1. Lecture 26 © slg CHM 151 Topics: 1. Ideal Gas Law calculations 2. Density, Molar Mass of gases 3. Stoichiometry involving gases

2. Combined Gas Law We can combine the three factors which describe a “confined gas” (constant n, number of particles) as follows: P  1 / V, P  T : P  T P =C c T V V PV = C c AND P 1 V 1 = C c =P 2 V 2 T T 1 T 2

3. If we hold the one factor constant for this “confined gas”, then : P 1 V 1 = P 2 V 2 T 1 T 2 V 1 = V 2 T 1 T 2 Constant T P 1 V 1 = P 2 V 2 Constant P Constant V P 1 = P 2 T 1 T 2

4. Units Utilized in Gas Law Problems: PRESSURE*: generally done in reference to a column of mercury immersed in a dish of mercury open to atmospheric pressure. (CD ROM) Pressure as a FORCE PER UNIT AREA: English SI = 14.7 lbs/ft 2 (psi) = 101.3 kilopascal (k Pa) 760 mm Hg = 760 Torr = 1 atmosphere (atm) 1 Pa, pascal = 1 Newton / m 2 1 Newton = 1 kg. m / s 2 (SI unit of force)

5. Volume*: Liters, L (most general) Temperature: Absolute or Kelvin Scale; K= o C +273.15 In all cases, K, Kelvin scale must be used! Where P, V = 0, T=0, absolute zero, - 273.15 o C * Any V or P unit OK for combined gas law, if used consistently Units Utilized in Gas Law Problems, continued:

6. Since gas law problems always include lots of data, it is a standard practice to set up a table of all given data before proceeding: For the combined gas law, an appropriate table is below:

7. A gas exerts a pressure of 735 mm Hg in volume of 2.50 L at a temperature of 73 o C. What pressure would it exert if the temperature were increased to 110 o C and the volume increased to 3.00 L?

8. data formula solve

9. A sample of CO 2 gas has a pressure of 56.5 mm Hg in a 125 mL flask. The gas is transferred to a new flask where it has a pressure of 62.3 mm Hg at the same temperature. What is the volume of the new flask?

11. GROUP WORK A sample of gas occupies 12.0 liters at 240. o C under a pressure of 80.0 kPa. At what temperature would the gas occupy 15.0 liters if the pressure were increased to 107 kPa?

12. KEY:

14. VARIATION OF P with n at constant V, T If we keep the volume of our container constant and also the temperature, the introduction of more particles into the container will mean more collision with the wall of the container. Accordingly, we can say: P  n P = C n n P 1 P 2 = C n = n 1 n 2

15. Relating P to n, V, T P  1 / V, P  T, P  n : P  n T P =R nT V V where R = gas constant used to interrelate the four factors. Rearranging: PV = nRT the “ideal gas equation” (No such thing as an “ideal gas” but if very low T’s and very high P’s are avoided, most gases can be described fairly accurately using the “ideal gas law”.)

16. Evaluation of value of R If P, V, n, and T for a gas are experimentally determined, the value for this constant can readily be calculated, which is appropriate for use with any gas (if the conditions are not too extreme): PV / nT =R We can calculate R for one mole of any gas using what is described as “Standard Conditions”, a conventional P and T adopted as the norm for gases (“STP”): Standard Pressure = 1 atmosphere or 760 Torr (mm Hg) Standard Temperature = 0 o C = 273 K

17. It was determined by Avogadro (1811) that equal volumes of any gas, at the same temperature and pressure, contain the same number of molecules. It has since been experimentally determined that at STP, 1.00 mole of any “ideal” gas occupies 22.414 L, called the standard molar volume. 1.00 mole gas, STP = 22.4 L

18. Calculation of R, STP: Let us use experimentally determined molar volume of 1.00 mole of gas at STP to evaluate R. Note that our table of values changes form when using PV= nRT :

