1520 mm Hg = ____ atms. Use your notes to find the equivalence line. Day 4 11-7.

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1520 mm Hg = ____ atms. Use your notes to find the equivalence line. Day 4 11- 7

description

1. Gases consist of large #s of tiny particles that are far apart relative to their size (each particle has mass) Kinetic-molecular Theory - Gases 2. Collisions between gas particles and between particles and container walls are elastic 3. Gas particles are in constant, rapid, random motion… they possess kinetic energy

Transcript of 1520 mm Hg = ____ atms. Use your notes to find the equivalence line. Day 4 11-7.

Page 1: 1520 mm Hg = ____ atms. Use your notes to find the equivalence line. Day 4 11-7.

1520 mm Hg = ____ atms.

Use your notes to find the equivalence line.

Day 4 11-7

Page 2: 1520 mm Hg = ____ atms. Use your notes to find the equivalence line. Day 4 11-7.

How does a cartesian diver work?

Day 4 11-7

Page 3: 1520 mm Hg = ____ atms. Use your notes to find the equivalence line. Day 4 11-7.

1. Gases consist of large #s of tiny particles that are far apart relative to their size (each particle has mass)

Kinetic-molecular Theory - Gases

2. Collisions between gas particles and between particles and container walls are elastic3. Gas particles are in constant, rapid, random motion… they possess kinetic energy

Page 4: 1520 mm Hg = ____ atms. Use your notes to find the equivalence line. Day 4 11-7.

Kinetic-molecular Theory - Gases

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

Page 5: 1520 mm Hg = ____ atms. Use your notes to find the equivalence line. Day 4 11-7.

cartesian divers

Day 4 11-7

Page 6: 1520 mm Hg = ____ atms. Use your notes to find the equivalence line. Day 4 11-7.

33,600 mL of Helium = ____ atoms at STP

Day 5 11-9

Page 7: 1520 mm Hg = ____ atms. Use your notes to find the equivalence line. Day 4 11-7.
Page 8: 1520 mm Hg = ____ atms. Use your notes to find the equivalence line. Day 4 11-7.

The Gas LawsBoyle’s Law: – Pressure-volume relationship – the volume of a fixed mass of gas varies inversely with the pressure at constant temperature

Comparing conditions:

P1V1 = P2V2Boyle’s Law:

Constant n and TP and V change

Page 9: 1520 mm Hg = ____ atms. Use your notes to find the equivalence line. Day 4 11-7.

The Gas LawsCharles’ Law: – Volume-temperature relationship – the volume of a fixed mass of gas varies directly with the Kelvin temperature at constant pressure

Comparing conditions:

V1/T1 = V2/T2

Temperature decreaseCharles’ Law: –

Constant n and PT and V change

Page 10: 1520 mm Hg = ____ atms. Use your notes to find the equivalence line. Day 4 11-7.

The Gas LawsGay-Lussac’s Law: – Pressure-temperature relationship – the pressure of a fixed mass of gas varies directly with the Kelvin temperature at constant volume

Comparing conditions:

P1/T1 = P2/T2

Low T - Slow moving

High T - Fast moving

Temperature Increase

Gay-Lussac’s Law: – Constant n and VT and P change

Page 11: 1520 mm Hg = ____ atms. Use your notes to find the equivalence line. Day 4 11-7.

http://mc2.cchem.berkeley.edu/Java/molecules/index.html

Increase pressure

Temperature __________

Volume __________

Increase temperature

Volume __________

Pressure __________

Increase volume Temperature __________

Pressure __________

increases decreases

increasesincreases

increases decreases

What happens if…?

Constant V Constant T

Constant VConstant P

Constant P Constant T

Page 12: 1520 mm Hg = ____ atms. Use your notes to find the equivalence line. Day 4 11-7.

The pressure exerted on a 3 L sample of gas is 600 mm Hg, if the pressure is decreased to 300 mm Hg what is the new volume?

1.Highlight important info.2.Choose a law3. Identify conditions as

beginning (1) or ending (2)4.Solve

Page 13: 1520 mm Hg = ____ atms. Use your notes to find the equivalence line. Day 4 11-7.

P1V1 = P2V2

DAY 6 11-101. What is Boyle’s law? _____________ 2. Write a Boyle’s law problem in your

notebook with the answer

3. Why are temperatures around zero kelvins bad for an ideal gas?

Page 14: 1520 mm Hg = ____ atms. Use your notes to find the equivalence line. Day 4 11-7.

