1. GENERAL CHEMISTRY SPRING 2010 Mr. Hoffman Mrs. Paustian The Behavior of Gases Unit 9

2. Air Pressure Pressure equation Newtons (N) Square meters (m 2 )

3. Unit for Air Pressure Pascals  1 N/m 2 = 1 Pascal (Pa)  Standard pressure is 101.325 kPa on Earth f P a Blaise Pascal (French philosopher and mathematician) Pascal’s Triangle 1 kPa = 1000 Pa!

4. Air Pressure Without changing the mass (force), how can you increase pressure? Without changing the mass (force), how can you decrease pressure? f P a Decrease the area Increase area

5. Practice Problem The mass of a brick is 2.0 kg (F=19.6N), the sides are 0.05m and 0.03m. What is the pressure exerted? f P a 19.6 N = 13,067 Pa (0.05 m x 0.03 m)

6. Pressure Units Ways to represent pressure Units of Gas Pressure Unit Standard Pressure Atmosphere1 atm (exactly) Inches of Hg29.9 in Hg cm of Hg76 cm Hg mm of Hg760 mm Hg Torr760 torr Pounds per sq. in.14.7 psi Kilopascal101 kPa Torr is named after Evangelista Torricelli

7. The Barometer Etymology of “barometer”  In Greek, “baros”= weight  Meter= measure  Literally means “measure the weight of air” or air pressure. Bariatric surgeory is weight loss surgeory $5 Footlong

8. The Barometer How a barometer works  Air presses down on an open tray of Hg  This downward pressure pushes the Hg up into a tube  Higher air pressure causes the mercury to go higher up in the tube (measured as height, mm Hg)

9. Compressibility A gas will expand to fill its container Compressibility  A measure of how much the volume of matter decreases under pressure.  Gases easily compress because of the space between the particles http://www.garagelibrary.com/images/airbag.jpg

10. Kinetic Molecular Theory Postulates the theory is based on… 1. Gases consist of tiny particles (of negligible mass) with great distances separating them 2. Gases are in constant random motion 3. Molecular collisions are elastic 4. Average kinetic energy is dependent only on temperature

11. The Kelvin Scale As T increases, so does kinetic energy Theoretically, kinetic energy can be zero, but it hasn’t been achieved and probably won’t ever be achieved Absolute zero- The temperature at which a substance would have zero kinetic energy The Kelvin Scale- a temperature scale directly related to kinetic energy  Zero on the Kelvin scale corresponds to zero kinetic energy

12. The Kelvin Scale Units are Kelvins (K), with no degree ( o ) sign Kelvin relates temperature to kinetic energy!

13. Temperature Conversions Easy to convert between Celsius and Kelvin  How do you think?  o C  K? Add 273  K  o C? Substract 273  25 o C  K?  (25+273)= 298 K  310 K  o C?  (310-273) = 37 o C Fahrenheit Celsius?  ( o F – 32 o F) x 5/9 = o C  ( o C x 9/5) + 32 o F = o F

14. Factors Affecting Gas Pressure Amount of gas Volume Temperature http://www.bmumford.com/photo/highspeed/Ted1.jpg

15. Boyle’s Law “Boyles inverse” States that the volume of a gas varies inversely (opposite) with pressure if temperature and amount are held constant Written:  P 1 V 1 = P 2 V 2 Robert Boyle

16. Boyle’s Law in motion http://www.grc.nasa.gov/WWW/K-12/airplane/Animation/gaslab/Images/chprmt.gif

17. Boyle’s Practice 1 A tank contains a volume of 3 L and a pressure of 4 atmospheres. What volume would the gas from this tank fill up at a pressure of 1 atmosphere?  P 1 = 4atm  V 1 = 3L  P 2 = 1 atm  V 2 = ? Substitute into Boyle’s Law equation and solve for V 2 P 1 V 1 = P 2 V 2 (4 atm)(3L) = (1atm)(V 2 ) 12 atm*L = V 2 1 atm V2= 12 L

18. Boyle’s Practice 2 Find the volume of a cylinder needed if you want to put 50 atmospheres of pressure with a volume of 3 L into a cylinder that can hold a pressure of no greater than 20 atmospheres.  P 1 = 50 atm  V 1 = 3 L  P 2 = 20 atm  V 2 = ? Substitute into Boyle’s Law equation and solve for V 2 P 1 V 1 = P 2 V 2 (50 atm)(3 L) = (20 atm)(V 2 ) 7.5 L = V 2

19. Charles’ Law “Charles direct” The volume of a gas varies directly with temperature if the pressure and amount remain constant Mathematically:

