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161-2007 CHEM CHAPTER 5 - GASES PRACTICE PROBLEMS DR. ED TAVSS
Pressure Gas laws Reaction stoichiometry Partial pressure (mixtures of gases) Kinetic molecular theory/Effusion and Diffusion/Real gases
PRESSURE
CHEM 161-2006 EXAM II HILL & PETRUCCI CHAPTER 5 GASES PRESSURE
14.
In the above drawing, the value of h, the height of the liquid in the tube above the liquid in the beaker, would depend on which of the following? X. Density of the liquid Y. Cross sectional area of the tube Z. Atmospheric pressure (a) Z only (b) Y and Z only (c) X and Y only
(d)
X and Z only (e) X, Y, and Z This is a barometer. The liquid is held in the tube by atmospheric pressure. If the atmospheric pressure is exactly 1 atmosphere (14.7 lb/in2), then the liquid would have to exert a pressure of 14.7 lb/in 2. (X) If the liquid was, let's say, less dense, then it would have to be a greater height to exert a pressure of 14.7 lb/in 2. If the liquid was more dense then it would have to be a shorter height.
(Y) The cross sectional area of the tube is irrelevant, since the pressure (14.7 lb/in 2) would be the same regardless of the cross-sectional area. (Z) If the atmospheric pressure decreased, then the height would decrease because it would require less liquid to be equal to a lower pressure. If the atmospheric pressure increased, then the height would need to increase.
14
Chem 161-2005 Hourly Exam II Chapter 5 - Gases Pressure Which of the following is a unit of pressure? A. B.
g cm 3 kg
m3 C. pounds D. N2 m E. g 2 cm
Pressure is force per unit area. A is force per unit volume. B is force per unit volume C is force D is force per unit area. E is force per unit area. The acceptable answer is "D". I disagree. "E" should also be acceptable. Some faculty members consider "N" as force, and "g" as mass. Since pressure is force per unit area then only "D" is allowed. In my opinion this is nit-picking semantics.
15.
Chem 162-2004 Hourly exam II Chapter 5 - Gases
Pressure Mercury, rather than water, is used as a liquid in barometers because:
A. B. C.
Mercury is more sensitive to pressure changes than water. Mercury is safer. Pressure is measured in mm of mercury. Mercury is denser. Mercury is metallic.
D.
E.
A barometer works by the pressure of the air (e.g., 1 atmosphere) being equal and opposite to the pressure exerted by the liquid in the barometer. The pressure exerted by the liquid in the barometer is due to its height and its mass. For mercury, this height is 760 mm. Since water has a density 13.6 times less than the density of mercury, it would require a height of 760 mm x 13.6 = 10336 mm = 33.9 feet, and therefore would be too cumbersome to be useful.
2. CHEM 161-2000 EXAM 2 CHAPTER 5 - GASES PRESSURE Which one of the following expressions represents pressure? A. B. C. distance time force x distance force volume
D.
E.
force
area mass volume
21. CHEM 161-2000 EXAM 2 CHAPTER 5 - GASES PRESSURE Mercury is used as a liquid in barometers rather than water because:
A.
B. C. D. E.
A water barometer would be impractically tall. Water does not respond to pressure differences as easily as mercury. Mercury is a metallic liquid and conducts electricity. A mercury barometer is safer to use. Torr is defined as mm of Hg.
CHEM 161-2001 EXAM II + ANSWERS CHAPTER 5 - GASES PRESSURE 20. The density of water is 1.00 g/cm3. The density of mercury is 13.6 g/cm3. What height of a water column is necessary to exert a pressure of 2.00 atm? A. 23.1 meters 20.7 meters 18.6 meters 11.2 meters 5.80 meters
B.
C. D. E.
1 atm = 760 mm Hg; 2 atm = 1520 mm Hg = 1.520 m Hg d1h1 = d2h2 13.6 x 1.520 = 1.00 x X X = 20.67 meters H2O 11. CHEM 161-2002 EXAM II + ANSWERS
CHAPTER 5 - GASES PRESSURE A lead brick has dimensions 10 cm x 20 cm x 30 cm. Under which face will the pressure underneath the brick be the greatest? A. B. C. under the 10 cm x 30 cm face under the 20 cm x 30 cm face the weight of the brick must be known in order to answer the question. under the 10 cm x 20 cm face the pressure is equal under all the faces Pressure equals force per unit area. The force of the brick is the mass of the brick, which is unaffected by the direction of the brick. The greatest pressure will therefore be one in which the brick will be positioned with the smallest face down, which is the 10 cm x 20 cm face. An analogy would be the pressure of a person with a flat heel stepping on your foot, versus a person in spikes stepping on your foot.
D.
E.
GAS LAWS
3 Chem 161-2006 Final Exam Chapter 5 Gases Gas laws A container of fixed volume contains 22.0 g of CO 2(g) exerting a pressure of 1.00atm. Some CH4(g) is introduced into the container, raising the pressure to 1.50atm. What mass of CH4(g) was introduced? A. 8.00g B
B.
4.00g C. 12.0g D. 16.0g E. 2.00g CO2(g) CH4(g) 22.0 g = 0.4999 mol ?g 1.00 atm V1 = V2 T1 = T2 P2 = 1.50 atm Condition 1 = just CO2 Condition 2 = Total gases P1V1/n1T1 = P2V2/n2T2 P1/n1 = P2/n2 1.00 atm/0.4999 mol = 1.50 atm/X mol X mol = 0.75 mol = total moles Therefore, 0.75 total mol 0.50 mol CO2 = 0.25 mol CH4 g = 0.25 mol x 16.05g/mol = 4.01 g CH4
CHEM 161-2006 EXAM II HILL & PETRUCCI CHAPTER 5 GASES GAS LAWS
7.
A high vacuum pump can produce a pressure of 1.0 x 10-6 Torr. How many molecules are in each mL of gas at 25 oC at this low pressure?
(a)
3.2 x 1010 (b) 2.5 x 1010 (c) 3.9 x 1011
(d) 2.5 x 1013 (e) 3.9 x 1013 Gas 1.0x10-6Torr x (1 atm/760 Torr) = 1.316 x 10-9 atm 0.001L 298K n = PV/RT n = ((1.316 x 10-9 atm) x 0.001L)/((0.08205Latmdeg-1mol-1) x 298K) = 5.382 x 10-14 mol 5.382x10-14 mol x (6.022 x 1023 molecules/mol) = 3.24 x 1010 molecules
CHEM 161-2006 EXAM II HILL & PETRUCCI CHAPTER 5 GASES GAS LAWS
1. A bicycle tire is filled with air to a pressure of 610 kPa, at a temperature of 19 oC. Riding the bike on asphalt on a hot day increases the temperature to 58 oC. The volume of the tire increases by 4.0%. What is the new pressure in the tire?
(a)
(b) (c) (d) (e)
665 kPa 517 kPa 868 kPa 434 kPa 1217 kPa Bike(condition 2) ?kPa = Y kPa 331 K 1.04X L
Bike(condition 1) 610 kPa 292 K XL
P1V1/n1T1 = P2V2/n2T2 n1 = n2 P1V1/T1 = P2V2/T2 (610kPa x XL)/(292K) = (YkPa x 1.04XL)/(331K) The X's and units cancel out.
(610/292) = (Y x 1.04)/(331) Y = 665 kPa
8
Chem 161-2005 Hourly Exam II Chapter 5 - Gases Gas laws Which graph is not a straight line? A. P vs T (same V, n) B. V vs n (same P, T)
C.
P vs V (same T, n) D. V vs T (same P, n) E. P vs n (same V, n) "same" presumably means that those variables are held constant. Assume ideal gas law: PV = nRT In the formula for a straight line, Y = mX + b, Y and X are variables, while m and b are constants. (A): If V and n are constant, then the ideal gas law may be written as P = (nR/V) x T + 0 This fits the straight line law of Y = mX + b, in which a plot of P vs T will provide a straight line. (B): If P and T are constant, then the ideal gas law may be written as V = (RT/P) x (n) + 0. This fits the straight line law of Y = mX + b, in which a plot of V vs n will provide a straight line. (C): If T and n are constant, then the ideal gas law may be written as P = (nRT) x (1/V) + 0. This fits the straight line law of Y = mX + b, in which a plot of P vs 1/V will provide a straight line. If a plot of P vs 1/V provides a straight line, then a plot of P vs V will not provide a straight line. (It will provide a parabola?) (D): If P and n are constant, then the ideal gas law may be written as V = (nR/P) x T + 0. This fits the straight line law of Y = mX + b, in which a plot of V vs T will provide a straight line. (E): This problem doesn't tell what "T" is. However, it says that one of the variables, n, is a constant. Regardless of what "T" is, if the variable P is plotted against a constant, n, the result is a straight line. That is, if any variable is plotted against a constant, the result would be a straight line.
9
Chem 161-2005 Hourly Exam II Chapter 5 - Gases Gas laws A gas has a volume of 25.0 mL at 20 C. When heated at constant pressure, its volume expands to 28.0 mL. What is the temperature of the heated gas?
A.
B. C. D. E.
55 C 45 C 65 C 85 C 75 C Gas2 28.0 mL ?K P1 = P2 n1 = n2
Gas1 25.0 mL 293K
P1V1/n1T1 = P2V2/n2T2 V1/T1 = V2/T2 25.0 mL/293K = 28.0 mL/X X = 328.2K = (328.2 273 =) 55.2oC
5 Chem 161-2005 Final exam CHAPTER 5 - GASES Gas laws A rigid container holds 160 g of methane, CH4(g), at a pressure of 780 Torr. What is the pressure if 16.0 g of methane leaks out of the container at constant temperature?
A.
B. C. D. E.
702 Torr 616 Torr 518 Torr 424 Torr 820 Torr
Condition I Condition II 160 g CH4 = 9.969 mol 144 g CH4 = 8.983 mol 780 torr ?torr T1 = T2 V1 = V2 P1V1/n1T1 = P2V2/n2T2 P1/n1 = P2/n2 780 torr/9.969 mol = X torr/8.983 mol X = 703 torr
31 Chem 161-2005 Final exam CHAPTER 5 - GASES Gas laws An ideal gas is contained in a container with a volume of 2.50 L at a temperature of 78.0 C and a pressure of 2.00 atm. What is the new volume the gas will occupy when the pressure is reduced to 1.50 atm and temperature is increased to 92.0 C?
A. 3.47 L
B. 3.21 L C. 1.95 L D. 1.80 L E. 2.10 L Condition I 2.50 L 78.0 + 273 = 351K 2.00 atm n1 = n2 Condition II ?L 92.0 + 273 = 365 K 1.50 atm
P1V1/n1T1 = P2V2/n2T2 P1V1/T1 = P2V2/T2 (2.00 atm x 2.50 L)/351K = (1.50 atm x X L)/365K X = 3.47 L
35 Chem 161-2005 Final exam CHAPTER 5 - GASES Gas laws A 2.85 g sample of gas X occupies a volume of 575 mL at a temperature of 25 C and a pressure of 0.95 atm. What is the molar mass of this gas?
g mol g B. 54.4 mol C. 128 g mol g D. 204 mol g E. 173 mol
A. 72.8 Gas 2.85 g 0.575 L 298K 0.95 atm ?MW PV = nRT = Ideal Gas Law PV = (g/MW)RT MW = gRT/PV MW = (2.85 g x 0.08205 Latmdeg-1mol-1 x 298 K)/(0.95 atm x 0.575 L) = 128 g/mol
2.
