The Gas Laws - Blended learning

Transcript The Gas Laws - Blended learning

The Gas Laws

Learning about the special behavior of gases

Objective #4

Section 21.5, Note pack pg. 11

Graham’s Law of Diffusion

• A key questions we need to be able to answer is, “How fast do gases travel in comparison to one another?”

• Diffusion The tendency for gases to travel from an area of higher concentration to lower concentration until an equilibrium is reached.

• Effusion – diffusion of gas molecules through a pin hole.

Effusion – diffusion of gas molecules through a pin hole.

Graham’s Law

• The ratio of velocity of a lighter gas to a heavier gas is equal to the square root of the inverse of their molar masses.

Graham’s Law

• The ratio of velocity of a lighter gas to a heavier gas is equal to the square root of the inverse of their molar masses.

Graham’s Law

• Show the mathematical proof that derives Graham’s Law : Kinetic energy = energy of movement

K.E. = ½ m v

(m) = Mass (v) = Velocity If 2 gas particles have the same K.E. (temp.), the smaller particle would be moving faster.

Restate Graham’s Law

• Graham’s law: Lighter gases travel faster than heavier gases at the same temperature and pressure.

• Example: Compare the velocities of Hydrogen and Oxygen.  We know that Hydrogen has a mass of 2 g/mol; Oxygen has a mass of 32g/mol.

Restate Graham’s Law

• Graham’s law: Lighter gases travel faster than heavier gases at the same temperature and pressure.

Hydrogen travels 4 x’s faster than oxygen.

Always list the lighter gas over the heavier one, then use the square root of the inverse to find the rate.

Graham’s law:

The ratio of velocity of a lighter gas to a heavier gas is equal to the square root of the inverse of their molar masses.

Lighter gas Heavier gas’ mass Reduce Heavier gas Lighter gas’ mass

Hydrogen travels 4 x’s faster than oxygen.

Please explain how to find this to your neighbor better than I did.

Graham’s Law: Example 1

pg. 12 • Place the following gases in order of increasing average molecular speed at 25 Celsius.

– Ne, HBr, SO 2 , NF 3 , CO

Graham’s Law: Example 1

• Place the following gases in order of increasing average molecular speed at 25 Celsius.

– Ne, HBr, SO 2 , NF 3 , CO We need to know the masses of each gas

Graham’s Law: Example 1

• Place the following gases in order of increasing average molecular speed at 25 Celsius.

– Ne, HBr, SO 2 , NF 3 , CO 20 81 64 71 28 (masses are g / mol )

Graham’s Law: Example 1

• Place the following gases in order of increasing average molecular speed at 25 Celsius.

– Ne, HBr, SO 2 , NF 3 , CO 20 81 64 71 28 (masses are g / mol ) So, the correct order, from smallest mass to largest, would be…

Graham’s Law: Example 1

• Place the following gases in order of increasing average molecular speed at 25 Celsius.

– Ne, HBr, SO Ne  CO  2 , NF SO …but in order of 2 3 , CO 20 81 64 71 28  NF (masses are 3  increasing speed g / HBr mol ) So, the correct order, from smallest mass to largest, would be… would be… HBr  NF 3  SO 2  CO  Ne

Graham’s Law: Example 2

• Compare the rate of diffusion of nitrogen gas to helium gas

Graham’s Law: Example 2

• Compare the rate of diffusion of nitrogen gas to helium gas Lighter Heavier

Graham’s Law: Example 2

• Compare the rate of diffusion of nitrogen gas to helium gas Lighter  Heavier  Rate He Rate N 2

Graham’s Law: Example 2

• Compare the rate of diffusion of nitrogen gas to helium gas Lighter  Heavier  Rate He Rate N 2 √28 √4 Graham’s law:

The ratio of velocity of a lighter gas to a heavier gas is equal to the square root of the inverse of their molar masses.

Graham’s Law: Example 2 • Compare the rate of diffusion of nitrogen gas to helium gas

Lighter  Heavier  Rate He √28 = 5.29

Rate N2 √4 = 2

Graham’s Law: Example 2

• Compare the rate of diffusion of nitrogen gas to helium gas Lighter  Rate He √28 = 5.29

Heavier  Rate N 2 √4 = 2 = 2.65

Therefore, Helium will diffuse 2.65 times faster than nitrogen

Graham’s Law: Example 3

• Compare the rate of diffusion of argon to neon gas

Graham’s Law: Example 3

• Compare the rate of diffusion of argon to neon gas Lighter  Neon Heavier  Argon

Graham’s Law: Example 3

• Compare the rate of diffusion of argon to neon gas Lighter  Neon √39.9

Heavier  Argon √20.2

Graham’s Law: Example 3

• Compare the rate of diffusion of argon to neon gas Lighter  Neon √39.9 = 6.32

Heavier  Argon √20.2 = 4.49

= 1.4

Graham’s Law: Example 3

• Compare the rate of diffusion of argon to neon gas Lighter  Neon √39.9 = 6.32

Heavier  Argon √20.2 = 4.49

= 1.4

Neon (lighter) will diffuse 1.4 times faster than Argon (heavier).

Example 4

• A gas of unknown molecular mass was allowed to effuse through a small opening under constant pressure conditions. It required 105 seconds for 1.0 Liter of the gas to effuse. Under identical conditions it required 31 seconds for 1.0 liter of oxygen to effuse. Calculate the molar mass of the unknown gas.

Example 4

Here’s what we know: Unknown gas Diffusion time = 105 seconds Heavier Oxygen Diffusion time = 31 seconds Lighter

Example 4

SO… Rate (oxy) √Mass (X) Rate (unknown) √Mass 32

Example 4

SO… Rate (oxy) √Mass (X)  Rate (unknown) √Mass 32  Invert the times to get the difference in rate between them sides.

3.39

11.49

105 31

=3.39

= √x_ √32 To get rid of the square root, we need to square both