Gases. Chapter 5 Chang & Goldsby. Modified by Dr. Juliet Hahn

Transcription

1 Gases Chapter 5 Chang & Goldsby Modified by Dr. Juliet Hahn Copyright McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. 1

2 Example 5.6 Argon is an inert gas used in lightbulbs to retard the vaporization of the tungsten filament. A certain lightbulb containing argon at 1.20 atm and 18 C is heated to 85 C at constant volume. Calculate its final pressure (in atm). End 10/4W class 9am Electric lightbulbs are usually filled with argon.

3 Example 5.6 (1) Strategy The temperature and pressure of argon change but the amount and volume of gas remain the same. What equation would you use to solve for the final pressure? What temperature unit should you use? Solution Because nn 1 = nn 2 and VV 1 = VV 2, Equation (5.9) becomes PP 1 TT 1 = PP 2 TT 2 which is Charles s law [see Equation (5.6)].

4 Example 5.6 (2) Next we write Initial Conditions PP 1 = 1.20 atm TT 1 = K = 291 K Final Conditions PP 2 =? TT 2 = K = 358 K The final pressure is given by PP 2 = PP 1 TT 2 TT 1 = 1.20 atm 358K 291K = 1.48 atm Check At constant volume, the pressure of a given amount of gas is directly proportional to its absolute temperature. Therefore the increase in pressure is reasonable.

5 Example 5.7 A small bubble rises from the bottom of a lake, where the temperature and pressure are 8 C and 6.4 atm, to the water s surface, where the temperature is 25 C and the pressure is 1.0 atm. Calculate the final volume (in ml) of the bubble if its initial volume was 2.1 ml. Error in question powerpoint originally gave 8 o C for temperature on the surface which did not match the powerpoint answer Used my example instead. End 10/6 Friday 9 am class

6 Example 5.7 (1) Strategy In solving this kind of problem, where a lot of information is given, it is sometimes helpful to make a sketch of the situation, as shown here: What temperature unit should be used in the calculation?

7 Example 5.7 (2) Solution According to Equation (5.9) PP 1 VV 1 nn 1 TT 1 = PP 2VV 2 nn 2 TT 2 We assume that the amount of air in the bubble remains constant, that is, nn 1 = nn 2 so that PP 1 VV 1 TT 1 = PP 2VV 2 TT 2 which is Equation (5.10).

8 Example 5.7 (3) The given information is summarized: Initial Conditions PP 1 = 6.4atm VV 1 = 2.1mL TT 1 = K = 281 K Final Conditions PP 2 = 1.0atm VV 2 =? TT 2 = K = 298 K Rearranging Equation (5.10) gives VV 2 = VV 1 PP 1 TT 2 PP 2 TT atm = 2.1 ml 1.0 atm 298 K 281K = 14 ml

9 Example 5.7 (4) Check We see that the final volume involves multiplying the initial volume by a ratio of pressures PP 1 PP 2 and a ratio of temperatures TT 2 TT 1. Recall that volume is inversely proportional to pressure, and volume is directly proportional to temperature. Because the pressure decreases and temperature increases as the bubble rises, we expect the bubble s volume to increase. In fact, here the change in pressure plays a greater role in the volume change.

10 Density and Molar Mass Density (d) Calculations dd = mm VV = PPM RRRR m is the mass of the gas in g M is the molar mass of the gas Molar Mass (M ) of a Gaseous Substance M = dddddd PP d is the density of the gas in g/l 10

11 Gas Stoichiometry 11

12 Example 5.11 Calculate the volume of O 2 (in liters) required for the complete combustion of 7.64 L of acetylene C 2 H 2 measured at the same temperature and pressure. 2C 2 H 2 gg + 5O 2 gg 4CO 2 gg + 2H 2 O ll The reaction of calcium carbide CCCCCC 2 with water produces acetylene CC 2 HH 2, a flammable gas.