Chem 162-2004 Hourly exam II Chapter 5 - Gases Gas Laws A 275 mL container holds a gas at a pressure of 1.50 atm and a temperature of 20 C. A valve is opened allowing some of the gas to escape. After closing the valve, the pressure of the remaining gas is 1.20 atm, at the same temperature. What amount of gas escaped? A. 7.7 x 10-3 mol
B. C. D.
2.6 x 10-3 mol 5.1 x 10-3 mol 1.7 x 10-3 mol 3.4 x 10-3 mol Condition 1 0.275 L 1.50 atm 293 K Condition 2 0.275 L 1.20 atm 293 K
E.
Since each of conditions 1 and 2 have three of the four ideal gas variables, then solve this problem using the ideal gas equation two times. Moles of gas at condition 1 : PV = nRT 1.50 atm x 0.275 L = n x 0.08205 Latmdeg -1 mol-1 x 293 K n = 0.01716 mol Moles of gas at condition 2: PV = nRT 1.20 atm x 0.275 L = n x 0.08205 Latmdeg -1 mol-1 x 293 K n = 0.01372 mol of gas at the second condition. Gas escaped: 0.01716 mol - 0.01372 mol = 0.00344 mol
11.
Chem 162-2004 Hourly exam II Chapter 5 - Gases Gas Laws A 1.00 L flask contains 0.20 g of H2 . A 2.00 L flask contains 8.0 g of gas X. Both gases are the same temperature and pressure. What is the molar mass of gas X?
A.
B. C. D. E.
g mol g 20 mol g 80 mol g 10 mol g 60 mol 40 H2 1.00 L X 2.00 L
0.20 g T1 = T 2 P1 = P2
8.0 g
?MW (P1V1MW1)/(g1T1) = (P2V2MW2)/(g2T2) Simplifying : (V1MW1)/(g1) = (V2MW2)/(g2) (1.00 L x 2.02 gmol-1)/(0.20 g) = (2.00 L x X)/(8.0 g) X = MW = 40.4 gmol-1
20.
Chem 162-2004 Hourly exam II Chapter 5 - Gases Gas laws A gas in a 250 mL container has a temperature of 25 C and a pressure of 1.00 atm. The volume of the container is decreased to 150 mL and the temperature increased to 50 C. What is the pressure of the gas inside the container?
A. B. C.
2.25 atm 1.25 atm 1.54 atm 1.81 atm 3.33 atm Conditions 1 0.250 L 298 K 1.00 atm Conditions 2 0.150 L 323 K ?P
D.
E.
P1V1/n1T1 = P2V2/n2T2 Simplify: Assume moles don't change. P1V1/T1 = P2V2/T2 (1.00 atm x 0.250 L)/298 K = (X x 0.150 L)/323 K X = 1.806 atm
30.
CHEM 161- 2004 FINAL EXAM + ANSWERS CHAPTER 5 - GASES GAS LAWS
A sample of gas has a pressure of 0.85 atm when placed in a temperature bath at 90 oC. It is placed in another temperature bath, and its pressure is decreased to 0.75 atm. The temperature of this second bath is: A. B. C. D. 102 oC 32 oC 79 oC 138 oC 47 oC
E.
P1V1/n1T1 = P2V2/n2T2 Gas condition 1 Gas condition 2 0.85 atm 0.75 atm o 90 C = 363K ?T Since the sample of gas doesn't change then n1 = n2. Since the pressure of the gas changes then it appears to be in a rigid container; therefore, P 1 = P2 . Simplifying: P1/T1 = P2/T2 0.85 atm/363 K = 0.75 atm/X X = 320.3 K = 47.3oC
31.
CHEM 161- 2004 FINAL EXAM + ANSWERS CHAPTER 5 - GASES GAS LAWS Which one of the following will always increase the density of a gas? A. decrease pressure and decrease temperature increase pressure and decrease temperature increase pressure and increase temperature none of the other choices would increase density, since density of a gas is independent of temperature and pressure. decrease pressure and increase temperature
B.
C. D. E.
Density = mass/volume Whatever will cause the volume to decrease will cause the density to increase. A. Decreasing the pressure will result in the volume increasing. B. Increasing pressure will result in the volume decreasing; decreasing temperature will result in the volume decreasing. Therefore, this combination will result in the gas density increasing. C. Increasing the temperature will result in the volume increasing. D. Incorrect. The density of a gas is dependent on temperature and pressure.
E. Decreasing the pressure will result in the volume increasing.
35.
CHEM 161- 2004 FINAL EXAM + ANSWERS CHAPTER 5 - GASES GAS LAWS Assume that each of the following reactions takes place in a container of fixed volume and that the temperature at the end of the reaction is the same as the initial temperature. X. Y Z. N2(g) + 3H2(g) 2NH3(g) 2H2O(g) 2H2(g) + O2(g) 2HCl(g) H2(g) + Cl2(g)
In which of the above reactions will the total pressure decrease.
A.
B. C. D. E.
X only Y only Y and Z only Z only X and Z only
P1V1/n1T1 = P2V2/n2T2 V1 = V2 T1 = T2 Therefore, P1/n1 = P2/n2 Since the moles within a reaction are proportional to the pressure, then when the moles decrease the pressure decreases. X: The moles decreases from 4 moles to 2 moles; therefore, the pressure decreases. Y: The moles increases from 2 moles to 3 moles; therefore, the pressure increases. Z: The moles begins and ends with two moles; therefore, the pressure doesn't change.
ZUMDAHL 5TH EDITION CHEM 161-2002 RECITATION 6TH WEEK CHAPTER 5 - GASES GAS LAWS
5-29
An aerosol can contains 400. mL of compressed gas at 5.20 atm pressure. When all the gas is sprayed into a large plastic bag, the bag inflates to a volume of 2.14 L. What is the pressure of gas inside the plastic bag? Assume temperature is constant. P V = n 5.20 atm 0.400 L X 2.14 L R T
Condition 1 Condition 2
Don't memorize; derive the equation when n and T are constant. (Put all of the constants on the same side of the equation as the constant "R", and all of the variables are the other side. Then, P 1V1 = k, and P2V2 = k. Therefore, P 1V1 = P2V2. Correspondingly, if V and T are constant, and P and n are variables, then P1/n1 = RT1/V1, and P2/n2 = RT2/V2. But since T and V are constants, then RT1/V1 = k = RT2/V2. Therefore P1/n1 = P2/n2.) At constant n and T: P1V1 = P2V2 5.20 atm x 0.400 L = X atm x 2.14 L X = 0.972 atm
ZUMDAHL 5TH EDITION CHEM 161-2002 RECITATION 6TH WEEK CHAPTER 5 - GASES GAS LAWS
43 An ideal gas is contained in a cylinder with a volume of 5.0 x 10 2 mL at a temperature of 30.oC and a pressure of 710. torr. The gas is then compressed to a volume of 25 mL, and the temperature is raised to 820.oC. What is the new pressure of the gas?
P V = (710/760 atm) 0.50 L X 0.025 L
n
R
T 303.15 K 1093.15 K
n and R are constant. Therefore, P1V1/T1 = P2V2/T2 ((710/760) atm x 0.50 L)/303.15 K) = (X atm x 0.025 L)/1093.15 K X = 67.4 atm 67.4 atm x (760 torr/1 atm) = 5.12 x 104 torr
ZUMDAHL 5TH EDITION CHEM 161-2002 RECITATION 6TH WEEK CHAPTER 5 - GASES GAS LAWS
37 A 2.50-L container is filled with 175 g argon. If the pressure is 10.0 atm, what is the temperature? P V = n R T -1 10.0 atm 2.50 L 175 g/39.95 gmole X 10.0 atm x 2.50 L = (175 g/39.95 gmole-1) x 0.08205 Latmdeg-1mol-1 x X X = 69.56 K
CHEM 161-1998 FINAL EXAM + ANSWERS CHAPTER 5 - GASES GAS LAWS 9. A container holds 3.0 g of hydrogen at a certain temperature and pressure. If the container is now evacuated and then filled with methane, CH4, at the same temperature and pressure, what mass of methane does it now hold? A. 16 g B. 19 g C. 13 g
D. 24 g
E. 40 g
P1V1/n1T1 = P2V2/n2T2 The problem explicity states that the reaction is done at constant temperature and pressure. Therefore, V1/n1 = V2/n2 But the reaction is done in the same container, i.e., the volume doesn't change. Therefore, V1 = V2. Therefore, 1/n1 = 1/n2 Therefore, n1 = n2; hence, everything drops out of the combined gas law except moles. Now the problem is easy. the number of moles of hydrogen = the number of moles of methane. moles of H2 = 3.0 g/2gmol-1 = 1.5 mol MW of CH4 = 16 gmol-1 16 gmol-1 x 1.5 mol = 24 g of CH4 This is kind of a neat problem.
CHEM 161-1998 FINAL EXAM + ANSWERS CHAPTER 5 - GASES GAS LAWS 20. Diallyl disulfide is the compound responsible for the odor of garlic. At 177 oC and 200 Torr a sample of diallyl disulfide vapor has a density of 1.04 g L-1. What is the molar mass of diallyl disulfide? A. 125 g mol-1 B. 176 g mol-1 C. 183 g mol-1 D. 104 g mol-1
E. 146 g mol-1
CHEM 161-1999 FINAL EXAM + ANSWERS CHAPTER 5 - GASES GAS LAWS 2. The density of oxygen is 1.43 g/L at STP. What is its density at 17C and 700 mm Hg? A. 1.12 g/L
B.
1.24 g/L C. 1.43 g/L D. 1.52 g/L E. 1.63 g/L
Since the conditions are changing, think combination gas law. However, we don't have a combination gas law involving density. We only have an ideal gas law involving density. "P = (D/MW)RT". So let's write a new combination gas law based on one of the modifications of the ideal gas law, "P = (D/MW)RT". PMW/DT = R P1MW 1/D1T1 = R P2MW 2/D2T2 = R P1MW 1/D1T1 = P2MW 2/D2T2 Conditions 1 Conditions 2 D = 1.43 g/L ?D T = 273K T = 290K P = 1 atm = 760 mm P = 700 mm The MW doesn't change since the gas is O 2 under both sets of conditions. MW 1 = MW 2 The new combination gas law is now simplified into: P1/D1T1 = P2/D2T2 760 mm/(1.43 g/L x 273 K) = 700 mm/(D 2 x 290 K) D2 = 1.24 g/L CHEM 161-1999 FINAL EXAM + ANSWERS CHAPTER 5 - GASES GAS LAWS 21. What is the molar mass of a gas if 4.50 grams of it occupies a volume of 4.00 L at 950 Torr and 182C ? A. 16.0 g/mol B. 26.5 g/mol 33.6 g/mol D. 30.2 g/mol E. 46.2 g/mol 1. CHEM 161-2000 EXAM 2 CHAPTER 5 - GASES
C.
