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Gas Laws and Gas Stoichiometry. Kinetic –Molecular Theory. Particles of matter (solid, liquid, or gas) are always in motion. This motion has consequences . Solid (defined shape and definite volume) Liquid (undefined shape and definite volume) Gases (undefined shape and volu)me.
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Kinetic –Molecular Theory • Particles of matter (solid, liquid, or gas) are always in motion. This motion has consequences . • Solid (defined shape and definite volume) • Liquid (undefined shape and definite volume) • Gases (undefined shape and volu)me
Kinetic-Molecular Theory of Gases • Gases are large numbers of tiny particles • Particles always moving in straight lines in all directions. Therefore, they have “kinetic energy” • These particles have elastic collisions between other particles and the container • Elastic – no loss of KE after collision • There is no attraction or repulsion between the particles • Kinetic energy of the particles is proportional to the temperature of the matter
Ideal Gas vs. Real gas • Ideal Gas: • Completely follows kinetic molecular theory (K-M theory) of gases • Hydrogen, maybe • Helium, sometimes • Real Gas: • Does not behave by K-M theory
K-M theory and nature of gases • Expansion • Fluidity (liquids and gases) • Low Density • Compressibility • Diffusion
Characteristics • Expansion • Fluidity
Characteristics 3. Low density
Characteristics • Compressibility
Characteristics • Diffusion
4 quantities that can be measured • Volume – How much space it takes up • Pressure – Amount of collisions particles have with container • Temperature – Average kinetic energy of the gas particles • Quantity or number of molecules – moles • We use these quantities to “work” with gases
Pressure • Pressure = Force / Area • Units of Pressure • 1 atm = 760 mm Hg = 760 torr = 1.01 x 105 pascals
Temperature Scales • Absolute zero = -273.15oC = 0 K (not 0oK) • Therefore K = 273 + oC • 0 oC = 273 K • As temperature increases, the number of gas collisions increases.
Temp. conversions K = oC + 273 25.0 oC = ? K
STP • “Slap The Pupils” -- No • “Such Total Pigs” -- No • Standard Temperature and Pressure • The volume of a gas depends on temperature and pressure. In order to compare volumes of gases, need a standard. • 0 degrees Celsius, 273 K • 1 atm of pressure
Boyle’s Law • Volume off a fixed mass of gas varies inversely with pressure at a constant temperature
Boyle’s Law • P1V1 = k and P2V2 = k • Therefore P1V1 = P2V2
A sample of oxygen gas occupies a vol. of 150 mL at a pressure of 720 mmHg. What would the volume be at 750 mmHg press.?
France, early 1800’s • Hot air balloons were extremely popular • Scientists were eager to improve the performance of their balloons. Two of the prominent French scientists were Jacques Charles and Joseph-Louis Gay-Lussac,
Charles’ Law • The volume of a fixed mass of gas varies directly with the Kelvin temperature at constant pressure
Charles’ Law • V1/ T1 = V2/ T2
A sample of Ne gas has a vol. of 752 mL at 25.0 deg. C. What is the vol. at 50.0 deg. C? T1 = 25.0 deg. C = 298 K T2 = 50.0 deg C = 323 K V1 = 752 mL
Gay-Lussac’s Law • The pressure of a fixed mass of gas varies directly with the Kelvin temp. at constant volume • P1/ T1 = P2/ T2
A sample of N gas is at 3.00 atm of pressure at 25 deg C what would the pressure be at 52 deg C? P1 = 3.00 atm T1 = 25 deg C = 298 K T2 = 52 deg C = 325 K P2 = ? P1/ T1 = P2/ T2 or P2 = P1T2/ T1
Combined Gas Law • Relates Pressure, Volume, and Temperature • Notice it is a combination of Boyles, Charles, and Gay-Lussac’s Laws • P1V1/ T1 = P2V2/ T2
A helium filled balloon has a vol. of 50.0 L at 25 deg C and 820. mmHg of pressure. What would the vol. be at 650 mmHg pressure and 10. deg C? • P1V1/ T1 = P2V2/ T2 • V1 = 50.0 L • T1 = 25 deg C = 298 K • P1 = 820. mmHg • T2 = 10. deg C = 283 K • P2 = 650 mmHg • V2 = ?
Molar Volume of a Gas • One mole of a gas has the same volume at STP as any other gas • 22.4 L / mole at STP
Practice Problem What volume would 0.0680 mol of oxygen gas occupy at STP? What about 0.0680 mol of nitrogen gas at STP?
Ideal Gas Law PV = nRT n = Number of Moles R = Ideal Gas Constant 0.0821 atm*L / mol*K V = Volume (must be in liters) P = Pressure (must be in atmospheres) T = Temperature (must be in Kelvin)
Practice Problem What is the P (in atm) exerted by a 0.500 mol sample of nitrogen gas in a 10.0 L container at 298 K? Pressure = 1.22 atm
Gas Stoichiometry • Uses volume-volume calculations (e.g. L L) • Liters can be used just like mole to mole ratios in a factor label problem
Practice Problem How many L of oxygen are required for the complete combustion of 0.250 L of propane? C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) • 5 L O2 needs 1 L C3H8
Practice Problem How many grams of calcium carbonate must be decomposed to produce 2.00 L of CO2 at STP? CaCO3(s) CaO(s) + CO2(g) 2.00 L CO2x 1 mol CO2x 1 mol CaCO3 X 100.086 g CaCO3 22.4 L CO2 1 mol CO21 mol CaCO3 = 8.94 g CaCO3