Gases

Gases

• Gases are made of independently moving particles.

• These particles move in random directions at varying speeds until they collide with another particle or a barrier.

Gases

• A gas has no definite volume or shape.

• The Gas Laws pertain to an Ideal Gas, an imaginary gas that serves as a model for gas behaviors.

Is it real or ideal?Ideal Conditions/Ideal Gas -

• imaginary gas that fits the all assumptions of the kinetic-molecular theory

Real Gas –

• particles have size and intermolecular attraction to other particles, so do not conform to gas laws under very high pressure or very low temperatures.

Kinetic Theory

• based on idea that particles are always in motion

• At SAME temperature, all gases have the SAME kinetic energy (KE)

• But gases do not all have same speed

• KE = ½ m v2

• m = mass

• v = velocity

• Rearrange to fastest to slowest • O2 CO2 H2 N2

• _____________________

• Why??

– Well who’s faster, a 350 lb football player or a 120 lb runner?

• ALL gases at the SAME temperature have the same average KE

• If the KE = 20, note the speed for the gases .

• H2

• KE = ½ m v2

• 20 = ½ 2g v2

• 20 = 1 v2

• 20/1 = v2 • 20 = v2

• v2 = /20

• v = 4.472

Ar

KE = ½ m v2

20 = ½ 40g v2

20 = 20 v2

20/20 = v2

1 = v2

v2 = /1

v = 1

Kinetic molecular theory assumptions:

• 1.   Gases consist of large # of tiny molecules that are far apart relative to their size. Thus they have low density and lots of empty space between them.

Kinetic molecular theory assumptions:

• 2.    Lots of elastic collisions

– - don’t stick together after collisions.

– No net loss of kinetic energy

• 3.    NO attraction or repulsion for each other (ideal gases do not condense to a liquid or a solid as no attraction)

Kinetic molecular theory assumptions:(cont)

• 4.    Kinetic energy = energy of motion

Gas particles are in constant, rapid motion that is random. Therefore, passes “kinetic energy” = energy of motion.

• 5.    Average kinetic energy depends on the temperature of the gas.

The Nature of Gases for Ideal gases

1. Expand to fit container - no definite shape or volume

2. Fluidity - slide past each other – fluid as attraction forces not significant

3. Low density - particles far apart4. Compressibility – particles can be

pressed close together.

Nature cont.5. Diffusion - random mixing caused by

random motion

• Rate of Diffusion depends on:

a.   Temperature (hot is faster)

b.   Size (small is faster)

c. Attractive forces between particles

(no attraction is faster)

– O2 vs H2O

Nature cont.6.Effusion - gas particles under pressure

pass through tiny openings

• Smaller molecule → faster effusion

(Does a balloon stay full forever?

Why not?)

To describe gases,

4 measurable qualitites are used

•           Volume (V)

•          Temperature (T)

•        Number of molecules (moles) (n)

•         Pressure (P) PV = nRT(R = constant)

Pressure

• Pressure = force/unit area • (Force is in Newtons (N))• Force is caused by gas molecules hitting wall -

their collisions• Air around us exerts a pressure on us. • What would happen to your eyeballs if you were to

go out into outer space??• Where is the most pressure - sea level, ocean

bottom, mountain top?

Barometer

• Note air pushing down

• Vacuum has nothing inside, so is not pushing down

A simple manometer.

Standard Temperature & Pressure (STP) -- must be standard to compare things

STP = 1 atmosphere (atm) and 0 oC at sea level, pressure = 1 atm = 760 mm Hg = 760 torr

Conversions

• Standard Pressure with different units

• 1 atmosphere (atm)

=760 mm Hg (millimeter of mercury)

=760 torr =14.7 lb/in2=76 cm Hg

=29.9 in Hg = 1013 millibar (mbar)

= 101,325 Pascal = 101.325 KPa (SI unit)

• Pascal - the SI unit = 1N/ m2

• These all equal each other,

Therefore can use as conversion factors

• Given 700 mm Hg, how many atm?

700 mm Hg ___1 atm_______ =

760 mm Hg

0.92 atm

• Covert the following

• 860 mm Hg = ______ atm

• 720 torr = ______ in Hg

• 0.89 atm = ______ mm Hg

• 1.2 atm = ______ in Hg

1.13

28.3

676

35.9

The Gas Laws

• Boyles Law -the volume of a fixed mass of gas varies INVERSELY with the pressure at constant temperature.

• So as P increases V decreases !• http://

preparatorychemistry.com/Bishop_Boyles_Law_audio.htm

• As pressure increases, the volume decreases

• Temperature and number of moles are constant

The effects of decreasing the volume of a sample of gas at constant temperature. Molecules hit more often, so Pressure Increases

Boyle’s Law

• P1V1 = P2V2

initial = final

• Inversely proportional

Boyle’s law

Ex. A sample of O2 has a volume of 100 ml and a pressure of 200 torr. What will it’s volume be if the pressure is increased to 300 torr (T is kept constant)

First think – what is the relationship?

