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Unit 6: Gases & The Kinetic Molecular TheoryPowerPoint Presentation

Unit 6: Gases & The Kinetic Molecular Theory

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### Unit 6:Gases & The Kinetic Molecular Theory

CHM 1045: General Chemistry and Qualitative Analysis

Dr. Jorge L. Alonso

Miami-Dade College – Kendall Campus

Miami, FL

- Textbook Reference:
- Module # 8

Characteristics of Gases

- Unlike liquids and solids, gases . . . .
- Are highly compressible.
- Expand to fill their containers.
- Have extremely low densities.

Condensed phases

Characteristics of Gases

- Variables affecting the behavior of gases
- Amount = number of moles ()
- Pressure (P)
- Volume (V)
- Temperature (T in Kelvin)

{PropGases*}

A

P =

PressureForce = mass x acceleration

Newton = 1kg . m/sec2

105 Newtons

= (104kg)(10 m/sec2)

- Pressure is the amount of force applied to an area.

Approx. 12 miles

105 Newtons

meter2

= 101.325 kPa

=

- Atmospheric pressure is the weight of air per unit of area.

Units of Pressure

Torricelli’s

- Atmosphere
- 1.00 atm
- = 760 mm Hg (torr)
- = 101.325 kPa

760 mm Hg = weight of equal surface area of the atmosphere

(Normal atmospheric pressure at sea level).

Barometer

33 ft H2O = weight of equal surface area of the atmosphere

Manometer

instrument used to measure the difference in pressure between atmospheric pressure and that of a gas in a vessel.

{Manometer}

Manometer

Used to measure the difference in pressure between atmospheric pressure and that of a gas in a vessel.

Pgas = 760 - 6 torrs

Pgas = 760 torr

Pgas = 760 + 6 torrs

Gas Laws

Variables affecting gases: moles (η), pressure (P), volume (V) and Temperature (T)

- Boyle’s Law
- Compared: P versus V ( & T are held constant).

- Charles’s Law
- Compared: V versus T ( & P are held constant).

- Avogadro’s Law
- Compared: Vversusη(P & T are held constant).

- Combined Gas Law
- Compared: P vs Vvs. T ( is held constant).

- Ideal Gas Law
- Compared: P vs Vvs. ηvsT (no variable held constant).

- Dalton’s Law of Partial Pressure
- Compared: individual pressures of gases in a mixture

P

V

Boyle’s Law: Pressure-Volume Relationship( & T are held constant).

2 x

The volume of a fixed quantity of gas at constant temperature is inversely proportional to the pressure.

V ?

V x ½

{Boyle’s Law}

P

k

P

V

V=

P ↑V ↓ = k

- Also,

OR

- This means a plot of V versus 1/P will be a straight line.

{PV.Graphs}

V T

Charles’s Law: Temp. – Volume Relationship( & P are held constant).

- The volume of a fixed amount of gas at constant pressure is directly proportional to its absolute temperature.

V ?

V 2 x

T 2 x

{*Charles’s Law Liq N2}

T

= k

V = kT

VT

or

Charles’s Law- The volume of a gas is directly proportional to its absolute temperature.

A plot of V versus T will be a straight line.

Avogadro’s Law: Moles-Volume Relationship

(P & T are held constant).

Vn

V ?

V 2 x

2 x

- The volume is directly proportional to the number of moles of the gas.

{Avogadro’s Law}

Standard Temperature & Pressure (STP) and Molar Volume

- Standard Temperature: 00C or 273K
- Standard Pressure: 760 torr (1 atm)

At STP the Molar Volume of any gas is 22.4 L

(11.1 in)3 or (28.2cm)3

1 mole = 6.022 x 1023 part. = gMM = 22.4 L

Standard Temperature & Pressure (STP) and Molar Volume

At STP the Molar Volume of any gas is 22.4 L

1 mole = 6.023 x 1023 part. = gMM = 22.4 L

H2 = 2.0g

O2 = 32.0g

CO2 = 44.0 g

Problem: At STP, what volume in mL would 75g of CO2 occupy?

knT

P

nT

P V

nT

P

V

V=

k =

Ideal-Gas EquationThe Gas Laws:

V 1/P (Boyle’s law)

VT (Charles’s law)

Vn (Avogadro’s law)

Combining these, we get

or

or

Ideal-Gas Equation:

Useful for pure gas under one set of conditions.

