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Gas Laws
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Boyle's Law
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Charles' Law
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Combined
Gas Laws
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Ideal Gas
Equation
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Graham's Law of Diffusion
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Dalton's Law of Partial
Pressures
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Mass-Volume Problem
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Boyle's Law: The pressure and volume of
a gas are inversely proportional, while the temperature is constant.
P1V1 = P2V2 |
| There are two ways to solve Boyle's
Law probems: 1) Use the formula or 2) Apply the law with a pressure or volume ratio. |
| Examples: |
| 25 ml of a gas is at a pressure
of 745 mm Hg. Find the new volume at 715 mm Hg. The temperature
is constant. V2 = 25 ml (745 mm Hg/715 mm Hg) = 26 ml |
| Solve: |
| Correct 1.95 L of a gas at
25 o C & 98.7 Kpa to 25oC
& 102 . 7 Kpa. |
| Answer: |
| V2 = 1.95
L (98 . 7 Kpa/102 .7 Kpa) = 1.87 L |
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Charles' Law: The temperature in Kelvin
and volume of a gas are directly proportional while the pressure is constant.
V1/T1
= V2/T2 |
| There are two ways to solve Charles'
Law probems: 1) Use the formula or 2) Apply the law with a temperature, in Kelvin, or volume ratio. |
| Examples: |
| 75.8 ml of a gas is at a
temperature of 15 oC. Find the new volume at 38
oC. Pressure is constant. T1 = 15 + 273 = 288 K T 2 = 38 + 273 = 311 K V2 = 75.8 ml (311 K/288 K) = 81.9 ml |
| Solve: |
| Correct 54.95 dl of a gas
at 35 o C & .95 atm to -8oC
& . 95 atm. |
| Answer: |
| V2 = 54.
95 dl (265 K/308 K) = 47.28 dl |
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Combined Gas Laws: Work both Boyle's
& Charles' Laws together: P1
V1/T1 = P2V2
/T2 |
| Solve: |
| 275 ml of a gas at .85 atm
& 6oC is changed to 1.15 atm & 22o
C. Find the new volume. |
| Answer: |
| V2 = 275 ml (.85
atm/1.15 atm)(295 K/279 K) V2 = 215 ml |
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Dalton's Law of Partial Pressures: In
a mixture of gases the total pressure is the sum of the gases pressure. Ptotal = P 1 + P 2 + P3 + .... |
| Examples: A vessel contains hydrogen, oxygen, and carbon dioxide gases. Find the pressure of the vessel if the Phydrogen = 84 .7 Kpa, Poxygen = 26.2 Kpa, and Pcarbon dioxide = 4 .62 Kpa. PTotal = Phydrogen + Poxygen + Pcarbon dioxide PT = 84.7 Kpa + 26.2 Kpa + 4 .62 Kpa = 115.52 = 115.5 Kpa |
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Water Vapor Pressure
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Temp.
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Pressure
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Temp.
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Pressure
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(oC)
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(mm of Hg)
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(oC)
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(mm of Hg)
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0.0
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4.6
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20.0
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17.5
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5.0
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6.5
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21.0
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18.6
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10.0
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9.2
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24.0
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22.4
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| Solve: 1) 250 ml of a gas was collected over water at 10 oC and 715 mm of Hg. Find the volume of the dry gas at STP. 2) 475 ml of a gas was collected over water at 21 oC and 1.4 atm. Determine the volume of the dry gas at 740 mm Hg and 35oC. |
| Answer: |
| 1) 715 mm Hg
Pgas + Pwater V2
= 250 ml(705.8 mm Hg/760 mm Hg)(273 K/283 K) - 9.2 mm Hg Pwater V 2 = 223.967 ml 705.8 mm Hg Pgas V 2 = 220 ml 2) 1064 mm Hg Pgas + Pwater 1.4 atm (760 mm Hg/1 atm) = 1064 mm Hg - 18.6 mm Hg Pwater V2 = 475 ml(1045.4 mm Hg/740 mm Hg)(308 K/294K) 1045.4 mm Hg Pgas V 2 = 703 ml |
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Graham's Law of Diffusion:
The velocity of a gas varies inversely with its mass. |
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Ideal Gas Equation: PV = nRT
P = pressure, V = volume, n = moles, T = temperature, in Kelvin, R = .082 L-atm/mol-K |
| Solve: 1) What pressure is exerted by .675 mol of a gas in a 45.8 L container at -24.0oC? 2) At what celsius temperature will 11.8 moles of a gas exert 592 Kpa of pressure in a container whose volume is 32 .8 L? Answers: 1) PV = nRT P = .675 mol(.082 L-atm/mol-K) 249K = .301 atm P = nRT 45.8 L V 2) PV = nRT P = 592 Kpa(1 atm/101.3 Kpa) = 5.844 atm T = PV nR T = 5.844 atm(32.8 L) = 198 K = -75 oC 11.8 mol(.082 L-atm/mol-K) |
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Mass-Volume Problems: They are worked liked
Mass-Mass problems except: 1 mole of any gas at
STP = 22.4 L |
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Steps:
1. Write & Balance the chemical equation.2. Change grams or liters of A to moles of A . 3. Change moles of A to moles of B. [Use molar ratio: Coeffients in the equation] 4. Change moles of B to grams or liters of B . |
| Solve: |
| 1) How many liters of CO2
are produced when 125 g of propane, C3H8 is
burned? C3H8 +O2
--> CO 2 + H 20 2) 750 ml of chlorine gas reacts with an excess of Al to produce how many grams of AlCl3? |
| Answers: |
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