Chemistry Chapter 12. Characteristics of Gases. Characteristics of Gases 1/31/2012. Gases and Liquids

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Importance of Gases Chemistry Chapter 12 Gases and Liquids Airbags fill with N 2 gas in an accident. Gas is generated by the decomposition of sodium azide, NaN 3. 2 NaN 3 ---> 2 Na + 3 N 2 THREE STATES OF MATTER General Properties of Gases There is a lot of free space in a gas. Gases can be expanded infinitely. Gases fill containers uniformly and completely. Gases diffuse and mix rapidly. Characteristics of Gases Gases are Fluids They flow just as liquids do Molecules are in constant motion Can be explained by the Kinetic-molecular Theory States there is no attractive or repulsive forces between ideal gas molecules Allows collisions to be elastic Gases have low densities Gases are mostly empty space Based on the distance between gas molecules Characteristics of Gases Gases are highly compressible It does not take much pressure to compress a gas It takes a tremendous amount of pressure to compress a liquid or solid Gases completely fill a container and exert pressure equally in all directions The temperature of a gas determines the average kinetic energy of its particles Expressed using the kinetic-energy equation KE = (1/2)mv 2 the higher the temperature the greater the kinetic energy 1

The Atmosphere is a Sea of Gases The atmosphere exerts pressure on the earth Force is measured in Newtons (N), and area is measured in Meters Squared (m 2 ) Therefore the SI unit for pressure is the Pascal (Pa) 1 pascal (Pa) = 1 N/m 2 Pressure Pressure of air is measured with a BAROMETER (developed by Torricelli in 1643) Hg rises in tube until force of Hg (down) balances the force of atmosphere (pushing up). (Just like a straw in a soft drink) Pressure of Hg pushing down related to Hg density column height Standard Temperature and Pressure We use a standard for comparison Standard Temperature and Pressure (STP) is the standard conditions at 0⁰ C and 1atm STP = 0⁰ C and 1atm Summary of units in table 12-1 on page 428 Pressure Column height measures Pressure of atmosphere 1 standard atmosphere (atm) * = 760 mm Hg (or torr) * = 29.92 inches Hg = 14.7 pounds/in 2 (psi) * = 101.3 kpa(si unit is PASCAL) * = about 10m (34 ft) of water! * Memorize these! Pressure Conversions A. What is 475 mm Hg expressed in atm? 1 atm 475 mm Hg x 760 mm Hg = 0.625 atm B. The pressure of a tire is measured as 29.4 psi. What is this pressure in mm Hg? 760 mm Hg 29.4 psi x = 1.52 x 10 14.7 psi 3 mm Hg Earth s Climate is Changing Radiant energy from the sun warms the Earth and in turn warms the atmosphere Greenhouse gases trap radiant energy All objects above 0 K radiate energy Greenhouse gases trap radiant energy and send it back to Earth Causes the Greenhouse Effect 2

Earth s Climate is Changing Earth is becoming warmer Primary is believed to be increases levels of CO 2 Over the last 100 years the average temperature has risen about 0.6 ⁰ C Chlorofluorocarbons (CFC s) also increase the greenhouse effect Once used in refrigeration and air conditioning Now banned in the US but not other places Earth s Climate is Changing The ozone is depleting Ozone (O 3 ) protects against ultraviolet radiation This can cause mutations in DNA and can cause skin cancer CFC molecules produce free radicals which deplete the ozone layer Produces a chain reaction that can destroy many thousand ozone molecules There is now a ozone hole over the south pole Pressure Conversions A. What is 2 atm expressed in torr? B. The pressure of a tire is measured as 32.0 psi. What is this pressure in kpa? Gas Laws Use specific variables for gas laws P = pressure exerted by the particles T = temperature in kelvins of the particles V = volume occupied by the particles n = number of moles of the particles 3

