3Section 2.2, Page 68Compressibility and Expansion are the properties of gas that are used to operate the ballast tanks of the submarine. When water is pumped in, the air in the tank is compressed. When water is pumped out, the compressed air expands to fill the empty tanks.Compressed gases must be kept away from heat sources and the containers must be handled with care. Increased temperature would cause the gas particles to move faster, and that would increase the pressure inside the container. Damaging a gas container could cause it to rupture or explode.
4Page 68, continuedThe compressibility of gases allows the gas spring to absorb the shock of opening or closing the hatchback.At the same temperature, two gases will have the same average kinetic energy. Their velocity will vary in proportion to their respective masses.As the balloon rises, the pressure outside will decrease, but the number of particles inside stays the same. The particles will continue to hit the side of the balloon pushing outwards. As the balloon rises, the gas inside will expand, filling all the space of the balloon.
5Page 68, concludedIn diffusion, a gas mixes gradually with another gas in the container, until the particles are evenly distributed. In effusion, a gas escapes from a container though one or more small openings in the container, or through a porous barrier.Fastest NH3, HF, CO, NO2, HI slowestAnswer: the molar mass of the unknown gas is 24.4 g/mol*The molar mass of the unknown gas is 146 g/lCurve 1 represents the gas with the greatest mass. (the gas with the lowest rate should be the one with the greatest mass.)*full answer for this question on the next slide
6Question 8, complete answer The effusion rate of an unknown gas is 43.0 mL/min. The effusion rate of CO2 is 32.0 mL/min. What is the molar mass of the unknown gas?Data:vx = 43.0 mL/minvCO2 = 32.0 mL/minTo find:MCO2 =Mx =Step 1: find molar mass of CO2 MCO2 = 12.0 g/mol + 2(16.0 g/mol) = 44.0 g/mol44.0 g/molStep 2: Use Graham’s Law24.4 g/molAnswer: The molar mass of the unknown gas is about 24.4 g/mol
7Question 9, complete answer It takes 192 s for an unknown gas to effuse. Nitrogen gas (N2) diffuses across the same barrier in 84s. What is the molar mass of the unknown gas.Data:tx = 192 stN2 = 84 sTo find:MN2 =Mx =Step 1: find molar mass of N2 MN2 = 2(14.0 g/mol) = 28.0 g/molStep 2: Use Graham’s Law variation28.0 g/mol146 g/molAnswer: The molar mass of the unknown gas is about 146 g/mol
8Section 2.3, Page 741. Molecules striking the sides of a container cause pressure.2. the pressure is atm or 91.7 kPa.3. a) 450 mmHg or 60.0 kPab) 99.9 kPa or 749 mmHgc) 160 kPa or 1199 mmHgd) kPa or4. The pressure is 18mmHg or 2.4 kPa**I think the book author intended a different answer, but I have answered what the question asked, not what I think it should have asked.
9Section 2.4.1, Page 97V2= 240 mLV2= 97.9 mLP2= 140 kPa (or 1.4x102 kPa)P2= kPaV2= 2.3 LP2= 1.2 x 102 kPa (or 120 kPa)V= 290 L (or 2.9 x 102 LP= 1.5 x 102 kPa (or 150 kPa)(a) P1= 108 kPa(b) The pressure was very high (good weather) but it is now falling (bad weather coming)10. V= 8.32 L
10Section 2.4.2, Page 97, 98 a) 298 K b) 310 K c) 423 K a) 100°C b) -175°C c) 152°CThe volume occupied by the gas will be 33L.The volume of the balloon will be 2.1 LThe final temperature will be 606°CThe variation will be 660°CThe volume of CO2 in the dough is 0.12LThe volume of the chlorine is 33 mLThe gas must be heated to 55°CThe nitrogen will occupy 386 mLThe temperature is 369 K or 96°C
11Section 2.4.3, Page 98 Answer: The temperature will be 97°C. The thermometer will show -3.7°C.The pressure will be 759 mmHg.The manometer will show a pressure of 109 kPa.The temperature in the freezer is -16°C.The gas must be heated to 313°C.The valve will open when the butane is 240°C.The gas will exert a pressure of 271 kPaThe pressure exerted by the helium is 18.9 atm
12Section 2.4.4, Page 98 The volume of the container is 100 L (1x102 L) The volume isThe balloon originally contained ≈4 mol of He. You add 0.5 mol of oxygen, making the new n≈4.5 mol So V1≈72 LAnswer: 0.57 mol of air escaped from the tire.
