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## Chapter 13 Gases

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**Some Gases in Our Lives**Air: oxygen O2 nitrogen N2 ozone O3 argon Ar carbon dioxide CO2 water H2O Noble gases: helium He neon Ne krypton Kr xenon Xe Other gases: fluorine F2 chlorine Cl2 ammonia NH3 methane CH4 carbon monoxide CO nitrogen dioxide NO2 sulfur dioxide SO2**13.1 The Nature of Gases**• Gases are compressible Why can you put more air in a tire, but can’t add more water to a glass full of water? • Gases have low densities Dsolid or liquid = 2 g/mL Dgas 2 g/L**Nature of Gases**1. Why does a round balloon become spherical when filled with air? 2. Suppose we filled this room halfway with water. Where would pressure be exerted?**Nature of Gases**• Gases fill a container completely and uniformly • Gases exert a uniform pressure on all inner surfaces of their containers**Barometers**760 mmHg atm pressure**Unit of Pressure**One atmosphere (1 atm) • Is the average pressure of the atmosphere at sea level • Is the standard of pressure • P = Force Area 1.00 atm = 760 mm Hg = 760 torr**Types of Pressure Units**Pressure Used in 760 mm Hg or 760 torr Chemistry 14.7 lb/in.2 U.S. pressure gauges 29.9 in. Hg U.S. weather reports 101.3 kPa (kilopascals) Weather in all countries except U.S. 1.013 bars Physics and astronomy**Check Yourself**A.The downward pressure of the Hg in a barometer is _____ than (as) the weight of the atmosphere. 1) greater 2) less 3) the same B.A water barometer has to be 13.6 times taller than Hg barometer (DHg = 13.6 g/mL) because 1) H2O is less dense 2) Hg is heavier 3) air is more dense than H2O**Solutions**A.The downward pressure of the Hg in a barometer is 3) the same (as) the weight of the atmosphere. B.A water barometer has to be 13.6 times taller than Hg barometer (DHg = 13.6 g/mL) because 1) H2O is less dense**Check Yourself**A. What is 475 mm Hg expressed in atm? 1) 475 atm 2) 0.625 atm 3) 3.61 x 105 atm B. The pressure of a tire is measured as 29.4 psi. What is this pressure in mm Hg? 1) 2.00 mm Hg 2) 1520 mm Hg 3) 22,300 mm Hg**Solutions**A. What is 475 mm Hg expressed in atm? 485 mm Hg x 1 atm = 0.625 atm (B) 760 mm Hg B. The pressure of a tire is measured as 29.4 psi. What is this pressure in mm Hg? 29.4 psi x 1.00 atm x 760 mmHg = 1.52 x 103 mmHg 14.7 psi 1.00 atm (B)**13.2 P and V Changes – Boyles Law**P1 P2 V1 V2**Boyle's Law**• The pressure of a gas is inversely related to the volume when T does not change • Then the PV product remains constant P1V1 = P2V2 P1V1= 8.0 atm x 2.0 L = 16 atm L P2V2= 4.0 atm x 4.0 L = 16 atm L**PV Problem**Freon-12, CCl2F2, is used in refrigeration systems. What is the new volume (L) of a 1.6 L sample of Freon gas initially at 50 mm Hg after its pressure is changed to 200 mm Hg at constant T?**PV Calculation**Prepare a list of given information: Initial conditions Final conditions P1 = 50 mm Hg P2 = 200 mm Hg V1 = 1.6 L V2 = ? ?**Find New Volume (V2)**Solve for V2: P1V2 = P2V2 V2 = V1 x P1 /P2 V2 = 1.6 L x 50 mm Hg = 0.4 L 200 mm Hg**Check Yourself**A sample of nitrogen gas is 6.4 L at a pressure of 0.70 atm. What will the new volume be if the pressure is changed to 1.40 atm? (T constant) Explain. 1) 3.2 L 2) 6.4 L 3) 12.8 L**Solution**A sample of nitrogen gas is 6.4 L at a pressure of 0.70 atm. What will the new volume be if the pressure is changed to 1.40 atm? (T constant) 6.4 L x 0.70 atm = 3.2 L (1) 1.40 atm Volume must decrease to cause an increase in the pressure**Check Yourself**A sample of helium gas has a volume of 12.0 L at 600. mm Hg. What new pressure is needed to change the volume to 36.0 L? (T constant) Explain. 1) 200. mmHg 2) 400. mmHg 3) 1200 mmHg**Solution**A sample of helium gas has a volume of 12.0 L at 600. mm Hg. What new pressure is needed to change the volume to 36.0 L? (T constant) Explain. 600. mm Hg x 12.0 L = 200. mmHg (1) 36.0 L Pressure decrease when volume increases.**13.3 Charles’ Law – Volume & Temp.**V = 125 mL V = 250 mL T = 273 K T = 546 K Observe the V and T of the balloons. How does volume change with temperature?**Charles’ Law: V and T**At constant pressure, the volume of a gas is directly related to its absolute (K) temperature V1 = V2 T1 T2**Check Yourself**Use Charles’ Law to complete the statements below: 1. If final T is higher than initial T, final V is (greater, or less) than the initial V. 2. If final V is less than initial V, final T is (higher, or lower) than the initial T.