1. Gases and their Properties

2. Nature of Gases 1. Gas particles have mass. If you fill up a basketball with air, its mass will be more than the mass of a deflated ball. 2. Gases are easy to compress: If you squeeze a gas, its volume decreases 3. Gases fill their containers completely: When a balloon is filled with air, the air is distributed evenly. There is no place around you that is not filled with air (exception: sealed vacuum).

3. Nature of Gases Cont. 4. Different gases move through each other quite rapidly; this is called diffusion. Examples gases diffusing through air are: smelling perfume left by a person, smelling a skunk nearby. 5. Gases exert pressure: you experience air pressure when your ears pop, or the balloon inflates. Pressure = force/unit area 6. The pressure of a gas depends on its temperature: the higher the temperature, the higher the air pressure and vice versa. In the summer the tire pressure in your car increases due to heat.

4. Kinetic Molecular Theory Ludwig Boltzmann Rudolf Claussius James Maxwell Kinetic Molecular Theory is based on six postulates: 1.Gases are small particles that have a mass 2.Since distances separating gas particles are large, the volume of individual gas molecules is negligible 3. Gas particles are in constant, rapid, random motion

5. Kinetic Molecular Theory Cont. 4.Collisions between gas particles or gases and their containers are perfectly elastic (in a perfectly elastic collision, no energy of motion is lost ex. bouncing ball) 5.The average kinetic energy of gas particles depends only on the temperature of the gas. The higher the temp. the greater the kinetic energy (energy of motion) 6.Gases exert no force on one another. The attractive forces between gas particles is negligible. No hydrogen bonding.

6. Measuring Gases In order to describe a gas, one must take into consideration four factors: 1.amount of gas 2.volume 3.temperature 4.pressure Amount of gas is measured in moles of gas (n) Volume of gas is measured in Liters = 1000 cm 3

7. Measuring Temperature of a Gas Temperature of a gas is usually measured in Celsius. However the gas laws are written using the Kelvin Scale. The zero point on the Celsius scale equals the melting/freezing point of water at 1 atmosphere of pressure. The boiling point of water at 1atmosphere of pressure is set to 100 degrees on the Celsius scale. The interval between the melting point and the boiling point of water is divided into 100 intervals, each equal to 1 degree. This interval is the same on the Kelvin scale. 1 Kelvin unit = 1 o Celsius. The zero point on the Kelvin scale is called absolute zero, the lowest possible temperature. This is the point at which all molecular motion stops. It is equal to - 273 o C. When working with the Kelvin scale, the term "degrees" is not used. Zero degrees Celsius is written as 273 K and read as "273 Kelvins". The zero point on the Fahrenheit scale is based on the freezing point of a mixture of salt and water and scientifically meaningless. The Fahrenheit scale is not used in science.

8. Conversions: Converting from Celsius to Kelvin: Example: convert 37 o Celsius to Kelvin. C + 273 = K or C = K -273 37 + 273 = 310K Check out the picture to verify this answer. To convert from Kelvin to Celsius subtract 273. Converting from Celsius to Fahrenheit: Example: convert 37 o Celsius to Fahrenheit. 9/5C + 32 = F ( 9/5 X 37 ) + 32 = 98.6 o F. 9/5 = 1.8, meaning that 1 degree on the Celsius scale = 1.8 degrees on the Fahrenheit scale. Look at the picture: a five degree interval on the Celsius scale equals a nine degree interval on the Fahrenheit scale. Remember, 37 o C equals 98.6 o F, because it is the body temperature. Converting from Fahrenheit to Celsius: C = (F - 32) / 9/5 (98.6 - 32) / 9/5 = 37 o C. First subtract 32, then divide by 1.8. Instead of dividing by 1.8, you may also multiply by 5/9. http://www.howe.k12.ok.us/~jimaskew/ptemps.htm

9. Calculating Pressure: Standard Pressure is the air pressure taken at sea level. It can be measured using a variety of barometers that use different scales with different values. The SI unit for pressure is Pascals The different units for standard pressure are as follows: 101,325Pa = 101.325 KPa = 1atm = 760 torr = 760 mm Hg = 14.70 lbs/in 2 The Standard pressures written above are all equal to one another. To convert between them, you can use their equivalencies to make conversion factors. Image from http://www.williamsclass.com/EighthScienceWork/Atmosphere/barometerTori. jpg

10. How to Convert Between Pressure Scales The standard pressures are all equal to one another. 101,325Pa = 101.325 KPa = 1 atm = 760 torr = 760 mm Hg = 14.70 lbs/in 2 One can convert between two different pressures by creating a conversion factor. For example: If a barometer reads 5 atm pressure, what is the pressure in torr? The standard pressures indicate that 760 torr = 1 atm. By turning that equivalency into a conversion factor (760 torr/1atm or 1atm/760 torr), one can use the conversion factor to convert from atm to torr. 5 atm x 760 torr =3,800 torr 1 atm

11. Sample Problems What is 780 torr in mm Hg? What is 15 atm in lbs/in 2 ? What is 3400 mm Hg in KPa? What is.02 atm in Pascal?