Review section 14.2 and complete #s 17-19, 22, 24, + 25 on page 463

Page 15: 1520 mm Hg = ____ atms. Use your notes to find the equivalence line. Day 4 11-7.

Whose LawP1/T1 = P2/T2

P1V1 = P2V2

V1/T1 = V2/T2

Gay-Lussac’s

Boyle’s

Charles’

Page 16: 1520 mm Hg = ____ atms. Use your notes to find the equivalence line. Day 4 11-7.

At 100 K a sample exerts a pressure of 2.026e5 Pa, if the pressure is decreased to 1.013e5 Pa what is the new temperature?

What law do you use? __________________

Solve:

Gay-Lussac’s: P1/T1 = P2/T2

Page 17: 1520 mm Hg = ____ atms. Use your notes to find the equivalence line. Day 4 11-7.

1. At 100 K and 1 atm a sample has a volume of 2L, if the gas expands to 6L what is the new temperature?

Not in notes

Page 18: 1520 mm Hg = ____ atms. Use your notes to find the equivalence line. Day 4 11-7.

Observe what happens.

How could we get the egg back out without breaking it?

EXPLAIN why the egg enters the flask.

Page 19: 1520 mm Hg = ____ atms. Use your notes to find the equivalence line. Day 4 11-7.

Ideal gas law – mathematical relationship of pressure, volume, temperature, and the # of moles of a gas

PV = nRT

R = the ideal gas constant = 8.314

Page 454 table 13.2

L * kPamol * K

http://en.wikipedia.org/wiki/Gas_constant

Page 20: 1520 mm Hg = ____ atms. Use your notes to find the equivalence line. Day 4 11-7.

What pressure in atms. is exerted by 0.325 mol of hydrogen gas in a 4.08 L container at 35°C?

1. Convert to Kelvins

2. Choose R

3. Solve

0.0821 atm-L/mol-K

8.314 J/mol-K

PV = nRT

Page 21: 1520 mm Hg = ____ atms. Use your notes to find the equivalence line. Day 4 11-7.

DAY 1 11-11A sample of gas that contains 4.0 moles at 400K and 2 atms. takes up how much space?

Page 22: 1520 mm Hg = ____ atms. Use your notes to find the equivalence line. Day 4 11-7.

1. If a balloon is in this room (about 1 atm. and 300 K) and its volume is 0.40 L, how many moles of gas does it contain?

On notes page 6

Page 23: 1520 mm Hg = ____ atms. Use your notes to find the equivalence line. Day 4 11-7.

What mass of chlorine gas in grams is contained in a 10.0 L tank at 27°C and 3.50 atm of pressure?

101 gOn notes page 5

Page 24: 1520 mm Hg = ____ atms. Use your notes to find the equivalence line. Day 4 11-7.

What is the volume occupied by 4.67 moles of H2 at 27.0 °C and 2.00 atm.?

On notes page 6

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Boyle’s Law

Ideal Gas

Kinetic Energy

Mole

Standard Temperature and Pressure

Attract

Volume-temperature relationship

Absolute zero

Charles’ Law

Avogadro’s Law

Pascal

Precipitate

Synthesis reaction

Page 26: 1520 mm Hg = ____ atms. Use your notes to find the equivalence line. Day 4 11-7.

Homework # 1 Unit 4

Page 27: 1520 mm Hg = ____ atms. Use your notes to find the equivalence line. Day 4 11-7.

At 300 K a sample of gas occupies 5 L, if the volume is decreased to 2 L what is the new temperature?

V1/T1 = V2/T2

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1. Explain what happens at very low temperatures in terms of kinetic molecular theory.

2. List one of the 5 assumptions of kinetic molecular theory.

5-6

Page 29: 1520 mm Hg = ____ atms. Use your notes to find the equivalence line. Day 4 11-7.

Read Section 12.1 (Kinetic Molecular Theory) and

complete #s 8-11 + 13 on page 410

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HomeworkPractice problems 1 + 2 on page 443

Practice Problems 4-6 on page 446

Practice Problems 8 + 9 on page 448

Periods 3 & 6 5-5

Page 31: 1520 mm Hg = ____ atms. Use your notes to find the equivalence line. Day 4 11-7.

1. Highlight important info.

2. Choose a law

3. Identify conditions as beginning (1) or ending (2)

4. Solve

Boyle’s Law: Constant n and TP and V change