two cartoons in honor of my husband’s birthday…

Two cartoons in honor of my husband’s birthday…

Friday, Jan. 25th: “A” DayMonday, Jan. 28th: “B” Day

Agenda“Charles’ Law” lab – finish/collectBegin Section 12.3: “Molecular Composition of

Gases”Ideal gas law, diffusion, effusion, Graham’s law of

diffusionHomework:

“Chemistry: Practice Problems for the Gas Laws”Concept Review – the 1st 2 sections should be done…

We will finish section 12.3 next time…

“Charles’ Law: The Effect of Temperature on Volume”

By now, you should have completed the data table and graphed your data.

Make sure your graph has a title and the axis’ are labeled, including units.

With your lab partner, take about 15 – 20 minutes to complete the conclusion questions before the lab will be due.

Remember to use complete sentences when answering all lab questions.

Review of Basic Gas Laws

Ideal GasIdeal gas: an imaginary gas whose particles

are infinitely small and do not interact with each other.

An ideal gas:1.Does not condense to a liquid at low

temperatures2.Does not have forces of attraction or repulsion

between the particles3.Is composed of particles that have no volume

Ideal Gas LawIdeal gas law: the law that states the

mathematical relationship of pressure (P), volume (V), temperature (T), the gas constant (R), and the number of moles of a gas (n).

PV = nRT“Pivnert”

Remember: temperature is in Kelvins!

Ideal Gas LawR is a proportionality constantThe value of R used in calculations depends on

the units used for pressure.

For Pressure in kPa: R = 8.314 L kPa∙ mol K∙

For Pressure in atm: R = 0.0821 L atm∙ mol K∙

Ideal Gas Law

Real gases deviate somewhat from an ideal gas and more so at very high pressures.

Sample Problem E Pg. 435How many moles of gas are contained in 22.41 liters at 101.325 kPa and 0°C?

Use the ideal gas law: PV = nRTP = 101.325 kPaV = 22.41 litersn = ?R = 8.314 L kPa∙ mol K∙T = 0˚C + 273 = 273 K

n = 1.00 mole

Additional PracticeWhat is the volume of 4.35 moles of a gas at a

pressure of 85.6 kPa and 26.0˚C?Use the ideal gas law: PV = nRT

P = 85.6 kPaV = ?n = 4.35 molesR = 8.314 L kPa∙ mol K∙T = 26.0˚C + 273 = 299 K

V = 126 Liters

DiffusionWhen you open a bottle of household

ammonia, the odor of ammonia gas doesn’t take long to fill the room. Why?

Diffusion: the movement of particles from regions of higher density to regions of lower density.

The process of diffusion involves an increase in entropy (S).

EffusionEffusion: The passage of a gas under pressure

through a tiny opening.

Scottish scientist Thomas Graham found that at constant temperature and pressure, the rate of effusion of a gas is inversely proportional to the square root of the gas’s molar mass, M.

Graham’s law of diffusion

VA = MB

VB MA

V = velocity (molecular speed) of gas A and BM = molar mass of gas A and B

Particles with lower molar mass travel faster than heavier particles.

Sample Problem F, pg 438Oxygen molecules have an average speed of about

480 m/s at room temperature. At the same temperature, what is the average speed of molecules of sulfur hexafluoride, SF6?

Gas A = SF6 VA = ? MA: 146 g/mol

Gas B = Oxygen VB = 480 m/s MB: 32 g/mol

VA__ = 32 - do division under and

480 146 - cross multiply to find VA

VA = 225 m/s

Additional PracticeThe average velocity of CO2 molecules at room temperature is

409 m/s. What is the molar mass of a gas whose molecules have an average velocity of 322 m/s under the same conditions?

Gas A = CO2 VA = 409 m/s MA = 44 g/mol

Gas B = X VB = 322 m/s MB = ?

409 m/s = MB 1.27 = MB

322 m/s 44 g/mol 44 g/mol- Square each side to solve for MB

MB = 71 g/mol

Homework

“Chemistry: PracticeProblems for the GasLaws” worksheet

DO NOT do #4 on the“Ideal Gas Law” sideOf the worksheet…

We will finish this section next time…