CHEMISTRY WKST: Gases Review p. 1 1) a) b) c) V directly proportional to Kelvin temperature V inversely proportional to pressure V directly proportional to moles 2) Air pressure is caused by the collisions of gas particles against a surface. 3) barometer, manometer 4) Gas particles are so small, compared to the distances between them, the volume of individual particles is assumed to be negligible (zero volume). Gas particles are in constant random motion, colliding with the walls of the container to cause air pressure. The collisions are perfectly elastic. Gas particles are assumed not to attract or to repel each other. The average kinetic energy of the gas particles is directly proportional to the Kelvin temperature of the gas. 5) Gases approach ideal behavior at low pressure and high temperature. 6) The mole ratio of N2:H2 is equal to the volume ratio of the gases. In this case both are 1 N2: 3 H2 7) 22.4 L 8) Average kinetic energy is directly proportional to the Kelvin temperature of the gas sample. 9) a) increases b) increases c) increases d) same e) same 10) a) same b) increases c) increases d) increases e) increases 11) The particles of Xe in the 100 C sample will have more kinetic energy and, therefore, be moving faster than the particles of Xe in the 25 C sample. 12) The He in both samples will be moving at the same speed. Since both samples have the same temperature, both samples will have the same average kinetic energy. Therefore, the particles in both samples will have the same velocity. 13) H2 2.02 g/mol CH4 16.05 g/mol N2O3 76.02 g/mol O2 32.00 g/mol NH3 17.04 g/mol The lighter the gas the faster it moves. N2O3 < O2 < NH3 < CH4 < H2 (slowest) (fastest)
CHEMISTRY WKST: Gases Review p. 2 14) An ideal gas is one in which its volume is: directly proportional to its Kelvin temperature directly proportional to its moles inversely proportional to its pressure 15) PHe = 0.75 atm ( ) = 76 kpa 1 atm P O2 = 225 torr ( ) = 30.0 kpa 760 torr P N2 = 12.0 psi ( ) = 82.7 kpa 14.7 psi Ptotal = 76 kpa + 30.0 kpa + 82.7 kpa = 189 kpa 16) Pvap of water at 25 C = 23.76 torr P1V1 T1 = P2V2 T2 V1 = 250.0 ml P1 = 750.0 torr 23.76 torr = 726.2 torr T1 = 25 C = 298 K V2 =? P2 = 760 torr T2 = 273.15 K P1V1T2 = P2V2T1 V2 = P1V1T2 V2= P2T1 (726.2 torr)(250.0 ml)(273.15 K) (760 torr)(298 K) V2 = 219 ml 17) V1 = 2.250 L T1 =? V2 = 4.115 L T2 = 0 C = 273 K V1 = V2 T1 T2 V1T2 = V2T1 T1 = V1T2 V2 T1 = (2.250 L)(273 K) (4.115 L) = 149 K T1 = 149 K 273.15 = 124 C 18) Since T goes up 4, V goes up 4. Since P goes up 8, V goes down 8. Therefore, V ends up going down 2.
CHEMISTRY WKST: Gases Review p. 3 19) C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) 5 mol O2 a) 2.25 L C3H8 ( ) = 11.25 L O2 = 11.2 L O2 1 mol C3H8 1 mol C3H8 4 mol H2O 18.02 g H2O b) 0.5000 L C3H8 ( ) ( ) ( ) = 1.61 g H2O 22.4 L C3H8 1 mol C3H8 1 mol H2O c) Need molar V at 725 torr and 15 C 725 torr ( 1 atm ) = 0.954 atm 760 torr PV = nrt V = nrt P L atm (1 mol)(0.08206 = mol K )(288 K) 0.954 atm = 24.8 L 1 mol O2 mol CO2 24.8 L CO2 75.0 g O2 ( ) (3 ) ( ) = 34.9 L CO2 32.00 g O2 5 mol O2 1 mol CO2 20) P1 = 1.000 atm T1 = 28.25 C = 296.40 K T2 =? P2 = 1.500 atm T 1 = P 2 T 2 T 2 = P 2 T 1 T 2 = P 2T 1 T 2 = (1.500 atm)(296.40 K) 1.000 atm = 444.6 K T2 = 444.6 K 273.15 = 171.45 C = 171.4 C 21) V1 = 22.50 ml P1 = 92.55 kpa V2 =? P2 = 105.50 kpa P1V1 = P2V2 V2 = P1V1 V2= P2 (92.55 kpa)(22.50 ml) (105.50 kpa) V2 = 19.74 ml
CHEMISTRY WKST: Gases Review p. 4 22) n1 = 0.750 mol = 685.2 mmhg ( ) = 91.35 kpa 760 mmhg P2 = 120.3 kpa n2 =? Both P s need to be the same unit doesn t matter which. If you made both mmhg, the 120.3 kpa would become 902.3 mmhg. Final answer is the same regardless. n 1 = P 2 n 2 n 2 = P 2 n 1 n 2 = P 2n 1 n 2 = (120.3 kpa)(0.750 mol) 91.35 kpa n2 = 0.988 mol 23) n1 = 0.00225 mol V1 = 62.00 ml V2 =? n2 = 0.500 mol V1 = V2 n1 n2 V1n2 = V2n1 V2 = V1n2 n1 V2= (62.00 ml)(0.500 mol) (0.00225 mol) V2 = 13800 ml 24) g = 32.0 g NH3 V = 17.4 L T = 41 C = 314 K P =? PV = grt (MM) P = grt V(MM) L atm (32.0 g)(0.08206 )(314 K) mol K = (17.4 L)(17.04 g ) mol = 2.78 atm 25) g =? SO3 V = 250.0 ml = 0.2500 L P = 745 torr ( 1 atm ) = 0.980 atm 760 torr T = 15 C = 258 K PV = grt (MM) g = PV(MM) RT = g (0.980 atm)(0.2500 L)(80.07 mol ) (0.08206 L atm = )(258 K) mol K 0.927 g 26) rate O2 rate Cl2 MM Cl2 = MM O2 rate O2 rate Cl2 70.90 g/mol Cl2 = 32.00 g/mol O2 = 1.488 O2 is 1.488 times faster than Cl2
