the gas laws learning about the special behavior of gases (section 12.3) note pack page 3, bottom...
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The Gas LawsThe Gas LawsThe Gas LawsThe Gas Laws
Learning about the special Learning about the special behavior of gases behavior of gases (section 12.3)(section 12.3)
Note Pack page 3, bottomNote Pack page 3, bottom
Objective #1Objective #1
Boyle’s Law• State Boyle’s Law in words:
– When temperature and the number of particles are held constant, pressure and volume are inversely related
• Equation: P1V1 = P2V2
• Graph Boyle’s Law:
Example 1• A high-altitude balloon contains 30.0 L of helium gas at 103
kPa. What is the volume when the balloon rises to an altitude where the pressure is only 25.0 kPa? (Assume that the temperature remains constant.)
When dealing with gases, we’re going to start by using values of:
Pressure
Volume
Temperature (K)
In the problem above, please identify what you are given, and what you want.
Example 1• A high-altitude balloon contains 30.0 L of helium gas at 103
kPa. What is the volume when the balloon rises to an altitude where the pressure is only 25.0 kPa? (Assume that the temperature remains constant.)
P1 V1 = P2 V2 (103 kPa)(30L) = (25 kPa) (x)
Example 1• A high-altitude balloon contains 30.0 L of helium
gas at 103 kPa. What is the volume when the balloon rises to an altitude where the pressure is only 25.0 kPa? (Assume that the temperature remains constant.)
P1 V1 = P2 V2 (103 kPa)(30L) = (25 kPa) (x)
3090 = 25x
Example 1• A high-altitude balloon contains 30.0 L of helium gas at 103
kPa. What is the volume when the balloon rises to an altitude where the pressure is only 25.0 kPa? (Assume that the temperature remains constant.)
P1 V1 = P2 V2 (103 kPa)(30L) = (25 kPa) (x)
3090 = 25x 25 25
123.6 = x
Example 1• A high-altitude balloon contains 30.0 L of helium gas at 103
kPa. What is the volume when the balloon rises to an altitude where the pressure is only 25.0 kPa? (Assume that the temperature remains constant.)
P1 V1 = P2 V2 (103 kPa)(30L) = (25 kPa) (x)
3090 = 25x 25 25
123.6 = x
So, the answer is 123.6 L (we’ll use the same unit for
Volume as V1, unless otherwise instructed.
Example 2The pressure on 2.5 L of anesthetic gas changes
from 105 kPa to 40.5 kPa . What will be the new volume if the temperature remains constant?
In the problem above, please identify what you are given, and what you want:
Pressure
Volume
Temperature (K)
Example 2The pressure on 2.5 L of anesthetic gas changes
from 105 kPa to 40.5 kPa . What will be the new volume if the temperature remains constant?P1 V1 = P2 V2 (105 kPa)(2.5L) = (40.5 kPa)(X)
Example 2The pressure on 2.5 L of anesthetic gas changes
from 105 kPa to 40.5 kPa . What will be the new volume if the temperature remains constant?P1 V1 = P2 V2 (105 kPa)(2.5L) = (40.5 kPa)(X)
262.5 = 40.5x
Example 2The pressure on 2.5 L of anesthetic gas changes
from 105 kPa to 40.5 kPa . What will be the new volume if the temperature remains constant?P1 V1 = P2 V2 (105 kPa)(2.5L) = (40.5 kPa)(X)
262.5 = 40.5x 40.5 40.5
X = 6.48L
Example 3A gas with a volume of 4.0 L at a pressure of 205 kPa
is allowed to expand to a volume of 12.0 L. What is the pressure in the container if the temperature remains constant?
In the problem above, please identify what you are given, and what you want:
Pressure
Volume
Temperature (K)
Example 3A gas with a volume of 4.0 L at a pressure of 205 kPa
is allowed to expand to a volume of 12.0 L. What is the pressure in the container if the temperature remains constant?
