 = T1 = T2 = V2 = V1 n2 n1 ----- ----- 2.0L ? V2 = (2.0L)(0.34 moles) (5.6 moles) 5.6 moles 0.34 moles V2 = 0.12 L ----- -----

Gas Law Mystery Game Directions: You will be given three TIMED gas law problems. (Exactly 3 minutes to do each problem…so work together). Your job is to find an answer to each mystery problem. The sum of the answers to all three mystery problems is: -189.088 (ignoring sig fig rules)

a. You have 60.2 mL of gas. You increase the volume to 701 mL by increasing the temperature to 502°C while holding pressure constant. What was the starting temperature of the gas (in °C)? b. A 22.0 cm3 sample of oxygen gas expands to 78.0 cm3. If the original pressure was 570. mmHg, what is the new pressure (in atm)? c. A 1.2 mole sample of acetylene gas has an initial pressure of 699 mm Hg at 40.0°C. When 0.6 moles of acetylene gas are added to the glass container, the volume increases to 25.0 ml. What was the initial volume of acetylene? Assume constant temperature and pressure.

Add your answers to all 3 mystery problems! (volunteers to show work???)  a + b + c = ANSWER (no units) -206 °C 0.212 atm 16.7 ml + + = -189.088