19. Calculation of R, changing pressure unit to atmospheres:

20. Useful Gas Law Knowledge, to date: P 1 V 1 = P 2 V 2 T 1 T 2 V 1 = V 2 T 1 T 2 Constant T P 1 V 1 = P 2 V 2 Constant P Constant V P 1 = P 2 T 1 T 2 Combined Gas Law, constant n:

21. PV = nRT the “ideal gas equation” 1.00 mole gas, STP = 22.4 L STP: Standard Pressure = 1 atmosphere or 760 Torr (mm Hg) Standard Temperature = 0 o C = 273 k R = 62.4 Torr L / K mol = 62.4 Torr L K -1 mol -1 R =.0821 atm L / K mol =.0821 atm L K -1 mol -1

22. n = ? 25.0 g N 2 gas = ? moles 28.0 g N 2 = 1 mole N 2 25.0 g N 2 1 mol =.893 mol N 2 28.0 g 25.0 g =.893 mol 28.0 g mol -1 More convenient form, gas laws

23. Problems using the ideal gas equation: A hot air balloon holds 30.0 kg of helium. What is the volume of the balloon if the final pressure is 1.20 atm and the temperature 22 o C?

24. Convert formula first!

25. Group Work The nitrogen gas in an air bag, with a volume of 65 L, exerts a pressure of 829 mm Hg at 25 o C. How many moles and how many g of N 2 are in the air bag? (Solve for moles using ideal gas equation, then solve for g) PV =nRT n=

27. Another handy “formula” to utilize: The ideal gas law can be used to calculate molar mass if grams of gas, and P,V, T are known: 1. Calculate moles of gas, n = PV / RT 2. Use n, moles and mass, g to calculate M n, moles = mass, g molar mass, M = mass, g molar mass, M n, moles Widely used

28. Gas Laws : Density, Molar Mass Chloroform is a common liquid which vaporizes readily. If the pressure of the vaporized liquid is 195mm Hg at 25 o C, and the density of the gas is 1.25 g/L, what is the molar mass of the chloroform? V= 1.00 L mass= 1.25 g P= 195 mm Hg T= 25 o C = 298 K R = 62.4 L mm Hg K -1 mol -1 M, molar mass = ? g/mol Procedure: 1) solve for n, # mol 2) M = #g / #mol density

29. V= 1.00 L mass= 1.25 g P= 195 mm Hg T= 25 o C = 298 K R = 62.4 L mm Hg K -1 mol -1 M, molar mass = ? g/mol CHCl 3 = 12 + 1 + 3(35.4) = 119.2 g/mol

30. Group Work What is the molar mass of a gas which has a density of 1.83 g/L measured at 27.0 o C and 0.538 atm? R =.0821 L atm K -1 mol -1

32. Gas laws, Stoichiometry Problem 49, text: A self contained breathing apparatus uses canisters containing potassium superoxide. The superoxide consumes the CO 2 exhaled by the person and replaces it with O 2. What mass of solid KO 2, in grams, is required to react with 8.90 L of CO 2 at 22.0 o C and 767 mm Hg?

33. When gases are included in a balanced equation for a reaction, the number of moles or grams of the gas used can be computed by the Ideal gas equation, PV = nRT. Use Ideal gas equation in equation situation to compute moles of gas, then proceed normally to do problem. Word of Warning: never use grams of solid or liquid in gas law equation: it won’t give a correct answer...

34. ? grams V= 8.90 L T= 22.0 o C = 296.1 K P= 767 mm Hg R= 62.4 L mm Hg K -1 mol -1 n =? FIRST

35. ? grams 1 K =39.1 2 O= 32.0 71.1 g/mol n =.369 moles

36. Group Work What volume of O 2, collected at 22.0 o C and 728 mm Hg would be produced by the decomposition of 8.15 g KClO 3, M = 122.5 g/mol? 2 KClO 3 (s) 2 KCl (s) + 3 O 2(g) 1. Calculate moles of O 2 from equation 2. Calculate V of O 2 from ideal gas law