P increases, then V decreases

Boyle’s law

• P1 x V1 = P2 x V2

200 torr x 100 ml = 300 torr x V2

V2 = 66.67 ml

• Plotting Boyle's data from Table 5.1. (a) A plot of P versus V shows that the volume doubles as the pressure is halved. (b)

• A plot of V versus 1/P gives a straight line. The slope of this line equals the value of the constant k.

A plot of PV versus P for several gases at pressures below 1 atm.

Charles Law

deals with Volume and temperature

(Pressure constant).

V and T (in Kelvin) are directly proportional

As you INCREASE the temperature - molecules MOVE faster

KE (Kinetic energy) increases as molecules jump around more

http://preparatorychemistry.com/Bishop_Charles_Law_audio.htm

Charles Law

Temperature MUST be in Kelvin (K)Found gases had “zero” volume at -273 oCSo named this Absolute Zero, which equals 0 Kelvin = 0 K

 K = 273 + ____ oCoC = ___ K - 273

 

Charles Law

V1 = T1

V2 T2

Can rewrite: V1T2 = V2T1

(Watch 1's & 2's)

Directly proportional

The effects of increasing the temperature of a sample of gas at

constant pressure.Molecules are moving FASTER

so they hit the container harder, thus size must increase to keep P same

Charles

Ex. A sample of neon gas occupies a volume of 752 ml at 25 oC. What volume will it occupy at 50 oC.

Pressure is constant.

Think first - as Temperature increases, Volume will ____________

Charles

• MUST change oC to Kelvin

• Given : V1 = 752 ml

• T1 = 25  oC + 273 = 298 K

• V2 = ?

• T2 = 50 oC + 273 = 323 K

Charles

V1 x T2 = V2 x T1

752 ml x 323 K= V2 x 298

K

V2 = 815 ml

Gay- Lussac’s Law the pressure of a fixed mass of gas at constant

volume varies directly with the Kelvin temperature

• P1 = T1

P2 T2 direct proportion!

Or P1 x T2 = P2 x T1

http://preparatorychemistry.com/Bishop_Gay_Lussac_Law_audio.htm

The effects of increasing the temperature of a sample of gas at constant volume. Molecules move faster, so pressure increases

Gay-Lussac’s Law in real life

• Ex. A gas in an aerosol can is at a pressure of 3.00 atm at 25 ° C. Directions warn the user not to keep the can in a place where temperature exceeds 52 ° C. What would the pressure in the can be at 52 oC?

Given

P1 = 3.00 atm

P2 = ?

T1 = 25 oC or 298 K

T2 = 52 oC or 325 K

• Ex. A gas in an aerosol can is at a pressure of 3.00 atm at 25 ° C. Directions warn the user not to keep the can in a place where temperature exceeds 52 ° C. What would the pressure in the can be at 52 oC?

Given

P1 = 3.00 atm 3.00 atm = 298 K

P2 = ? P2 325 K

T1 = 25 oC or 298 K

T2 = 52 oC or 325 K P1 x T2 = P2 x T1

P2 = 3.27 atm

Combined Gas LawExpresses the relationship between pressure, volume, and temperature of a fixed amount of gas.

• P1V1 = P2V2

T1 T2

or

P1V1 T2 = P2V2 T1

Ex. A helium balloon has a volume of 50.0 L at 25 ° C and 1.08 atm. What volume

will it have at 10 oC and 0.855 atm?

Given:

V1 = 50.0L

V2 = ?

P1 = 1.08 atm

P2 = 0.855 atm V2 =

T1 = 25 ° C = 298 K

T2 = 10 ° C = 283 K

Ex. A helium balloon has a volume of 50.0 L at 25 ° C and 1.08 atm. What volume

will it have at 10 oC and 0.855 atm?

Given:

V1 = 50.0L (1.08 atm)(50.0L) = (0.855 atm)(V2)

V2 = ? 298 K 283 K

P1 = 1.08 atm

P2 = 0.855 atm V2 = 60.0 L He

T1 = 25 ° C = 298 K

T2 = 10 ° C = 283 K

Ideal Gas Equation

• Avogadro’s Law - equal volumes of gases at the same temperature and pressure contain equal numbers of molecules

• http://preparatorychemistry.com/Bishop_Avogadros_Law_audio.htm

• Thus, the volume occupied by one mole of gas at STP = standard molar volume of a gas = 22.4 L

Ideal Gas Equation

• Ex. A chemical reaction produces 0.0680 mol of O2. What volume will it occupy at STP?

Given: 0.0680 mol O2

1 mol = 22.4 L at STP

Ideal Gas Equation

• Ex. A chemical reaction produces 0.0680 mol of O2. What volume will it occupy at STP?

