1. A Review Gases

2. Elements that exist as gases at 25 0 C and 1 atmosphere

3. Gases assume the volume and shape of their containers. Gases are the most compressible state of matter. Gases will mix evenly and completely when confined to the same container. Gases have much lower densities than liquids and solids. Physical Characteristics of Gases

4. Properties of Gases Gas properties defined using only 4 varibles V = volume of the gas (L)V = volume of the gas (L) T = Kelvin temperature (K)T = Kelvin temperature (K) n = number of moles (mol)n = number of moles (mol) P = pressure in eitherP = pressure in either (atm), (mm Hg) or (kPa) (atm), (mm Hg) or (kPa)

5. Units of Pressure 1 pascal (Pa) = 1 N/m 2 1 atm = 760 mmHg = 760 torr 1 atm = 101,325 Pa Barometer Pressure = Force Area (force = mass x acceleration)

6. Sea level1 atm 4 miles0.5 atm 10 miles0.2 atm

7. Atmospheric pressure Barometer Manometer

8. Boyle’s Law If n and T are constant, then PV = (nRT) = k This means, for example, that P goes up as V goes down. Robert Boyle (1627-1691). Son of Early of Cork, Ireland.

9. Boyle’s Law A bicycle pump is a good example of Boyle’s law. As the volume of the air trapped in the pump is reduced, its pressure goes up, and air is forced into the tire.

10. P  1/V P x V = constant P 1 x V 1 = P 2 x V 2 Boyle’s Law Constant temperature Constant amount of gas

11. A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL? P 1 x V 1 = P 2 x V 2 P 1 = 726 mmHg V 1 = 946 mL P 2 = ? V 2 = 154 mL P 2 = P 1 x V 1 V2V2 726 mmHg x 946 mL 154 mL = = 4460 mmHg

12. Charles’s Law If n and P are constant, then V = (nR/P)T = kT V and T are directly related. Jacques Charles (1746- 1823). Isolated boron and studied gases. Balloonist.

13. Charles’s original balloon Modern long-distance balloon

14. As T increasesV increases

15. Variation of gas volume with temperature at constant pressure. V  TV  T V = constant x T V 1 /T 1 = V 2 /T 2 T (K) = t ( 0 C) + 273.15 Charles’ & Gay-Lussac’s Law Temperature must be in Kelvin

16. A sample of carbon monoxide gas occupies 3.20 L at 125 0 C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? V 1 = 3.20 L T 1 = 398.15 K V 2 = 1.54 L T 2 = ? T 2 = V 2 x T 1 V1V1 1.54 L x 398.15 K 3.20 L = = 192 K V 1 /T 1 = V 2 /T 2 T 1 = 125 ( 0 C) + 273.15 (K) = 398.15 K

17. Avogadro’s Law Equal volumes of gases at the same T and P have the same number of molecules. V = n (RT/P) = kn V and n are directly related. twice as many molecules

18. Avogadro’s Law V  number of moles (n) V = constant x n V 1 /n 1 = V 2 /n 2 Constant temperature Constant pressure

19. Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of NO are obtained from one volume of ammonia at the same temperature and pressure? 4NH 3 + 5O 2 4NO + 6H 2 O 1 mole NH 3 1 mole NO At constant T and P 1 volume NH 3 1 volume NO

23. Ideal Gas Equation Charles’ law: V  T  (at constant n and P) Avogadro’s law: V  n  (at constant P and T) Boyle’s law: V  (at constant n and T) 1 P V V  nT P V = constant x = R nT P P R is the gas constant PV = nRT

24. The conditions 0 0 C and 1 atm are called standard temperature and pressure (STP). PV = nRT R = PV nT = (1 atm)(22.414L) (1 mol)(273.15 K) R = 0.082057 L atm / (mol K) Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L.

25. What is the volume (in liters) occupied by 49.8 g of HCl at STP? PV = nRT V = nRT P T = 0 0 C = 273.15 K P = 1 atm n = 49.8 g x 1 mol HCl 36.45 g HCl = 1.37 mol V = 1 atm 1.37 mol x 0.0821 x 273.15 K Latm molK V = 30.6 L

26. What is the volume (in liters) occupied by 49.8 g of HCl at STP?

27. Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0 C is heated to 85 0 C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? PV = nRT n, V and R are constant nR V = P T = constant P1P1 T1T1 P2P2 T2T2 = P 1 = 1.20 atm T 1 = 291 K P 2 = ? T 2 = 358 K P 2 = P 1 x T2T2 T1T1 = 1.20 atm x 358 K 291 K = 1.48 atm

28. Density (d) Calculations d = m V = PMPM RT m is the mass of the gas in g M is the molar mass of the gas Molar Mass ( M ) of a Gaseous Substance dRT P M = d is the density of the gas in g/L

29. A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and 27.0 0 C. What is the molar mass of the gas? dRT P M = d = m V 4.65 g 2.10 L = = 2.21 g L M = 2.21 g L 1 atm x 0.0821 x 300.15 K Latm molK M = 54.6 g/mol

30. Gas Stoichiometry You can apply the discoveries of Gay -Lussac and Avogadro to calculate the stoichiometry of chemical reaction. Step 1  Write a balanced chemical equation. Step 2  Convert all units to common units Step 3  Start with given and Use factor-label skills (Via MOLES) to get to desired units. Step 4  If gas laws are needed use appropriately at this point or prior to factor- label step.