GAS LAWS o A sample of gas has a volume of 50.0 mL at 20 C. To what temperature must this gas sample be heated in order to increase its volume to 55.0 mL? Assume the pressure of the gas remains constant. A. B. C. D. 87 o C 74 o C 63 o C 22 o C 49 o C
E.
8. CHEM 161-2000 EXAM 2 CHAPTER 5 - GASES GAS LAWS A 2.67 gram sample of gas X occupies 548 mL at a pressure of 710 Torr and a temperature of 300 K. What is the molar mass of gas X? A. B. C. 144 g mol-1 64.2 g mol-1 96.3 g mol-1 128 g mol-1 193 g mol-1
D.
E.
18. CHEM 161-2000 EXAM 2 CHAPTER 5 - GASES GAS LAWS A sample of gas has a volume of 60.0 mL. If the pressure of this gas is tripled and its Kelvin temperature is doubled, what is the new volume of the gas?
A. B. C. D.
360 mL 10.0 mL 160 mL 80.0 mL 40.0 mL
E.
Conditions 1 Conditions 2 V = 0.060 L ?V P 3P T 2T Assume moles of gas haven't changed, i.e., n1 = n2 Use the combination gas law when the conditions of the gas changes. P1V1/n1T1 = P2V2/n2T2 At constant n, this reduces to: P1V1/T1 = P2V2/T2 (P x 0.060 L)/T = (3P x XL)/2T X = 0.040 L
CHEM 161-2000 FINAL EXAM + ANSWERS CHAPTER 5 - GASES GAS LAWS 24. The vapor pressure of water at 25.0C is 23.8 Torr. Water is placed in a 1000 L container originally having no water vapor. What mass of water must evaporate in order to reach liquid/vapor equilibrium? (Assume the water vapor occupies the entire volume of the container) A. B. C. 1.28g 2.30 x 10-4 g 5.75 g 23.0 g 1.28 x 10-5 g
D.
E.
This problem is talking about water in the gas form. Consider this as an ideal gas. PV = (g/MW)RT (23.8 Torr/760 Torr) x 1000 L = (g/18.02gmol-1) x 0.08205 Latmdeg-1 mol-1 x (273 + 25 K) g = 23.1
CHEM 161-2000 FINAL EXAM + ANSWERS CHAPTER 5 - GASES GAS LAWS 43. A sample of 1.90 moles of a gas is at pressure (P) in a constant-volume container at 30oC. If the container is heated to 60oC, how many moles of gas must be added to the
container in order to double the original pressure? A. B. 3.80 mol 1.90 mol 1.56 mol 4.62 mol 5.70 mol Gas ?moles 2P 333K
C.
D. E. Gas 1.90 moles P 303K
Combination gas law: Combination gas law: (P1V1/n1T1) = (P2V2/n2T2) V1 = V2 (P1/n1T1) = (P2/n2T2) (P/(1.90 x 303)) = (2P/(X x 333)) X = 3.46 moles 3.46 - 1.90 = 1.56 moles needs to be added to go from original 1.90 to final 3.46 moles.
CHEM 161-2001 EXAM II + ANSWERS CHAPTER 5 - GASES GAS LAWS 6. Box A contains a mass of argon; box B contains an equal mass of neon. Both boxes have the same volume and are at the same temperature. Which of the following statements is (are) true? Assume Ar = 39.95g; therefore 1 mole Ar. 39.95g Ne approx. equal to 2 mole Ne. X. Both gases have the same pressure. Not true; 1 mole Ar and 2 moles of Ne provide twice the pressure for Ne. Y. The molecules in both boxes have the same average velocity. Not true; argon is a smaller molecule. Based on the effusion rule, argon would have a much greater velocity than neon. Z. Both gases have the same number of molecules. Not true. They would have the same number of molecules if they had the same number of moles, but the number of moles is different. Ne has twice as many molecules as Ar.
A. B.
Z only all are true
C.
D. E.
none are true Y only X only
CHEM 161-2001 EXAM II + ANSWERS CHAPTER 5 - GASES GAS LAWS 11. The molar volume of an ideal gas at a temperature of 273 K and a pressure of 1.00 atm is 22.4 Liters. Under which circumstances would the molar volume of an ideal gas always be less than 22.4 Liters?
A.
B. C. D. E.
T<273K, P>1.00atm T>273K, P< 1.00 atm the molar volume of an ideal gas is always 22.4 liters T>273K, P> 1.00 atm T<273K, P< 1.00 atm
At a temperature of <273K a gas is reduced in volume; >273 a gas expands. At a pressure of less than 1.00 atm a gas expands in volume; >1.00 atm a gas is reduced in volume. The only combination of conditions in which you can be certain that the gas will be reduced in volume would be <273K and >1 atm.
CHEM 161-2001 EXAM II + ANSWERS CHAPTER 5 - GASES GAS LAWS
21. An 11.0 gram sample of CO2 exerts a pressure of 50.0 kPa. What mass of CH4 must be added to the CO2 in order to increase the pressure to 75.0 kPa? Assume no CO2 escapes. A. B. 4.00 g 5.00 g 2.00 g 3.00 g 1.00 g
C.
D. E.
Gas 1 (CO2) 11.0 g (= 0.25 mol) 50.0 kPa P1V1/n1T1 = P2V2/n2T2 T1 = T2 V1 = V2 Therefore, P1/n1 = P2/n2
Gas 2 (CO2 + CH4) ?g CH4 0.25 mol + ?mol CH4 75.0 kPa
50.0 kPa/0.25 mol CO2 = 75.0 kPa/(X + 0.25 mol) X = 0.125 mol CH4 0.125 mol x 16 g/mol = 2.00 g CH4
CHEM 161-2001 FINAL EXAM WITH ANSWERS CHAPTER 5 - GASES GAS LAW 15. A mixture of 22 g of CO2 and 16 g of O2 and an undetermined quantity of H2 occupies 30 L at 1.0 atm and 27C. What is the mass of H2 present? A. 0.58 g B. 0.99 g C. 0.76 g
D. 0.44 g
E. 0.15 g CO2 O2 H2
22g 0.5 mol
16g 0.5 mol
Xg ? mol
n = PV/RT n = (1 x 30)/(0.08205 x 300) = 1.219 mol Therefore 0.219 mol H2 = 0.438 g H2
CHEM 161-2001 FINAL EXAM WITH ANSWERS CHAPTER 5 - GASES GAS LAWS 25. The density of a gas is 1.76 g/L at 1.00 atm and 10C. What is its density at 1.50 atm and 20C? A. 4.65 g/L B. 3.23 g/L C. 1.52 g/L D. 2.15 g/L
E. 2.55 g/L
Gas D=1.76 g/L P = 1 atm T = 283.15 K Gas D=? P = 1.5 atm T = 293.15 K
P = DRT/MW MW = DRT/P D1R1T1/P1 = D2R2T2/P2 D1T1/P1 = D2T2/P2 1.76 x 283.15/1 = X x 293.15/1.5 X = 2.55
CHEM 161-2001 SUMMER FINAL EXAM + ANSWERS CHAPTER 5 - GASES GAS LAWS 28. A rigid container holds 4.00 atm of SO2 and 3.00 atm of O2. When the following reaction is allowed to proceed to completion, what is the final pressure? 2SO2(g) + O2(g) -> 2SO3(g)
A. 5.00 atm
B. 7.00 atm C. 4.00 atm
D. 8.00 atm E. 2.00atm
In a lot of gas problems there's a direct relationship between pressure and moles. This applies in the present case and can be proven as follows: P1V1/n1T1 = P2V2/n2T2 This reaction takes place in a rigid container. Hence, the volume at the beginning equals the volume at the end, or V1 = V2. Since it doesn't say anything about the temperature, you can assume that the beginning and the end of the reaction are at equilibrium with the surroundings, and therefore there is no change in temperature. Hence, T1 = T2. Therefore, P1/n1 = P2/n2. Note that this means that there's a direct relationship between pressure and moles, so we can convert from one to the other. Since the reaction is 2SO2 + O2 -> 2SO3, let's say that it is 2 atm SO2 + 1 atm O2 -> 2 atm SO3 Let's convert this to: 4 atm SO2 + 2 atm O2 -> 4 atm SO3 Hence, 4 atm SO2 + 2 atm O2 will get converted into 4 atm SO3. But you started with 4 atm SO2 + 3 atm O2, i.e., 1 atm O2 in excess. Hence, at the end of the reaction you have 4 atm SO3 + 1 atm O2 = 5 atm.
CHEM 161-2001 SUMMER-EXAM I + ANSWERS ZUMDAHL CHAPTER 5 - GASES GAS LAWS 4. A sample of 1.90 moles of a gas is in a constant-volume container at 30oC and pressure P. How many moles of additional gas must be added to this container if the pressure doubles when the container is heated up to 60oC? A. 3.80 mol B. 1.90 mol
C. 1.56 mol
D. 4.62 mol E. 5.70 mol
CHEM 161-2001 SUMMER-EXAM I + ANSWERS ZUMDAHL CHAPTER 5 - GASES GAS LAWS 14. A balloon filled with helium gas at 20oC occupies 2.91 L at 1.00 atm. The balloon is immersed in liquid nitrogen at -196oC that applies an external pressure of 5.20 atm to it. What is the volume of the balloon in the liquid nitrogen?
A. 0.15 L
B. 2.1 L C. 5.5 L
D. 6.8 L E. 2.24 L The gas changes conditions. Hence, this is a typical combination gas law problem. He (cond. 1) He (cond. 2) 293K 273-196 = 77 K 2.91 L ?L 1.00 atm 5.20 atm Assume no moles of He were lost. Hence, n1 = n2 P1V1/n1T1 = P2V2/n2T2 P1V1/T1 = P2V2/T2 (1.00 atm x 2.91 L)/293 K = (5.20 atm x X)/77 K X = 0.147 L
2.
CHEM 161-2002 EXAM II + ANSWERS CHAPTER 5 - GASES GAS LAWS A gas has a volume of 50.0 mL at a temperature of 19 o C. What temperature is necessary to expand the gas to a volume of 60.0 mL? Assume pressure and amount remain the same. A. B. 92 o C 63 o C 77 o C 46 o C 23 o C
C.
D. E.
P1V1/n1T1 = P2V2/n2T2 At constant pressure and moles: V1/T1 = V2/T2 50 mL/292 K = 60 mL/X X = 350.4K 350.4K - 273K = 77.4K Common student error: Whereas you get the same answer by plugging in mL or converting to L, you must use Knot oC. oC won't give the correct answer.
10. 15. CHEM 161-2002 EXAM II + ANSWERS CHAPTER 5 - GASES GAS LAWS When NCl3(l) is placed in a previously evacuated container and then heated, it decomposes into N2(g) and Cl2(g), with a final total pressure of 2.00 atm. What is the partial pressure of N 2 in the container?
A.
1.00 atm 0.50 atm 1.25 atm 1.75 atm 1.50 atm
B.
