Apparatus for Studying the Relationship Between Pressure and Volume of a Gas

Transcription

1 The Gas Laws

2 Apparatus for Studying the Relationship Between Pressure and Volume of a Gas As P (h) increases V decreases

3 Boyle s Law P x V = constant P 1 x V 1 = P 2 x V 2 Constant temperature Constant amount of gas

4 Example 1 A helium balloon was compressed from 4.0L to 2.5L at a constant temperature. If the pressure of the gas in the 4.0L balloon is 210 kpa, what will the pressure be at 2.5L? Given: V 1 = 4.0L V 2 = 2.5L P 1 = 210 kpa P 2 =? P 1 V 1 = P 2 V 2 (210 kpa) (4.0L) = P 2 (2.5L) P 2 = 336 kpa 340 kpa

5 Example 2 A sample of neon gas occupies 0.200L at atm. What will be its volume at 29.2 kpa pressure? Given: V 1 = 0.200L V 2 =? P 1 = atm P 2 = 29.2 kpa **Units must match for each variable (doesn t matter which one is converted) P 1 V 1 = P 2 V 2 (87.1 kpa) (0.200L) = (29.2 kpa) V 2 V 2 = 0.597L atm kpa 1 atm = 87.1 kpa

6 Variation in Gas Volume with Temperature at Constant Pressure As T increases V increases

7 Charles s Law **Temperature must be in Kelvin K = 0 C Constant pressure Constant amount of gas

8 Example 1 A gas ample at 40.0 C occupies a volume of 2.32L. If the temperature is raised to 75.0 C, what will the volume be, assuming the pressure remains constant? Given: T 1 = 40.0 C = 313K T 2 = 75.0 C = 348K V 1 = 2.32L V 2 =? V 1 V 2 = T 1 T L V 2 = V 2 = 2.58L

9 Example 2 A gas ample at 55.0 C occupies a volume of 3.50L. At what new temperature in kelvin will the volume increase to 8.00L? Given: T 1 = 55.0 C = 328K T 2 =? V 1 = 3.50L V 2 = 8.00L V 1 V 2 = T 1 T L 8.00 = 328 T 2 T 2 = 750K

10 Gay-Lussac s Law Constant volume Constant amount of gas **Temperature must be in Kelvin K = 0 C + 273

11 Example 1 The pressure of a gas in a tank is 3.20 atm at 22.0 C. If the temperature rises to 60.0 C, what will be the gas pressure in the tank? Given: P 1 = 3.20 atm P 2 =? T 1 = 22.0 C = 295K T 2 = 60.0 C =333K P 1 P 2 = T 1 T atm P 2 = 295K 333K P 2 = 3.61 atm

12 Example 2 A rigid container has a gas at constant volume at 665 torr pressure when the temperature is 22.0C. What will the pressure be if the temperature is raised to 44.6C? Given: P 1 = 665 torr P 2 =? T 1 = 22.0 C = 295K T 2 = 44.6 C =317.6K P 1 P 2 = T 1 T torr P 2 = 295K 317.6K P 2 = 716 torr

13 What is the same about these gases? What is different about these gases?

14 What is the relationship between number of moles and volume?

15 Avogadro s Law Constant temperature Constant pressure 1 mol of gas at STP = 22.4L

16 Gas Stoichiometry When gases are involved, the coefficients in a balanced chemical equation represent not only molar amounts (mole ratios) but also relative volumes (volume ratios).

17 Example L of a gas is known to contain mol. If the amount of gas is increased to 1.80 mol, what will be the new volume? Assume constant temperature and pressure. Given: V 1 = 5.00 L V 2 =? n 1 = mol n 2 = 1.80 mol V 1 V 2 = n 1 n L V = mol 1.80 mol V 2 = 9.33 L

18 Example 2 Calculate the volume that mol of gas at STP will occupy mol 22.4 L 1 mol = 19.7 L

19 Example 3 How many grams of carbon dioxide gas are in a 0.75 L balloon at STP? 0.75 L CO 2 1 mol CO g CO L CO 2 1 mol CO 2 = 1.5 g CO 2