Boyle s Law P α1/v This means Pressure and Volume are INVERSELY PROPORTIONAL if moles and temperature are constant (do not change). For example, P goes up as V goes down. P 1 V 1 = P 2 V 2 Robert Boyle (1627-1691). 1691). Son of Earl of Cork, Ireland. A bicycle pump is a good example of Boyle s law. As the volume of the air trapped in the pump is reduced, its pressure goes up, and air is forced into the tire. Boyle s Law Boyle s Law Sample Problem Sample Problem Cont. The gas in a balloon has a volume of 7.5 L at 100 kpa. The balloon is released into the atmosphere, and the gas in it expands to volume of 11 L. What is the pressure on the balloon at the new volume? V 1 = 7.5L P 1 = 100 kpa V 2 = 11 L P 2 =? kpa Boyle s Law at Work Boyle s Law: Summary Pressure * Volume = Constant Doubling the pressure reduces the volume by half. Conversely, when the volume doubles, the pressure decreases by half. p 1 * V 1 = p 2 * V 2 With constant temperature and amount of gas, you can use these relationships to predict changes in pressure and volume. By knowing 3 variables you can solve for the fourth 4

Partial Pressure In containers with more than one kind of gas each will exert a pressure The pressures of all gases added together is the total pressure Dalton s Law of Partial Pressures 2 H 2 O 2 (l) ---> 2 H 2 O (g) + O 2 (g) 0.32 atm 0.16 atm What is the total pressure in the flask? P total in gas mixture = P A + P Therefore, P total = P H + P 2O O = 0.48 atm 2 + P B +... Dalton s Law: total P is sum of PARTIAL pressures. Dalton s Law Mole Fractions and Partial Pressures The air we breath contains 0.78 mol fraction of N2, 0.21 of O2 and 0.01 mol argon. The mixture has a total pressure of 0.97 atm. What is the partial pressure of each gas? John Dalton 1766-18441844 Mole Fractions and Partial Pressures Health Note When a scuba diver is several hundred feet under water, the high pressures cause N 2 from the tank air to dissolve in the blood. If the diver rises too fast, the dissolved N 2 will form bubbles in the blood, a dangerous and painful condition called "the bends". Helium, which is inert, less dense, and does not dissolve in the blood, is mixed with O 2 in scuba tanks used for deep descents. 5

Collecting a gas over water Gases, since they mix with other gases readily, must be collected in an environment where mixing can not occur. The easiest way to do this is under water because water displaces the air. So when a gas is collected over water, that means the container is filled with water and the gas is bubbled through the water into the container. Thus, the pressure inside the container is from the gas AND the water vapor. This is where Dalton s Law of Partial Pressures becomes useful. Table of Vapor Pressures for Water A student collects some hydrogen gas over water at 20 degrees C and 768 torr. What is the pressure of the H 2 gas? Solve This! 768 torr 17.5 torr = 750.5 torr Charles s Law If n and P are constant, then V αt V and T are directly proportional. V 1 V 2 T 1 = T 2 If one temperature goes up, the volume goes up! Jacques Charles (1746-1823). Isolated boron and studied gases. Balloonist. Charles Law at Work Application of Charles Law As the temperature increases, the volume increases. Conversely, when the temperature decreases, volume decreases. V 1 / T 1 = V 2 / T 2 V 1 = initial volume T 1 = initial temperature V 2 = final volume T 2 = final temperature If you know three of the four, you can calculate the fourth. 6

