Unit 10
Gas Laws
I. Kinetic Theory
Particles in an ideal gas…
1.gases are hard, small, spherical
particles
2.don’t attract or repel each other.
3.are in constant, random,
straight-line motion.
4.indefinite shape and volume.
5.have “perfectly” elastic collisions.
A. Graham’s Law
• Diffusion
–The tendency of molecules to move
toward areas of lower concentration.
•Ex: air leaving tire when valve is
opened
• Effusion
–Passing of gas molecules
through a tiny opening in
a container
A. Graham’s Law
Tiny opening
Diffusion
Effusion
Which one is Diffusion and which one is Effusion?
II. Factors Affecting Gas Pressure
A. Amount of Gas
Add gas - ↑ pressure
Remove gas - ↓ pressure
Ex: pumping up a tire
adding air to a balloon
aerosol cans
II. Factors Affecting Gas Pressure
B. Volume
Reduce volume - ↑ pressure
Increase volume - ↓ pressure
Ex: piston in a car
II. Factors Affecting Gas Pressure
C. Temperature
Increase Temp. - ↑ pressure
Decrease Temp. - ↓ pressure
Ex: Helium balloon on
cold/hot day, bag of chips
Gas Pressure- collision of gas
molecules with the walls of the
container
Atmospheric Pressure- collision
of air molecules with objects
Atmospheric pressure is measured with
a barometer.
Vacuum- empty space
with no particles and no
pressure Ex: space
Increase altitude – decrease pressure
Ex. Mt. Everest – atmospheric pressure is 253 mm Hg
Gas Pressure (Cont.)
-- 3 ways to measure pressure:
»atm (atmosphere)
»mm Hg
»kPa (kilopascals)
U-tube Manometer
III. Variables that describe a gas
Variables
Units
Pressure (P) –
kPa, mm Hg, atm
Volume (V) –
L , mL , cm3
Temp (T) –
°C , K
(convert to Kelvin)
K = °C + 273
Mole (n) -
mol
Draw on the Left Side of
Your Spiral
Pressure Volume
kPa
Temperature
Mole
How pressure units are
related:
1 atm = 760 mm Hg = 101.3 kPa
How can we make these into
conversion factors?
1 atm
760 mm Hg
101.3 kPa
1 atm
Guided Problem:
1. Convert 385 mm Hg to kPa
385 mm Hg x 101.3 kPa
760 mm Hg
= 51.3 kPa
2. Convert 33.7 kPa to atm
1 atm
33.7 kPa x
= .33 atm
101.3 kPa
STP
Standard Temperature and Pressure
Standard pressure – 1 atm, 760 mmHg,
or 101.3 kPa
Standard temp. – 0° C or 273K
Gases (cont.)
Kelvin Temperature scale is directly
proportional to the average kinetic
energy
IV. Gas Laws
A. Boyle’s Law
• The pressure and volume of
a gas are inversely related
-at constant mass & temp
• P1 × V1 = P 2 × V 2
P
V
Example Problems
pg 335 # 10 &11
10. The pressure on 2.50 L of anesthetic gas
changes from 105 kPa to 40.5 kPa. What will
be the new volume if the temp remains
constant?
P1 = 105 kPa
V1 = 2.5 L
P2 = 40.5 kPa
V2 = ?
P1 × V1 = P2 × V2
(105) (2.5) = (40.5)(V2)
262.5 = 40.5 (V2)
6.48 L = V2
Example Problems
pg 335 # 10 &11
11. A gas with a volume of 4.00L at a pressure
of 205 kPa is allowed to expand to a volume of
12.0L. What is the pressure in the container if
the temp remains constant?
P1 = 205 kPa
P2 = ?
V1 = 4.0 L
V2 = 12.0 L
P1 × V1 = P2 × V2
(205) (4.0) = (P2)(12)
820 = (P2) 12
68.3 L = P2
B. Charles’ Law
• The volume and temperature
(in Kelvin) of a gas are directly
related
– at constant mass & pressure
• V1 =
V2
T1
T2
***Temp must be in Kelvin
K = °C + 273
V
T
Example Problems
pg. 337 # 12 & 13
12. If a sample of gas occupies 6.80 L at
325°C, what will be its volume at 25°C if
the pressure does not change?
