Plan for Mon, 13 Oct 08 Exp 2 post-lab question Lecture –KMT (5.6) –Effusion and Diffusion (5.7) –Real gases and the van der Waal equation (5.8) –The nature.

Plan for Mon, 13 Oct 08

• Exp 2 post-lab question

• Lecture– KMT (5.6)– Effusion and Diffusion (5.7)– Real gases and the van der Waal equation (5.8)– The nature of energy?? (6.1)

• Quiz 2 returned

What can KMT do for you?• The main ideas you should take from KMT is

that we can describe T and P from a molecular perspective.

• Pressure: arises from molecules banging into the container walls.

• Temperature: arises from the kinetic energy of the gas molecules. The more KE they have, the faster they can move around, the “hotter” they are.

Temperature according to KMT

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Kinetic energy: The energy an object has by virtue of its motion.

Basically, the energy you must apply to an object to accelerate it from rest to a given velocity (u):

221 muEk

The average Ek of a molecule is directly proportional to the absolute temperature in K.

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Dr. Villarba subs. QUIZ 5

Dr. Schultz subs in lab

EXAM 2: Ch 5,6,10

Metathesis Lab (Exp3)

QUIZ 4

Vol. Anal. FORMAL RPT (Exp4)

We are here. QUIZ 3

A Closer Look at Molecular KE• Let’s consider the average KE per molecule and

see how it determines molecular speed.

3

2RTTotal KE in a mole of gas:

21

2muAverage KE per molecule:

According to KMT, and beyond the scope of this course.

Where u is an average of molecular velocity, and m is the mass of one molecule.

21 1 3

2 2A

mu RTN

We are apportioning the total KE in the mole of gas among all the molecules in an average fashion.

A Closer Look at Molecular KE

21 1 3

2 2A

mu RTN

We are apportioning the total KE in the mole of gas among all the molecules in an average fashion.

2 3

A

RTu

mN Note that mNA = M.

3rms

RTu

M

“Root-mean-square” speed, one kind of average molecular speed.

urms is the speed of a molecule that

has the average KE.

urms gives us a formal connection

between average gas speed, T, and M.

Distribution of Molecular Speeds

Petrucci, Fig 6.17

“Maxwell-Boltzmann” curve (a statistical distribution)

um – most probable speed

uav – average speed

urms – the speed of a molecule with the average molecular kinetic energy

urms Dependence on T & M

Petrucci, Fig. 6.18

E&G, Fig. 5.25M(O2) = 32 g/mol

M(H2) = 2 g/mol

M

RTurms

3

Trends

• Increased T increased average KE increased urms

– Maximum of curve shifts to higher u, and distribution spreads out.

• Increased M decreased urms

– Heavier molecules have lower average speed than lighter molecules.

• At a given T, are there more molecules at low speeds (u < urms) or high speeds (u > urms)? – There are more molecules at lower speeds than high.

M

RTurms

3

Calculating urms

• What is urms at 25oC for He(g) and N2(g)?

M(He) = 4.003 g/mol

M(N2) = 28.01 g/mol

• Already we know that urms(N2) < urms(He).• We also need T and R. What units do we need

for these values?

M

RTurms

3

Calculating urms

• What is urms at 25oC for He(g) and N2(g)?

M(He) = 4.003 g/mol

M(N2) = 28.01 g/molT = (25oC + 273) K = 298 K

• urms is in units of m/s• Will 0.08206 L atm/mol K work?• How about 8.314 J/mol K?

M

RTurms

3

Calculating urms

• How about 8.314 J/mol K?

• What’s a J (joule)?

1 J = 1 N m

• What’s a N (newton)?

1 N = 1 kg m/s2

1 J = 1 kg m2/s2

M

RTurms

3

Calculating urms for He(g)M(He) = 4.003 g/mol

T = 298 K

R = 8.314 J/mol K = 8.314 kg m2/mol K s2

M

RTurms

3

2

2

kg m3 8.314 298 K

mol K s

g 1 kg4.003

mol 1000 g

31.36 10 m/s

Calculating urms for N2(g)M(N2) = 28.01 g/mol

T = 298 K

R = 8.314 J/mol K = 8.314 kg m2/mol K s2

M

RTurms

3

2

2

kg m3 8.314 298 K

mol K s

g 1 kg28.01

mol 1000 g

515 m/s

Comparison• At 25oC,

urms(N2) = 515 m/s M(N2) = 28 g/molurms(He) = 1.36 x 103 m/s M(He) = 4 g/mol

• A car travelling at 60 mph,

ucar = 26.8 m/s

If gases travel so fast, why does it take so long for you to smell a bottle of perfume from

across the room?

Diffusion• Gas molecules travel in a

straight line only until they collide with a container wall or another gas molecule.

• Gaseous perfume molecules do not have an uninterrupted path in from of them.

• They are constantly colliding with gas molecules in the air.

Diffusion

Show Z Fig 5.24 p. 207

Diffusion is the process of mixing gases.

This is analogous to solution formation.

Diffusion of gases

In a closed container, diffusion will eventually lead to a homogeneous mixture.




Effusion

Petrucci, Fig 6.21b

Effusion is a special case of diffusion, which exploits the difference in velocities of lighter gas molecules.

This process was used during the Manhatten Project to separate 235U and 238U isotopes.

Effusion of a Gas

Rate of effusion is proportional to urms. So lighter particles will have a higher rate of effusion.




Real Gases

• Generally speaking, there is no such thing as an “Ideal Gas.”

• There are conditions under which a gas will behave ideally…– low P – moderate to high T

• van der Waal developed some corrections to the Ideal Gas law, based on a molecular picture, to explain these observed deviations.

Real Gases: Molecules have volume

• At high P, the volume of the individual gas molecules becomes non-negligible.

• Macroscopic gas is compressible, individual gas molecules are not.

• Under high P conditions, the space available for a gas molecule to move through is decreased by its neighbors, so the volume of the system is reduced relative to the ideal case.

Volume correction

• vdW corrected the volume available to a gas:

• P’ is a “corrected” ideal pressure. • What is the result of this volume correction, a

higher or lower pressure relative to ideal?

V V nb nRTP

V nb

Number of moles of gas

Empirical constant… different for each gas

Real Gases: Molecules attract each other• Under high P, gas molecules

get very close to each other, so intermolecular forces become significant.

• At low T, molecular Ek is reduced, and molecular speed drops, so the molecules become “trapped” by attractions to other molecules.

• Under these conditions, the molecules don’t collide with the container as frequently, so the pressure of the system is reduced relative to the ideal case.

Pressure Correction• vdW corrected for pressure, by including a mutual

attraction term.• Molecular attractions are proportional to concentration,

n/V.

• What is the result of the pressure correction?

2n

P P aV

Intermolecular

attraction termEmpirical constant… different for each gas

vdW Equation

• b generally increases with the size of the molecule• a seems to depend on the strength of intermolecular

forces.

2

obs

nP P a

V

2

obs

nRT nP a

V nb V

2

obs

nP a V nb nRT

V

vdW Equation

vdW equation corrects two major flaws in ideal gas theory:

• Gas molecules have finite volume which becomes important at high P.

• Gas molecules have nontrivial attractions that become important at low T and high P.

2

obs

nP a V nb nRT

V

• Although nonideal behavior is evident at all temperatures, the deviation is less at higher temperatures.

• Why do you think that could be?

N2(g)

Real Gases

Plan for Mon, 13 Oct 08 Exp 2 post-lab question Lecture –KMT (5.6) –Effusion and Diffusion (5.7) –Real gases and the van der Waal equation (5.8) –The nature.

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