C. D. E.
2NCl3 N2 + 3Cl2 These are two gases with changed/different conditions (different number of moles and different number of atmospheres). Hence, use the Combination gas law equation. (P1V1)/(n1T1) = (P2V2)/(n2T2) N2 PN2+Cl2 = 2.00 atm ?P V1 = V2 T1 = T2 X mol Combination gas law: P1/n1 = P2/n2 n1/n2 = P1/P2 Xmol/3Xmol = P1/P2 1/3 = P1/P2 Let Y = pressure of N2 ; therefore 3Y = pressure of Cl2. Y + 3Y = 2.00 atm Y = N2 = 0.50 atm 3X mol Cl2
16. CHEM 161-2002 EXAM II + ANSWERS CHAPTER 5 - GASES GAS LAWS A 542 mL sample of a gas has a mass of 1.50 g at a pressure of 1.20 atm and a temperature of 25 o C. What is the molar mass of this gas? A. B. C. 78.2 g/mol 234 g/mol 156 g/mol 56.4 g/mol 117 g/mol
D.
E.
PV = nRT n = g/MW PV = gRT/MW MW = gRT/PV = (1.50 g x 0.08205 Latmdeg -1 mol-1 x (273.15 + 25))/(1.20 atm x 0.542 L) = 56.42 g/mol
CHEM 161- 2002 FINAL EXAM + ANSWERS CHAPTER 5 - GASES GAS LAWS 32. The pressure of a sample of gas halves while its Kelvin temperature triples. The volume of this gas will: A. increase to 3/2 of its original volume C. decrease to 2/3 of its original volume
B.
D. E.
increase to 6 times its original volume decrease to 1/6 of its original volume remain the same
Use the combination gas law. P1V1/n1T1 = P2V2/n2T2 n1 = n2 P1V1/T1 = P2V2/T2 Let P1 = X; therefore, P2 = 0.5X Let T1 = Y; therefore, T2 = 3Y
1XV1/1Y = 0.5XV2/3Y V2/V1 = 6 CHEM 161- 2002 FINAL EXAM + ANSWERS CHAPTER 5 - GASES GAS LAWS 36. What is the density of SF6 gas at a temperature of 20.0 C and a pressure of 925 mmHg? A. 6.88 g/L B. 1.72 g/L C. 3.44 g/L
D.
E.
7.39 g/L 4.71 g/L
PV = nRT PV = (g/MW)RT D= g/L PV = (DL/MW)RT P = (D/MW)RT D = (PxMW)/RT MWSF6 = 146.07 D = ((925 mm/760 mm atm-1) x 146.07 gmol-1)/(0.08205 Latmdeg-1 mol-1 x (273.15 + 20.0 deg)) D = 7.39 g/L
CHEM 161- 2002 FINAL EXAM + ANSWERS CHAPTER 5 - GASES GAS LAWS 40. How many molecules are in 1.00 cm3 of CO2 gas at a temperature of 20.0C and a pressure of 775 mmHg? A. 1.42 x 1021 molecules B. 5.80 x 1017 molecules C. 3.47 x 1020 molecules D. 1.94 x 1022 molecules
E.
2.55 x 1019 molecules
Strategy: Use the ideal gas equation to calculate moles of CO 2; then convert moles of CO2 to molecules of CO2. PV = nRT n = PV/RT n = ((775 mm/760 mmatm-1) x 0.001 L)/(0.08205 Latmdeg -1 mol-1 x (273.15 + 20.0 deg) n = 4.24 x 10-5 mol 4.24 x 10-5 mol x (6.022 x 1023 molecules/mol) = 25.5 x 1018 = 2.55 x 1019 molecules
CHEM 161- 2002 FINAL EXAM + ANSWERS CHAPTER 5 - GASES GAS LAWS 46. A 32.0 g sample of CH4 gas in a container exerts a pressure of 1.00 atm. What mass of CO2 gas must be added to the container in order to increase the total pressure to 1.50 atm? Assume no CH4 escapes. A. 33.0 g 44.0 g 11.0g 55.0 g 22.0 g
B.
C. D. E.
Since the conditions are changing, use combination gas law. P1V1/n1T1 = P2V2/n2T2 Since the volume of the container is constant, then volume is constant. Since there was no mention of T, assume T is constant. P1/n1 = P2/n2 n1 = 32 g CH4/16gmol-1 = 2 mol CH4 1.00 atm/2 mol gas = 1.50 atm/X mol gas X = 3 mol total gas Since there was already 2 mol of gas, then 1 more mol of gas is needed. One mol CO2 = 44 g.
2.
CHEM 161-2003 EXAM II + ANSWERS CHAPTER 5 - GASES GAS LAWS A container with NH4CO2NH2(s) decomposes into NH3(g) and CO2(g). After decomposition is complete, the total pressure in the container is 0.600 atm. What is the
partial pressure of the NH3? NH4CO2NH2(s) 2NH3(g) + CO2(g) A. B. C. D. E. NH3 ?atm V1 = V2 2X mol T1 = T2 P1V1/n1T1 = P2V2/n2T2 "1" NH3; "2" is CO2. P1/n1 = P2/n2 P1/P2 = n1/n2 Therefore, P1/P2 = 2X/1X = 2/1 According to the law of partial pressures, the sum of the individual pressures = the total pressure. 2X + 1X = 0.600 atmospheres Therefore X = 0.200 atm. 2X = 0.400 atm. 1X mol 0.500 atm 0.300 atm 0.400 atm 0.200 atm 0.100 atm CO2 P1+P2 = 0.600 atm
11.
CHEM 161-2003 EXAM II + ANSWERS CHAPTER 5 - GASES GAS LAWS The pressure of a gas is doubled while its Kelvin temperature is tripled. The volume of this gas will:
A.
B.
increase to 3/2 of its original volume go down to 1/6 of its original volume
C. D. E.
remain the same go up 6 times decrease to 2/3 of its original volume
P1V1/n1T1 = P2V2/n2T2 P1 = X; P2 = 2X T1 = Y; T2 = 3Y n1 = n2 P1V1/T1 = P2V2/T2 XV1/Y = 2XV2/3Y V1 = 2V2/3 V2/V1 = 3/2
21.
CHEM 161-2003 EXAM II + ANSWERS CHAPTER 5 - GASES GAS LAWS A 2.25 g sample of gas X has a volume of 534 mL at 20 o C and 0.950 atm. What is the molar mass of this gas? A. B. g mol g 176 mol g 107 mol g 75.0 mol g 151 mol 50.2
C.
D. E.
PV = nRT PV = (g/MW)RT MW = gRT/PV MW = (2.25 g x 0.08205 Latmdeg -1 mol-1 x 293 K)/(0.950 atm x 0.534 L) = 106.6 gmol-1
5.
Chem 161-2003 Final exam Chapter 5 - Gases
Gas laws/stoichiometry A container at 500 K holds Cl2(g) and Ar(g), each with a partial pressure of 0.10 atm. The temperature is increased to 5000 K, and at that high temperature, all the Cl2 breaks down into Cl atoms. What is the total pressure in the container at 5000 K? The volume of the container is unchanged. A. B. 5.0 atm 2.0 atm 3.0 atm 1.0 atm 4.0 atm Conditions 1 500 K Cl2(g) Ar(g) 0.10 atm 0.10 atm V1 = Conditions 2 5000 K Cl(g) Cl(g) Ar(g) ?Total atm V2
C.
D. E.
This is a combination gas law problem. P1V1/n1T1 = P2V2/n2T2 P1 = 0.2 atm; P2 = X atm V1 = V2 T1 = 500; T2 = 5000 n1 = ?; n2 = ? Therefore, 0.2/(n1x500) = X/(n2x5000) But there are too many unknowns (P2, n1 and n2) to solve this equation. So can we find the moles of gas in condition 1 and the moles of gas in condition 2 and then solve for P2? (If we realize that in the first container, in which the Cl2 and Ar are at the same temperature, volume and pressure, then the moles of Cl2 and Ar must be identical, then we can easily determine n1 and n2, but if we don't realize this then we have to solve for the moles of Cl2 and Ar.) Let's use the combination gas law to get information on the relative moles of Cl2 and Ar gas under condition 1. P1V1/n1T1 = P2V2/n2T2 Call "1" the Cl2 and "2" the Ar T1 = 500 K; T2 = 500 K P1 = 0.1 atm; P2 = 0.1 atm Same container; therefore V1 = V2 Therefore the pressure, volume and temperature cancel out and 1/n1 = 1/n2 Therefore n1 = n2 Therefore, the moles of Cl2 = the moles of Ar Now let's set up the combination gas law for conditions 1 and conditions 2.
P1V1/n1T1 = P2V2/n2T2 In conditions 2, we assume that the Ar doesn't dissociate because it's an inert gas, but the Cl2 does. Since there were equal moles of gas before Cl2 dissociation (i.e., X + X), then after dissociation there are twice as many moles of Cl as Ar (i.e., 2X + X), so the ratio of moles is 3:2. n1 = 2X; n2 = 3X V1 = V2. T1 = 500 K; T2 = 5000 K P1 = 0.1 atm + 0.1 atm = 0.2 atm; P2 = Y P1/n1T1 = P2/n2T2 0.2 atm/2Xx500 = Y/3Xx5000 Y = P2 = 3.0 atm
17.
Chem 161-2003 Final exam Chapter 5 - Gases Gas laws What volume of O2, measured at a pressure of 1.00 atm and a temperature of 25 o C, would react completely in the combustion of 15.0 g C4H10? The reaction is: 2 C4H10(g) + 13 O2(g) C (g) + 10 H2O(l) g Molar mass of C4H10 = 58.1 mol A. B. C. D. 6.32 L 12.6 L 3.16 L 25.3 L 41.1 L C (g) + 10 H2O(l)
E.
2 C4H10(g) + 13 O2(g) 15.0 g ?L MW=58.1 1.00 atm 298 K
Plan: gC4H10 molC4H10 molO2 LO2 15.0 g/58.1gmol-1 = 0.2582 mol C4H10 0.2582 mol C4H10 x (13 mol O2/2 mol C4H10) = 1.6783 mol O2 V = nRT/P V = (1.6783 mol x 0.08205 Latmdeg -1 mol-1 x 298 K)/1.00 atm V = 41.04 L
20.
Chem 161-2003 Final exam
Chapter 5 - Gases Gas laws Box A contains a mass of argon; box B contains an equal mass of neon. Both boxes have the the same volume and are at the same temperature. Which of the following is true? X. Both gases have the same pressure. Y. The molecules in both boxes have the same average velocity. Z. Both gases have the same density. A. B. none are true all are true Z only X only Y only Box A Xg Ar X/39.948 mol A V1 = T1 = Box B X g Ne X/20.180 mol Ne V2 T2
C.
D. E.
P1V1/n1T1 = P2V2/n2T2 P1/n1 = P2/n2 P1/(X/39.948) = P2/(X/20.180 mol) 39.948 P1 = 20.180 P2 A. Both gases don't have the same pressure. B. Since the average velocity is RMS = (3RT/MW) , and the temperature of the two gases is identical but the MW's are different, then the average velocities will be different. C. Density = g/V. Since the masses are the same, and the volumes are the same then the densities are the same.