PV= nRT

(torr) (L) = (mol) R (K)

Units:

{PV= nRT RapVideo}

RapVideoLinkYouTube

Ideal Gas Law Problems

What volume (in mL) would a 2.20 g sample of hydrogen gas (H2) at 50.00C occupying at 443 torr?

PV= nRT

V= nRT

P

R

V =

= 50.0 L

Ideal-Gas Equation: Densities of Gases

PV = nRT

For Ideal Gas Equation:

Since

Then

and

Dividing both sides of the equation on the left by V we get

( )

Where d = Density of Gas

If we solve the equation for density, we get……..

Ideal-Gas Equation: Densities and Molecular Weigh of Gases

Problem: What is the density of the oxygen in a tank in an AC room (25°C) and whose pressure gauge reads 25.0 atm

Problem: A gas whose density is 0.0131 g/mL and is in a container at room temperature and whose pressure gauge reads 1.9 x 104 mmHg. What is its MW?

Ideal-Gas Equation: Densities & Molecular Weigh of Gases Problems

What is the density (in g/mL) of SO2 at STP?

PV = nRT

( )

= 2.62 g/L

=

Ideal-Gas Equation: Densities & Molecular Weigh of Gases Problems

What is the molecular weight of a gas whose density @ STP is 7.78 g/L?

PV = nRT

( )

=

2006 A Problems

Combining, we can get Problems

P2V2

T2

P1V1

T1

Combined Gas Law EquationThe Gas Laws:

V 1/P (Boyle’s law)

VT (Charles’s law)

The Combined Gas Law

k

=

Useful for a constant amount of a pure gas under two different conditions.

P Problems1V1

T1

P2V2

T2

=

Combined Gas Law ProblemA scuba diver takes a gas filled 1.0 L balloon from the surface where the temperature is 34 0C down to a depth of 66 ft (33 ft H2O = 1 atm). What volume will the gas balloon have at that depth if the temperature is 15 0C?

V2

V2

Dalton’s Law of ProblemsPartial Pressures

- The total pressure of a mixture of gases equals the sum of the pressures that each would exert if it were present alone.

- In other words,
- Ptotal = P1 + P2 + P3 + …
- Pair = P N2 + PO2 + PH2O + …

Partial Pressures Problems

P of gas

P of atm

- When one collects a gas over water, there is water vapor mixed in with the gas.

- To find only the pressure of the desired gas, one must subtract the vapor pressure of water from the total pressure.

Vapor Pressure of Water Problems

- Daltons Law:

Ptotal = Pgas + PH2O

- To find only the pressure of the desired gas, one must subtract the vapor pressure of water from the total pressure.

Pgas = Ptotal - PH2O

{Press on can}

Evaporation vs Boiling in terms of Vapor Pressure Problems

Patm

Patm

+

+

Pvap

Pvap

Vapor Pressure (v.p. or Pvap)

Patm

- Caused by the tendency of solids & liquids to evaporate to gaseous form. It is temperature (K.E.) dependent.

=

Pvap

Stoichiometry with Gases Problems

Mg (s) + 2HCl (aq) MgCl2 (aq) + H2 (g)

Problem: If 2.0 g of Mg are reacted with excess HCl, what volume of H2 will be produced at 250C and 775 torr? At STP?

PV = nRT

Kinetic-Molecular Theory Problems

A model that aids in our understanding of what happens to gas particles as environmental variables change.

Main Tenets:

- Gases consist of large numbers of molecules that are in continuous, random motion.

2.Collisionsbetween gas molecules and between gas molecules and the walls of the container mustbe completely elastic(energy may be transferred between molecules, but none is lost).

Kinetic-Molecular Theory Problems

Main Tenets:

3.Attractive and repulsive forces between gas molecules are negligible.

4. The combined volumeof all the molecules of the gasis negligible (excluded volume) relative to the total volume in which the gas is contained.

Kinetic-Molecular Theory Problems

Main Tenets:

5. Energy can be transferred between molecules during collisions, but the average kinetic energy of the molecules does not change with time, as long as the temperature of the gas remains constant.

@ 100 0C

6. The average kinetic energy (KE=½mv2) of the molecules is proportional to the absolute temperature.

{KE T(K)}

Diffusion Problems

EffusionMovement of molecules from an area of high concentration to an area of low concentration until equilibrium is reached (homogeneity).

