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## Properties of Gases

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**Density of air (outside) = 1.2 mg/mL**Density of air (inside) = 0.88 mg/mL Properties of Gases**Pressure: the pressure of a gas is caused by the gas**particles colliding with the walls of the container. Pressure is always a force exerted over an area. (P = F/A) The SI unit of pressure is the Pascal (Pa). Other units of pressure are: 1 atm = 760 mm Hg = 760 Torr = 101.325 kPa = 750 bar Reading manometers: Pext = 760 Torr PA = 760 -520 = 240 Torr PB = 760 + 67 = 827 Torr PC = 103 Torr**heat**Heating a gas causes the gas particles to move more rapidly and increases the distance between them. Volume #1 (V1) Temperature #1 (T1) Volume #2 (V2) Temperature #2 (T2) Charles’ Law: For a fixed number of gas particles, the volume of the gas will increase as the temperature increases.**heat**Volume #1 (V1) Temperature #1 (T1) Volume #2 (V2) Temperature #2 (T2) V1 V2 V2 = T1 T2 T2 V1 = T1 Ex: A 2.0 L balloon at 0°C is heated to 20°C. What is its new volume? V2 = ? 2.0 L V1 = 2.0 L 0°C 0°C 20°C T2 = 20°C T1 = 0°C**What's wrong with**heat Volume #1 (V1) Temperature #1 (T1) Volume #2 (V2) Temperature #2 (T2) this problem?? V1 V2 V2 = T1 T2 T2 V1 = T1 Ex: A 2.0 L balloon at 0°C is heated to 20°C. What is its new volume? V2 = ? 2.0 L V1 = 2.0 L 0°C 0°C 20°C T2 = 20°C T1 = 0°C**V1 =**2.0 L V2 = ? T1 = 0°C T2 = 20°C 2.0 L V2 0.00733 = V2 = 273 K 293 K 293 K K = °C + 273 Kelvin temperature scale Ex: Convert 0°C into Kelvin. K = °C + 273 0°C + 273 K = 273 K Looking back at our balloon problem: = 273 K = 293 K V2 = 2.1 L**Rapidly cooling off a gas causes the gas particles to move**more slowly and exert less pressure on the walls of their container. Gay-Lussac’s Law: For a fixed number of gas particles, the pressure of the gas will increase as the temperature increases. P1 P2 = T1 T2 In this movie, the can collapses when cooled because the pressure outside of the can is much greater than the pressure inside of the can. Click picture to start movie**P1 = 1 atm**P2 = ? T2 = 0C = 273 K T1 = 100C = 373 K 1 atm P2 = 373 K 273 K P1 P2 = T2 T1 Let’s calculate what the final pressure inside the can was. P2 = 0.73 atm No wonder the can collapsed with less than ¾ of an atmosphere’s worth of pressure inside!**P1V1 = P2V2**Pressure #2 (P2) Volume #2 (V2) Pressure #1 (P1) Volume #1 (V1) Boyle’s Law: For a fixed amount of gas, the pressure exerted by a gas decreases as the volume increases.**The Combined Gas Law: can be used to calculate T, P and V**changes for a fixed number of particles. = (1 atm)(2.2L) (2.5 atm)V2 P2V2 P1V1 = 303K 277K T2 T1 Example: A diver with a lung volume of 2.2 L takes a breath at the surface of the water (P = 1 atm, T = 30C) and dives down to the bottom of a deep lake where the pressure is 2.5 atm and the temperature is 4C. What is the volume of his lungs at this depth? P1 = V1 = T1 = P2 = V2 = T2 = 1 atm 2.5 atm ? 2.2 L 4C = 277K 30C = 303K V2 = 0.80 L**Q: What if the mass of the gas is changing?**Ah, that’s where that number comes from! ! A: Whenever mass or number of particles is involved in a chemical problem, that means that moles are involved. The Ideal Gas Law: PV = nRT P = pressure in atm n = # of moles R = gas constant = 0.0821 L·atm V = volume in Liters T = temperature in Kelvin Ex: What is the volume of 1.0 mol of He at 0C and 1.0 atm? P = 1.0 atm V = ? n = 1.0 mol R = 0.0821 L·atm T = 0C + 273 = 273 K (1.0 atm)(V) = (1.0 mol)(0.0821 L·atm)(273 K) V = 22.4 L**Avogadro’s Law: The volumes and partial pressures of gases**in the same container and at the same temperature will depend upon their stoichiometric coefficients. 3x x 2x T = 877 K 3 H2(g) + N2(g) 2 NH3(g) Vtot = 180 L 7.5 mol 2.5 mol 5.0 mol 90 L 30 L 60 L 2 atm 3 atm 1 atm Ex A: 5.0 mol of ammonia were formed. How many moles of hydrogen and nitrogen were used? Ex B: 90 L of hydrogen were used. How many liters of nitrogen were also used? How many liters of ammonia were formed? Ex C: The final partial pressure of ammonia is 2 atm. What were the partial pressures of the hydrogen and nitrogen that reacted?**Rearrange**Another application of the Ideal Gas Law: Gas Densities and Molar Masses (MM) PV = nRT Multiply both sides by MM As n(MM) = mass and D = M/V The units on density in this case are g/L**Dalton’s Law of Partial Pressures: The total gas pressure**in a container of a mixture of gases is the sum of the individual (partial) pressures of each gas. Ex: The Earth’s atmosphere contains 78% nitrogen, 21% oxygen and 1% argon. What is the partial pressure of each gas if the total pressure is 1.0 atm? 78% of 1.0 atm = 0.78 atm N2 21% of 1.0 atm = 0.21 atm O2 1% of 1.0 atm = 0.01 atm Ar Ptot = PN2 + PO2 + PAr Math check: Ptot = 0.78 atm + 0.21 atm + 0.01 atm Ptot = 1.0 atm**Collecting gases over water: an application of Dalton’s**Law Vapor: a substance that is in its gas phase at temperatures when it is normally a liquid or a solid 2 KClO3(s) 2 KCl(s) + 3 O2(g) Ptot = Patm = PO2 + PH2O**Average speed of gaseous molecules/atoms**The internal energy, U, of ALL gases depends only upon temperature: In ideal gases, kinetic energy is the ONLY form of energy present *Using molar mass, MM, in place of mass of a single particle and u, the root mean square of the particle velocities Combining the two expressions: The rms of gas particles depends only upon Temperature and the molar mass of the gas.**Graham’s Law of Effusion:**• Based on the principle that heavier gas molecules travel more slowly than the lighter ones. u1 = rms of Gas 1 MM1 = molar mass of Gas 1 u2 = rms of Gas 2 MM2 = molar mass of Gas 2 Ex: How much faster will He travel than CH4? MM2 = 16 g/mol MM1 = 4 g/mol Helium will travel 2 x faster than methane, CH4**Van der Waals Gas Law: For Nonideal Gases**a = constant to correct for molecular attractions (affects pressure of the gas) b = constant to correct for the volume of the molecules (V 0 when T = 0 K) The van der Waals’ constants, a and b, have to be experimentally determined for each gas.