The Gas Laws
Chemistry
Dr. May
Gaseous Matter
Indefinite volume and no fixed shape
 Particles move independently of each
other


Particles have gained enough energy to
overcome the attractive forces that held
them together as solids and liquids
Avogadro’s Number

One mole of a gas contains Avogadro’s
number of molecules

Avogadro’s number is 6.02 x 1023 or
602,000,000,000,000,000,000,000
Diatomic Gas Elements
Gas





Hydrogen (H2)
Nitrogen (N2)
Oxygen (O2)
Fluorine (F2)
Chlorine (Cl2)
Molar Mass





2 grams/mole
28 grams/mole
32 grams/mole
38 grams/mole
70 grams/mole
Inert Gas Elements
Gas






Helium
Neon
Argon
Krypton
Xenon
Radon
Molar Mass






4 grams/mole
20 grams/mole
40 grams/mole
84 grams/mole
131 grams/mole
222 grams/mole
Other Important Gases






Gas
Formula
Molar Mass
Carbon Dioxide
Carbon Monoxide
Sulfur Dioxide
Methane
Ethane
Freon 14
CO2
CO
SO2
CH4
CH3CH3
CF4
44
28
64
16
30
88
g/mole
g/mole
g/mole
g/mole
g/mole
g/mole
One Mole of Oxygen Gas (O2)

Has a mass of 32 grams

Occupies 22.4 liters at STP
273 Kelvins (0oC)
 One atmosphere (101.32 kPa)(760 mm)


Contains 6.02 x 1023 molecules
(Avogadro’s Number)
Mole of Carbon Dioxide (CO2)

Has a mass of 44 grams

Occupies 22.4 liters at STP

Contains 6.02 x 1023 molecules
One Mole of Nitrogen Gas (N2)

Has a mass of 28 grams

Occupies 22.4 liters at STP

Contains 6.02 x 1023 molecules
Mole of Hydrogen Gas (H2)

Mass

2.0 grams

Volume at STP

22.4 liters

Molecules

6.02 x 1023
Standard Conditions (STP)

Molar Volume


Standard
Temperature


Standard Pressure

22.4 liters/mole
0 oC
 273 Kelvins
1 atmosphere
 101.32 kilopascals
 760 mm Hg
Gas Law Unit Conversions








liters  milliliters
milliliters  liters
o C  Kelvins
Kelvins  o C
mm  atm
atm  mm
atm  kPa
kPa  atm








Multiply by 1000
Divide by 1000
Add 273
Subtract 273
Divide by 760
Multiply by 760
Multiply by 101.32
Divide by 101.32
Charles’ Law

At constant pressure, the volume of a
gas is directly proportional to its
temperature in Kelvins
V1
T1
=
V2
T2
As the temperature goes up , the volume goes up 
Boyle’s Law

At constant temperature, the volume of
a gas is inversely proportional to the
pressure.
P1V1 = P2V2
As the pressure goes up , the volume goes down 
Combined Gas Law
P1V1
T1
=
P2V2
T2

Standard Pressure (P) = 101.32 kPa, 1 atm,
or 760 mm Hg

Standard Temperature (T) is 273 K

Volume (V) is in liters, ml or cm3
Charles’ Law Problem
A balloon with a volume of 2 liters and a
temperature of 25oC is heated to 38oC.
What is the new volume?
1. Convert oC to Kelvins
25 + 273 = 298 K
38 + 273 = 311 K
2. Insert into formula
Charles’ Law Solution
V1
T1
V1 = 2 liters
T1 = 298 K
=
V2
T2
V2 = Unknown
T2 = 311 K
2 = V2
298 311
Charles’ Law Solution
2
298
=
V2
311
298 V2 = (2) 311
V2 = 622
298
V2 = 2.09 liters
Charles’ Law Problem Answer
A balloon with a volume of 2 liters and a
temperature of 25oC is heated to 38oC.
What is the new volume?
V2 = 2.09 liters
Boyle’s Law Problem
A balloon has a volume of 2.0 liters at
743 mm. The pressure is increased to 2.5
atmospheres (atm). What is the new
volume?
1. Convert pressure to the same units
743  760 = .98 atm
2. Insert into formula
Boyle’s Law Solution
P1V1 = P2V2
P1 = 0.98 atm
V1 = 2.0 liters
P2 = 2.5 atm
V2 = unknown
0.98 (2.0) = 2.5 V2
Boyle’s Law Solution
P1V1 = P2V2
0.98 (2.0) = 2.5 V2
V2 = 0.98 (2.0)
2.5
V2 = 0.78 liters
Boyle’s Law Problem Answer
A balloon has a volume of 2.0 liters at
743 mm. The pressure is increased to
2.5 atmospheres (atm). What is the
new volume?
V2 = 0.78 liters
Combined Gas Law Problem
A balloon has a volume of 2.0 liters at a
pressure of 98 kPa and a temperature of 25
oC. What is the volume under standard
conditions?
1.
2.
3.
4.
Convert 25 oC to Kelvins 25 + 273 = 298 K
Standard pressure is 101.32 kPa
Standard temperature is 273 K
Insert into formula
Combined Gas Law Solution
P1V1
T1
P1 = 98 kPa
V1 = 2.0 liters
T1 = 298 K
=
P2V2
T2
P2 = 101.32 kPa
V2 = unknown
T2 = 273 K
Combined Gas Law Solution
P1V1
=
T1
98 (2.0)
298
P2V2
T2
=
101.32 V2
273
(298) (101.32) V2 = (273) (98) (2.0)
Combined Gas Law Solution
P1V1
T1
=
P2V2
T2
(298) (101.32) V2 = (273) (98) (2.0)
V2
=
273 (98) (2.0)
(298) (101.32)
Combined Gas Law Solution
P1V1
T1
V2
V2
=
=
=
P2V2
T2
273 (98) (2.0)
(298) (101.32)
53508 = 1.77 liters
30193
Combined Gas Law Problem Answer
A balloon has a volume of 2.0 liters at a
pressure of 98 kPa and a temperature
of 25 oC. What is the volume under
standard conditions?
V2 = 1.77 liters
Combined Gas Law – V2
P1V1
=
T1
P2V2
T2
P1V1T2 = P2V2T1
P1V1T2
P2T1
=
V2
The End