13 Example 5.11 (1) Strategy Note that the temperature and pressure of O 2 and C 2 H 2 are the same. Which gas law do we need to relate the volume of the gases to the moles of gases? Solution According to Avogadro s law, at the same temperature and pressure, the number of moles of gases are directly related to their volumes. From the equation, we have 5 mol O 2 2 mol C 2 H 2 ; therefore, we can also write 5 L O 2 2 L C 2 H 2. The volume of O 2 that will react with 7.64 L C 2 H 2 is given by volume of O 2 = 7.64 L C 2 H 2 5 L O 2 2 L C 2 H 2 = 19.1L

14 Example 5.12 Sodium azide NaN 3 is used in some automobile air bags. The impact of a collision triggers the decomposition of NaN 3 as follows: 2NaN 3 ss 2Na ss + 3N 2 gg The nitrogen gas produced quickly inflates the bag between the driver and the windshield and dashboard. Calculate the volume of N 2 generated at 80 C and 823 mmhg by the decomposition of 60.0 g of NaN 3. An air bag can protect the driver in an automobile collision.

15 Example 5.12 (1) Strategy From the balanced equation we see that 2 mol NaN 3 3 mol N 2 so the conversion factor between NaN 3 and N 2 is 3 mol N 2 2 mol NaN 3 Because the mass of NaN 3 is given, we can calculate the number of moles of NaN 3 and hence the number of moles of N 2 produced. Finally, we can calculate the volume of N 2 using the ideal gas equation.

16 Example 5.12 (2) Solution First we calculate number of moles of N 2 produced by 60.0 g NaN 3 using the following sequence of conversions so that grams of NaN 3 moles of NaN 3 moles of N 2 moles of N 2 = 60.0 g NaN 3 1 mol NaN g NaN 3 3 mol N 2 2 mol NaN 3 = 1.38molN 2 The volume of 1.38 moles of N 2 can be obtained by using the ideal gas equation: VV = nnnnnn PP = 36.9 L 1.38 mol L atm K mol K = atm

17 Dalton s Law of Partial Pressures V and T are constant PP 1 PP 2 PP total = PP 1 + PP 2 17

18 Partial Pressure Consider a case in which two gases, A and B, are in a container of volume V. PP AA = nn AARRRR VV PP BB = nn BBRRRR VV nn AA is the number of moles of A n B is the number of moles of B PP TT = PP AA + PP BB XX AA = nn AA nn AA + nn BB XX BB = nn BB nn AA + nn BB PP AA = XX AA PP TT PP BB = XX BB PP TT PP ii = XX ii PP T mole fraction XX ii = nn ii nn TT 18

19 Example 5.14 A mixture of gases contains 4.46 moles of neon Ne, 0.74 mole of argon Ar, and 2.15 moles of xenon Xe. Calculate the partial pressures of the gases if the total pressure is 2.00 atm at a certain temperature.

20 Example 5.14 (1) Strategy What is the relationship between the partial pressure of a gas and the total gas pressure? How do we calculate the mole fraction of a gas? Solution According to Equation (5.14), the partial pressure of Ne PP Ne is equal to the product of its mole fraction XX Ne and the total pressure PP T

21 Example 5.14 (2) Using Equation (5.13), we calculate the mole fraction of Ne as follows: XX Ne = nn Ne nn Ne + nn Ar + nn Xe = 4.46 mol 4.46 mol mol mol = Therefore, PP Ne = XX Ne PP T = atm = 1.21 atm

22 Example 5.14 (3) Similarly, PP Ar = XX Ar PP T = atm = 0.20 atm and PP Xe = XX Xe PP T = atm = atm Check Make sure that the sum of the partial pressures is equal to the given total pressure; that is, atm = 2.00 atm.

23 Collecting a Gas over Water 2KClO 3 ss 2KCl ss + 3O 2 (gg) PP T = PP O2 + PP H2 O 23

24 Copyright McGraw-Hill Education. Permission required for reproduction or display. Vapor of Water and Temperature Table 5.3 Pressure of Water Vapor at Various Temperatures Temperature ( CC) Water Vapor Pressure (mmhg)

25 Example 5.15 Oxygen gas generated by the decomposition of potassium chlorate is collected as shown in Figure The volume of oxygen collected at 24 C and atmospheric pressure of 762 mmhg is 128 ml. What is the pressure of the oxygen. The pressure of the water vapor at 24 C is 22.4 mmhg. End 10/9Monday 10 am class 25