Gases

Properties of Gases • takes shape of container • takes volume of container • exerts pressure on its surroundings • particles are very far apart • exert no forces on each other • move in straight line random paths

Factors or measurements that affect gases: Pressure- occurs from the force and number of collisions with the walls of the container • Temperature- a measure of the average kinetic energy of the particles therefore indirectly relates to the motion of the particles • Volume- space occupied by a gas • Number of moles- amount of gas

Pressure Atmospheric pressure measured by a barometer; usually contains mercury • units of pressure: mm of Hg, torr, atmosphere, kPa • defined as force/unit area; in SI Newtons/m2 = pascal • 1 atmosphere= 760 mm Hg = 760 torr= 101,325 pascals= 101.325 kPa=14.7 lbs/in2; standard pressure; average pressure at sea level • use these to convert from one to another Ex. 723 mm Hg= _______ kPa

Pressure Manometer- instrument used to measure gas pressure Two types of manometers- Open- open to the atmosphere Closed-closed to the environment; barometer is a special type of closed manometer that measures atmospheric pressure is called a barometer.

Using Liquids Other than Mercury in Manometers Use formula based upon density: hb = ha x da db

Using Liquids Other than Mercury in Manometers Ex. A liquid with a density of 1.15 g/mL was used in an open-ended manometer. In a particular experiment, a difference in the heights of the levels in the two arms was measured to be 14.7 mm, with the level in the arms connected to an apparatus containing a trapped gas lower than the level open to the atmosphere. The barometric pressure had been measured to be 756.00 torr. What was the pressure of the trapped gas?

: Gas Laws: Boyle’s Law Boyle’s Law- relates pressure and volume; inverse relationship; gives hyperbola graph; true under low pressures such as atmospheric pressure; used to predict the new volume of a gas when the pressure is changed PV = k ;true for any gas with constant temperature and amount Therefore at different volumes for same gas PV= k P1V1 = P2V2

Gas Laws: Charles’ Law Charles’ Law- relates volume of a gas to temperature of a gas; direct relationship; gives straight line; based upon absolute temperature which is the kinetic energy of particles; must use Kelvin temperature ( K = oC + 273) V = k T   V1= V2 T1 T2

Gas Laws: Gay-Lussac’s Law Relates pressure and temperature; direct relationship; gives straight line; use any pressure units and Kelvin temperature P1 = P2 T1 T2

Gas Laws: Combined Gas Law Combined Gas Law- combines all three effects P1V1 = P1V1 T1 T2

Gas Laws: Avogadro’s Law Avogadro’s Law- involves volume and amount of gas in moles V = kn V is volume; k is constant; n is symbol for # of moles

Ideal Gases Ideal Gases- name given to gases that follow the above laws at all temperatures and pressures. At room temperature and pressure, most real gases follow those of ideal. At high pressures and low temperatures, we must make adjustments in our formula to account for attractions between molecules

Ideal Gas Law Ideal Gas Law- incorporates all factors into one formula  PV= nRT P is pressure in atmospheres, V is volume in L, n is moles, and T is temperature in K. R is a constant; known as the gas constant; can have different values based upon units used in equation. R= 0.0821 L atm/mol K R= 62.4 mm Hg L/mol K R= 8.31 kPa dm3/mol K

Molar Gas Volume When one mole of any gas is at STP the volume will be 22.4 L. Known as molar gas volume.

Gas Stoichiometry If conditions differ from STP use Combined Gas Laws equations or Ideal Gas Law to determine the new volume. Ex. Calculate the volume of CO2 at STP produced from the decomposition of 152 g of CaCO3 according to the reaction: CaCO3 (s)  CaO (s) + CO2(g)

Gas Stoichiometry Ex. A sample of methane gas having a volume of 2.80 L at 25oC and 1.65 atm was mixed with a sample of oxygen having a volume of 35.0 L at 31oC and 1.25 atm. The mixture was ignited to form carbon dioxide and water. Calculate the volume of CO2 formed at a pressure of 2.50 atm and a temperature of 125 oC.

Since D = m/V substitute MM = D R T Molar Mass of a Gas Molar Mass of a Gas   Moles(n) = m___ MM ( molar mass) PV=nRT P = nRT V P= (m/MM) R T   MM = mRT PV Or Since D = m/V substitute MM = D R T P

Molar Mass of a Gas Ex. A student measured the density of a gaseous compound to be 1.34 g/L at 25oC and 760 torr and was told that the compound was composed of 79.8% carbon and 20.0% hydrogen. A) What is the empirical formula of the compound? B) What is its molecular mass? C) What is the molecular formula of the compound?

Gas Density To determine the density of a gas at STP simply take the molar mass/ 22.4 L.

Dalton’s Law of Partial Pressure For a mixture of gases in a container, the total pressure exerted is the sum of the pressures that each gas would exert if it were alone in the container. PT = P1 + P2 + P3 + … Each P represents the partial pressure of each gas PT = P1 + P2 + P3 … = n1RT + n2RT + n3RT = V V V   (n1 + n2 + n3 +…) RT V

Ideal Gases Since the pressure of the ideal gas is not affected by the identity of the gas particles, tells us that ideal gases the volume of the individual gas particles must not be important and the forces among the particles must not be important.

