Examples

3.50 L

O2

1.50 L

N2

2.70 atm• When these valves are opened, what is

each partial pressure and the total pressure?

4.00 L

CH4

4.58 atm 0.752 atm

Vapor Pressure

• Water evaporates!• When that water evaporates, the

vapor has a pressure.• Gases are often collected over water

so the vapor. pressure of water must be subtracted from the total pressure.

• Water vapor pressure must be given.

Example• N2O can be produced by the following

reaction

• NH4NO3 N2O + 2H2O

• what volume of N2O collected over water at a total pressure of 94 kPa and 22ºC can be produced from 2.6 g of NH4NO3? ( the vapor pressure of water at 22ºC is 21 torr)

1) Gases are composed of molecules whose size is negligible compared with the average distance between them. (Ideal gas molecules are point masses.)

2) Gas molecules are in constant motion, in random directions, and at various speeds. (Properties of a gas that depend on motion like pressure are the same in all directions.)

Kinetic Molecular Theory

3) Ideal gas molecules exert neither an attractive force nor repulsive force on each other.

4) When molecules collide with one another, the collisions are elastic. What does this mean?

5) The average kinetic energy of the molecules is proportional to the temperature on the Kelvin scale. What does this mean?

All gases at the same temperature on the K scale have the same average kinetic energy.

K.E. = 1/2 mv2 or 1/2 mu2 where v is velocity and is a vector quantity, and u is speed which is a scaler quantity.

Two gases at the same T have the same average K.E.If one gas is heavier than the other, what must be different if the average K.E. must be the same?

This is the basis for Graham's Law.avg. KEA = avg. KEB

1/2 mAu2A = 1/2 mBu2

B

mAu2A = mBu2

B

Using force factor of gases

• (KE)avg = 3/2 RT

• This the meaning of temperature.• Finding root mean squre velocity: u is the average particle velocity.• u 2 is the average particle velocity

squared.

• the root mean square velocity is

u 2 = urms

Combine these two equations

• (KE)avg = NA(1/2 mu 2 )

• (KE)avg = 3/2 RT

Combine these two equations

• (KE)avg = NA(1/2 mu 2 )

• (KE)avg = 3/2 RT

Where M is the molar mass in kg/mole, and R has the units 8.3145 J/Kmol.

• The velocity will be in m/s

u = 3RT

Mrms

Dalton’s Law

Ptotal= P1+ P2+ P3+ …

= UF6

= H2 (2.01 g/mol)

(352 g/mol)

KE = ½ mu2 m = massu = average velocity

root mean square velocity urms

urms = 3RT/M

M = molar mass (kg)

R= 8.314 J/K mol

urms =

urms = 145 m/s

1926 m/s

Example

• Calculate the root mean square velocity of carbon dioxide at 25ºC.

• Calculate the root mean square velocity of hydrogen at 25ºC.

• Calculate the root mean square velocity of chlorine at 25ºC.

Boltzman-Maxwell Distribution of gases

Starting in the 1850's, Boltzmann in Germany and Maxwell in England found that

properties of gases could be satisfactorily explained in terms of the motion of the individual gas molecules.

1) in a given sample of gas, the speeds of the molecules vary over a wide range of values - a few are moving very slow, a few are moving very fast, most have intermediate speeds.

2) Graphically we represent it as:

3) If we raise the temperature, then the curve moves to the right and lowers.

Velocity

• Average increases as temperature increases.

• Spread increases as temperature increases.

Effusion - a process in which a gas flows through a small hole.

Theoretically simpler than diffusion since we don't have to be concerned with molecules hitting each other and zigzagging across the room.

Graham’s Law of Effusion

escape of gases through a small hole

rate1 = rate2

M2 / M1

diffusion mixing of gases

mean free path

1 atm 6 x 10-9 mspace 3 x 1010 m

Graham's Law of Effusion - Under the same conditions of pressure and temperature, the rates of effusion of gases are inversely proportional to the square roots of their molar masses.

How many times faster does H2 effuse than O2 at the same temperature and pressure?

The rate of effusion of H2 is 4 times faster than the rate of effusion of O2.

If it takes 205 seconds for 1.50 L of an unknown gas to effuse through a porous cup, and 95 seconds for the same volume of N2 at the same T and P, what is the approximate molar mass of the gas?

Examples• A compound effuses through a porous

cylinder 3.20 time faster than helium. What is it’s molar mass?

• If 0.00251 mol of NH3 effuse through a

hole in 2.47 min, how much HCl would effuse in the same time?

• A sample of N2 effuses through a hole in

38 seconds. what must be the molecular weight of gas that effuses in 55 seconds under identical conditions?

Diffusion

• The spreading of a gas through a room.

• Slow considering molecules move at 100’s of meters per second.

• Collisions with other molecules slow down diffusions.

• Best estimate is Graham’s Law.

Real Gases

• Real molecules do take up space and they do interact with each other (especially polar molecules).

• Need to add correction factors to the ideal gas law to account for these.

Volume Correction

• The actual volume free to move in is less because of particle size.

• More molecules will have more effect.• Corrected volume Vreal = V - nb• b is a constant that differs for each

gas.

• P’ = nRT (V-nb)

Pressure correctionPressure correction Because the molecules are attracted Because the molecules are attracted

to each other, the pressure on the to each other, the pressure on the container will be less than idealcontainer will be less than ideal

depends on the number of molecules depends on the number of molecules per liter.per liter.

since two molecules interact, the since two molecules interact, the effect must be squared.effect must be squared.

Pobserved = P’ - a

2

( )Vn

Altogether• Pobs= nRT - a n 2

V-nb V

• Called the Van der Wall’s equation if rearranged

• Corrected Corrected Pressure Volume

( )

P + an

V x V - nb nRTobs

2

Where does it come from

• a and b are determined by experiment.

• Different for each gas.• Bigger molecules have larger b.• a depends on both size and

polarity.• once given, plug and chug.

Example

• Calculate the pressure exerted by 0.5000 mol Cl2 in a 1.000 L container at 25.0ºC

• Using the ideal gas law.• Van der Waal’s equation

• a = 6.49 atm L2 /mol2 • b = 0.0562 L/mol

Boyle’s LawV 1/P

Charles’ LawV T

Avogadro’s LawV n

Dalton’s Law

Ptotal= P1+ P2+ P3+ …

3. Attractive and repulsive forces negligible.

Real Gases

n = PVRT

need to correct V for Vgas

need to correct P for interactions

P + n2aV2

V - nb = nRT

P + n2aV2

V - nb = nRT

van der Waals Equation

a related to Intermolecular Forcesmolecular complexity

b related to molecular volume

a, b experimental