Gases

Ideal Gases Ideal gases are imaginary gases that perfectly fit all of the assumptions of the kinetic molecular theory.  Gases consist of tiny particles that are far apart relative to their size.  Collisions between gas particles and between particles and the walls of the container are elastic collisions  No kinetic energy is lost in elastic collisions

Ideal Gases Ideal Gases (continued)  Gas particles are in constant, rapid motion. They therefore possess kinetic energy, the energy of motion  There are no forces of attraction between gas particles  The average kinetic energy of gas particles depends on temperature, not on the identity of the particle.

The Nature of Gases  Gases expand to fill their containers  Gases are fluid – they flow  Gases have low density  1/1000 the density of the equivalent liquid or solid  Gases are compressible  Gases effuse and diffuse

Pressure  Is caused by the collisions of molecules with the walls of a container  is equal to force/unit area  SI units = Newton/meter 2 = 1 Pascal (Pa)  1 atmosphere = 101,325 Pa  1 atmosphere = 1 atm = 760 mm Hg = 760 torr

Measuring Pressure The first device for measuring atmospheric pressure was developed by Evangelista Torricelli during the 17 th century. The device was called a “barometer” Baro = weight Meter = measure

An Early Barometer The normal pressure due to the atmosphere at sea level can support a column of mercury that is 760 mm high.

Standard Temperature and Pressure “STP” P = 1 atmosphere, 760 torr T =  C, 273 Kelvin The molar volume of an ideal gas is liters at STP

Converting Celsius to Kelvin Gas law problems involving temperature require that the temperature be in KELVIN! Kelvin =  C °C = Kelvin - 273

Are You Ready For Some Practice Time? Please complete handout 2-2 Practice Problems #1-10 & 13-2 Practice Problems (on back) #1-8 only!!! 1 atm = 760mm Hg 760 torr 101,325 Pa 29.2 in Hg 14.7 lb/in 2 Units of Pressure

The Combined Gas Law The combined gas law expresses the relationship between pressure, volume and temperature of a fixed amount of gas. Boyle’s law, Gay-Lussac’s law, and Charles’ law are all derived from this by holding a variable constant.

Boyle’s Law Pressure is inversely proportional to volume when temperature is held constant.

Charles’s Law The volume of a gas is directly proportional to temperature, and extrapolates to zero at zero Kelvin. (P = constant)

Gay Lussac’s Law The pressure and temperature of a gas are directly related, provided that the volume remains constant.

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

Ideal Gas Law PV = nRT  P = pressure in atm  V = volume in liters  n = moles  R = proportionality constant  = L atm/ mol·   T = temperature in Kelvin Holds closely at P < 1 atm

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

Practice Time! Please do problems: pp #26, &

Gas Density … so at STP…

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

Gas Stoichiometry #1 If reactants and products are at the same conditions of temperature and pressure, then mole ratios of gases are also volume ratios. 3 H 2 (g) + N 2 (g)  2NH 3 (g) 3 moles H mole N 2  2 moles NH 3 3 liters H liter N 2  2 liters NH 3

Gas Stoichiometry #2 How many liters of ammonia can be produced when 12 liters of hydrogen react with an excess of nitrogen? 3 H 2 (g) + N 2 (g)  2NH 3 (g) 12 L H 2 L H 2 = L NH 3 L NH

Gas Stoichiometry #3 How many liters of oxygen gas, at STP, can be collected from the complete decomposition of 50.0 grams of potassium chlorate? 2 KClO 3 (s)  2 KCl(s) + 3 O 2 (g) = L O g KClO 3 1 mol KClO g KClO 3 3 mol O 2 2 mol KClO L O 2 1 mol O

Gas Stoichiometry #4 How many liters of oxygen gas, at 37.0  C and atmospheres, can be collected from the complete decomposition of 50.0 grams of potassium chlorate? 2 KClO 3 (s)  2 KCl(s) + 3 O 2 (g) = “n” mol O g KClO 3 1 mol KClO g KClO 3 3 mol O 2 2 mol KClO mol O 2 = 16.7 L

Dalton’s Law of Partial Pressures For a mixture of gases in a container, P Total = P 1 + P 2 + P This is particularly useful in calculating the pressure of gases collected over water.

Try This! Mixtures of helium and oxygen can be used in scuba diving tanks to help prevent “the bends.” For a particular dive, 46 L He at 25 o C and 1.0 atm and 12 L O 2 at 25 o C and 1.0 atm were pumped into a tank with a volume of 5.0 L. Calculate the partial pressure of each gas and the total pressure in the tank at 25 o C. Wanna hint? Calculate the number of moles of each gas Good Luck!

And the answer is……….. P He = 9.3 atm P O2 = 2.4 atm P T = 11.7 atm Mole Fraction: The ratio of the number of moles of a given component in a mixture to the total number of moles in the mixture. Χ = n1n1 n TOTAL = n2n2 n 1 + n 2 + n 3 +…… And…..the mole fraction of each component in a mixture of ideal gases is directly related to its partial pressure! Χ 1 = n1n1 n TOTAL = P1P1 P TOTAL So….. P 1 = χ1χ1 X P TOTAL

Kinetic Energy of Gas Particles At the same conditions of temperature, all gases have the same average kinetic energy.

The Meaning of Temperature  Kelvin temperature is an index of the random motions of gas particles (higher T means greater motion.)

Kinetic Molecular Theory  Particles of matter are ALWAYS in motion  Volume of individual particles is  zero.  Collisions of particles with container walls cause pressure exerted by gas.  Particles exert no forces on each other.  Average kinetic energy  Kelvin temperature of a gas.