REACTION STOICHIOMETRY
38 Chem 161-2006 Final Exam Chapter 5 Gases Reaction stoichiometry Nitrogen trichloride gas exerting a pressure of 5.40 atm is added to a 1.50L container. After a period of time, 35% of the nitrogen trichloride decomposes to N2(g) and Cl2(g). Calculate the total pressure in the container. A. B. C. D. 3.24atm 1.62atm 4.38atm 5.75atm 7.29atm + 3Cl2(g)
E
E.
2NCl3(g) N2(g) 5.40 atm 35% decomposed V = 1.50L T = 303K
NCl3 left = 0.65 x 5.40 atm = 3.51 atm P1V1/n1T1 = P2V2/n2T2 P1/n1 = P2/n2 At constant V and T, P is directly related to moles. This is one part of Dalton's law. Therefore, atmospheres of N2 formed = (0.35 x 5.40) x (1mol N2/2molNCl3) = 0.945 atm Atmospheres of Cl2 formed = (0.35 x 5.40) x (3molCl2/2molNCl3) = 2.835 atm Total atmospheres = 3.51 + 0.945 + 2.835 = 7.29 atm
CHEM 161-2006 EXAM II HILL & PETRUCCI CHAPTER 5 GASES REACTION STOICHIOMETRY
25.
Element X reacts with hydrochloric acid to produce hydrogen, according to the following full-formula equation: 2X(s) + 6HCl(aq) 2XCl3(aq) +3H2(g)
If 1.27 g of X reacts with excess HCl, and produces 400.0 mL of H 2 at 1.00 atm pressure and temperature of 20 oC, calculate the molar mass of X.
g mol g (b) 45.0 mol (c) 115 g mol g (d) 27.9 mol g (e) 139 mol
(a) 69.7
2X(s) + 6HCl(aq) 1.27g XS ?MW
2XCl3(aq) +
3H2(g) 0.400L 1.00 atm 293K
Plan: PVTH2 molH2 molX MWX PV = nRT n = PV/RT n = (1.00atm x 0.400L)/(0.08205Latmdeg -1 mol-1 x 293K) = 0.01664 mol H2 0.01664 mol H2 x (2molX/3molH2) = 0.01109 mol X MW = g/mol = 1.27g/0.01109mol = 114.5 g/molX
CHEM 161-2006 EXAM II HILL & PETRUCCI CHAPTER 5 GASES REACTION STOICHIOMETRY
20.
Consider the reaction: 2NO(g) + O2(g)
2NO2(g)
If 0.176 mol NO(g) and 0.176 mol O2(g) are placed in a 5.00 L container at 298 K, and the reaction proceeds as far as possible, calculate the final pressure in the container. (a) 0.861 atm (b) 1.72 atm (c) 2.58 atm
(d)
2NO(g)
1.29 atm (e) 3.44 atm + O2(g) 2NO2(g)
0.176mol 5.00L 298K
0.176mol 5.00L 298K
Limiting reactant: NO, because 0.176 mol NO requires 0.088 mol O2, so O2 is in excess. Since the moles of NO2:NO is 2:2, or 1:1, then 0.176 mol NO2 will form, in 5.00L and at 298K. PV = nRT P = nRT/V P = (0.176mol x 0.08205Latmdeg -1mol-1 x 298deg)/5.00L = 0.861 atm of NO2. Note that the excess O2 also contributes to the final pressure of the container. Moles of excess O2 = 0.176 moles - 0.088 mol = 0.088 mol P = nRT/V P = (0.088mol x 0.08205Latmdeg -1mol-1 x 298deg)/5.00L = 0.431 atm of excess O2. Total final pressure = 0.861 atm NO2 + 0.431 atm O2 = 1.29 atm
12
Chem 161-2005 Hourly Exam II Chapter 5 - Gases Reaction stoichiometry Consider the reaction: 2KClO3(s) 2KCl(s) + 3O2(g) If 0.0500 mol of KClO3 reacts, what volume of O2, measured at 20 C and 1.00 atm, will be collected? A. 1.20 L
B.
1.80 L C. 2.40 L D. 3.00 L E. 3.60 L 2KClO3(s) 0.0500 mol 2KCl(s) + 3O2(g) ?V 293K 1.0 atm
0.0500 mol KClO3 x (3 mol O2/2 mol KClO3) = 0.075 mol O2 PV = nRT 1.0 atm x X = 0.075 mol x 0.08205 Latmdeg -1 mol-1 x 293K V = 1.80 L
34.
CHEM 161- 2004 FINAL EXAM + ANSWERS
CHAPTER 5 - GASES REACTION STOICHIOMETRY Assume that 1.20 g of Ne is added to a 1.50 L container holding some Ar gas, and no Ar gas escapes. The total pressure in the container (from both the Ne and Ar) is 2.00 atm at 20o C. Calculate the mass of argon in the bulb. Assume no change of volume.
A.
B. C. D. E.
2.61 g 3.49 g 2.07 g 4.14 g 1.42 g
Ne Ar 1.20 g x (1 mol/20.18 g) = 0.0595 mol X mol 1.50 L 1.50 L 293K 293K PNe + PAr = 2.00 atm Ne: PV = nRT P = nRT/V P = 0.0595 mol x 0.08205 Latmdeg -1 mol-1 x 293K/1.50 L P = 0.954 atm PAr = 2.00 - 0.954 = 1.046 atm nAr = PV/RT n = (1.046 atm x 1.50 L)/(0.08205 Latmdeg -1 mol-1 x 293K) = 0.0653 mol Ar g = 0.0653 mol x 39.95 g/mol = 2.61 g Ar This problem probably also could be done by using the combination gas law.
ZUMDAHL 5TH EDITION CHEM 161-2002 RECITATION 6TH WEEK CHAPTER 5 - GASES REACTION STOICHIOMETRY
49 Calculate the volume of O2, at STP, required for the complete combustion of 125 g octane (C8H18) to CO2 and H2O.
C8H18 + O2 CO2 + H2O C8H18 + 12.5O2 8CO2 + 9H2O 2C8H18 + 25O2 16CO2 + 18H2O 125 g C8H18 x (1 mol C8H18/114.26 g C8H18) x (25 mol O2/2 mol C8H18) = 13.68 mol O2 P V = n R T 1 atm X 13.68 mol 273.15 K 1 atm x X L = 13.68 mol O2 x 0.08205 Latmdeg-1mol-1 x 273.15 K X = 307 L O2
CHEM 161-1999 FINAL EXAM + ANSWERS CHAPTER 5 - GASES REACTION STOICHIOMETRY 10. When 21.7 grams of MgCO3.xH2O was treated with excess acid, 3.51L of CO2 were collected at STP. What is the value of (x) ? ( Molar mass of MgCO3 = 84 g/mol ) A. 2
B.
3 C. 4 D. 5 E. 6 MgCO3.xH2O + 21.7 g H2CO3 2H+ XS H2CO3 H2O + Mg2+ + xH2O
CO2 + 3.51 L 273K 1 atm
Plan: LCO2 molCO2 molMgCO3.xH2O MWMgCO3.xH2O PV = nRT n=PV/RT = (1 atm x 3.51 L)/(0.08205Latmdeg -1 mol-1 x273K) = 0.1567 mol CO2 0.1567 mol CO2 x (1 mol MgCO3.xH2O/mol CO2) = 0.1567 mol MgCO3.xH2O MW MgCO3.xH2O = 21.7 g MgCO3.xH2O /0.1567 mol MgCO3.xH2O = 138.48g MgCO3.xH2O MW MgCO3 = 24.30 + 12.01 + 48.00 = 84.31 g Therefore, mass of H2O = 138.48 - 84.31 = 54.17 g MW H2O = 18.02 g. 54.17 g/18.02 gmol-1 = 3.01 mol H2O Therefore, X = 3 mol H2O
CHEM 161-2000 FINAL EXAM + ANSWERS CHAPTER 5 - GASES REACTION STOICHIOMETRY 12. The law of combining volumes states that at a given temperature and pressure, the reaction of 50.0 mL of N2(g) with 150 mL of H2(g) to form NH3(g) will produce A. B. C. D. 250 mL of NH3(g) 50.0 mL of NH3(g) 200 mL of NH3(g) 150 m Lof NH3(g) 100 mL of NH3(g) 3H2 2NH3
E.
N2 +
50 ml
100 ml
?ml
Combination gas law: (P1V1/n1T1) = (P2V2/n2T2) At constant pressure and temperature: (V1/n1) = (V2/n2) V1 and n1 can either be 50 ml and 1 mole (for nitrogen) or 150 ml and 3 moles (for ammonia). (50 ml/1 mol) = (X ml/2 mol) X = 100 ml
CHEM 161-2000 FINAL EXAM + ANSWERS CHAPTER 5 - GASES REACTION STOICHIOMETRY 22. Consider the following reaction: 2NO(g) + O2(g) -> 2NO2(g) A 0.020 moles of NO(g) is mixed with 0.050 moles of O2(g) in a 1.71 L container at 25 C. What is the partial pressure of NO2(g) in the container after the reaction goes to completion? A. 0.333 atm 0.286 atm 0.500 atm 0.400 atm 1.00 atm + O2(g) -> 0.050 mol 1.71L 298K 2NO2(g)
B.
C. D. E.
2NO(g) 0.020 mol 1.71L 298K
?P Limiting reactant: Stoichiometric ratio of NO to O2 = 2:1 Actual ratio of NO to O2 ==0.020 to 0.050 = 0.4:1 Limiting reactant = 0.020 mol NO Based on the stoichiometry, 0.020 mol NO will form 0.020 mol NO 2. Use the combination gas law for two different gases or one gas under two different conditions: P1V1/n1T1 = P2V2/n2T2 Since 0.020 mol NO will form 0.020 mol NO2, then n1 = n2. T1 = T2
V1 = V2 Therefore, the combination gas law simplifies to P 1 = P2 P1, which is the pressure of the NO, can be found from the ideal gas law: PV = nRT P = nRT/V P = (0.020 mol x 0.08205 Latmdeg -1mol-1 x 298K)/1.71 L P = 0.286 atm
CHEM 161-2000 FINAL EXAM + ANSWERS CHAPTER 5 - GASES REACTION STOICHIOMETRY 33. A 100.0 g sample of aqueous hydrogen peroxide solution decomposes over time producing 2.17 liter of O2(g) at 25oC and 755 Torr. 2H2O2(aq) 2H2O(l) + O2(g) What must have been the mass percent of H2O2 in the solution? A. 3.00 % H2O2 6.00 % H2O2 9.00 % H2O2 12.0 % H2O2 15.0% H2O2 2H2O2(aq) 100g solution 2H2O(l) + O2(g) 2.17 L 298K (755/760) atm
B.
