GAS LAWS
BOYLE’S LAW
DEMO
Bell Jar and Marshmallow
-
The marshmallow is getting bigger (expanding – volume increases). Why?
How do volume and pressure relate to each other?
SHAVING CREAM DEMO
WHAT IS BOYLE’S LAW?
• The volume of a given amount of gas varies
inversely with its pressure if the temperature remains
constant
• Inverse Relationship – as pressure increases, the
volume decrease by the same factor
• Example: If pressure doubles, the volume decreases
by ½
• Example: If the pressure decreases by a factor of 4,
Quadruple
the volume will _____________.
BOYLE’S LAW
Formula:
P1V1 = P2V2
• Remember, if pressure increases from P1 to P2, then
Decrease from V1 to V2.
volume must _________
• So, if pressure decreases from P1 to P2, then volume
Increase from V1 to V2.
must _________
BOYLE’S LAW GRAPH
BOYLE’S LAW EXAMPLE
• A sample of helium gas in a balloon is compressed
from 4.0L to 2.5L, at a constant temperature. If the
pressure of the balloon started at 210kPa, what will
the final pressure (P2) be?
BOYLE’S LAW EXAMPLE ANSWER
• Formula: P1V1 = P2V2
• What do we know?
•
•
•
•
P1 = 210 kPa
V1 = 4.0 L
P2 = ______
V2 = 2.5 L
(210kPa)(4.0L) = (P2)(2.5L)
P2 = 336 kPa
CHARLES’ LAW
DEMO
Erlenmeyer Flask and a balloon
Why does the balloon expand?
How do temperature and volume relate to each other?
WHAT IS CHARLES’ LAW?
• The volume of a given mass of gas is directly
proportional to its Kelvin temperature when held at
constant temperature
• Direct Relationship – as temperature increases, the
volume increases by the same factor
CHARLES’ LAW GRAPH
CHARLES’ LAW
Formula:
V1 = V2
T1 T2
• Remember, if pressure increases from P1 to P2, then
Increase from V to V .
volume must _________
1
2
• So, if pressure decreases from P1 to P2, then volume
Decrease from V to V .
must _________
1
2
CHARLES’ LAW EXAMPLE
• A gas sample at 40.0oC occupies a volume of 2.32L.
If the temperature is raised to 75.0oC, what will its
volume be? (always assume the 3rd variable is
constant if it is not mentioned)
(K = oC + 273)
What is absolute zero?
Why do we use Kelvin instead of Celsius degrees when working with gases?
Temperature must be in Kelvin!
K = oC + 273
CHARLES’ LAW EXAMPLE ANSWER
Formula: V1 = V2
T1 T2
• What do we know?
•
•
•
•
V1 = 2.32 L
T1 = 40.0o C
V2 = ______
T2 = 75.0o C
We must convert from degrees
Celsius to Kelvin
(K = oC + 273)
• T1 = 40.0o C + 273 = 313 K
• T1 = 75.0o C + 273 = 348 K
2.32 L = V2
313 K
348 K
V2 = 2.58 L
GAY-LUSSAC’S LAW
DEMO
Egg in a bottle
WHAT IS GAY-LUSSAC’S LAW
• The pressure of a given mass of gas varies directly
with the Kelvin temperature when the volume
remains constant.
• Direct Relationship – as temperature increases, the
volume increases by the same factor
GAY-LUSSAC’S LAW
Formula: P1 = P2
T1 T2
• Remember, if pressure increases from P1 to P2, then
Increase from T1 to T2.
temperature must _________
• So, if pressure decreases from P1 to P2, then
Decrease from T to T .
temperature must _________
1
2
GAY-LUSSAC’S LAW GRAPH
Temperature
(K)
150
300
450
Pressure (atm)
1.0
2.0
3.0
GAY-LUSSAC’S LAW
Real World Example
Pressure cooker is sealed so that the volume is
constant. Pressure increases in the cooker as
temperature increases
GAY-LUSSAC’S LAW EXAMPLE
The pressure of a gas tank is 3.20atm at 22oC.
If the temperature raises to 60oC, what will be
the pressure in the gas tank?
Temperature must be in Kelvin!
K = oC + 273
GAY-LUSSAC’S LAW EXAMPLE ANSWER
Formula: P1 = P2
T1 T2
• What do we know?
•
•
•
•
P1 = 3.20 atm
T1 = 22.0o C
P2 = ______
T2 = 60.0o C
We must convert from degrees
Celsius to Kelvin
(K = oC + 273)
• T1 = 22.0o C + 273 = 295 K
• T1 = 60.0o C + 273 = 333 K
3.20 atm = P2
295 K
333 K
P2 = 3.61 atm