1335. Answer: 0. 0098 mol of CO2 must be added to the balloon. 36 35. Answer: mol of CO2 must be added to the balloon. 36. There are 6.34x10-3 g of oxygen in the swim bladder of the fish. 37. The total volume of argon is 77.3 L 38. a) 0.21 mol, b) 9.24x10-4 mol c) 3.7x10-3 mol 39. The volume occupied by the hydrogen is about 180 L 40. The amount of sulphur dioxide is 2.0x10-3 mol. 41. Neon will occupy 50.4 L 42. Hydrogen volume is L
14The volume of hydrogen sulphide is 1400 L The volume of hydrogen is 24.5 L45. There are 3.01x1023 molecules of heliumSeven or Eight envelopes (seven is the closest answer, but if you needed all the air, you would need eight envelopes)SATP is a higher temperature than STP (25°C vs. 0°C). According to kinetic theory at a higher temperature the molecules would move faster, and try to expand to a greater volume.The water is a liquid. It does not have a standard molar volume (only gases do), and liquid is a condensed form of matter. It will have less volume.
15The air pressure in the tire is 190 kPa It would take 24 mol of gas Since the rubber ball is elastic, the volume can increase. If the volume increases the pressure may not. In the propane tank the volume is constant, so the pressure must increase.Fifty twoP1n1P2n2c) 71.4 kPae) 175 mola) 150 kPab) 3.7 mold) 158 kPa
16Page 104P=104 kPa, V=2L, T=295K so n=PV / RT so n=0.085mol (for all three containers)The containers hold the same moles (0.085mol), and therefore molecules of gasThe mass of the gases are: 0.34g, 2.7g, and 3.7g*The pressure exerted by the methane is 971 kPa.sample contains 4.10x10-2 mol of methane.The temperature will be 497 K or 224°C*It could be argued that the answers should be rounded to just one significant digit, since the volume was given as 2L, rather than 2.0L, but that would render the answers (0.3g, 3g and 4g) somewhat meaningless.
17Page 104 continuedWhat quantity of methane gas is found in a sample with a volume of 500mL at 35C and 210 kPaP= 210 kPa PV=nRTV= L n = PV / RTn= ? n= (210 x 0.5) / (8.31 x 308)T= = 308K n= molThe amount of methane is moles (or 0.66g)6. Determine the pressure exerted in a 50L cylinder if it contains 30 mol of air and is heated to 40C?P=? PV=nRTV= 50L P= nRT / Vn= 30 mol P= (30 x 8.31 x 313) / 50T= =313K P = kPaThe pressure is 1.6 x 103 kPa
18Page 104 continued9. The molar mass of the gas is 242 g/mol (2.4x102 g/mol)10. The mass of sulphur dioxide is 99g11. The molar mass of the gas is 1.3 x 102 g/mol12. The molar mass of the gas is 44.2 g/mol. The gas could be Carbon dioxide or propane13. The sample contains moles14. The volume of the balloon is 2.5x102 L15. The pressure of the argon in the bulb is 78 kPa16. The temperature of the chlorine gas is 23°C
19Page 104 continuedWhat volume does 50kg of oxygen gas occupy at a pressure of 150 kPa and a temperature of 125°CP=150 kPa n= 50000g / 32g/mol= 1562 molV=? PV=nRTn= 1562 mol V= (nRT) / PT= =398K V = (1562 x8.31x398) / 150V= 34462LThe volume is 3.4 x104 LAt what temperature does 10.5 g of ammoina (NH3) gas exert a pressure of 85 kPa in a 30 L container?P=85kPa n= 10.5g / 17 g/mol = molV=30L PV = nRTm=10.5g T = PV / nRn=0.618 mol T= (85x30) / (0.618x8.31)T=? T=496.5 kThe temperature would be about 497 K or 224° C
20Page 107The gas must be brought to a temperature of K or °C (1.0x102 K or -173°C)The volume of the gas would be L (67 L)The volume of the gas would be 5.8LThe final temperature will be 231°C (or 504K)The new volume of the balloon is 1.0 LThe missing quantities are:88C b C c. 3.8x102 kPa d. 5 L* e. 1.03x103L* L rounded to 1 SF. (a nasty trick to play on you!)