C. D. E.
PV = nRT n = PV/RT n = ((755/760) x (2.17))/(0.08205 x 298) = 0.08817 moles O2 Therefore, 0.08817 moles O2 x (2mole H2O2/1moleO2) = 0.1763 moles H2O2 0.1763 moles x 34 g/mole = 5.995 g H2O2 5.995g/100g = 6.0% H2O2
CHEM 161-2001 EXAM II + ANSWERS CHAPTER 5 - GASES REACTION STOICHIOMETRY 10. When ethane gas, C2H6, burns, the products of the combustion are carbon dioxide and water. What volume of O2 is required to react completely with 4.00 mL of ethane? Assume all gases are at the same temperature and pressure. A. 24.0 mL B. 18.0mL
C
D. E.
14.0 mL 9.00 mL 6.00 mL
C2H6 + 7/2 O2 -> 2CO2 + 3H2O At the same temperature and pressure, one mole of C2H6 and one mole of O2 occupy equal volumes. Hence, if one mole of C2H6 occupies 4.00 mL, then one mole of O2 would occupy 4.00 mL. But there are 3.5 moles O2. Therefore, O2 occupies 4.00 x 7/2 = 14.00 mL.
CHEM 161-2001 EXAM II + ANSWERS CHAPTER 5 - GASES REACTION STOICHIOMETRY 16. Metal M reacts with HCl to produce hydrogen gas and a solution of MCl3. If 0.724 grams of M reacts with excess HCl, producing 375 mL of pure H 2 at 1.00 atm and 20oC, what is the molar mass of M? 2M(s) + 6HCl(aq) ----> 3H2(g) + 2MCl3(aq) A. B. C. D. 27.0 g/mol 139 g/mol 95.9 g/mol 45.0 g/mol 69.7 g/mol
E.
Plan: Find moles of H2; convert that into moles M; then MW find M. PV = nRT; n = PV/RT; n = (1 x 0.375)/(0.08205 x 293) = 0.0156 moles H 2. 0.0156 moles H2 x (2 moles M/3 moles H2) = 0.0104 moles M MW = (g/moles) = 0.724/0.0104 = 69.6 g/mole
CHEM 161-2001 SUMMER EXAM II + ANSWERS CHAPTER 5 - GASES REACTION STOICHIOMETRY 9. What will the final pressure be (in Torr) if 1.00 mol of O 2 is allowed to react with excess C6H6(l) in a rigid 1.000 L container at 25oC? (The original question was written as, "What will the final pressure be (in Torr) if 1.20 mol of O2 is allowed to react with excess C6H6(l) in a rigid 1.000 L container at 25oC?", but the answer to this problem is inconsistent with the answer key. I changed the text to something that is consistent with "D".) A. 22,302 Torr B. 13,621 Torr C. 19,563 Torr
D. 14,868 Torr
E. 33,452 Torr C6H6(l) + 15/2 O2(g) 6CO2(g) + 3H2O(l) 2C6H6(l) + 15 O2(g) 12CO2(g) + 6H2O(l) XS 1.00 mol ?P 1L 298K Strategy: moles O2 moles CO2; then P = nRT/V 1.00 mol O2 x 12 mol CO2/15 mol O2 = 0.80 mol CO2 P P V = 1.000 L n 0.8 mol R 0.08205 T 273.15 + 25 K
P = 19.57 atm 19.57 atm x (760 torr/atm) = 14868 Torr
6.
CHEM 161-2002 EXAM II + ANSWERS CHAPTER 5 - GASES REACTION STOICHIOMETRY When an automobile's airbag is activated, sodium azide, NaN 3(s), decomposes into sodium and N2 gas. What mass of NaN3 is needed to produce 50.0 L of N2 at a pressure of 1.00 atm and a temperature of 20 o C? Always begin problems with a balanced equation. 2NaN3 2Na + 3N2 MW NaN3 = 23 + 42 = 65 PV = nRT n = PV/RT = (1 atm x 50 L)/((0.08205 Latmdeg -1mol-1) x (293.15K)) = 2.08 mol 2.08 mol N2 x (2 mol NaN3/3 mol N2) x (65 g NaN3/1 mol NaN3) = 90.1 g NaN3 A. 135 g
B.
C. D. E.
90.1 g 241 g 172 g 202 g
CHEM 161-1998 FINAL EXAM + ANSWERS CHAPTER 3 - STOICHIOMETRY REACTION STOICHIOMETRY 4. Calculate the mass of H2O produced from the reaction of 600 mL of O2(g) at 27.0oC and 2.00 atm with excess ammonia assuming 80.0% yield. 4 NH3(g) + 5 O2(g) A. 0.235 g
Pt
4 NO(g) + 6 H2O(g)
B. 0.842 g
C. 0.662 g D. 0.753 g E. 0.987 g 4 NH3(g) XS + 5 O2(g) 0.600 L 300 K 2.00 atm
Pt
4 NO(g)
+
6 H2O(g) ?g 80% yield
Strategy: LO2 molO2 molH2O gH2O %yield H2O PV = nRT n = PV/RT n = 2.00 atm x 0.600 L/0.08205Latmdeg -1mol-1 x 300K n = 0.04875 mol O2 0.04875 mol O2 x (6 mol H2O/5 mol O2) = 0.0585 mol H2O 0.0585 mol H2O x (18.02 g H2O/mol H2O) = 1.054 g H2O But only 80% yield: 0.80 x 1.054 = 0.843 g H2O
CHEM 161-1998 FINAL EXAM + ANSWERS CHAPTER 3 - STOICHIOMETRY REACTION STOICHIOMETRY 42. Consider the following reaction: X + 2 HCl XCl2 + H2 If 0.0729 g of X reacts with excess HCl to produce 73.9 mL of H 2 at 1.00 atm and 300 K, the element X is
A. Mg
B. Cu C. Zn D. Fe E. Ni X + 0.0729g 2 HCl XS XCl2 + H2 0.0739 L 1.00 atm 300 K
Plan: LH2 molH2 molX MWX PV = nRT n = PV/RT n = (1.00 atm x 0.0739 L)/(0.08205Latmdeg -1 mol-1 x 300K) = 0.003002 mol H2 0.00302 mol H2 x (1 mol X/1 mol H2) = 0.00302 mol X 0.0729 g X/0.00302 mol X = 24.28 g X/mol X = Mg
CHEM 161-2001 FINAL EXAM WITH ANSWERS CHAPTER 3 - STOICHIOMETRY REACTION STOICHIOMETRY 20. What mass of oxygen will have the same volume as 4.0 g of hydrogen gas at the same temperature and pressure?
A.64 g
B.32 g C.16 g D.4.0 g E.8.0 g moles = 4.0/2 = 2 mol 2 mol O2 = 64 g
CHEM 161-2001-HOURLY EXAM I + ANSWERS ZUMDAHL CHAPTER 3 - STOICHIOMETRY REACTION STOICHIOMETRY 9. The direct high temperature reaction of H2 (g) and N2 (g) produces NH3 (g). What mass of H2(g) is required to form 5.00 x 103 mol of NH3 ? A.7.00x104 g H2
B.1.51x104 g H2
C.7.50x103 g H2 D.1.01xl04 g H2 E.5.04xlO3 g H2 3H2 + N2 2NH3 Strategy: molNH3 molH2 g H2 5 x 103 mol NH3 x (3 mol H2/2 mol NH3) x (2.02 g H2/mol H2) = 1.52 x 104 g H2
16.
CHEM 161-2003 EXAM II + ANSWERS CHAPTER 5 - GASES REACTION STOICHIOMETRY When heated, potassium chlorate, KClO3, decomposes into KCl and O2. 2KClO3(s) 2KCl(s) + 3O2(g)
If 50.0 mL of O2 is collected at 25 o C and 1.00 atm, what amount of KClO3 decomposed? A. B. C. D. 0.00307 mol 0.00152 mol 0.00102 mol 0.00204 mol 0.00136 mol 2KCl(s) + 3O2(g) 0.050 L 298 K 1.00 atm
E.
2KClO3(s) ?n
Plan: L O2 mol O2 mol KClO3 PV = nRT n = PV/RT n = (1.00 atm x 0.050 L)/(0.08205 Latmdeg -1 mol-1 x 298 K) = 0.002045 mol O2 0.002045 mol O2 x (2 mol KClO3/3 mol O2) = 0.00136 mol KClO3
CHEM 161-1998 FINAL EXAM + ANSWERS CHAPTER 3 - GASES REACTION STOICHIOMETRY 7. Consider the following reaction: CO(g) + 3 H2(g) CH4(g) + H2O(g)
If 2.00 mol of CO and 2.00 mol of H2 are mixed in a 30.0 L sealed vessel and the reaction proceeds to completion, what is the pressure inside the container if the final temperature is 400 K? A. 1.86 atm
B. 2.92 atm
C. 3.55 atm D. 4.78 atm E. 5.95 atm CHEM 161-1999 FINAL EXAM + ANSWERS CHAPTER 5 - GASES REACTION STOICHIOMETRY 6. Methane is burned in oxygen according the reaction below: CH4(g)+2O2(g) CO2(g)+2H2O(g) 2.00 moles of methane and 2.00 moles of oxygen are mixed and the reaction proceeds as far as possible. If the final pressure of the reaction mixture was 3.56 atm, what is the partial pressure of the H2O(g)?
A.
B. C. D. E.
1.78 atm 1.19 atm 7.12 atm 3.56 atm 2.38 atm
PARTIAL PRESSURE
CHEM 161-2006 EXAM II HILL & PETRUCCI CHAPTER 5 GASES PARTIAL PRESSURE (MIXTURES OF GASES)
4. A sample of NH3(g) is heated and completely decomposes to N2 and H2 in a closed container. When the reaction is complete, the total pressure in the container is 868 Torr. What is the partial pressure of N2 in the container? (a) 434 Torr
(b)
217 Torr (c) 289 Torr (d) 651 Torr (e) 326 Torr
2NH3 N2 + 3H2 0 Torr 868 Torr Dalton's law: At constant temperature and volume, (n1/P1) = (n2/P2) Also, Dalton's law is P1 + P2 + P3 . . . = PT Total pressure for 4 moles of gas is 868 torr. Therefore 1 mole provides 217 Torr. Since the ratio of N2 to H2 is 1:3, then N2 is 217 Torr, and H2 is 651 Torr.
25
Chem 161-2005 Hourly Exam II Chapter 5 - Gases Partial pressure A gaseous mixture contains 1.00 mol of O2 and 2.00 mol of CO2, exerting a total pressure of 1.50 atm. What amount of CH4(g) must be added to the mixture in order to increase the total pressure to 2.00 atm? Assume no gas escapes, and there is no change of total volume. A. 0.50 mol B. 0.75 mol
C.
1.00 mol D. 1.25 mol E. 1.50 mol O2 1 mol 1.50 atm 2.00 atm P1V1/n1T1 = P2V2/n2T2 The volume is constant, and assume that the temperature is constant. P1/n1 = P2/n2 Therefore, the pressure of each gas is directly related to the number of moles of the gas, respectively. Since there is 1 mol O2 and 2 mol CO2, and their total pressure is 1.50 atm, then the O2 must exert 0.50 atm pressure and the CO2 must exert 1.00 atm pressure. Hence, there is 0.50 atm per mole of gas. Since each gas behaves exactly like any other gas, then we have to determine how many moles of gas (CH4) is required for the additional (2.00 1.50 = ) 0.50 atm. Since there is 0.50 atm per mole of gas, then the additional 0.50 atm would require 1.0 mol CH4. CO2 2 mol CH4 ?X mol
6.
