gases laws notes. pressure pressure- force per unit area caused by particles hitting the walls of a...
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Pressure• Pressure- force per unit area
caused by particles hitting the walls of a container
• Barometer- Measures atmospheric pressure
• Atmospheric Pressure- Results from the weight of the air- mass of air being pulled toward the center of the earth by gravity.
Units for pressure
mm Hg, torr, Pascal (Pa), Kilopascal (kPa), atmospheres (atm), pounds square inch (psi)
1 atm = 760 mm Hg = 760 torr = 101325 Pa = 101.3 kPa = 14.69 psi
Ex: You have 28 psi. How many atm? Torr? Pascals?
=28 psi x ________
psi
atm
14.69
1 1.9 atm =28 psi x ________
psi
Pa
14.69
101325 1.9 x 105 Pa
=28 psi x ________
psi
torr
14.69
760 1400 torr
Boyle’s Law
-Relationship between pressure and volume (constant temp.)
-Inverse proportion P V
P1V1= P2V2
Given
V1= 1.5 L
P1= 56 torr
P2= 150 torr
V2= ?
V2 = P1V1
P2
____ V2 = (56 torr)(1.5 L)____________150 torr
V2 = 0.56 L
Ex: Consider a 1.5 L sample of CCl2F2 at a pressure of 56 torr. If pressure is changed to 150 torr at constant temperature, what is the new volume?
Learning Check
• Boyle’s Law variables:
• Is Boyle’s Law an inverse or direct relationship?
• Boyle’s Law constants:
• Boyle’s Law formula:
Charles’s Law-Relationship between volume and
temperature (constant pressure)
-Directly proportional V T-Temperature in Kelvin V1 = V2
T1 T2
Given
V1= 2.0 L
T1= 298 K
T2= 278 K
V2= ?
V2 = T2V1
T1
______ V2 = (278 K)(2.0 L)____________298 K
V2 = 1.9 L
°C + 273 = ____ K
Ex: A 2.0 L sample of air is collected at 298 K and cooled to 278 K, the pressure is held constant. What is the volume?
Learning Check
• Charles’ Law variables:
• Is Charles’ Law an inverse or direct relationship?
• Charles’ Law constants:
• Charles’ Law formula:
Gay-Lussac’s Law-Relationship between
pressure and temperature (constant volume)
-Directly proportional P T-Temperature in Kelvin
Given
P1= 107 kPa
T1= 22 °C + 273 = 295 K
T2= 45 °C + 273 = 318 K
P2= ?
P2 = T2P1
T1
______ P2 = (318 K)(107 kPa)______________295 K
P2 = 115 kPa
°C + 273 = ____ K
Ex: A mylar balloon is filled with helium gas to a pressure of 107 kPa when the temperature is 22 °C. If the temperature changes to is 45 °C, what will be the pressure of helium in the balloon?
P1 = P2
T1 T2
Learning Check
• Gay-Lussac’s Law variables:
• Is Gay-Lussac’s Law an inverse or direct relationship?
• Gay-Lussac’s Law constants:
• Gay-Lussac’s Law formula:
Avogadro’s Law -Relationship between volume
and moles (constant temperature and pressure)
-Directly proportional V n V1 = V2
n1 n2
Given
V1= 12.2 L
n1= 0.50 mol
n2= 0.50 mol O2
V2= ?
V2 = n2V1
n1
______ V2 = (0.33 mol)(12.2 L)______________0.50 mol
V2 = 8.1 Lx ________
mol O2
mol O3
3
2= 0.33 mol
Ex: 12.2L sample contains 0.50 moles O2. If O2 is converted to O3, what will the volume be?
3O22O3
n = # of moles
Learning Check
• Avogadro’s Law variables:
• Is Avogadro’s Law an inverse or direct relationship?
• Avogadro’s Law constants:
• Avogadro’s Law formula:
Combined Gas LawCombination of Boyle’s law, Charles’ law and Gay-Lussac’s Law
Given
V1= 3.5 L
P1= 6.32 atm
T1= 27 °C + 273 = 300. K
V2= 4.7 L
P2= 4.15 atm
T2= ?
P1 V1 = P2 V2
T1 T2
T2 = P2V2T1_______P1V1
T2 = (4.15 atm)(4.7 L)(300. K)_____________________(6.32 atm)(3.5 L)
T2 = 260 K
Ex: A 3.5 L sample of Argon exerts a pressure of 6.32 atm at 27 °C. When the volume is increased to 4.7 L and the pressure is decreased to 4.15 atm, what is the final temperature?