**Solution**V1 = V2 T1 T2 1. If final T is higher than initial T, final V is (greater) than the initial V. 2. If final V is less than initial V, final T is (lower) than the initial T.**V and T Problem**A balloon has a volume of 785 mL on a Fall day when the temperature is 21°C. In the winter, the gas cools to 0°C. What is the new volume of the balloon?**VT Calculation**Complete the following setup: Initial conditions Final conditions V1 = 785 mL V2 = ? T1 = 21°C = 294 K T2 = 0°C = 273 K V2 = _______ mL x __ K = _______ mL V1 K Check your answer: If temperature decreases, V should decrease.**Check Yourself**A sample of oxygen gas has a volume of 420 mL at a temperature of 18°C. What temperature (in °C) is needed to change the volume to 640 mL? 1) 443°C 2) 170°C 3) - 82°C**Solution**A sample of oxygen gas has a volume of 420 mL at a temperature of 18°C. What temperature (in °C) is needed to change the volume to 640 mL? 2) 170°C T2 = 291 K x 640 mL = 443 K 420 mL = 443 K - 273 K = 170°C**Gay-Lussac’s Law: P and T**The pressure exerted by a confined gas is directly related to the temperature (Kelvin) at constant volume. P (mm Hg) T (°C) 936 100 761 25 691 0**Check Yourself**Use Gay-Lussac’s law to complete the statements below: 1. When temperature decreases, the pressure of a gas (decreases or increases). 2. When temperature increases, the pressure of a gas (decreases or increases).**Solution**1. When temperature decreases, the pressure of a gas (decreases). 2. When temperature increases, the pressure of a gas (increases).**PT Problem**A gas has a pressure at 2.0 atm at 18°C. What will be the new pressure if the temperature rises to 62°C? (V constant) T = 18°C T = 62°C**PT Calculation**? P1 = 2.0 atm T1 = 18°C + 273 = 291 K P2 = ? T2 = 62°C + 273 = 335 K What happens to P when T increases? P increases (directly related to T) P2 = P1 x T2 T1 P2 = 2.0 atm x K = atm K**Check Yourself**Complete with 1) Increases 2) Decreases 3) Does not change A. Pressure _____, when V decreases B. When T decreases, V _____. C. Pressure _____ when V changes from 12.0 L to 24.0 L (constant n and T) D. Volume _____when T changes from 15.0 °C to 45.0°C (constant P and n)**Solution**A. Pressure 1) Increases, when V decreases B. When T decreases, V 2) Decreases C. Pressure 2) Decreases when V changes from 12.0 L to 24.0 L (constant n and T) D. Volume 1) Increases when T changes from 15.0 °C to 45.0°C (constant P and n)**13.4 Volume and Moles**How does adding more molecules of a gas change the volume of the air in a tire? If a tire has a leak, how does the loss of air (gas) molecules change the volume?**Avogadro’s Law**When a gas is at constant T and P, the V is directly proportional to the number of moles (n) of gas V1 = V2 n1 n2 initial final**Combined Gas Law**P1V1 = P2V2 T1 T2 Rearrange the combined gas law to solve for V2 P1V1T2 = P2V2T1 V2 = P1V1T2 P2T1**Combined Gas Law**P1V1 = P2V2 T1 T2 Isolate V2 P1V1T2 = P2V2T1 V2 = P1V1T2 P2T1**Check Yourself**Solve the combined gas laws for T2.**Solution**Solve the combined gas law for T2. (Hint: cross-multiply first.) P1V1 = P2V2 T1 T2 P1V1T2 = P2V2T1 T2 = P2V2T1 P1V1**Combined Gas Law Problem**A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29°C. What is the new temperature(°C) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm?**Data Table**List given information: P1 = 0.800 atm V1 = 0.180 L T1 = 302 K P2 = 3.20 atm V2= 90.0 mL T2 = ?? ??**Solution**Solve for T2 Enter data T2 = 302 K x atm x mL = K atm mL T2 = K - 273 = °C**Calculation**Solve for T2 T2 = 302 K x 3.20 atm x 90.0 mL = 604 K 0.800 atm 180.0 mL T2 = 604 K - 273 = 331 °C**Check Yourself**A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. What is the temperature in °C when the gas has a volume of 0.315 L and a pressure of 802 mm Hg?**Solution**T1 = 308 K T2 = ? V1 = 675 mL V2 = 0.315 L = 315 mL P1 = 0.850 atm P2 = 802 mm Hg = 646 mm Hg T2 = 308 K x 802 mm Hg x 315 mL 646 mm Hg 675 mL P inc, T inc V dec, T dec = 178 K - 273 = - 95°C**Check Yourself**True (1) or False(2) 1.___The P exerted by a gas at constant V is not affected by the T of the gas. 2.___ At constant P, the V of a gas is directly proportional to the absolute T 3.___ At constant T, doubling the P will cause the V of the gas sample to decrease to one-half its original V.**Solution**True (1) or False(2) 1. (2)The P exerted by a gas at constant V is not affected by the T of the gas. 2. (1) At constant P, the V of a gas is directly proportional to the absolute T 3. (1) At constant T, doubling the P will cause the V of the gas sample to decrease to one-half its original V.