217. This is a nasty question, since both the volume and pressure can change, there are too many variables, so it cannot be solved using the general gas lawa. No answer.b. No answer.The volume of methane is 127 mL (or L)The new volume of the gas is 163 mLThe new temperature of the gas is 1.5x102K or 1.2x102°CThe temperature of the bulb is 310°CThe final volume of the gas is 1.5 times the original volume.A pressure of 642 kPa must be applied.
22Page 113 The resulting pressure is 300.65 kPa The partial pressure of Ar is 50.2 kPaFind the percentages of the total pressure.Partial pressure of Ne is kPa(14kPa)Partial pressure of He is kPa(27kPa)Partial pressure of Rd is 75.4 kPa (75 kPa)The total pressure of mixture is 110 kPa (1.1x102 kPa)
23The total number of moles is 2.92 mol. The partial pressures are: Methane (CH4) is kPa ( 7.5x102 kPa)Oxygen (O2) is kPa ( 4.8x102 kPa)Nitrogen (N2) is kPa (1.1x103 kPa)Propane (C3H8) is kPa (2.4x102 kPa)The partial pressure of CO2 is 0.7 atm.The partial pressures of the gases are:Carbon dioxide = kPa (3.6 kPa)Oxygen = kPa ( 4.1 kPa)Water vapour= kPa (95 kPa)The container has 64% chlorine inside.
24Chapter 2. Physical Properties of Gases (lines are drawn through questions which may have misprints in the textbook)Answer is Pressure of the gas is 69.6 kPaPressure of CO2 is 103 kPaIt moves towards the open endSince the end is closed, atmospheric pressure does not exert force on the mercury, so the difference in height is equal to the gas pressure.(explanation pending)
25The pressure in the cylinder is 57 atmospheres The hydrogen will occupy 375 dm3 (or 375 L)The final temperature of the gas is 49.3°CAbout 1.7 L of water vapour will form in the cakeThe volume of chlorine is 170 LThe final temperature will be 74°CThe final pressure will be 3.0x102 kPaa)318 K b) 340 K c) 623 Ka) 200°C b) -165°C c) -48°C
26The volume of CO2 in the dough is 0.56 L The volume of argon is 2.6 LThe new volume of gas is 8.2 LThe heat increases the kinetic energy inside the balloon, this can increase the pressure until the balloon bursts.The nitrogen will occupy 0.50 LThe volume of dry air is 21 LThe volume of the balloon is 2.1 LThe final temperature is 606°CThe final temperature of the oxygen is -233°C
27The pressure in the tire is 252 kPa A pressure of 6.4x102 kPa must be exerted.The nitrogen will occupy 2.8 LThe volume of the balloon will be 2.2 LThe container can be safely heated to 374°CThe final temperature of argon is 273°CThe final temperature of neon is -67°CThe pressure of oxygen is 9.5x102 kPaThe pressure in the bulb will be 24 atm.The volume of the bubble will be 21 mLThe molar mass of the gas is 48.2 g/mol
28The molar mass of halothane is 197 g/mol The molar mass of the gas is71.2 g/molThe oxygen occupies a volume of LThe nitrogen occupies a volume of 0.33 La) 2.41g b) 1.75 gThe temperature is 12°CThe gas will occupy 5.0 L at 0°CThe increase in pressure is 127%There are 0.18 mol of oxygenThe effusion rate of nitrogen is higher than the unknown gas, therefore, by Graham’s law, the molar mass of the unknown gas must be higher than nitrogen.