Learning Check
• Combined Gas Law variables:• Combined Gas Law constants:• Combined Gas Law formula:• A gas is heated from 263.0 K to 298.0 K and the volume is
increased from 24.0 liters to 35.0 liters by moving a large piston within a cylinder. If the original pressure was 1.00 atm, what would the final pressure be?
Ideal Gas LawP= pressure (atm)
V= volume (L)n= Mole Gas (mol) T= temperature (K)
R= universal gas constant
Ex: H2 has a volume of 8.56 L at 0oC and 1.5 atm. Calculate the moles of H2 present.
PV = nRT
R = 0.08206 L* atm mol * K
Given
V= 8.56 L
T = 0oC + 273 = 273 K
P = 1.5 atm
n = ?
R = 0.08206 atm L/mol K
n = PV
RT____ n = (1.5 atm)(8.56 L)_______________________
(0.08206 atm L/mol K)(273 K)
n = 0.57 mol
Dalton’s Law of Partial Pressures
Partial Pressure – pressure that a gas would exert if it alone in the container.
Ptotal = P1 + P2 + P3…Ex: A container holds three gases: oxygen, carbon dioxide,
and helium. The partial pressures of the three gases are 2.00 atm, 3.00 atm, and 4.00 atm, respectively. What is the total pressure inside the container?
Given
P1= 2.00 atm
P2= 3.00 atm
P3= 4.00 atm
Ptotal= ?
Ptotal = P1 + P2 + P3
Ptotal = 2.00 atm + 3.00 atm + 4.00 atm
Ptotal = 9.00 atm
A common method of collecting gas samples in the laboratory is to bubble the gas into a bottle filled with water and allow it to displace the water. When this technique is used, however, the gas collected in the bottle contains a small but significant amount of water vapor. As a result, the pressure of the gas that has displaced the liquid water is the sum of the pressure of the gas plus the vapor pressure of water at that temperature. The vapor pressures of water at various temperatures are given in Table.
PTotal - Pvapor = PGas
Given
T = 16°C
Pvapor =
V = 188 mL (doesn’t matter)
Ptotal = 92.3 kPa
Pgas= ?
PTotal - Pvapor = PGas
92.3 kPa – 1.82 kPa = PGas
Pgas = 90.5 kPa
Ex: A student collects oxygen gas by water displacement at a temperature of 16°C. The total volume is 188 mL at a pressure of 92.3 kPa. What is the pressure of oxygen collected?
1.82 kPa
Given
T = 17°C + 273 = 290. K
Pvapor =
V = 0.461 L
Ptotal = 0.989 atm
Pgas= ?
ngas =
PTotal - Pvapor = PGas
100. kPa – 1.94 kPa = PGas
Pgas = 98 kPa
Ex: Hydrogen gas is collected by water displacement. Total volume collected is 0.461 L at a temperature of 17°C and a pressure of 0.989 atm. What is the pressure of dry hydrogen gas collected? How many moles of hydrogen are present?
1.94 kPa
x ________
atm
kPa
1
101.3= 100. kPa
n = PV
RT____
n = (0.97 atm)(0.461 L)_______________________(0.08206 atm L/mol K)(290. K)
n = 0.019 mol
x ________
kPa
atm
101.3
1= 0.97 atm
Learning Check
• What does it mean to collect gas
over water?
• A sample of oxygen gas is collected over water. The total pressure is 98.56 kPa. The partial pressure of the dry oxygen calculated to be 95.70 kPa. What is the vapor pressure of water?
Gas Stoichiometry
STP = Standard Temperature and Pressure• Standard Temperature = 0oC• Standard Pressure = 1 atm• Molar Volume 1.00 molgas = 22.4 Lgas
Ex 1: A sample of N2 has a volume of 1.75 L at STP. How many moles of N2 are present?
x ________L
mol
22.4
1= 0.0781 mol1.75 L
or
use PV = nRT
Ex 3: Calculate the volume of CO2 produced at STP from 152g of CaCO3.
CaCO3 CaO + CO2
mol CaCO3
mol CO2
1
1=
34.0 L
x ___________
g CaCO3
mol CaCO3
100.1
1152 g CaCO3 x _______x __________ L
mol
22.4
1
or
use PV = nRT
Ex 2: Calculate the volume of O2 produced at 1.00 atm and 25 oC by the decomposition of 10.5g KClO3.
2KClO3 2KCl + 3O2
Given
V= ?
T = 25oC + 273 = 298 K
P = 1.00 atm
R = 0.08206 atm L/mol K
n = 10.5 g KClO3
V = nRT
P____
V = (0.128 mol) (0.08206 atm L/mol K)(298 K)_______________________________ 1.00 atm
V = 3.13 L
x __________mol KClO3
mol O2
2
3= 0.128 mol x ___________
g KClO3
mol KClO3
122.6
1