31. Gas Stoichiometry What is the volume of CO 2 produced at 37 0 C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C 6 H 12 O 6 (s) + 6O 2 (g) 6CO 2 (g) + 6H 2 O (l) g C 6 H 12 O 6 mol C 6 H 12 O 6 mol CO 2 V CO 2 5.60 g C 6 H 12 O 6 1 mol C 6 H 12 O 6 180 g C 6 H 12 O 6 x 6 mol CO 2 1 mol C 6 H 12 O 6 x = 0.187 mol CO 2 V = nRT P 0.187 mol x 0.0821 x 310.15 K Latm molK 1.00 atm = = 4.76 L

32. Gas Stoichiometry problems 1) Ammonia can react with Oxygen to produce nitrogen and water. If 1.78 L of O 2 reacts, what volume of nitrogen will be produce? (Temp and pressure remain constant) 2) Ethylene gas( C 2 H 4 ) combust in Air to form H 2 O and CO 2. If 13.8 L of C 2 H 4 measured at 21°C and 1.038 atm burns completely with O 2, calculate the volume of CO 2 produced assuming the CO 2 is measured at 44°C and 0.989 atm. 3) When Arsenic (III) sulfide is roasted in air, it reacts with oxygen to produce arsenic (III) oxide and sulfur dioxide. When 89.5 g As2S3 is roasted with excess O 2, what volume of SO 2 is produced? The gaseous product is measured at 20°C and 98.0 kPa. Ans.= 1.19 L of N 2 Ans.= 32.6 L of CO 2 Ans.= 27.1 L of SO 2

33. Dalton’s Law of Partial Pressures V and T are constant P1P1 P2P2 P total = P 1 + P 2

34. Consider a case in which two gases, A and B, are in a container of volume V. P A = n A RT V P B = n B RT V n A is the number of moles of A n B is the number of moles of B P T = P A + P B X A = nAnA n A + n B X B = nBnB n A + n B P A = X A P T P B = X B P T P i = X i P T mole fraction (X i ) = nini nTnT

35. A sample of natural gas contains 8.24 moles of CH 4, 0.421 moles of C 2 H 6, and 0.116 moles of C 3 H 8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C 3 H 8 )? P i = X i P T X propane = 0.116 8.24 + 0.421 + 0.116 P T = 1.37 atm = 0.0132 P propane = 0.0132 x 1.37 atm= 0.0181 atm

36. Chemistry in Action: Scuba Diving and the Gas Laws PV Depth (ft)Pressure (atm) 01 332 663 PV = nRT

37. Kinetic Molecular Theory of Ideal Gases 1.A gas is composed of molecules that are separated from each other by distances far greater than their own dimensions. The molecules can be considered to be points; that is, they possess mass but have negligible volume. 2.Gas molecules are in constant motion in random directions. Collisions among molecules are perfectly elastic. 3.Gas molecules exert neither attractive nor repulsive forces on one another. 4.The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy

38. Kinetic theory of gases and … Compressibility of Gases Boyle’s Law P  collision rate with wall Collision rate  number density Number density  1/V P  1/V “12m10an3.mov” Charles’ Law P  collision rate with wall Collision rate  average kinetic energy of gas molecules Average kinetic energy  T P  TP  T “12m10an2.mov”

39. Kinetic theory of gases and … Avogadro’s Law P  collision rate with wall Collision rate  number density Number density  n P  nP  n “12m10an1.mov” Dalton’s Law of Partial Pressures Molecules do not attract or repel one another P exerted by one type of molecule is unaffected by the presence of another gas P total =  P i

40. Deviations from Ideal Gas Law Real molecules have volume.Real molecules have volume. There are intermolecular forces.There are intermolecular forces. –Otherwise a gas could not become a liquid.

41. Deviations from Ideal Behavior 1 mole of ideal gas PV = nRT n = PV RT = 1.0 Repulsive Forces Attractive Forces

42. Deviations from Ideal Gas Law Account for volume of molecules and intermolecular forces with Van der Waal’s Equation. Measured V = V(ideal) Measured P intermol. forces vol. correction J. van der Waals, 1837-1923, Professor of Physics, Amsterdam. Nobel Prize 1910. nRT V - nb V 2 n 2 a P + ----- )(

43. Deviations from Ideal Gas Law Cl 2 gas has a = 6.49, b = 0.0562 For 8.0 mol Cl 2 in a 4.0 L tank at 27 o C. P (ideal) = nRT/V = 49.3 atm P (van der Waals) = 29.5 atm

44. Van der Waals equation nonideal gas P + (V – nb) = nRT n2an2a V2V2 () } corrected pressure } corrected volume During Calculations plug in a and b values into the equation and solve algebraically