20 Example 4 What volume of oxygen gas is needed for the complete combustion of 4.00 L of propane gas (C 3 H 8 )? Assume constant temperature and pressure L C 3 H 8 5 L O 2 1 L C 3 H 8 = 20.0 L

21 Gas Law Summary

22 Charles s Law Gay-Lussac s Law

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24 Combined Gas Law Combine all 4 to make one master equation:

25 Example 1 A gas at 110 kpa and 30.0 C fills a flexible container with an initial volume of 2.00L. If the temperature is raised to 80.0 C and the pressure increased to 440 kpa, what is the new volume? Given: P 1 = 110 kpa P 2 = 440 kpa T 1 = 30.0 C =303K T 2 = 80.0 C = 353K V 1 = 2.00L V 2 =? 110 (2.00) 440 (V 2 ) = V 2 = 0.583L 0.58L

26 Example 2 An unopened bottle of soda contains 46.0 ml of gas confined at a pressure of 1.30 atm and temperature of 5.00 C. If the bottle is dropped into a lake and sinks to a depth at which the pressure and temperature changes to 1.52 atm and 2.90 C, what will be the volume of gas in the bottle? Given: V 1 = 46.0 ml V 2 =? P 1 = 1.30 atm P 2 = 1.52 atm T 1 = 5.00 C = 278K T 2 = 2.90 C = 275.9K 1.30 (46.0) 1.52 (V 2 ) = V 2 = 39.0mL

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28 Ideal Gas Follows all gas laws under all conditions of temperature and pressure. Follows all conditions of the Kinetic Molecular Theory (KMT) An ideal gas does not exist in real life The Ideal Gas Law Real Gas Follows some gas laws under some conditions of temperature and pressure. Does not conform to the Kinetic Molecular Theory. Real gases have a volume and attractive and repulsive forces. A real gas differs from an ideal gas the most at low temperature and high pressure. Low T gas is moving less High P gas is more attractive

29 Ideal Gas law PV = nrt R is the universal gas constant R = PV nt = (1 atm)(22.4l) (1 mol)(273k)

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31 Example 1 Calculate the number of moles of gas contained in a 3.00L vessel at 298K with a pressure of 1.50 atm. V = 3.00L T = 298K P = 1.50atm n =? R = L atm mol K PV=nRT (1.50)(3.00) = n (0.0821) (298) n= mol

32 Example 2 What will the pressure (in kpa) be when there are mol of gas in a 5.00L container at 17.0 C? V = 5.00L T = 17.0 C = 290K P =? kpa n = mol R = L kpa mol K PV=nRT P(5.00) = (0.400)(8.314)(290) P= 193 kpa

33 Stoichiometry

34 Example 1 If 5.00L of nitrogen reacts completely with excess hydrogen at a constant pressure and temperature of 3.00 atm and 298K, how many grams of ammonia are produced mol N 2 2 mol NH g NH 3 1 mol N 2 1 mol NH 3 = g NH 3

35 Example 2 How many grams of calcium carbonate will be needed to form 6.75L of carbon dioxide at a pressure of 2.00 atm and 298K? CaCO 3 (s) CO 2 (g) + CaO(s) mol CO 2 1 mol CaCO g CaCO 3 1 mol CO 2 1 mol CaCO 3 = 55.2 g CaCO 3

36 Example 3 How many liters of chlorine will be needed to make 95.0 grams of C 2 H 2 Cl 4 at 3.50 atm and 225K? Cl 2 (g) + C 2 H 4 (g) C 2 H 2 Cl 4 (l) 95.0 g C 2 H 2 Cl 4 1 mol C 2 H 2 Cl 4 2 mol Cl g C 2 H 2 Cl 4 1 mol C 2 H 2 Cl 4 = 1.13mol Cl 2

37 Density (d) Calculations n V = P RT so m M V = P RT d = PM RT n = m M d = m V m is the mass of the gas in g M is the molar mass of the gas d is the density of the gas in g/l Molar Mass (M ) of a Gaseous Substance M = drt P 37

38 Example 1 What is the molar mass of a pure gas that has a density of 1.40g?L at STP?

39 Example 2 Calculate the density a gas will have at STP if its molar mass is 39.9g/mol.