Chem 162-2004 Hourly exam II Chapter 5 - Gases Parial pressure Consider the reaction: 2N2O5(g)
4NO2(g) + O2(g)
Some N2O5(g) is placed in a container, exerting an initial pressure of 0.50 atm before reaction. Some of the N2O5 reacts as shown in the above equation, reducing its partial pressure to 0.40 atm. What is the resultant total pressure in the container? Assume constant temperature.
A.
1.4 atm 0.65 atm 0.95 atm 0.80 atm 0.35 atm
B.
C. D. E.
0.40 atm of N2O5 remained untouched. 0.10 atm of N2O5 was converted into NO2 and O2. The question is, how much NO2 and O2 can come from 0.10 atm of N2O5 ? 2N2O5 4NO2 1O2 0.10 atm x atm y atm Volumes are equal Temperatures are equal Moles are in a ratio of 2:4:1 First do N2O5 vs NO2 P1V1/n1T1 = P2V2/n2T2 P1/n1 = P2/n2 Let 2Y = moles of N2O5, and 4Y = moles of NO2 0.10 atm/2Y = X atm/4Y X = 0.2 atm of NO2. The same thing should apply for N2O5 vs O2. P1V1/n1T1 = P2V2/n2T2 P1/n1 = P2/n2 0.10 atm/2Y = X atm/1Y X = 0.05 atm of O2. Total pressure: 0.4 atm N2O5 + 0.2 atm NO2 + 0.05 atm O2.
16.
Chem 162-2004 Hourly exam II
Chapter 5 - Gases Partial pressure N2(g) and H2(g) are in two separate containers as shown in the diagram below
The valve between the two containers is opened allowing the gases to mix. What is the partial pressure of nitrogen in the system? Assume the temperature remains constant. A. B. C. 1.00 atm 0.50 atm 2.25 atm 0.90 atm 1.50 atm
D.
E.
The H2 works independently of the N2. So the presence of the H2 is irrelevant. The conditions of N2 are changing. Consider the combination gas law. P1V1/n1T1 = P2V2/n2T2
Initial conditions Final conditions V = 3.00 L V = 5.00 L P = 1.50 atm ?P Ti = Tf ni = nf Simplifying the combination gas law: P1V1 = P2V2 1.50 atm x 3.00 L = X atm x 5.00 L X = 0.9 atm
ZUMDAHL 5TH EDITION
CHEM 161-2002 RECITATION 6TH WEEK CHAPTER 5 - GASES PARTIAL PRESSURE 65
The partial pressure of CH4(g) is 0.175 atm and that of O2(g) is 0.250 atm in a mixture of the two gases. What is the mole fraction of each gas in the mixture? PV = nRT At constant temperature and volume, P 1/n1 = P2/n2 Therefore, the ratio of the moles = 0.175 CH4 : 0.250 O2 Mole fraction CH4: 0.175 CH4/(0.175 CH4 + 0.250 O2) = 0.412 Mole fraction O2: 0.250 O2/(0.175 CH4 + 0.250 O2) = 0.588
12. CHEM 161-2000 EXAM 2 CHAPTER 5 - GASES PARTIAL PRESSURE When ammonium nitrate, NH4NO3(s), is heated, it decomposes to form N2O(g) and H2O(g) . A sample of ammonium nitrate is placed in a previously evacuated reaction vessel. After decomposition, the total pressure in the vessel is 600 Torr. What is the partial pressure of the water vapor?
A.
B. C. D. E.
400 Torr 500 Torr 300 Torr 200 Torr 600 Torr N2O(g) + 2H2O(g) ?P Total P = 600 Torr
NH4NO3(s) sample 0 atm
Solve this using the combination gas law. Let gas "1" = N2O and gas "2" = H2O. P1V1/n1T1 = P2V2/n2T2 What is equal? What is different? There is no information regarding the pressures, so we don't know that they are the same. Furthermore, since they are asking for the pressure of one of the gases, the pressures of the two gases probably aren't the same. The gases are in the same container, so V1 = V2. The temperatures of two gases in the same container must be the same, so T1 = T2. The stoichiometric mole ratio is 1:2, respectively, for N 2O and H2O, so if N2O is X, then H2O is 2X. Since the volumes and the temperatures are equal, then the combination gas law reduces to P 1/n1 = P2/n2. Substituting X and 2X for the moles: P1/X = P2/2X. Therefore, P2/P1 = 2:1. Let the pressure of the H2O = 2Y and the pressure of N2O = Y. Y + 2Y = 600 Torr Y = N2O = 200 Torr 2Y = H2O = 400 Torr The key in this problem is to know or to calculate (as we did above) that when two gases have any two parameters identical, then the other two parameters are directly related to each other. In this case the T and V are identical; hence the moles and pressure are directly related.
24. CHEM 161-2000 EXAM 2 CHAPTER 5 - GASES PARTIAL PRESSURE The explosion of nitroglycerine may be represented by:
4C3H5(NO3)3(l)
6N2(g) + O2(g) + 12CO2(g) + 10H2O(g)
If 10.0 grams of nitroglycerine (molar mass 227 g mol-1) explodes completely in a 1.50 liter container, what is the partial pressure of the N2 produced? Assume the gas is at 30 oC, and the volume of the container is unchanged by the reaction. A. B. C. 1.64 atm 1.46 atm 2.92 atm 1.10 atm 0.731atm 6N2(g) + O2(g) + 12CO2(g) + 10H2O(g)
D.
E.
4C3H5(NO3)3(l) 10.0 g MW = 227gmol-1 1.5 L 303K
1.5L ?P 303K
Plan: gC3H5(NO3)3 molC3H5(NO3)3 molN2 PN2 10.0 gC3H5(NO3)3 x (1 mol/227g) = 0.04405 mol C3H5(NO3)3 0.04405 mol C3H5(NO3)3 x (6 mol N2/4 mol C3H5(NO3)3) = 0.06608 mol N2 PV = nRT P = nRT/V P = (0.06608 mol x 0.08205 Latmdeg -1mol-1 x303K)/1.5L = 1.095 atm
CHEM 161-2001 SUMMER EXAM II + ANSWERS CHAPTER 5 - GASES PARTIAL PRESSURE 3. How many grams of Mg must react with excess HCl to produce 35.50 mL of H 2(g) at 40.0 oC when collected over water at a pressure of 775.0 Torr. The vapor pressure of water at 40 oC is 55.3 Torr.
A. 31.8 mg
B. 63.6 mg C. 34.2 mg D. 24.16 mg E. 248.9 mg Mg(s) + 2H+(aq) Mg2+(aq) + H2(g) Strategy: moles H2 moles Mg g Mg
P (775.0 - 55.3 Torr) x (1 atm/760 Torr) n = 0.001308 moles H2
V 0.03550 L
=
n n
R 0.08205
T 273.15 + 40.0 K
0.001308 mol H2 x (1 mol Mg/mol H2) x (24.30 g Mg/mol Mg) = 0.0318 g Mg CHEM 161-2001 SUMMER FINAL EXAM + ANSWERS CHAPTER 5 - GASES PARTIAL PRESSURE 19. 197.6 g of Na2CO3 are allowed to react with excess CH3COOH. The resulting gas was collected over water at 25.0 C and at a pressure of 745.0 Torr. What volume of gas is produced? The vapor pressure of water at 25 C is 23.8 Torr. A. 4.74 x l0-4 m3 B. 6.32 x 10-2 m3 C. 9.60 x l0-2m3 D. 5.48 x 1O-4m3
E. 4.80 x lO-2m3
Na2CO3 + 2CH3COOH H2CO3 + 2NaCH3COO H2CO3 H2O + CO2 Net reaction: Na2CO3 + 2CH3COOH 2NaCH3COO 197.6g XS
+
H2O
+
CO2 298 K
23.8/760 atm Total pressure = 745/760 atm ?V This problem is a very difficult one in that it has several concepts combined. For one thing one needs to know that the reaction of sodium carbonate and acetic acid forms carbonic acid, and that carbonic acid dissociates into the gas carbon dioxide. For a second thing one must be able to balance these equations so that the stoichiometry between sodium carbonate and CO2 can be determined. Then one must realize that since the carbon dioxide was collected over water, then the carbon dioxide is saturated with water vapor, and that the vapor pressure of the water must be subtracted from the total pressure to provide the pressure of CO2. Hence, since pressures operate independently of each other, then if the total pressure = 745 torr and the H2O pressure = 23.8 torr, then the CO2 pressure = 721.2 torr The strategy is gNa2CO3 mol Na2CO3 mol CO2 VCO2 197.6 g Na2CO3 x (1 mol Na2CO3/105.99 g) x (1 mol CO2/mol Na2CO3) = 1.864 mol CO2 PV = nRT (721.2 torr x (1 atm/760 torr) x V) = (1.864 mol CO2 x 0.08205 Latmdeg-1 mol-1 x 298 K) V = 48.02 L Then L has to be converted into m3. 1 cm = 0.01 m 1 cm3 = 1 x 10-6 m3 48.02 L x (1000 mL/L) x (1cm3/mL) (1x10-6 m/cm3 ) = 0.0480 m3 This is a ridiculously time-consuming problem for an exam in which one can only spend an average of
three minutes per problem.
KINETIC MOLECULAR THEORY/ EFFUSION AND DIFFUSION/REAL GASES
34 Chem 161-2006 Final Exam Chapter 5 Gases Kinetic molecular theory/Effusion and Diffusion/Real gases Which one of the following statements is true? A. For a real gas, PV is always < nRT. B. For a real gas, PV is always > nRT. C. Gases are likely to be more real and less ideal at high temperatures.
D
D.
Some real gases have lower pressures than that calculated for the ideal gas because of attraction between molecules. E. Some real gases have lower pressures than that calculated for the ideal gas because of the finite volume occupied by the molecules.
1.00 PV/nRT
Pressure PV = nRT for an ideal gas. Therefore, for an ideal gas PV/nRT = 1, which includes atmospheric pressure and room temperature. As the pressure rises (and the temperature decreases) PV/nRT decreases to below 1.00. This is because P and V decreases to below what would calculate based on it being an ideal gas. As the pressure rises further, the curve changes direction as PV/nRT rises to above 1.00. This is because at very high pressure P and V rises, now being above what one would calculate based on it being an ideal gas. A. For a real gas, PV is always < nRT. False. At very high pressures it is > nRT. B. For a real gas, PV is always > nRT. False. At moderately high pressures it is < nRT. C. Gases are likely to be more real and less ideal at high temperatures. False. Gases tend to be most ideal at high temperatures and low pressures. D. Some real gases have lower pressures than that calculated for the ideal gas because of attraction between molecules. True. At moderately high pressures, and low temperatures, the gas molecules come into close contact with each other. This slows down their journey toward a collision with the wall to provide pressure. Hence, the pressure decreases relative to an ideal gas. E. Some real gases have lower pressures than that calculated for the ideal gas because of the finite volume occupied by the molecules. False. They will have higher volumes (not lower pressures) than that calculated for the ideal gas, due to the volume occupied by the molecules.