• P1V1 = P2V2 (205 kPa)(4.0L) = (x)(12.0L)
68.3 kPa
Charles’s Law -• State Charles’s Law in Words:
– When pressure and amount of gas is held constant, Kelvin temperature and volume are directly related.
• Show this law in a formula:
• What would the graph for this look like? NOTE: Temp for all
gases needs to be in Kelvin!!
Example 1• A balloon inflated in a room at 24oC has a
volume of 4.0 L. The balloon is then heated to a temperature of 58oC. What is the new volume if the pressure remains constant?
Begin by changing the temp to Kelvin (oC + 273 = K)
24oC + 273 = 297 Kelvin
58oC + 273 = 331 Kelvin
Example 1• A balloon inflated in a room at 24oC has a
volume of 4.0 L. The balloon is then heated to a temperature of 58oC. What is the new volume if the pressure remains constant?
V1 = V2 T1 = T2
Example 1• A balloon inflated in a room at 24oC has a
volume of 4.0 L. The balloon is then heated to a temperature of 58oC. What is the new volume if the pressure remains constant?
V1 = V2 4.0L = X T1 = T2 297 = 331
Example 1• A balloon inflated in a room at 24oC has a
volume of 4.0 L. The balloon is then heated to a temperature of 58oC. What is the new volume if the pressure remains constant?
V1 = V2 4.0L = XT1 = T2 297 = 331
To solve, cross multiply and then divide…
X = 4.46L
Example 2
• If a sample of gas occupies 6.8 L at 325o C, what will be its volume at 25o C, if the pressure does not change?
Example 2
• If a sample of gas occupies 6.8 L at 325o C, what will be its volume at 25o C, if the pressure does not change?
V1 = V2 6.8L = X
T1 = T2 598 = 298
X = 3.39 L
To solve:
Cross-multiply & divide
(6.8L)(298K) = (598)(X)
Example 3
• Exactly 5.0 L of air at - 50.0o C is warmed to 100o C. What is the new volume if the pressure remains constant?
V1 = V2 T1 = T2
Change temp into Kelvin
Example 3
• Exactly 5.0 L of air at - 50.0o C is warmed to 100o C. What is the new volume if the pressure remains constant?
V1 = V2 5.0L = X .T1 = T2 223K = 373K
Example 3
• Exactly 5.0 L of air at - 50.0o C is warmed to 100o C. What is the new volume if the pressure remains constant?
V1 = V2 5.0L = X .T1 = T2 223K = 373K
To solve: Cross-multiply & divide
Example 3
• Exactly 5.0 L of air at - 50.0o C is warmed to 100o C. What is the new volume if the pressure remains constant?
V1 = V2 5.0L = X .T1 = T2 223K = 373K
X = 8.36 L
To solve: Cross-multiply & divide
Gay-Lussac’s LawState Gay-Lussac’s Law in words:
When the amount of gas and the volume of the gas are held constant, Kelvin temperature and pressure are directly related.
• Express this in a formula:
• What would the graph look like?
Example 1• A gas left in a used aerosol can is at a
pressure of 103 kPa at 25o C. If this can is thrown onto a fire, what is the pressure of the gas when its temperature reaches 928o C ?
P1 = P2T1 = T2
Example 1• A gas left in a used aerosol can is at a
pressure of 103 kPa at 25o C. If this can is thrown onto a fire, what is the pressure of the gas when its temperature reaches 928o C ?
P1 = P2 103 kPa = XT1 = T2 298 K = 1201K
Example 1• A gas left in a used aerosol can is at a
pressure of 103 kPa at 25o C. If this can is thrown onto a fire, what is the pressure of the gas when its temperature reaches 928o C ?
P1 = P2 103 kPa = XT1 = T2 298 K = 1201K
X = 415 kPa
Example 2• 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 = P2T1 = T2
Example 2• 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 = P2 6.58 kPa = XT1 = T2 539 K = 211K
Example 2• 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 = P2 6.58 kPa = XT1 = T2 539 K = 211K
X = 2.58 kPa
That concludes content for Obj. #1.
Practice makes permanent!