This presentation was created for the
benefit of our students by the Science
Department at Howard High School of
Technology

Please send suggestions and comments
to rmay@nccvt.k12.de.us
The Ideal Gas Law
Chemistry
Dr. May
Kinetic Molecular Theory
Molecules of an ideal gas
 Are dimensionless points
 Are in constant, straight-line motion
 Have kinetic energy proportional to
their absolute temperature
 Have elastic collisions
 Exert no attractive or repulsive forces
on each other
Ideal Gas Law
PV = nRT
P = pressure in kilopascals (kPa) or
atmospheres (atm)
V = volume in liters
n = moles
T = temperature in Kelvins
R = universal gas constant
Ideal Gas Law: PV = nRT
Pressure (P)
 Volume (V)
 Moles (n)
 Temperature (T)


The universal gas
constant (R)
Atm or kPa
Always liters
Moles
Kelvins
0.0821 ( P in atm) or
8.3 (P in kPa)
Universal Gas Constant
R = 0.0821 if P = atmospheres
R = 8.3 if P = kilopascals
R = PV
nT
Deriving R for P in Atmospheres
R = PV
nT
Assume n = 1 mole of gas
Standard P = 1 atmosphere
Standard V = molar volume = 22.4 liters
Standard T = 273 Kelvins
R Value When P Is In Atmospheres
R = PV
nT
R = (1) (22.4)
(1) 273
R = 0.0821 atm Liters
mole Kelvins
Deriving R For P In Kilopascals
R = PV
nT
Assume n = 1 mole of gas
Standard P = 101.32 kilopascals
Standard V = molar volume = 22.4 liters
Standard T = 273 Kelvins
R Value When P Is In Kilopascals
R = PV
nT
R = (101.32) (22.4)
(1) 273
R = 8.3 kPa Liters
mole Kelvins
Ideal Gas Law - Pressure
PV = nRT
P
=
nRT
V
Solves for pressure when moles,
temperature, and volume are known
Ideal Gas Law - Volume
PV = nRT
V
=
nRT
P
Solves for volume when moles, temperature,
and pressure are known
Ideal Gas Law - Temperature
PV = nRT
T
=
PV
nR
Solves for temperature when moles,
pressure, and volume are known
Ideal Gas Law - Moles
PV = nRT
n
=
PV
RT
Solves for moles when pressure,
temperature, and volume are known
Ideal Gas Law Problem
What is the mass of nitrogen in a 2.3 liter
container at 1.2 atmospheres, and 25 oC ?





V = 2.3 liters
P = 1.2 atmospheres
T = 25 oC = 298 Kelvins
R = 0.0821 since P is in atms.
Find moles (n), then grams
Ideal Gas Law Solution (moles)
PV = nRT
1.2 (2.3) = n (0.0821) (298)
n =
1.2 ( 2.3)
= 0.11 moles
(0.0821) (298)
Ideal Gas Law Solution (Grams)
Grams = moles x molecular weight (MW)
Moles = 0.11
 Molecular Weight of N2 = 28 g/mole
 Grams = 0.11 x 28 = 3.1 grams

Ideal Gas Law Answer
What is the mass of nitrogen in a 2.3
liter container at 1.2 atmospheres, and
25 oC ?
The answer is 0.11 moles and
3.1 grams
The End

This presentation was created for the
benefit of our students by the Science
Department at Howard High School of
Technology

Please send suggestions and comments
to rmay@nccvt.k12.de.us