1. Warmup A 40.0m 3 weather balloon at standard temperature encounters a sudden icy cool breeze during its journey into the Alaskan sky. If the weather balloon decreases to 37.9 m 3, calculate the temperature of the icy cool breeze. Assume pressure is held constant at 610 mmHg.

2. More Gas Laws and two demos

3. 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: 2C 20 H 42 (s) + 61O 2 (g)  40CO 2 (g) + 42H 2 O(l) There is a net loss of 21 moles of gas as O 2 is used and CO 2 is made. Also, CO 2 is more water soluble than O 2 (stays dissolved in the liquid rather than being present as gas). Avogadro’s Law: as moles of gas decreases, volume decreases.

4. Avogadro’s Law 1 L of Helium 1 L of Oxygen equal numbers of molecules (moles) of gas at the same temperature (with same KE and speed) V 1 = V 2 n 1 n 2 should exert the same pressure, therefore should have the same volume 1 mole of ANY gas at STP will occupy 22.4 L molar volume = 22.4 L/mole

5. 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? V1V1 n1n1 V2V2 InitialAfter P 1 =P 2 = V 1 =V 2 = n 1 =n 2 = T 1 =T 2 = 75.0 mL ? ----- n 2 = 102 moles V 1 = V 2 n 1 n 2 n2n2 ----- 1680 ml 4.55 moles n 2 = n 1 V 2 V 1 n 2 = (4.55 moles)(1680ml) (75.0ml)

6. 2. A 5.6 mole sample of air is inside a 2.0L soda bottle. What volume would 5.6 x 10 19 molecules of air occupy under the same conditions? V1V1 n1n1 V2V2 InitialAfter P 1 =P 2 = V 1 =V 2 = n 1 =n 2 = T 1 =T 2 = 9.3x10 -5 moles ? ----- V 2 = 3.3 x 10 -5 L (or 33 ul) V 1 = V 2 n 1 n 2 n2n2 ----- 2.0L 5.6 moles V 2 = V 1 n 2 n 1 V 2 = (2.0L)(9.3x10 -5 moles) (5.6 moles) (5.6 x 10 19 molec.) (1 moles) = (6.02 x 10 23 molec.)

7. P 1 V 1 = P 2 V 2 T 1 T 2 (101.3 kPa)(V 1 )=(99.0 kPa) (466 ml ) (273K) (-4.6°C+273) (101.3kPa)(V 1 )(268.4°K)=(99.0 kPa)(466 ml)(273K) (101.3 kPa)(268.4K) (101.3 kPa)(268.4K) V 1 = 463ml Ex1. A half-full water bottle (at STP) travels by car and is put in a freezer high up in the mountains, changing the temperature to -4.6° C. The volume changes to 466 ml and the pressure decreases to 99.0 kPa. What was the original volume of the gas inside the water bottle?

8. Ex 2: A helium balloon at standard, constant temperature is placed in a vacuum, decreasing the pressure to 0.843 atm. The volume expands from 1.78 L to 4.98 L. What was the pressure of the balloon in the beginning? Use Boyle’s Law Use Combined Gas Law same answer…. I swear!

9. P 1 V 1 = P 2 V 2 T 1 T 2 P 1 (1.78 L)=(0.843 atm) (4.98 L) (273K) (273K) P 1 (1.78 L) (273K) =(0.843 atm)(4.98 L)(273K) (1.78 L) (273K) (1.78 L) (273K) P 1 = 2.36 atm

10. He CO 2 H2OH2O 0.7 atm 0.8 atm 2.3 atm Dalton’s Law of Partial Pressures the total pressure of a mixture of gases is equal to the sum of the partial pressures of the individual gases P total = P 1 + P 2 + P 3 + …

11. Ex 1: A container holds three gases: oxygen, carbon dioxide, and helium. The partial pressures of the three gases are 2.00 atm, 3.00 atm, and 4.00 atm, respectively. What is the total pressure inside the container? P total = P 1 + P 2 + P 3 + … P total = P O2 + + P CO2 + P He P total = 2.00 atm + 3.00 atm + 4.00 atm P total = 9.00 atm

12. Gas Collected Over Water: you need to account for the additional vapor pressure : P total = P gas + P H 2 O Note: most of the time, the total pressure is equal to the pressure inside the room!

13. Ex 2: Helium gas is collected over water at 25 °C. What is the partial pressure of the helium, given that the barometric pressure is 750.0 mm Hg? P total = P He + P H 2 O 750.0 mm Hg = P He + 23.8 mm Hg P He = 726.2 mm Hg

14. Ex 3: A student has stored 100.0 mL of neon gas over water on a day when the temperature is 28.0°C. If the barometer in the room reads 743.3 mm Hg, what is the pressure of the neon in its container? P total = P Ne + P H 2 O 743.3 mmHg = P Ne + 28.3 mm Hg P Ne = 715.0 mm Hg

15. Ex 4: A container has two gases, helium and argon. Suppose that the mixture contains 30% helium by volume. Calculate the partial pressure of helium AND argon if the total pressure of the container is 4.00 atm. Total pressure = 4.00 atm. 30% of 4.00 is 1.2 atm (0.30 X 4.00 atm = 1.2 atm) 70 % of 4.00 is 2.8 atm (0.70 X 4.00 atm = 2.8 atm) P He = 1.2 atm AND P Ar = 2.8 atm Double Check: P total = P He + + P Ar P total = 1.2 atm + 2.8 atm = 4.00 atm

16. Ex 5: A certain mass of oxygen was collected over water when potassium chlorate was decomposed by heating. The volume of the oxygen sample collected was 720 mL at 25°C and an atmospheric pressure of 750 mmHg. What would the volume of the oxygen be at STP? P 1 V 1 = P 2 V 2 T 1 T 2 P 1 (720ml) = (760mmHg)V 2 (273 + 25 ⁰ C) (273K) P total = P O2 + P H 2 O 750 mmHg = P O2 + 23.8 mmHg P O2 = 726.2 mmHg Answer: 630 mL (726.2 mmHg)(720ml) = (760mmHg)V 2 (298K) (273K)

17. Ex 6: An 80.0 mL xenon gas sample is collected over water at 27.0°C and 0.200 atm. What volume will the same gas sample occupy at standard temperature and standard pressure? (volunteer?)

18. Molar Volume Class Activity I need an assistant to: 1.Measure the mass of magnesium. 2.Measure acid volume and pour it in. 3.Read graduated cylinder volume. 4.Report out the room temperature. Finish this for HW (packet).