Embed Size (px)
Citation preview
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:
2
22
1
11
T
VP
T
VP
This is for a gas undergoing changing conditions of temp, pressure, and volume.
Combining Boyle’s law (pressure-volume) with Charles’ Law (volume-temp):
122211 TVPTVP
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?
P1 = 1148 mm Hg
V1 = 30.0 L
T1 = 40.0°C = 313 K
P2 = 1 atm = 760 mm Hg
V2 = ?
T2 = 0°C = 273 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?
P1 = 1148 mm Hg
V1 = 30.0 L
T1 = 40.0°C = 313 K
P2 = 1 atm = 760 mm Hg
V2 = ?
T2 = 0°C = 273 K
P1V1T2 = P2V2T1
(1148 mm Hg)(30.0 L)(273 K) = (760mm Hg)(V2)(313K)
V2= 39.5 L
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?
P1 = 985 torr
V1 = 105 L
T1 = 27.0°C = 300 K
P2 = 1 atm = 760 torr
V2 = ?
T2 = 0°C = 273 K
P1V1T2 = P2V2T1
(985 torr)(105 L)(273K) = (760torr)(V2)(300K)
V2= 124 L
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?
P1 = 80.0 kPa
V1 = 10.0 L
T1 = 240.°C = 513 K
P2 = 107 kPa
V2 = 20.0 L
T2 = ?
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 1.30 atm. At what pressure (in mm Hg) would the gas occupy 11.6 L if the temperature were lowered to 12.5 °C?
P1 = 1.30 atm = 988 mmHg
V1 = 23.2 L
T1 = 22.2 °C = 295.2 K
P2 = ?
V2 = 11.6 L
T2 = 12.5 °C = 285.5 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 = nRTP = 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.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 8.96 L sample of methane, CH4, measured at standard conditions.
P = 1.00 atm V = 8.96 L n = ? T = 273 K
PV = nRT
(1 atm)(8.96 L) = (n)(0.0821 L·atm/mol·K)(273 K)
n = 0.400 mol
(a)
Example #3: Calculate (a) the # moles in, and (b) the mass of an 8.96 L sample of methane, CH4, measured at standard conditions.
Convert moles to grams…
0mol40.0
mol 1
g 0.164CH g 40.6
(b)
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
