Chapter 11 Gases 11.1 Gases & Pressure

Number of Particles Increases Defining Gas Pressure How are number of particles and pressure related? Pressure –force per unit area that particles exert on walls of their container Gas particles collide with walls = greater pressure Pressure is directly proportional to number of particles. Number of Particles Increases Pressure Increases

Temperature & Pressure Higher temperature results in more kinetic energy! IF the volume of container remains constant and IF the amount of gas remains constant: the pressure of a gas increases in direct proportion to the Kelvin temperature. (Kelvin Temp = Celsius Temp + 273) Volume of a gas at constant pressure is directly proportional to Kelvin temp. Pressure of Gas Increases Kelvin Temperature Increases

Temperature Conversions Kelvin & Celsius Tk = (Tc + 273) Tc = (Tk - 273) Fahrenheit & Celsius Tf = (9/5 Tc) + 32 Tc = (Tf - 32) 5/9

Devices to Measure Pressure Barometer: an instrument that measures pressure exerted by the atmosphere. Invented in 1600’s by an Italian scientist, Evangelista Torricelli Height of column of mercury shows the atmospheric pressure. (atm)

Atmospheric Pressure We live at the bottom of an ocean of air; highest pressure occurs at the lowest altitudes! Standard Atmosphere is pressure that supports a 760 mm column of mercury. 1.00 atm = 760 mm Hg

Devices to Measure Pressure Pressure Gauge: instrument used to measure pressure inside a tire or oxygen tank. Tire Pressure Blood Pressure

Absolute Pressure When measuring tire pressure; you measure pressure ABOVE atmosphere pressure. Recommended pressures for tires are gauge pressures. Absolute pressure – the TOTAL pressure of all gases including the atmosphere. Q: How would you figure it for an inflated tire? A: Add barometric pressure to the gauge pressure.

Pressure Units SI unit for measuring pressure is the pascal (pa) after the French physicist Blaise Pascal (1600s) A kilopascal (kPa) is 1000 pascals and is more commonly used. Equivalent Pressures 1.00 atm 101300 Pa 101.3 kPa 760 mm Hg 760 Torr 14.7 psi

Sample Calculations Express 1.56 atm in kPa. Convert 801 mm Hg to Pa. How many psi are equivalent to 95.6 kPa?

Answers 101.3 kPa 1.56 atm 801 mm Hg 95.6 kPa X = 158 kPa 1.00 atm 14.7 psi X = 13.9 psi 101.3 kPa

Dalton’s Law of Partial Pressures The pressure exerted by each gas in a mixture The total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases Dalton’s Law: PT = P1 + P2 + P3 + …

Practice Calculate the partial pressure in mm Hg exerted by the four main gases in air at 760 mm Hg: nitrogen, oxygen, argon and carbon dioxide. Their abundance by volume is 78.08%, 20.95%, 0.934% and 0.035%, respectively. N2= 593.4 mm Hg O2= 159.2 mm Hg Ar = 7.10 mm Hg CO2= 0.27 mm Hg

Gases Collected by Water Displacement Gases produced in the lab are often collected by the displacement of water in a collection bottle Water vapor will be present in the collected gas, and it exerts a pressure Water vapor pressure = PH20 Water vapor pressure increases with temperature (Appendix A, Table-8) Pressure of the dry gas P atm = P gas + P H20 so… P gas = P atm – P H2O

Practice A student has stored 100.0 mL of neon gas over water on a day when the temperature if 27.0 °C. If the barometer in the room reads 743.3 mm Hg, what is the pressure of the neon gas in its container? P atm = P Ne + P H2O P Ne = P atm – P H2O P Ne = 743.3 mm Hg – 26.7 mm Hg =716.6 mm Hg

Chapter 11 Gases 11.2 The Gas Laws

Pressure & Volume are Inversely Proportional! In the 1600s, Robert Boyle did many experiments involving gases. He did these experiments at constant temperature. if pressure increases, volume decreases if pressure decreases, volume increases Pressure & Volume are Inversely Proportional!

Boyle’s Law Graph

Boyle’s Law V1P1=V2P2 Where: V1 = initial volume P1 = initial pressure V2 = final volume P2 = final pressure

Kinetic Explanation of Boyle’s Law As volume is reduced, number of particles and temperature remains constant but number of collisions with the walls of the container increases. There is a smaller area of space for the same number of particles to move around, so pressure increases.

