Properties of Gases

1. Density of air (outside) = 1.2 mg/mL Density of air (inside) = 0.88 mg/mL Properties of Gases

2. 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

3. 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.

4. 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

5. 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

6. 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

7. 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

8. P1 = 1 atm P2 = ? T2 = 0C = 273 K T1 = 100C = 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!

9. 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.

10. 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 = 30C) and dives down to the bottom of a deep lake where the pressure is 2.5 atm and the temperature is 4C. What is the volume of his lungs at this depth? P1 = V1 = T1 = P2 = V2 = T2 = 1 atm 2.5 atm ? 2.2 L 4C = 277K 30C = 303K V2 = 0.80 L

11. 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 0C and 1.0 atm? P = 1.0 atm V = ? n = 1.0 mol R = 0.0821 L·atm T = 0C + 273 = 273 K (1.0 atm)(V) = (1.0 mol)(0.0821 L·atm)(273 K) V = 22.4 L

12. 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?

13. 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

14. 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

15. 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

16. 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.

17. 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

19. 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.