NOTES: 14.2 - 14.3 (Combined and Ideal Gas Laws)

Pressure-Volume-Temperature Relationship: Combined Gas Law • Pressure and volume are inversely proportional • Both pressure and volume are directly proportional to temperature

COMBINED GAS LAW: Combining Boyle’s law (pressure-volume) with Charles’ Law (volume-temp): OR This is for a gas undergoing changing conditions of temp, pressure, and volume.

Combined Gas Law Example #1: The volume of a gas filled balloon is 30.0 L at 40.0oC and 1148 mm Hg of pressure. What volume will the balloon have at STP? P2 = 1 atm = 760 mm Hg V2 = ? T2 = 0°C = 273 K P1 = 1148 mm Hg V1 = 30.0 L T1 = 40.0°C = 313 K P1V1 = P2V2 T1 T2

Combined Gas Law Example #1: The volume of a gas filled balloon is 30.0 L at 40.0oC and 1148 mm Hg of pressure. What volume will the balloon have at STP? P2 = 1 atm = 760 mm Hg V2 = ? T2 = 0°C = 273 K P1 = 1148 mm Hg V1 = 30.0 L T1 = 40.0°C = 313 K P1V1T2 = P2V2T1 (1148 mm Hg)(30.0 L)(273 K) = (760mm Hg)(V2)(313K) V2= 39.5 L

(985 torr)(105 L)(273K) = (760torr)(V2)(300K) Example #2: A sample of neon gas occupies 105 L at 27.0°C under a pressure of 985 torr. What volume would it occupy at standard conditions? P2 = 1 atm = 760 torr V2 = ? T2 = 0°C = 273 K P1 = 985 torr V1 = 105 L T1 = 27.0°C = 300 K P1V1T2 = P2V2T1 (985 torr)(105 L)(273K) = (760torr)(V2)(300K) V2= 124 L

(80.0kPa)(10.0L)(T2) = (107kPa)(20.0L)(513K) Example #3: A sample of gas occupies 10.0 L at 240.°C under a pressure of 80.0 kPa. At what temperature would the gas occupy 20.0 L if we increased the pressure to 107 kPa? P2 = 107 kPa V2 = 20.0 L T2 = ? P1 = 80.0 kPa V1 = 10.0 L T1 = 240.°C = 513 K P1V1T2 = P2V2T1 (80.0kPa)(10.0L)(T2) = (107kPa)(20.0L)(513K) T2= 1372K

Example #4: A sample of oxygen gas occupies 23. 2 L at 22. 2 °C and 1 Example #4: A sample of oxygen gas occupies 23.2 L at 22.2 °C and 1.30 atm. At what pressure (in mm Hg) would the gas occupy 11.6 L if the temperature were lowered to 12.5 °C? P2 = ? V2 = 11.6 L T2 = 12.5 °C = 285.5 K P1 = 1.30 atm = 988 mmHg V1 = 23.2 L T1 = 22.2 °C = 295.2 K P1V1T2 = P2V2T1 (988mm Hg)(23.2L)(285.5K) = (P2)(11.6L)(295.2K) P2= 1938 mm Hg ≈ 1940 mmHg

IDEAL GASES: • If a gas follows the following assumptions, it is considered to have ideal behavior:  Particles take up no volume  Random, straight line trajectories  Elastic collisions  No attractive or repulsive forces

Ideal Gases continued… • An ideal gas can be described using the following variables:  Pressure (P)  Volume (V)  Temperature (T)  Number of Moles (n) • If any three of these variables can be measured, the fourth can be determined using the ideal gas law.

Ideal Gas Law • PV = nRT P = pressure (in atm or kPa) V = volume (Liters) n = moles T = temperature (Kelvin) R = ideal gas constant R = 0.0821 (L•atm)/(mol•K) or R= 8.314 (L•kPa)/(mol•K) **use whichever value matches your pressure units!

Ideal Gas Law Example #1: You fill a rigid metal container with a volume of 20.0 L with nitrogen gas to a pressure of 197 atm at 28.0oC. How many moles of gas are present in the container? P = 197 atm V = 20.0 L n = ? T = 28.0°C + 273 = 301 K PV = nRT (197 atm)(20.0 L) = (n)(0.0821 L·atm/mol·K)(301 K) n = 159 mol

Example #2: What volume would 50. 0 g of ethane, C2H6, occupy at 140 Example #2: What volume would 50.0 g of ethane, C2H6, occupy at 140.ºC under a pressure of 1820 torr? P = (1820 torr)(1 atm/760 torr) = 2.39 atm V = ? n = (50.0 g)(1 mol / 30.0 g) = 1.67 mol T = 140.°C + 273 = 413 K PV = nRT (2.39 atm)(V) = (1.67 mol)(0.0821 L·atm/mol·K)(413 K) V = 23.7 L

Example #3: Calculate (a) the # moles in, and (b) the mass of an 8 Example #3: Calculate (a) the # moles in, and (b) the mass of an 8.96 L sample of methane, CH4, measured at standard conditions. P = 1.00 atm V = 8.96 L n = ? T = 273 K (a) PV = nRT (1 atm)(8.96 L) = (n)(0.0821 L·atm/mol·K)(273 K) n = 0.400 mol

(b) Convert moles to grams… Example #3: Calculate (a) the # moles in, and (b) the mass of an 8.96 L sample of methane, CH4, measured at standard conditions. (b) Convert moles to grams…

Example #4: Calculate the pressure exerted by 50 Example #4: Calculate the pressure exerted by 50.0 g ethane, C2H6, in a 25.0 L container at 25.0ºC? P = ? V = 25.0 L n = (50.0 g)(1 mol / 30.0 g) = 1.67 mol T = 25 °C + 273 = 298 K PV = nRT (P)(25.0 L) = (1.67 mol)(0.0821 L·atm/mol·K)(298 K) P = 1.63 atm