20. Charles’ Law in Motion

21. Charles’ Law Problem 1 Always convert temperature to Kelvin (K) A tank contains a volume of 3 L and a temperature of 100 o C. What volume would the gas from this tank fill up at a temperature of 200 o C?  V 1 = 3 L  T 1 = 100 o C = 373K  V 2 = ?  T 2 = 200 o C= 473K Substitute into equation and solve for V2 3L V2 = 373K 473K (3L)(473K) = (V 2 )(373K) 4L = V 2

22. Charles’ Law Problem 2 A 275 L helium balloon is heated from 20 o C to 40 o C. Calculate the final volume assuming pressure remains constant.  V 1 = 275 L  T 1 = 20 o C (+273K)= 293K  V 2 = ?  T 2 = 40+273= 313K 275L = V 2 293K 313K (293)(v 2 ) = (275L)(313K) V 2 = 294 L

23. Temperature-Pressure Relationships Gay-Lussac’s Law  The pressure of a gas varies directly w/ temperature if volume and amount remain constant.  Mathematically: Joseph Louis Gay-Lussac 1778-1850 Tire inflation…. Summer vs. winter…hmmm how do these two differ?

24. Gay-Lussac Problem 1 Temperature in Kelvin (K) A tank contains a pressure of 3 atm and a temperature of 100 o C. What pressure would the gas from this tank be at a temperature of 200 o C?  P 1 = 3 atm  T 1 = 100 o C (+273K)= 373K  P 2 = ?  T 2 = 200 o C= 473K 3 atm = P 2 373 K 473 K P 2 = 4 atm

25. Gay-Lussac Problem 2 Find the pressure needed if you wanted to put gas at 50 o C and 75 atm into a vessel that is at 65 o C.  P 1 = 75 atm  T 1 = 50 o C= 323 K  P 2 = ?  T 2 = 65 o C= 338 K 75 atm = P 2 323 K 338 K P 2 = 78 atm

26. Combined Gas Law Puts together several scientists’ work on how gases behave when conditions are changes. You can change three things about a gas  Amount of gas- this stays the same for now  Temperature  Volume  Pressure changes in response to these STP?? Temperature= 273K Pressure= 1 atm

27. CBL Problem 1 A 50 mL sample of hydrogen gas is collected at 772 mm Hg and 21 o C. Calculate the volume of hydrogen at STP.  P 1 = 772 mm Hg  atm  V 1 = 50 mL  L  T 1 = 21 o C= 294 K  P 2 = 1 atm  V 2 = X  T 2 = 273K Since your final conditions are at STP, you need to convert initial conditions to atm and L

28. CBL Problem 1 solution (1.016 atm)(0.05L) = (1atm)(V 2 ) 294 K 273 K Cross multiply, then solve for V 2 (1.016 atm)(0.05L)(273 K) = (1atm)(294K)(X) 0.05 L = V 2 P 1 = 772 mm Hg x 1 atm/760 mm Hg= 1.016 atm V1= 50 mL / 1000 mL= 0.05 L

29. Ideal Gases Ideal gases are…  Gases that behave under all conditions as predicted by the kinetic molecular theory Gases are not ideal…  When they don’t behave as predicted by kinetic molecular theory Kinetic energy is…  Energy associated with motion

30. The Ideal Gas Law “piv-nert” PV=nRT  P= Pressure (atm)  V= Volume (L)  n= # moles (mol)  R= Ideal gas constant= 0.0821 (L*atm)/(mol*K)  T= Temperature (ALWAYS Kelvin)

31. IGL Problem 1 If a container has a volume of 3 L and is at a temperature of 60 o C and a pressure of 6 atm, how many moles of gas are in the container?  V= 3L  T= 60 o C= 333K  P= 6 atm  R= 0.0821 L*atm/mol*K  n= ?

32. IGL Problem 1 Solution Substitute in PV=nRT and solve for n (6 atm)(3 L) = (n)(0.0821)(333 K) n= 0.7 mol The unit for R is L*atm mol*K Be sure units match this, and units will cancel

33. IGL Problem 2 If a container has 50 moles of O 2 and is at a temperature of 40 o C and a pressure of 3 atm, how many liters are in this container?  n= 50 mol O2  T= 40 o C = 313 K  P= 3 atm  R= 0.0821 L*atm/mol*K  V= ?

34. IGL Problem 2 Solution Substitute in PV=nRT and solve for V (3atm)(V) = (50 mol)(0.0821)(313 K) V = 428 L