The escape (diffusion) of gas molecules through a tiny hole into an evacuated space.

Effect of Molecular Mass on Rate of Effusion and Diffusion Problems

Thomas Graham (1846): rate of diffusion is inversely proportional to the square root of its molar mass

Kinetic Energy per individual molecule:

Dropper with Br Problems(l)

Rate of Diffusion & EffusionThomas Graham (1846): rate of diffusion is inversely proportional to the square root of its molar mass

{BrDiffusion}

Comparing the rates of two gases:

Graham’s Law of Diffusion and Effusion of Gases

{GasDiff}

Effusion and Diffusion Problems

- This is the most widespread uranium enrichmentmethod. Uranium is reacted with fluorine to make uranium hexafluoride gas: 235UF6 & 238UF6

- The physical principle is that the diffusion speed of a gas molecule depends on the mass of the molecule: the lighter ones diffuse faster and get through a porous material easier.

- In gas diffusion units, uranium-hexafluoride gas diffuses through an etched foil made of either an aluminum alloy or teflon, due to artificially maintained difference in pressure. The lighter molecules(i.e. those containing 235U) get through easier to the other side, therefore the gas accumulating there will be richer in 235U.

Gas Centrifugation Problems

The gas centrifuge is essentially a bowl, in which there is a rotor spinning round at a very high speed. The gas (UF6) directed to the centrifuge is forced to spin by the rotor. Due to the centrifugal force the heavier molecules (those which contain 238U) will accumulate near the wall of the bowl, while the lighter molecules containing 235U will stay closer to the center of the centrifuge.

Boltzmann Distributions Problems

The Maxwell–Boltzmann distribution is the statistical distribution of molecular speeds in a gas. It corresponds to the most probable speed distribution in a collisionally-dominated system consisting of a large number of non-interacting particles.

{Boltzman Plot}

Kinetic Energy of Gas Molecules Problems

Kinetic Energy per individual molecule:

☺

☺

Kinetic Energy per mole:

Combining above equations and solving for velocity we get:

- The root-mean square velocity of gases is a very close approximation to the average gas velocity.

☺

- To calculate this correctly:
- The value of R = 8.314 kg m2/s2 K mol
- Mm= molar mass, and it must be in kg/mol.

The Kinetic-Molecular Theory Problems

- Example: What is the root mean square velocity of N2 molecules at room T, 25.0oC?

- To calculate this correctly:
- The value of R = 8.314 kg m2/s2 K mol
- And M must be in kg/mol.

The Kinetic-Molecular Theory Problems

Problem: What is the root mean square velocity of He atoms at room T, 25.0oC?

You do it!

- To calculate this correctly:
- The value of R = 8.314 kg m2/s2 K mol
- And M must be in kg/mol.

GasMW

He 4

N2 28

SF6 146

- Can you think of a physical situation that proves He molecules have a velocity that is so much greater than N2 molecules?
- What happens to your voice when you breathe He(g) or SF6 (g)?

Ideal Problemsvs.Real Gases

In the real world, the behavior of gases only conforms to the ideal-gas equation at relatively high temperature and low pressure.

Deviations from Ideal Behavior Problems

Two particular assumptions made in the kinetic-molecular model break down at high pressure and/or low temperature:

(1) attractive forces and (2) excluded volume.

2003 A Problems

Corrections for Non-ideal Gas Behavior Problems

- The ideal-gas equation can be adjusted to take these deviations from ideal behavior into account.

- The corrected ideal-gas equation is known as the van der Waals equation.

Real Gases: ProblemsDeviations from Ideality

- van der Waals’ equation accounts for the behavior of real gases at low temperatures and high pressures.

a

b

- The van der Waals constants a and b take into account two things:
- a accounts for intermolecular attraction
- b accounts for volume of gas molecules

- At large volumes a and b are relatively small and van der Waal’s equation reduces to ideal gas law at high temperatures and low pressures.

Real Gases: ProblemsDeviations from Ideality

- Example: Calculate the pressure exerted by 84.0 g of ammonia, NH3, in a 5.00 L container at 200. oC using the ideal gas law.
You do it!

Real Gases: ProblemsDeviations from Ideality

- Example: Calculate the pressure exerted by 84.0 g of ammonia, NH3, in a 5.00 L container at 200. oC using the van der Waal’s equationYou do it!

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