(5.5 L) x (1.6 atm) = (x L) x (1.2 atm) Practice If you have 5.5 L of gas at a pressure of 1.6 atm, and the pressure changes to 1.2 atm, what is your new volume? V1P1 = V2P2 (5.5 L) x (1.6 atm) = (x L) x (1.2 atm) x = 7.3 L

Temperature & Volume are Directly Proportional! Jacques Charles did experiments concerning gases held at constant pressure, while varying temperature. As Kelvin temperature increases, volume increases. As Kelvin temperature decreases, volume decreases. Temperature & Volume are Directly Proportional!

Charles’s Law Graph

Charles’s Law V1 V2 T1 T2 = Where: V1 = initial volume T1 = initial temperature V2 = final volume T2 = final temperature =

Kinetic Explanation of Charles’s Law When a gas is heated, its temperature increases, which means the kinetic energy of the particles has increased. Then the particles begin to move faster, which causes its volume to increase. The reverse occurs as the temperature begins to fall.

Practice 3.0 L of Helium gas is in a balloon at 22 C and a pressure of 760 mm Hg. If the temperature rises to 31 C and the pressure remains constant, what will the new volume be? (remember to convert any temperatures to KELVIN!!!) V1 V2 T1 T2 3.0 L V2 (273 + 22C) (273 + 31C ) V2 = (3.0 L x 304 K) / 295 K V2 = 3.1 L = =

Pressure & Temperature From the prior relationships of volume & pressure, and temperature & volume, it could be concluded that a relationship exists between pressure & temperature. For a given mass of a dry gas, if the volume is constant, the pressure is directly proportional to the Kelvin temperature Pressure & Temperature are Directly Proportional!

Gay-Lussac’s Law Graph

Gay-Lussac’s Law P1 P2 T1 T2 = Where: P1 = initial pressure T1 = initial temperature P2 = final pressure T2 = final temperature =

Practice At 27 C, Helium gas is in a balloon at pressure of 760 mm Hg. If the temperature rises to 31 C, what will the new pressure be? (remember to convert any temperatures to KELVIN!!!) P1 P2 T1 T2 760 mm Hg P2 (273 + 27C) (273 + 31C ) P2 = (760 mm Hg x 304 K) / 300 K P2 = 770 mm Hg = =

Combined Gas Law All 3 Gas Laws require one variable to be held constant. How can we solve a problem when all 3 variables; volume, pressure & temperature change? Since 2 out of the 3 laws always have a variable in common, there should be a way to relate these laws into one formula. This new formula is called the Combined Gas Law.

Combined Gas Law P1 V1 = P2 V2 T1 T2 Gay-Lussac’s Law P1 V1 = P2 V2 T1 T2 Where: P1, V1 & T1 are initial values P2, V2 & T2 are final values *0C & 1 atm = Standard Temperature & Pressure, or STP Boyle’s Law Charles’s Law

Practice 154 mL of Carbon Dioxide gas is at a pressure of 121 kPa and a temperature of 117C. What volume would this gas occupy at STP? (Remember to convert your temps to Kelvin!!!) 1 atm = 101.3 kPa P1V1/ T1 = P2V2/ T2 (154 mL)(121 kPa) = (101.3 kPa)(V2) (117C + 273) (0C + 273) V2 = (154 mL)(121 kPa)(273 K) (390 K)(101.3 kPa) V2 = 129 mL

11.3 Gas Volumes & the Ideal Gas Law Chapter 11 Gases 11.3 Gas Volumes & the Ideal Gas Law

The Law of Combining Gas Volumes If one volume of water, H2O, is decomposed, one volume of oxygen will be formed and 2 volumes of hydrogen will be formed. How can 3 volumes be formed from only 1 initial volume? 1 L H2O 1 L O2 1 L H2 1 L H2 + +

The Law of Combining Gas Volumes The law of combining volumes states that in chemical reactions involving gases, the ratio of the gas volumes is a small whole number. All of the gases are at the same temperature & pressure, each of the identical flasks contains the same number of molecules. Notice how the combining ratio: 2 volumes H2 : 1 volume O2 : 2 volumes H2O leads to a result in which all the atoms present initially are accounted for in the product.

The Law of Combining Gas Volumes Avogadro was the first to study this and concluded a water molecule is composed of particles. We now know that a water molecule is composed of 2 hydrogen atoms & 1 oxygen atom. When a molecule of water breaks down, it breaks down according to the ratio of particles that compose it; 2 volumes of H2 & 1 volume of O2 from 1 volume of H2O.