CHEMISTRY WKST: Gases Review p. 5 27) rate NH3 rate unk = MM unk MM NH3 rate unk rate NH3 MM NH3 = MM unk 42.00 ml/min 22.75 ml/min = 17.04 g/mol NH3 22.75 ml/min 42.00 ml/min = 17.04 g/mol NH3 1.846 = 17.04 g/mol NH3 17.04 g/mol NH3.5417 = 3.408 = 17.04 g 0.2934 = 17.04 g mol = 58.07 g/mol = 17.04 g mol 0.2934 = 58.08 g mol 28) a) When heated, the gas particles inside the can gain KE and will move faster. This increases the number of collisions against the surface of the can, causing an increase in pressure. Eventually the pressure will become too much for the can to hold, causing the can to eplode. b) When you inhale the air in the straw you are creating a vacuum inside the straw and your mouth. The higher atmospheric pressure pushing on the surface of the liquid in the container will force the liquid up the straw. c) The higher you go into the atmosphere the lower the pressure becomes (so does temperature but not as fast as pressure). Therefore, as the balloon rises and the pressure lowers, the balloon s volume increases. d) As the football cools the gas particles inside the ball loses KE and slows down. This causes fewer collisions against the side of the ball. Thus, the pressure decreases and ball gets soft. e) As more air goes into the tire, the number of collisions against the tire s surface increases. Therefore, the pressure inside the tire increases. f) When the air inside the can is removed there will be less collisions against the can s surface, causing the pressure to decrease. The higher outside pressure will then force the walls of the container to collapse. g) When the container is first closed, gravity pulls out a small amount of water through the hole causing the volume of air inside the container to epand and lower its pressure. Then the higher pressure outside the container will keep the water inside the container since the air inside has less pressure. When the lid is opened, the pressures equalize and water will pour out the hole. h) When the hot soup is put in the freezer the gas particles inside the container lose KE and slow down. This lowers the number of collisions of particles against the surface of the container and thereby lowers the pressure inside the container. When going to open the container you then must overcome the higher atmospheric pressure on the outside of the container. i) A popcorn kernel has a small amount of water inside. As the kernel is heated, the water gets superheated well above its boiling point. This causes etreme pressure inside the kernel as the water molecules begin to move faster with the increased KE. Eventually the pressure gets to the point where the kernel s hull can t contain it any longer and the kernel pops.
CHEMISTRY WKST: Gases Review p. 6 j) As the air inside the balloon is heated the gas particles gain KE and move faster. This causes the particles to spread out and increase the volume of the balloon. This in turn decreases the density of the air inside the balloon. The more dense air outside is pulled down by gravity causing the less dense air inside the balloon to rise. 29) Gas particles do have a volume and that space is NOT available to other gas particles as assumed. Gas particles do have attraction forces between them when they get close to each other, causing the particles to slow down a little. This would cause the pressure to be slightly less than epected. 30) The vapor pressure of water increases as temperature increases.