Application of Charles Law V 1 / T 1 = V 2 / T 2 V 1 = 2.5 liters T 1 = 250 K V 2 = 4.5 liters T 2 =? Solving for T 2, the final temperature equals 450 K. So, increasing the volume of a gas at constant pressure from 2.5 to 4.5 liters results in a temperature increase of 200 K. Gay-Lussac s Law If n and V are constant, then P αt P and T are directly proportional. P 1 P 2 T 1 = T 2 If one temperature goes up, the pressure goes up! Joseph Louis Gay- Lussac (1778-1850) 1850) Combined Gas Law The good news is that you don t have to remember all three gas laws! Since they are all related to each other, we can combine them into a single equation. BE SURE YOU KNOW THIS EQUATION! P 1 V 1 P 2 V 2 = T 1 T 2 Combined Gas Law If you should only need one of the other gas laws, you can cover up the item that is constant and you will get that gas law! P 1 V 1 T 1 P 2 V 2 T 2 Boyle s Law Charles Law Gay-Lussac s Law 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? Set up Data Table P 1 = 0.800 atm V 1 = 180 ml T 1 = 302 K P 2 = 3.20 atm V 2 = 90 ml T 2 =?? Calculation P 1 = 0.800 atm V 1 = 180 ml T 1 = 302 K P 2 = 3.20 atm V 2 = 90 ml T 2 =?? P 1 V 1 P 2 V 2 = P 1 V 1 T 2 = P 2 V 2 T 1 T 1 T 2 T 2 = P 2 V 2 T 1 P 1 V 1 T 2 = 3.20 atm x 90.0 ml x 302 K 0.800 atm x 180.0 ml T 2 = 604 K -273 = 331 C = 604 K 7

And now, we pause for this commercial message from STP Standard Pressure = 1 atm (or an equivalent) Standard Temperature = 0 deg C (273 K) OK, so it s really not THIS kind of STP STP in chemistry stands for Standard Temperature and Pressure STP allows us to compare amounts of gases between different pressures and temperatures IDEAL GAS LAW P V = n R T Brings together gas properties. Can be derived from experiment and theory. BE SURE YOU KNOW THIS EQUATION! P = Pressure V = Volume T = Temperature N = number of moles Using PV = nrt R is a constant, called the Ideal Gas Constant Instead of learning a different value for R for all the possible unit combinations, we can just memorize onevalue and convert the units to match R. R = 0.0821 L atm Mol K R = 0.0821 L atm Mol K Using PV = nrt How much N 2 is required to fill a small room with a volume of 960 cubic feet (27,000 L) to 745 mm Hg at 25 o C? Solution 1. Get all data into proper units V = 27,000 L T = 25 o C+ 273 = 298 K P = 745 mm Hg (1 atm/760 mm Hg) = 0.98 atm And we always know R, 0.0821 L atm/ mol K Using PV = nrt How much N 2 is req dto fill a small room with a volume of 960 cubic feet (27,000 L) to P = 745 mm Hg at 25 o C? Solution 2. Now plug in those values and solve for the unknown. PV = nrt GAS DENSITY Low density RT n = RT (0.98 atm)(2.7 x 10 4 L) (0.0821 L atm/k mol)(298 K) n = 1.1 x 10 3 mol (or about 30 kg of gas) 22.4 L of ANY gas AT STP = 1 mole High density 8

Gases and Stoichiometry 2 H 2 O 2 (l) ---> 2 H 2 O (g) + O 2 (g) Decompose 1.1 g of H 2 O 2 in a flask with a volume of 2.50 L. What is the volume of O 2 at STP? Bombardier beetle uses decomposition of hydrogen peroxide to defend itself. Gases and Stoichiometry 2 H 2 O 2 (l) ---> 2 H 2 O (g) + O 2 (g) Decompose 1.1 g of H 2 O 2 in a flask with a volume of 2.50 L. What is the volume of O 2 at STP? Solution 1.1 g H 2 O 2 1 mol H 2 O 2 1 mol O 2 22.4 L O 2 34 g H 2 O 2 2 mol H 2O 2 1 mol O 2 = 0.36 L O 2 at STP GAS DIFFUSION AND EFFUSION diffusionis is the gradual effusionis is the mixing of molecules of movement of molecules through a different gases. small hole into an empty container. Vapor Pressure The vapor pressure of a liquid equals 1 atm at the boiling point of the liquid Therefore water s vapor pressure equals 1 atm at 100⁰C Phase Diagram 9