Diffusion: describes the mixing of gases. The rate of diffusion is the rate of gas mixing. Diffusion

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

Effusion: Diffusion: Graham’s Law Rates of Effusion and Diffusion

Real Gases  corrected pressure corrected volume P ideal V ideal Must correct ideal gas behavior when at high pressure (smaller volume) and low temperature (attractive forces become important). “van der Waals equation”

Warmup Time Before the Practice Test Ideal gases vary from real gases at conditions of: A.High temperature and low pressure B.High temperature and high pressure C.Low temperature and low pressure D.Low temperature and high pressure E.Both high density and low pressure ANSWER: D At low temperature and high pressure, the molecules are closer together and therefore the forces between the molecules become more important as they are stronger. The actual (finite) volume of individual molecules also becomes more important as more of the total space is actually occupied by finite molecular volume.

What is the volume of 3.00 mol of STP? A.22.4 L B.3 x 22.4 L C.3 x 22.4 L x 760 D.3 x 22.4 x 273/760 E.It cannot be determined without knowing which gas is involved. ANSWER: B The molar volume of all gases at STP is about 22.4 L, so three moles would occupy 22.4 L / mol x 3 moles.

An ideal gas of volume 189. mL is collected over water at 30 o C and 777 torr. The vapor pressure of water is o C. What pressure is exerted by the dry gas under these conditions? A.320 torr B.745 torr C.777 torr D.32/77 torr E.32 x 777 torr ANSWER : B The total pressure = 777 torr. Of this, 32 torr is due to the water vapor, hence = 745 torr of pressure are allocated to the dry gas.

A 14.0-L cylinder contains 5.60 g N 2, g Ar and 6.40 g O 2. What is the total pressure in atm at 27 o C? (R = the ideal gas constant.) A.20 R B.26 R C.30 R D.60 R E.120 R ANSWER : C Use P total = N total RT/V to determine the pressure. From the mass of each of the gases you can find mol of N 2, 1.00 mol of Ar, and mol of O 2 to give a total number of moles (n) of 1.4 mol. Therefore: P total = R x 1.40 mol x 300 K / 14.0 L = 30 R.

In a closed inflexible system, 7.0 mol CO 2, 7.0 mol Ar, 7.0 mol N 2 and 4.0 mol Ne are trapped, with a total pressure of 10.0 atm. What is the partial pressure exerted by the neon gas? A.1.6 atm B.4.0 atm C.10.0 atm D.21.0 atm E.29.0 atm ANSWER : A The total number of moles is 25 ( = 25), hence the Ne is 4/25 of the total amount of gas and exerts 4/25 of the total pressure, or 4/25 x 10 atm = 1.6 atm.

Consider the reaction C 2 H 6 (g) + 7/2 O 2 (g) 2CO 2 (g) + 3H 2 O(g). If 6.0 g ethane, C 2 H 6 (g) burn (as shown above), what volume of CO 2 (g) will be formed at STP? A.0.20 L B.0.40 L C.2.2 L D.9.0 L E.22.4 L ANSWER: D 6.0 g of ethane / 30. g/mol yields 0.20 mol of ethane, which forms twice that number of moles of carbon dioxide. Since one mole of gas occupies 22.4 L at STP, 22.4 L/mol x 0.20 mol x 2 CO 2 / C 2 H 6 = 8.96 L of carbon dioxide (rounded to two sigfigs = 9.0 L)

Cl 2 and F 2 combine to form a gaseous product; one volume of Cl 2 reacts with three volumes of F 2 yielding two volumes of product. Assuming constant conditions of temperature and pressure, what is the formula of the product? A.ClF B.Cl 2 F 2 C.ClF 2 D.Cl 2 F E.ClF 3 ANSWER: E This problem assumes you understand Avogadro’s Law, “At constant temperature and pressure, the volume of a gas is directly proportional to the number of moles of the gas”. To apply that to this problem, 1 volume Cl volumes F 2 2 volumes Cl x F x. The simplest formula for the product becomes ClF 3

Decreasing the temperature of an ideal gas from 80 o C to 40 o C causes the average kinetic energy to A.Decrease by a factor of two B.Decrease by a factor of four C.Increase by a factor of two D.Increase by a factor of four E.Decrease by less than a factor of two. ANSWER: E Be careful here to note the difference between the kinds of temperature scales and what they mean. Even though the Celsius temperature is half as much, the average kinetic energy is proportional to the Kelvin temperature. In this case that ratio is only 353/313 = 1.13 so the average kinetic energy decreases only by a factor of 1.13

The effusion rate of helium gas, He, compared with that of Methane gas, CH 4 is A.Four times greater for helium B.Four times less for helium C.Twice as great for helium D.Twice as great for methane E.Sixteen times as great for helium ANSWER: C Effusion rates are inversely proportional to the square root of the molecular masses of the gases.

The average speed of the molecules of a gas is proportional to the A.Volume of the container B.Reciprocal of absolute temperature C.Absolute temperature D.Square root of the absolute temperature E.Square of the absolute temperature ANSWER: D in this case the question is about the average speed (not the energy) of the molecules.

Do Now: Please remove everything from your workstation except for a PENCIL and calculator. Also set up dividers in preparation for today’s exam. Objective: Test on Chapter 5 “GASES” HW: Read pp Chapter 6 Thermochemistry: Introduction, The Nature of Energy & Chemical Energy