Gases Laws Notes

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.

Manometer- Measures pressure of a gas in a container

Two types – Open and Closed

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

Learning Check

• How many mmHg are 8.92atm?

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

Learning Check

• How many moles of gas are contained in a 2.00L flask at 98.8 kPa and 25.0°C?

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

CHALLENGE!

• What would the value of the Gas Constant, R, be with the following units?

1) kPa*L

mol*K

2) atm*L

kmol*K