12. Boyle’s Law Pressure versus Volume ► Robert Boyle (1627-1691) was the chemist who first proposed a law to describe the behavior of gases at differing volumes and pressures. ► Boyles law is: P 1 V 1 = P 2 V 2 ► As Pressure increases volume must decrease and vice versa. ► Boyles law example example

13. Boyle’s Law is based on an Inverse Relationship

14. Solving Boyle’s Law ► Sample problem: A marshmallow has a volume of 10mL at the initial pressure of.89 atm. If the pressure changes to 1.02 atm, what is the new volume of the marshmallow? ► Step 1: Write out the known values. P 1 = initial pressure V 1 = initial volume P 2 = final pressure V 2 = final volume Step 1 continued: P 1 = 0.89 atm V 1 = 10 ml P 2 = 1.02 atm V 2 = x Step 2: Plug the values into the formula. 0.89 atm x 10 ml = 1.02 atm x X Step 3: Solve for X. 8.9 atm ml = 1.02 atm X Divide both sides by 1.02 atm to isolate X. 8.9 atm ml = 1.02 atm X 1.02 atm 8.725 ml = X Note: Due to their inverse relationship, when pressure increased, volume had to decrease

15. Example Problems ► If the initial pressure of a gas is 5 atm with a volume of 6 L, what is the new pressure of the gas if the volume increases to 15 L? ► If the pressure of a tire is 30 psi with a volume of 15 L, what is the pressure of the tire if the volume increases to 18 L?

16. Charles’s Law ► In 1787 French physicist Jacques Charles studied the effect of temperature on volume when pressure is held constant ► Volume and temperature have a direct relationship: when temperature increases, increases, so does volume.ncreases Formula is: V 1 = V 2 T 1 T 2 or it can be re- written as V 1 T 2 = V 2 T 1 ► Jacques Alexandre César Charles (November 12, 1746,– April 7, 1823) was a French inventor, scientist, mathematician, and balloonist. ► Charles and the Robert brothers launched the world's first (unmanned) hydrogen-filled balloon in August 1783, then in December 1783, Charles and his co- pilot Nicolas-Louis Robert ascended to a height of about 1,800 feet (550 m) in a manned balloon. Their pioneering the use of hydrogen for lift led to this type of balloon being named a CharlièreRobert brothersballoonNicolas-Louis Robert Quote from: http://en.wikipedia.org/wiki/Jacques_Charles

17. Charles’s Law ► V 1 /T 1 = V 2 /T 2 Pressure and moles are held constant. Volume and temperature are directly related As temperature increases, the molecules move faster creating more pressure on the walls of the container. The pressure is relieved by pushing the walls of the container outward causing an increase in volume.

18. Volume verses Temperature Graph ► Graph of Charles Law. Volume is proportional to absolute temperature.

19. How to solve Charles’ Law Problems ► CONVERT ALL temperature to KELVIN!!! ► Never use Celsius or Fahrenheit in the gas law formulas. Sample Problem: A balloon has a volume of 5L at 25 o C. What happens to the volume of the balloon if the temperature goes up to 35 o C? Formula V 1 /T 1 = V 2 /T 2 V 1 = initial volume T 1 = initial temperature V 2 = final volume T 2 = final temperature ► Step 1: Write out your know values. V 1 = 5L T 1 = 25 o C +273 = 298K V 2 = X T 2 = 35 o C + 273= 308K ► Step 2: Plug the values into the formula: 5L/298K = X/308K ► Step 3: Solve for X Cross multiply:5Lx308K = X x 298K 1540 LK = 298K X Divide both sides by 298K to isolate X 1540 LK = 298K X 298K X = 5.17L Note: Because they are directly related, when temperature increased, so did volume

20. Sample Problems ► A gas is collected and found to fill a 2.85 L container at 25 o C. What will its new volume be if temperature decreases to standard temperature? ► A 5.00 L sample of a gas is collected at 100 K and then allowed to expand to 20 L what must the new temperature be in order to maintain the same pressure (as required by Charles’ Law)?

21. Joseph Gay-Lussac’s Law ► 1778-1850  French Physicist interested in the affect that temperature has on pressure ► Discovered that temperature and pressure have a direct mathematical relationship: ► Whenever temperature increases, so does pressure (and vice-versa) ► His formula is as follows : P 1 /T 1 = P 2 /T 2 P 1 /T 1 = P 2 /T 2 Remember: Temperature is measured in Kelvin

22. How does Temperature affect pressure? ► If temperature is increased, the gas molecules will move faster (remember: kinetic molecular theory defines temperature as the average speed of molecules). The particles will collide with the walls of the container more often. If the walls of the container are rigid (volume stays constant), then the pressure inside the container must increase.

23. Pressure v. Temperature Graph

24. Lussac’s Sample Problem If the air inside a car has an initial temperature of 30 o C, and a pressure of 1 atm, what will the new pressure be if the temperature increases to 45 o C? P 1 /T 1 = P 2 /T 2 P 1 = initial pressure T 1 = initial temperature P 2 = final pressure T 2 = final temperature ► Step 1: Write down all your known values. P 1 = 1 atm T 1 = 30 o C + 273 = 303K P 2 = X T 2 = 45 o C + 273 =318K ► Step 2: Plug the values into the formula: 1atm/303K = X/318K ► Step 3: Solve for X Cross-multiply: 318K atm = 303KX Divide both sides by 303K to isolate X: 318K atm = 303KX x = 1.05 atm 303K Note: Because they are directly related, when temperature increase so did pressure.

25. Sample Problems ► What is the pressure in a tire, if the temperature of the tire increased from 32 o C to 40 o C and its original pressure was 30 psi? to 40 o C and its original pressure was 30 psi? ► What is the new temperature of gas in a rigid container, when pressure is increased from 1 atm to 10 atm and the original temperature was 37 o C?

26. Combined Gas Law