Mole Fraction Concentration unit Mole Fraction (C)- moles of solute = n1 Total moles of solution n1 + n2 + n3 C = n2 = P1 n T PT P1= C x PT

Example Ex. Mixtures of helium and oxygen are used in scuba diving tanks to help prevent “the bends.” For a particular dive 46 L of O2 at 25oC and 1.0atm was pumped along with 12L of He at 25oC and 1.0 atm 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 25oC.

Example Ex. The partial pressure of oxygen in air was observed to be 156 torr when the atmospheric pressure was 743 torr. Calculate the mole fraction of O2 present.

Example Ex. The mole fraction of nitrogen in the air is 0.7808. Calculate the partial pressure of N2 in air when the atmospheric pressure is 760 torr.

Collecting gases by the displacement of water Water vapor is always present when a gas is collected over water. The water vapor becomes a constant since equilibrium is obtained. Therefore to find the pressure of the “dry gas” (gas pressure without water vapor pressure present) subtract the water vapor pressure from the total. Pgas + PH20 = PT PT - PH2O = Pgas

Collecting Gases over Water The pressure of the water vapor is a constant depending on the temperature. Usually this information is given or on a table of water vapor pressures by temperature. When calculating the pressure of the gas, the pressure of the dry gas only must always be used. The volume of the dry gas is the total volume of the container.

Kinetic Molecular Theory of Gases explains the properties of an ideal gas specifically for behavior of the individual gas particles the volume of the individual particles can be assumed to be negligible. particles are in constant motion; collisions of the particles with the walls of the container cause the pressure of the gas particles exert no forces on each other average kinetic energy of the gas particles is assumed to be directly proportional to the Kelvin temperature

Kinetic Molecular Theory of Gases Real gases do not necessarily conform to these assumptions; however at low pressures (atmospheric) and high temperatures (normal temperatures) they are very close.

Root Mean Square Velocity urms- root mean square velocity Derived: urms = 3RT/M T- Temperature in Kelvins M- molar mass in kilograms R- 8.31 J/K mol if velocity is to come out in m/s J (Joules- unit of energy = k m2/s2) Path of individual gas particles is very erratic Average distance a particle travels between collisions is very small and is called the mean free path. Large number of collisions produces large range of velocities. Therefore even if the root mean square velocity is given this is not the velocity of the entire group of molecules.

Effects of temperature on velocity As temperature is increased, the curve peak moves toward higher values and the range of velocities becomes much larger. The peak of the curve reflects the most probable velocity. Called Maxwell- Boltzmann Distribution.

Effusion and Diffusion Diffusion- describes the mixing of gases (The movement of a substance from an area of high concentration to an area of lower concentration) Effusion- the passage of a gas through a tiny orifice into an evacuated chamber (vacuum)

Effusion and Diffusion Rate of diffusion is the rate at which gases mix.   Rate of effusion is the speed at which a gas is transferred into the chamber. 1/2 m 1v 12 = 1/2 m 2v22 for two particles at same temperature m1v12 = m2v22 v12/v22 = m2/m1 This is Graham’s Law of Effusion

Example Ex. The mole fraction of nitrogen in the air is 0.7808. Calculate the partial pressure of N2 in air when the atmospheric pressure is 760 torr.

Real Gases Corrections for real gases can be made using van der Waal’s equation. It contains corrections for pressure and volume.

Corrections for Read Gases For pressure: In real gases there are some attractions. These attractions will decrease the frequency with which the molecules hit the walls of the container. Therefore a correction has been made. Pobs = Pcor pres - a (n/V)2 Therefore Pcor pres = Pobs + a(n/V)2

Corrections for Real Gases For volume: In real gases molecules have volume and therefore reduces the overall volume of the container. A correction has been made. V – nb is volume correction

Corrections for Real Gases The variables a and b are constants which depend upon the kind of molecule; specifically the attractions between the molecules and the size of the molecules. They can be found in a table in your book.

Example Ex. For any given gas, the values of the constants a and b can be determined experimentally. Indicate which physical properties of a molecule determine the magnitudes of the constants a and b. Which of the two molecules, H2 or H2O, has a higher value for a and which has the higher value for b? Explain.

Example Ex. Three volatile compounds X, Y and Z each contain element Q. The percent by weight of element Q in each compound was determined. Some of the data obtained are given in the next slide. a) The vapor density of compound X at 27o C and 750 mm Hg was determined to be 3.53 g/L. Calculate the molecular weight of compound X. b) Determine the mass of element Q contained in 1.00 moles of each of the three compounds. c) Calculate the most probable value of the atomic weight of element Q. d) Compound Z contains carbon, hydrogen, and element Q. When 1.00 g of compound Z is oxidized and all of the carbon and hydrogen are converted to oxides, 1.37 g of CO2 and 0.281 g of water are produced. Determine the most probable molecular formula.

Example Compound Percent by Weight of Element Q Molecular Weight X 64.8% ? Y 73.0% 104 Z 59.3% 64.0