the gas laws learning about the special behavior of gases (section 12.3) note pack page 3, bottom...

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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!