V1= 6.8L
T1 = 325°C = 598 K
6.8 = V2
598 298
598 × V2 = 2026.4
598
598
V2 = 3.39 L
V2 = ?
T2 = 25°C = 298 K
Example Problems
pg. 337 # 12 & 13
13. Exactly 5.00 L of air at -50.0°C is
warmed to 100.0°C. What is the new
volume if the pressure remains constant?
V1= 5.0L
T1 = -50°C = 223 K
5
= V2
223
373
(223) V2 = 1865
223
223
V2 = 8.36 L
V2 = ?
T2 = 100°C = 373 K
C. Gay-Lussac’s Law
• The pressure and absolute
temperature (K) of a gas are
directly related
– at constant mass & volume
P1
T1

=
P2
T2
***Temp must be in Kelvin
K = °C + 273
P
T
Example Problems
1. The gas left in a used aerosol can is at
a pressure of 103 kPa at 25°C. If this
can is thrown onto a fire, what is the
pressure of the gas when its
temperature reaches 928°C?
P1= 103 kPa
T1 = 25°C = 298 K
103 = P2
298 1201
298 × P2 = 123,703
P2 = 415 kPa
P2 = ?
T2 = 928°C = 1201 K
Example Problem
pg. 338 # 14
14. A gas has a pressure of 6.58 kPa at
539 K. What will be the pressure at 211
K if the volume does not change?
P1= 6.58 kPa
T1 = 539 K
6.58 = P2
539 211
539 × P2 = 1388
539
539
P2 = 2.58 kPa
P2 = ?
T2 = 211 K
D. Combined Gas Law
Combines the 3 gas laws as follows:
P1V1
T1
=
P2V2
T2
•The other laws can be obtained from this law by
holding one quantity (P,V or T) constant.
•Use this law also when none of the variables are
constant.
How to remember each Law!
P
Boyles
Cartesian Divers
V
Gay-Lussac
Charles
Fizz Keepers
T
Balloon and flask Demo
E. Ideal Gas Law
• The 4th variable that considers the
amount of gas in the system
• Equal volumes of gases contain equal
numbers of moles (varies directly).
P1V1
T1 n
=
P2V2
T2 n
E. Ideal Gas Law
•You can calculate the # of n of gas at
standard values for P, V, and T
PV
=R
Tn
(1 atm)(22.4L)
(273K)(1 mol)
UNIVERSAL GAS CONSTANT
R= 0.0821 atm∙L/mol∙K
You don’t need to memorize this value!
=R
E. Ideal Gas Law
PV=nRT
P= pressure in atm
V = volume in liters
n = number of moles
R= 0.0821 atm∙L/mol∙K
T = temperature in Kelvin
E. Example Problems
1. At what temperature will 5.00g of Cl2
exert a pressure of 900 mm Hg at a
volume of 750 mL?
2. Find the number of grams of CO2 that
exert a pressure of 785 mm Hg at a
volume of 32.5 L and a temperature of
32 degrees Celsius.
3. What volume will 454 g of H2 occupy
at 1.05 atm and 25°C.
F. Dalton’s Partial Pressure
Law
• The total pressure of a
mixture of gases equals the
sum of the partial pressures
of the individual gases.
Ptotal = P1 + P2 + P3 + ...
F. Dalton’s Law
• Example problem:
1. Air contains oxygen, nitrogen, carbon
dioxide, and trace amounts of other
gases. What is the partial pressure of
oxygen (PO2) if the total pressure is
101.3 kPa. And the partial pressures of
nitrogen, carbon dioxide, and other
gases are 79.10 kPa, 0.040 kPa, and
0.94 kPa.
PO2 = Ptotal – (PN2 + PCO2 + Pothers)
= 101.3 kPa – (79.10 kPa + 0.040 kPa + 0.94 kPa)
= 21.22 kPa
F. Dalton’s Law
2. 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 of the
container?
3. A gas mixture contains oxygen, nitrogen and carbon
dioxide. The total pressure is 50.0 kPa. If the carbon
dioxide has a partial pressure of 21 kPa and the
nitrogen has a partial pressure of 15 kPa, what is
the partial pressure of the oxygen?
4. A container contains two gases – helium and argon,
at a total pressure of 4.00 atm. Calculate the partial
pressure of helium if the partial pressure of the
argon is 1.5 atm.