Gas Law Applications Edward A. Mottel Department of Chemistry Rose-Hulman Institute of Technology

Gas Law Applications  Reading Assignment: Zumdahl Chapter 5.4,  This lecture concludes the topic of gas laws by describing the kinetic theory of gases and applying gaseous relationships to solve a variety of problems.

Gas Law Applications  Molecular weight determination  Pressure measurements  Isotope separation  Stoichiometric reactions

Dumas Method of Molecular Weight Determination  The weight of a vapor is used to determine the approximate molecular weight of the compound.

Dumas Method of Molecular Weight Determination A liquid is placed in an empty weighed retort. The liquid is boiled until is just completely evaporates. The tip of the glass retort is sealed with a flame. The mass of the trapped gas is used in the Ideal Gas equation to calculate the MW

Dumas Method of Molecular Weight Determination  A gaseous sample was found to have the following composition: 1.6% H, 39.7% C, 58.7% Cl  At 400. K and 870. torr, a 3.17 gram sample occupies liters. Diagram an approach to determine the empirical formula and the molecular formula of this compound.

Dumas Method of Molecular Weight Determination Percentage composition data Approximate molecular weight Molecular formula Empirical formula Gas Data 1.6% H, 39.7% C, 58.7% Cl 400. K and 870. torr, 3.17 grams occupies liters

Pressure Measurements Barometer mercury vacuum What forces determine the height of the mercury in the glass tube?

Pressure Measurements Manometer mercury P atm = 740. mm Hg 15.0 cm 10.0 cm What is the pressure of the gas in the bulb? open to atmosphere

Kinetic Theory of Gases  Gas is composed of discrete molecules.  Molecules are in continuous motion.  Molecular collisions are elastic.  Molecules are small.  The absolute temperature is proportional to the average kinetic energy.

Kinetic Energy of Molecules  All gases at the same temperature have the same average kinetic energy. KEmv  If an oxygen molecule has a velocity of m·s -1, what will be the velocity of a nitrogen molecule at the same temperature?

Boltzmann Distribution Maxwell Speed Distribution Law  Pv MW RT ve MW vRT           / velocity or energy number of molecules The same gas at a given average temperature has a range of different velocities.

Boltzmann Distribution Maxwell Speed Distribution Law velocity or energy number of molecules average Gas velocity is a measure of energy (temperature) KEmv  T 2 > T 1

Graham's Law of Effusion  Isotope separation evacuated chamber mixed gases pinhole leak

Graham's Law of Effusion  Isotope separation RateofeffusionofgasA RateofeffusionofgasB MWofgasB MWofgasA  Derive this equation from KEmv  1 2 2

Graham's Law of Effusion evacuated chamber mixed gases pinhole leak Which gas has a lower molecular weight?

Graham's Law of Effusion  Uranium hexafluoride (UF 6 ) is a gas that has been used as a method to enrich the amount of uranium-235 used in nuclear reactions.  Uranium has two principle isotopes, uranium-235 and uranium-238.

Graham's Law of Effusion How many enrichment cycles will be needed to raise the uranium-235 content from the natural abundance 0.3% to 5%? If the effusion method is used to separate 235 UF 6 (MW = 349) from 238 UF 6 (MW = 352) what will be the percentage enrichment per cycle?

Graham's Law of Effusion Rate of effusion of 235 UF 6 Rate of effusion of 238 UF 6 MW of 238 UF 6 = MW of 235 UF = 349 = If the effusion method is used to separate 35 UF 6 (MW = 349) from 238 UF 6 (MW = 352) what will be the percentage enrichment per cycle?

Graham's Law of Effusion each enrichment cycle increase x increase first cycle second cycle (1.0043) n increasen th cycle

Graham's Law of Effusion 0.3% natural abundance of U % required purity of U-235 for nuclear applications 5.0%0.3% x (1.0043) n 5.0% / 0.3% = 16.7(1.0043) n ln (16.7)n ln (1.0043) n = 657 cycles

Non-ideal Behavior van der Waal Equation  intermolecular force correction (a) collisions are not perfectly elastic  molecular volume correction (b) molecules are not point masses  Pa n V VbnnRT obs         2

Gas particles are molecules  Gas is composed of discrete particles of matter called molecules. All molecules of the same substance are the same.

Molecules are in continuous motion  Collide with each other and the walls that contain them. The pressure of a gas is due to the collision of molecules with the wall.

Molecular collisions are elastic  There is no net loss of kinetic energy.  A perfectly insulated vessel will maintain the same total kinetic energy (the temperature will remain constant).

Molecules are small  Molecules are small compared to the volume containing them. Molecules can be treated as point masses. Gases are compressible because there is a large distance between molecules.

Kinetic energy is proportional to absolute temperature  The absolute temperature of a gas is directly proportional to the average kinetic energy of the molecules. The translational velocity of a molecule (a measure of its kinetic energy) is proportional to its temperature.