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The Chemistry of Gases. Unit 4 Chapter 12. 12.1 Characteristics of Gases. Read Section 12.1 p. 416-422. Title: Crushed Bottle
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The Chemistry of Gases Unit 4 Chapter 12
12.1 Characteristics of Gases Read Section 12.1 p. 416-422
Title: Crushed Bottle This bottle was opened at the top of Mount San Jacinto. After it was resealed, it was then brought back down to the bottom of the mountain. Discuss with a neighbor and then write a paragraph explaining why it is crushed. Pre-assessment
See p. 422 Write a paragraph explaining what this diagram is telling us.
Gases Discuss and write down the names of three chemicals which are gases. Demo: Marble + acid Zinc + acid
Properties of gases: • Gases are fluids • Gases have low density • Gases fill whatever container they are in • Gases can be compressed • Why? In a gas, the molecules are moving around quickly and randomly
Pressure: caused by molecules striking the wall of a container Pressure = Force / Area The SI unit of pressure is the Pascal (Pa) See Fig 6 p. 419 Barometer Video clip: “Barometer – device to calculate air pressure” Notes
Writing assignment Title: Barometer See Fig 6 p. 419 Write a paragraph explaining how a barometer works to measure air pressure. Draw a diagram.
Some different units of pressure p. 420 See Chem Ref Sheet: Define units
Pressure unit conversion Convert a pressure of 225 kPa to atmospheres. ANSWER = 2.22 atm Wanted: bring an empty soda can to class
Practice question Convert a pressure of 2 atmospheres to Pascals. (hint: first convert atm to KPa, then convert KPa to Pa) ANS = 202 650 Pa (Extras: Convert the local pressure to Pascals)
Warm-up: pressure unit conversion 1. Convert 426 mm mercury pressure to Pascals (Hint: first convert mm Hg to KPa (using conversion factor on Chem Ref Sheet), then convert KPato Pa) ANS = 56795 Pa 2.Convert 12 444 Pascals pressure to mm of mercury. ANS = 93.338 mm Hg
Pressure unit conversion Classwork p. 421 #1,2,3,4* Be prepared for a quiz on pressure unit conversions. Suggested practice extras: p. 422 # 8,9,10
Worksheet Kinetic Molecular Theory
CA Chemistry Standard 4b Students know the random motion of molecules explains the diffusion of gases.
Diffusion Read p. 436 Diffusion KWS: Theprocesswhereparticlesmixastheresultoftheirrandommovement The process where particles mix as the result of their random movement Homogenous mixture: same throughout Draw it!! Sim: PhET gas properties
Standard Temperature and Pressure (STP) standardtemperatureandpressure(stp)isastandardsetofconditionsforexperimentalmeasurementstoallowcomparisonstobemadebetweendifferentsetsofdata. STPis1atmospherepressureand0°C
STP Standard Temperature and Pressure (STP) is a standard set of conditions for experimental measurements, to allow comparisons to be made between different sets of data. STP is 1 atmosphere pressure and 0°C. p. 420
CA Content Standard 4c. Students know how to apply the gas laws to relations between the • Pressure (P) • Temperature (T) • Volume(V) of any amount of an ideal gas or any mixture of ideal gases. Combined Gas Law preview
Warm-up In the following sentences, all the words have been joined together and the punctuation has been removed. Rewrite the definition into correct English in your notes: ‘in1662theenglishscientistrobertboylestudiedtherelationshipbetweenthevolumeandthepressureofagashefoundthatasthepressureonagasincreasesinaclosedcontainerthevolumeofthegasdecreases’
Warm-up ‘In 1662, the English scientist Robert Boyle studied the relationship between the volume and the pressure of a gas. He found that as the pressure on a gas increases in a closed container, the volume of the gas decreases.’
Think about it….. • What happens to the pressure of a gas when the volume is decreased? (reminder: volume is the amount of space something takes up) The pressure increases.
Boyle’s Law Pressure and volume are inversely(directly/inversely) proportional
Boyle’s Law Why? Think Kinetic Molecular Theory Sim: gas properties
P1= initial pressure V1= initial volume P2= final pressure V2= final volume Boyle’s Law
Boyle’s Law Practice question A sample of gas occupies 523 mL at 1.00atm. The volume of the gas is decreased to 265 mL, while the temperature remains the same. What is the new pressure of the gas? GIVEN: P1 = 1.00 atm V1 = 523 mL V2 = 265 mL T1 = T2 UNKNOWN: P2 = ? atm Prediction: P2 < 1.00 atm OR P2> 1.00 atm
Boyle’s Law Equation Start with Combined Gas Law When temperature is constant T1 = T2 P1V1 = P2V2
Boyle’s Law Practice question A sample of gas occupies 523 mL at 1.00atm. The volume of the gas is decreased to 265 mL, while the temperature remains the same. What is the new pressure of the gas? P1 = 1.00 atm V1 = 523 mL V2 = 265 mL T1 = T2 UNKNOWN: P2 = ? atm Now, rearrange, substitute and solve….. ANSWER = 1.97 atm
Boyle’s Law Practice questions Practice: p. 425 # 1-4 Then, complete Kinetic Molecular Theory worksheet Then, can you figure out Charles’s Law on your own??? P. 428 #1-4
Learning Objective Students know how to apply the gas laws to relations between the • Pressure (P) • Temperature (T) • Volume(V) of any amount of an ideal gas or any mixture of ideal gases.
Charles’s Law: Think about it….. • What happens to the volume of a balloon if the temperature is increased? Volume increases
Charles’s Law T↑ V↑ T↓ V↓ (IF pressure is held constant) V and T are directly proportional.
Can be derived from Combined Gas Law, if P1 = P2 T must be converted to Kelvin. Charles’s Law
Rearrange eqn to solve for: V2 = ? T2 = ? Charles’s Law(Can be derived from Combined Gas Law)
Chem Joke Why shouldn’t you believe in atoms? Because they make up everything
Charles’s Law Practice Problem • A balloon is inflated to 665 mL volume at 27°C. It is immersed in a dry-ice bath at −79°C. What is its volume, assuming the pressure remains constant? First, need to convert the temperatures to Kelvin T1 = 27oC = ?K V1 = 665 mL T2 = -79oC = ?K V2 = ? mL ANS = 430 mL
Charles’s Law Practice Classwork p.428 #1-4
Think about it…… • What happens to the pressure of a gas in a balloon if the temperature is increased? Think Kinetic Molecular theory
Gay-Lussac’s Law: T↑ P↑ T↓ P↓ P is directly proportional to T (IF volume is held constant)
Can be derived from Combined Gas Law Temperature must be in Kelvin Mathematically…..
Practice question A sealed can containing gas at 101 kPa and 22oC is heated to 55oC. Calculate the pressure in the heated can (assume volume is constant) Use G.U.E.S.S. Remember: convert oC to K T1 = 22oC = ?K P1 = 101 kPa T2 = 55oC = ?K ANSWER = 112 kPa
G-L Law Practice p. 431 # 1-4 (Gay-Lussac's Law) Early finishers: p. 432 #5-9