Grahm’s Law of Effusion

Grahm’s Law of EffusionEffusionEquation

Application

Review• Kinetic Molecular Theory

Gases have low density Gas molecules interact in elastic collisions Gas molecules are always in motion There are no forces of attraction between gas

molecules The speed of gas molecules is proportional to the

temperature• Pressure = Force / Area

Force - Newtons (N) ; Area - m2 ; Pressure = N / m2 = Pa• Units of Pressure

1 atm = 760 mm of Hg = 760 torr = 101.3 kPa = 1.013 x 105 Pa

Review• Partial Pressure

Dalton’s Law of Partial Pressure• Total Pressure = Pressure1 + Pressure2

• Stoichiometry of Gases Avogadro’s Law

• Volume, molecules, and mols of gases are all proportional to the coefficients of the balanced chemical equation

Standard Molar Volume• 22.4 L / mol of STP

Introduction• Grahm’s law of effusion compares the rate at which

two gases go through the same opening at the same temperature and pressure

• This relationship can be used to determine the identity of unknown gases

• Throughout the lecture form an answer to the following question Why is there an inverse relationship between the molar mass of a gas and the rate at which it effuses?

Diffusion & Effusion

• Diffusion - spontaneous mixing of gas particles from a high concentration to a low concentration

• Effusion - a process by which gas particles pass through a tiny openingWhat causes the particles to go through the opening?

Effusion Simulation

• http://phet.colorado.edu/en/simulation/gas-properties• http://phet.colorado.edu/en/contributions/view/3237

Kinetic Energy

• Grahm’s law can be derived from comparing the kinetic energy of two gases at the same temperature

• What is Kinetic energy?• Energy of motion• KE = 1/2 mv2

• Temperature determines the kinetic energy At the same temperature, two gases will

have the same KE

What did we see?

• Why did we only open the lid a little bit?

• Why did we hold the temperature constant?

• Which size of particle left the container more quickly?

Derivation

• KEA = KEB

• 1/2 mA(vA)2 = 1/2 mB(vB)2 • mA(vA)2

=mB(vB)2

• mB mB

• mA/ mB(vA)2 = (vB)2

• (vA)2

(vA)2• mA/ mB =

(vB)2 /(vA)2

• √ mA / √mB = vB / vA

Grahm’s Law of Effusion

• “the rates of the effusion of gases at the same temperature and pressure are inversely proportional to the square roots of their molar masses”

VA

VB=

√MB

√MA

What can we do with this formula?

• Compare the rates of effusion of different gasesWhich will gas will leave a container first

Which gas will fill a room first• Determine the identity of an

unknown gas

Basic Steps for Effusion Problems

• Determine the givens and the unknowns

• Rearrange equation to solve for unknown

• Plug in known numbers• Calculate square roots first• Finish Solving Problem

Examples

• How much faster does Neon gas effuse than Xenon gas?

• Molar Mass of Neon = 20.17 g/mol

• Molar Mass of Xenon = 131.29 g/mol

VA

VB=

√MB

√MA

VNe

VXe=

√MXe

√MNe

VNe

VXe=√131.29 g/mol

√20.17g/mol= 2.56 times

Examples• If a sample of Helium gas effuses 6 times faster

than an unknown sample of gas, what is the molar mass of the unknown gas?

• Molar Mass of Helium = 4.00 g/molVA

VB=

√MB

√MA

√MBVHe

Vunknown√MA= x

6VHe

Vunknown√4.00g/mol= x =144.00g/mol