KMT and Gas Laws Characteristics of Gases Gases expand to fill any container. –random motion, no attraction Gases are fluids (like liquids). –no attraction

KMT and

Gas Laws

Characteristics of GasesCharacteristics of Gases

• Gases expand to fill any container.– random motion, no attraction

• Gases are fluids (like liquids).– no attraction

• Gases have very low densities.– no volume = lots of empty space

Characteristics of GasesCharacteristics of Gases• Gases can be compressed.

– no volume = lots of empty space

• Gases undergo diffusion & effusion.Diffusion is the tendency of molecules and ions to move from areas of high concentration to areas of low concentration.

The diffusion of gas through a small opening in an otherwise closed container is called effusion.

Kinetic-Molecular Theory (KMT)

• Based on the idea that particles of matter are always in motion

• Provided a model of “ideal gas”– An imaginary gas that perfectly fits all the

assumptions of the KMT.

• FIVE (5) assumptions on KMT

K.E. = (½) mv2

Assumptions1. Gases consist of large number of tiny

particles that are far apart relative to their size.

2. Collisions between gas particles and between particles and container walls are elastic collisions (no loss in K.E.).

3. Gas particles are in continuous, rapid, random motion; K.E. = (½)mv2

Assumptions

4. There are no forces of attraction or repulsion between gas particles.

5. The average kinetic energy of gas particles depends on the temperature of the gas.

Temperature : Kinetic Energy

2

2

1.. mvEKeTemperatur

PressurePressure

area

forcepressure

Which shoes create the most pressure?

Atmospheric PressureAtmospheric Pressure

• Barometer– measures atmospheric pressure

Mercury Barometer

Aneroid Barometer

Atmospheric pressure is measured with a barometer.

Pressure Units and Conversions

• Kilopascal = kPa; just 1000 times greater than a Pascal

• Relationships between pressure units: **important for conversions!!!

1 atm = 14.7 psi = 760 mm Hg = 29.9 in Hg = 760 Torr = 101,300 Pa = 101.3 kPa

TemperatureTemperature

ºF

ºC

K

-459 32 212

-273 0 100

0 273 373

32FC 95 K = ºC + 273

• Always use absolute temperature (Kelvin) when working with gases.

Boyle’s LawBoyle’s Law

• The pressure and volume of a gas are inversely related – at constant mass & temp

P

V

P1V1 =P2V2

Example:• A 5.0 L container of nitrogen gas is at a pressure of 1.0 atm. What

is the new pressure if the volume is decreased to 500 mL, and the

temperature remains constant?

Formula needed

Rearrange for unknown

Plug-in to formulaInclude units and cancellation!

Final answer Unit

Unit conversion(s) needed:

kn

VV

n

Avogadro’s LawAvogadro’s Law

• Equal volumes of gases contain equal numbers of moles– at constant temp & pressure– true for any gas

Avogadro’s Law

• Avogadro’s Law says that equal volumes of gases at the same temperature and pressure contain equal numbers of moles.

• This means that the coefficients in the balanced equation

stand for volume as well as moles, but only for the gases.

twice as many twice as many moleculesmolecules

Example

5.00 L of a gas is known to contain 0.965 mol. If the amount of gas is increased to 1.80 mol, what new volume will result (at an unchanged temperature and pressure)?

Answer: this time I'll use V1n2 = V2n1

(5.00 L) (1.80 mol) = (x) (0.965 mol)

x =

ExampleConsider the chemical equation: 2NO2 (g) → N2O4 (g)

If 25 mL of NO2 gas is completely converted to N2O4 gas,under the same conditions, what volume will the N2O4 occupy?

Solution:Here, from the equation given, 2 moles of NO2 (g) forms 1 mole of N2O4 (g).So, V1 = 25 mL n1 = 2 moleV2 =? n2 = 1mole

V2 = (V1 n2) / n1, V2 = (25 mL) (1 mole) / (2 moles) = 12.5 mL

ExampleA cylinder with a movable piston contains 2.00 g of helium, He, at room

temperature. More helium was added to the cylinder and the volume was adjusted so that the gas pressure remained the same. How many grams of helium were added to the cylinder if the volume was changed from 2.00 L to 2.70 L? (The temperature was held constant.)

Solution:1. Convert grams of He to moles: 2.00 g / 4.00 g/mol = 0.500 mol2. Use Avogadro's Law: V1/n1 = V2/n2 2.00 L / 0.500 mol = 2.70 L / x x = 0.675 mol 3. Compute grams of He added: 0.675 mol - 0.500 mol = 0.175 mol 0.175 mol x 4.00 g/mol = 0.7 grams of He added

1 2

1 2

V V

T TV

T

Charles’ LawCharles’ Law

• The volume and absolute temperature (K) of a gas are directly related – at constant mass & pressure

Example:• A container of helium gas at 25°C in an expandable 500 mL

container is heated to 80°C. What is the new volume if the pressure remains constant?