Given: 0.0680 mol O2

1 mol = 22.4 L at STP

0.0680 mol O2 22.4 L = 1.52 L O2

1 mol

• Ex. A chemical reaction produces 98.0 mL of SO2 at STP. What mass in grams was produced?

Given: Volume SO2 at STP 98.0 mL = 0.098 L

0.098 L SO2 1 mol = .004375 mol SO2

22.4L

0.004375 mol SO2 64.07 g = 0.28 g SO2

1 mol

OR 0.098 L SO2 1 mol 64.07 g = 0.28g SO2

22.4L 1 mol

Ideal Gas Law

• the mathematical relationship of pressure, volume, temperature, and the number of moles of gas.

PV = nRTpressure x volume = # of moles x constant x temperature

the constant, R, is the ideal gas constant has a value of .0821 L atm

mol KWATCH units !!

The effects of increasing the number of moles of gas particles at constant

temperature and pressure.

Plots of PV/nRT versus P for several gases (200 K).

Plots of PV/nRT versus P for nitrogen gas at three temperatures.

• Ex. What is the pressure in atm exerted by 0.50 mol sample of N2 in a 10.0L container at 298 K?

Given: V of N2 = 10.0L n of N2 = 0.50 mol T = 298 K

P = ? PV = nRT therefore, P = nRT V

• Ex. What is the pressure in atm exerted by 0.50 mol sample of N2 in a 10.0L container at 298 K?

Given: V of N2 = 10.0L n of N2 = 0.50 mol T = 298 K

P = ? PV = nRT therefore, P = nRT V

P = (0.50 mol)(0.0821 L atm /mol K)(298 K) 10.0 L = 1.22 atm

Combining the ideal gas law with density - to find molar mass or gas density.

Ex. What is the density of ammonia gas at

63 oC and 705 mm Hg?

Given: P = 705 mmHg / 760 = 0.928 atm

V = ?

n = assume 1 mole

R = 0.0821 L atm /mol K

T = 63 oC + 273 K = 336 K

Combining the ideal gas law with density - to find molar mass or gas density.

V = 1mol(0.0821 L atm / mol K) ( 336 K) .928atm

= 29.73 LD = mass mass of 1

volume mol NH3 = 17g (from Periodic table)D = 17g 29.73L = 0.572 g / L

Gas Stoichiometry

Stoichiometry – the mass relationship between reactants and products in a chemical reaction. In general, follow these rules:

a. Use stoichiometry to do mole or mass conversions.

b. Use the Ideal Gas Law (PV=nRT) to convert between moles and volume.

CaCO3 → CaO + CO2

How many grams of calcium carbonate must be decomposed to produce 5.00 L of carbon dioxide at STP?

Use PV = nRT first, solve for n

0.223 mol CO2 → 22.3 g CaCO3

Gas Stoichiometry Practice

Gas Stoichiometry Practice• Calculate the volume of H2 gas that can

be obtained under laboratory conditions of temperature and pressure of 25 C and 0.900 atm when 5.00 g of sodium is reacted with water:

2 Na + 2H2O H2 + 2 NaOH

→ 0.109 mol H2 → 2.95 L

Gas Mixtures and Partial Pressures

Many gases are a mixture of gases; air being a prime example. We define the partial pressure of a gas which is the pressure a component of a gas mixture would have if it were all by itself in the same container.

Dalton’s Law

John Dalton (atomic theory) extended the gas laws simply as:

Ptotal = P1 + P2 + P3 + …

This simply says that the sum of the partial pressures of the gases will add up to the total pressure. This way we can treat a mixture of gases just like a pure gas.

Partial Pressure Practice• What would be the final pressure of the

mixture of gases for the processes depicted in each of the following illustration?

400 Torr N2 200 Torr O2

600 Torr N2 + O2

Effusion and Diffusion

Graham’s law of effusion - the rates of effusion of gases at the same temperature and pressure are inversely proportional to the square roots of their molar masses.

Rate of effusion A = Mb

Rate of effusion B Ma

The effusion of a gas into an evacuated chamber.

• Ex. Compare the rates of effusion of H2 and O2 at same temperature and pressure.

Given:

H2 mol weight 2

O2 mol weight 32

• Rate of effusion H2 = √ Mass O2

Rate of effusion O2 √ Mass H2

• Ex. Compare the rates of effusion of H2 and O2 at same temperature and pressure.

Given:

H2 mol weight 2

O2 mol weight 32

• Rate of effusion H2 = √ Mass O2

Rate of effusion O2 √ Mass H2

• √ 32 = 4

√ 2

• Therefore H2 effuses 4 times faster than O2

Important Gas relationships• As volume increases, pressure decreases at

constant temperature• As temperature increases, pressure increases at

constant volume• As temperature increases, volume increases at

constant pressure • Standard Temperature and Pressure -to

compare gases use (standard temperature & Pressure STP)

• Standard temperature = 0 ° C• Standard Pressure = 760 mm Hg = 1 atm

average barometric pressure at sea level