The Law of Combining Gas Volumes My principle states that equal volumes of gases at the same temp & pressure contain equal numbers of particles. He reasoned that the volume of a gas depends on the number of gas particles, provided the temperature & pressure are constant.

The Law of Combining Gas Volumes Under the same conditions of temperature and pressure, the volumes of reacting gases and their gaseous products are expressed in ratios of small whole numbers 2 L H2 + 1 L O2 → 2 L H2O (g) 2 volumes H2 + 1volume O2 → 2 volumes H2O (g) 1 volume H2 + 1 volume Cl2 → 2 volumes HCl 1 volume HCl + 1 volume NH3 → NH4Cl (s)

Avogadro’s Law For a gas at constant temperature and pressure, the volume is directly proportional to the number of moles of gas (at low pressures). V = an a = proportionality constant V = volume of the gas n = number of moles of gas

Standard Molar Volume Equal volumes of all gases at the same temperature and pressure contain the same number of molecules. - Amedeo Avogadro

Standard Molar Volume

Practice You are planning an experiment that requires 0.0580 mol of nitrogen monoxide gas at STP. What volume would you need? 0.0580 mol x 22.4 L = 1.30 L 1 mol

Gas Stoichiometry Volume-Volume Calculations Assume: All products and reactants are at the same temp and pressure Unless otherwise stated, assume STP Solve by normal stoichiometric processes Volume ratios are the same as mole ratios

Volume-Mass and Mass-Volume Calculations Order of Calculations You are given a gas volume and asked to find a mass: gas volume A →moles A →moles B → mass B You are given a mass and asked to find a gas volume: mass A → moles A →moles B →gas volume B

Ideal Gas Law PV = nRT P = pressure in atm V = volume in liters n = moles R = proportionality constant = 0.08206 L∙ atm/ mol·K For units of kPa, L & K: R = 8.31 kPa ∙ L Mol ∙ K T = temperature in Kelvin

Calculate the Value of R Use all standard values! P = 1 atm V = 22.4 L n = 1 mole T = 273 K Try substituting different standard pressures to obtain different values of R

= 2.88 mol x 0.0821 (atm∙L/mol∙K) x 295 K Practice A 2.07 L cylinder contains 2.88 mol of helium gas at 22.0 °C. What is the pressure in atmospheres of the gas in the cylinder? PV = nRT P = nRT V = 2.88 mol x 0.0821 (atm∙L/mol∙K) x 295 K 2.07 L = 33.7 atm

Gas Density so at STP…

Variations on the Ideal Gas Law n = mass (m) molar mass (M) So replace n with m/M If PV = nRT then PV = mRT M So rearrange for M M = mRT PV

Variations on the Ideal Gas Law D = mass (m) volume (V) So replace m / V with D If M = mRT then VP M = DRT P

Density and the Ideal Gas Law Combining the formula for density with the Ideal Gas law, substituting and rearranging algebraically: M = Molar Mass P = Pressure R = Gas Constant T = Temperature in Kelvin

Practice At 28°C and 0.974 atm, 1.00 L of a gas has a mass of 5.16 g. What is the molar mass of this gas? What is the molar mass of a gas if 0.427g of the gas occupies a volume of 125 mL at 20.0°C and 0.980 atm? What is the density of a sample of ammonia gas if the pressure is 0.928 atm and the temp is 63.0°C? The density of a gas was found to be 2.0 g/L at 1.50 atm and 27°C. What is the molar mass of the gas? What is the density of argon gas at a pressure of 551 torr and a temp of 25°C? 131 g/mol 83.8 g/mol 0.572 g/L 33 g/mol 1.18 g/L

Chapter 11 Gases 11.4 Diffusion & Effusion

Effusion Effusion: describes the passage of gas into an evacuated chamber.

Graham’s Law Rates of Effusion and Diffusion

Graham’s Law Density can replace molar mass in Graham’s formula, since density is directly proportional to molar mass. Isotopes of elements can be separated by vaporizing the element, and allowing it to effuse. The heavier isotope effuses more slowly than the lighter isotope

Practice At 25 °C, the average velocity of oxygen molecules is 420 m/s. What is the average velocity of helium atoms at the same temperature? Rate of O2 is 420 m/s = √MHe Rate of He √MO2 420 m/s = √4 g/mol x √32 g/mol 420 m/s = 0.3536 x 1 x = 1188 m/s ≈ 1200 m/s