Formula needed



Final answer Unit


1 2

1 2

P P

T TP

T

Gay-Lussac’s LawGay-Lussac’s Law

• The pressure and absolute temperature (K) of a gas are directly related – at constant mass & volume

Example:• A tank of propane gas at a pressure of 3.0 atm is cooled from 90°C

to 30°C. What is the new pressure if the volume remains constant?

Formula needed



Final answer Unit


= kPVPTVT

PVT

Combined Gas LawCombined Gas Law

P1V1

T1

=P2V2

T2

P1V1T2 = P2V2T1

Example:

• A helium-filled balloon at sea level has a volume of 2.1 L at 0.998 atm and 36°C. If it is released and rises to an elevation at which the pressure is 0.900 atm and the temperature is 28°C, what will be the

new volume of the balloon?

Formula needed



Final answer Unit


Ideal Gas Law

• An ideal gas is one whose particles take up no space and have no attractive forces for each other.

• In reality, gases do not always behave

ideally, especially under conditions of high pressure and low temperature.

A. Ideal Gas LawA. Ideal Gas Law

UNIVERSAL GAS CONSTANT

R=0.0821 Latm/molK

PV=nRT

You don’t need to memorize these values!

ExamplePV=nRT, R=0.0821 Latm/molK

1. 32.0 g of oxygen gas is at a pressure of 760 mm Hg and a temperature of 0°C. What is the volume of the gas?

2. How many moles of oxygen will occupy a volume of 2.5 liters at 1.2 atm and 25°C?

3. What pressure will be exerted by 25 g of CO2 at a temperature of 25°C and a volume of 500mL?

4. At what temperature will 5.00g of Cl2 exert a pressure of 900 torr at a volume of 750 mL?

YOU MUST CONVERT:YOU MUST CONVERT:

• P into atm– Use pressure conversions

• V into L– Divide by 1000 if mL

• n into moles – divide by molecular weight

• T into K– Add 273 to Celcius

Example:

• 32.0 g of oxygen gas is at a pressure of 760 mm Hg and a temperature of 0°C. What is the volume of the gas?

Formula needed



Final answer Unit


Dalton’s Law of Partial Pressures

V and T are

constant

P1 P2 Ptotal = P1 + P2

Consider a case in which two gases, A and B, are in a container of volume V.

PA = nART

V

PB = nBRT

V

nA is the number of moles of A

nB is the number of moles of B

PT = PA + PB XA = nA

nA + nB

XB = nB

nA + nB

PA = XA PT PB = XB PT

Pi = Xi PT

Pi ----- Partial pressurePT ------- Total pressureXi -------- Mole fraction

GAS DIFFUSION AND GAS DIFFUSION AND EFFUSIONEFFUSION

• diffusiondiffusion is the is the gradual mixing of gradual mixing of molecules of different molecules of different gases.gases.

• effusioneffusion is the is the movement of movement of molecules through a molecules through a small hole into an small hole into an empty container.empty container.


Graham’s law governs Graham’s law governs effusion and diffusion effusion and diffusion of gas molecules.of gas molecules.

Thomas Graham, 1805-1869. Thomas Graham, 1805-1869. Professor in Glasgow and London.Professor in Glasgow and London.

Rate of effusion is Rate of effusion is inversely proportional inversely proportional to its molar mass.to its molar mass.

Rate of effusion is Rate of effusion is inversely proportional inversely proportional to its molar mass.to its molar mass.

M of AM of B

Rate for B

Rate for A


Molecules effuse thru holes in a rubber Molecules effuse thru holes in a rubber balloon, for example, at a rate (= balloon, for example, at a rate (= moles/time) that ismoles/time) that is

• proportional to Tproportional to T• inversely proportional to M.inversely proportional to M.

Therefore, He effuses more rapidly than Therefore, He effuses more rapidly than OO22 at same T. at same T.

HeHe

Gas DiffusionGas Diffusionrelation of mass to rate of diffusionrelation of mass to rate of diffusion

Gas DiffusionGas Diffusionrelation of mass to rate of diffusionrelation of mass to rate of diffusion

• HCl and NH3 diffuse from opposite ends of tube.

• Gases meet to form NH4Cl

• HCl heavier than NH3

• Therefore, NH4Cl forms closer to HCl end of tube.

• HCl and NH3 diffuse from opposite ends of tube.

• Gases meet to form NH4Cl

• HCl heavier than NH3

• Therefore, NH4Cl forms closer to HCl end of tube.

Documents

KMT and Gas Laws Characteristics of Gases Gases expand to fill any container. –random motion, no attraction Gases are fluids (like liquids). –no attraction