11 Chem 161-2006 Final Exam Chapter 5 Gases Kinetic molecular theory/Effusion and Diffusion/Real gases HCl(g) and NH3(g) react to form NH4Cl, a white solid. If NH3(g) and HCl(g) are introduced at opposite ends of a 60.0cm tube, how far from the HCl end will the white ring of NH4Cl form?
A
A.
B. C. D. E.
24.3cm 35.7cm 29.6cm 19.1cm 40.9cm
HCl(g) + NH3(g) NH4Cl(s) 60 cm tube Effusion rate: Rate1/Rate2 = (M2/M1) Rate 1 = NH3 Rate 2 = HCl Rate1/Rate2 = (36.46/17.04) = 1.463 That is, the NH3 travels 1.463 times faster than HCl. This also means that in the same time, the NH3 covers 1.463 times the distance that HCl covers. Let X = the distance traveled by HCl. Let 1.463X = the distance traveled by NH3. X + 1.463X = 60 cm X = 24.36 cm traveled by the HCl, and 35.64 cm traveled by the NH 3.
8 Chem 161-2006 Final Exam Chapter 5 Gases Kinetic molecular theory/Effusion and Diffusion/Real gases The average kinetic energy of UF6(g) is 4.98 kJ/mol at some temperature. What is the average kinetic energy of SF6(g) at that same temperature? A. 7.73 kJ/mol B. 3.21 kJ/mol
C
C.
4.98 kJ/mol D. 28.9 kJ/mol E. 12.0kJ/mol (KE)avg = (3/2)RT The KE only depends on the temperature, not on the MW. Since the temperature is the same for both molecules, then there is no difference in KE.
CHEM 161-2006 EXAM II HILL & PETRUCCI CHAPTER 5 GASES KINETIC MOLECULAR THEORY/EFFUSION AND DIFFUSION/REAL GASES
18.
At what temperature will the average velocity of an O2 molecule be 515 m/s? (a) 298 K
(b) 340 K
(c) 620 K (d) 535 K (e) 787 K RMS = (3RT/MW) 515 meter/sec = ((3 x 8.314Jdeg-1 mol-1 x Tdeg)/0.032 kgmol-1) Square both sides: 2.652x105 m2/s2 = ((3 x 8.314Jdeg-1 mol-1 x Tdeg)/0.032 kgmol-1) (Note conversion factor if you wish to cancel the units: 1J = 1 kgm2s-2) T = 340 K
CHEM 161-2006 EXAM II HILL & PETRUCCI CHAPTER 5 GASES KINETIC MOLECULAR THEORY/EFFUSION AND DIFFUSION/REAL GASES
13.
At 25oC, the atoms of helium move approximately six times faster than the molecules of an unknown gas. Identify the gas. (a) N2 (b) SO2
(c) SF6
(d) Cl2 (e) N2O5 Rate1/Rate2 = (M2/M1) Let He = gas 1. Let unknown gas = gas 2. Let X = rate of gas 2. Since He moves approximately six times faster than the unknown gas, then 6X = rate of helium. 6X/X = (Y/4) 6 = (Y/4) 36 = Y/4 Y = 144g/mol (a) N2: MW = 28.02 (b) SO2: MW = 64.06 (c) SF6: MW = 146 (d) Cl2: MW = 70.90 (e) N2O4: MW = 92.02
CHEM 161-2006 EXAM II HILL & PETRUCCI CHAPTER 5 GASES KINETIC MOLECULAR THEORY/EFFUSION AND DIFFUSION/REAL GASES
6.
Which of the following observations indicates that attractive forces may exist between gas molecules? X. Doubling the pressure on H2(g) results in halving of its volume. Y. The measured molar volume of Cl2(g) is < 22.4 L at STP. Z. When NH3(g) is placed under 10.0 atm pressure at 25 oC, the gas condenses into a liquid. (a) X only (b) X and Y only (c) Z only
(d)
Y and Z only (e) X, Y, and Z
X. Ideal gases have no attractive forces between molecules. Ideal gases follow the ideal gas law, PV = nRT, and all of its corollaries, such as (P1/P2) = (V2/V1). Hence, if the pressure were doubled, and the volume was halved, then this would be an ideal gas, not a real gas. In a real gas that has attractive forces between molecules, if the pressure doubled then the volume would be less than half. So this is not true. Y. If there is an attractive force between gas molecules then that would tend to decrease the rate at which the gas molecules would hit the wall, which means that the pressure would decrease. If the system was under conditions of constant pressure, then the volume would decrease instead of the pressure. So this is true. Another way of looking at this is as follows: Attractive forces tend to convert a gas into a liquid. Liquids, or molecules that are about to be converted into liquids have less volume than gases. Hence, the gas volume should decrease from the ideal gas calculated volume. Z. Ideal gases have no attractive forces between molecules. Liquids have attractive forces between molecules. That's why they are liquids and not gases. In this case, the high pressure of 10.0 atm apparently forced the molecules of NH3 to be close enough to each other so that the attractive forces between the NH3 gas molecules resulted in conversion of NH3(g) to NH3(l). This is true.
2
Chem 161-2005 Hourly Exam II Chapter 5- Gases Kinetic molecular theory/Effusion and Diffusion/Real gases Under identical conditions, gas X escapes through a small hole 1.48 times faster than CO 2 ? Which one of the following is gas X? A. NH3
B.
Ne C. N2 D. O2 E. He Gas X 1.48 Y M1 CO2 Y 44.01
Velocity: MW:
v1/v2 = (M2/M1) 1.48Y/Y = (44.01/M1) M1 = 20.09 MWNe = 20.18
4
Chem 161-2005 Hourly Exam II Chapter 5 - Gases Kinetic molecular theory/Effusion and Diffusion/Real gases A gas has a volume of 1.00 L at a pressure of 1.00 atm. When its volume is decreased, at 1 constant temperature, to 0.010 L ( of its original volume), the pressure required is measured to 100 be 120 atm (120 times its original pressure), a result which does not follow Boyle's law. The reason for this is: A. Molecules are attracting each other B. Kinetic energy per molecule is increasing Molecular size is significant compared to the size of the container D. Molecules are in constant random motion E. Molecules have elastic collisions. Boyle's law can be derived from the combination gas law at constant T and n. P1V1/n1T1 = P2V2/n2T2 P1V1 = P2V2 Gas1 1.00L 1.00 atm Gas 2 0.010L 120 atm
C.
If Boyle's law was followed: 1.0 atm x 1.00L = X atm x 0.010 L X = 100 atm Since the actual pressure is greater than the calculated pressure for an ideal gas, then this is a positive deviation from the ideal gas law. This can be looked at as the numerator in the equation, PV/nrT = 1, being larger than what it should be in the ideal gas law. If the numerator is too large than either the pressure of the gas or the volume of the gas is too large. (At 120 atm of pressure the volume should be 0.0083 L, not the observed 0.010L.) (A) Molecules attracting each other would give a negative deviation from the ideal gas law, because the pressure would be lower than for an ideal gas.. (B) The kinetic energy per molecule is not increasing since the system is at constant temperature, and kinetic energy is a function of temperature. (C) Positive deviations from the ideal gas law can be explained by the molecular size being significant compared to the size of the container. That is, when the piston is pushed down on this gas, the total volume won't decrease as much as it would like to because the gas molecules occupy a significant volume preventing the piston from moving down. (D) Gas molecules always are in constant random motion. This is irrelevant. (E) Gas molecules always have elastic collisions. This is irrelevant.
15 Chem 161-2005 Final exam CHAPTER 5 - GASES Kinetic molecular theory/Effusion and Diffusion/Real gases The rate of effusion of an unknown gas was measured and found to be 34.8 mL/min. Under the same conditions, the rate of effusion of O2 is found to be 51.8 mL/min. Which of the following could be the unknown gas?
A.
B. C. D. E.
Cl2 C4H10 H2 O N2O NH3
v1/v2 = (M2/M1) Unknown v = 34.8 mL/min M=X Oxygen v = 51.8 mL/min M = 32
(34.8/51.8) = (32/X)0.5 X = 70.90 MWCl2 = 35.45 x 2 = 70.90 MWC4H10 = 58.14 MWH2O = 18.02 MWN2O = 44.02 MWNH3 = 17.04
25 Chem 161-2005 Final exam CHAPTER 5 - GASES Kinetic molecular theory/Effusion and Diffusion/Real gases Which of the following is not an assumption of the kinetic-molecular theory of gases?
A.
B. C. D. E.
Gas molecules do not react. Volume of gas molecules is insignificant. The kinetic energy of the particles is proportional to the temperature of the gas. The collisions between gas molecules is perfectly elastic, with no loss of momentum. Collisions with the walls only change the direction, not the speed, of the particles.
A. False. Although reactions in the gas phase are very slow, some gases do react, e.g., NH3 and HCl to form NH4Cl. B. True. The volume of ideal gas molecules is considered to be insignificant. C. True. (KE)avg = (3/2)RT D. True. As the particles collide they exchange kinetic energy. Therefore, as one molecule slows down, another one speeds up, maintaining constant momentum in the system. E. True. RMS = v = (3RT/MW)
ZUMDAHL 5TH EDITION CHEM 161-2002 RECITATION 6TH WEEK CHAPTER 5 - GASES KINETIC MOLECULAR THEORY/EFFUSION AND DIFFUSION/REAL GASES 5. Chem 162-2004 Hourly exam II Chapter 5 - Gases Kinetic molecular theory/Effusion and Diffusion/Real Gases
. Based on this diagram, which statements are true? X Y Z. Of the four gases, CO2 has the strongest intermolecular attractions. Of the four gases, H2 shows the weakest intermolecular attractions. At 600 atm, the volume of the molecules is significant compared to the volume of the container.
A.
Y only X, Y, and Z X only Z only X and Y only
B.
C. D. E.
X. Negative deviations from the ideal gas law can only be explained by strong intermolecular attractions. The results show that CO 2 has the strongest intermolecular attraction. Y. The fact that H2 doesn't show any negative deviation from the ideal gas law demonstrates that it has no intermolecular attraction. Z. The fact that at 600 atm all of the molecules show a positive deviation from an ideal gas demonstrates that the volume of the molecules is significant compared to the volume
of the container. This would result in a higher than calculated V or P.
10.
Chem 162-2004 Hourly exam II Chapter 5 - Gases Kinetic Molecular Theory/Effusion and Diffusion/Real Gases 1.0 mol of H2 and 1.0 mol of CO2 are placed in a 5.0 L container at 298 K. Which of the following properties will be the same for the H2 and the CO2 gases, assuming ideal behavior? X. Y. Z. A. B. C. Average kinetic energy of the H2 and CO2 molecules Average speed of the H2 and CO2 molecules Partial pressure of the H2 and CO2 molecules
X, Y, and Z X and Y only Z only X and Z only X only
D.
E.
X. KE = (3/2)RT Since T is the same, the kinetic energies are the same. Y. RMS = (3RT/MW) Although T is the same, the MW's differ, so the speed will differ. Z: H2 CO2 1.0 mol 1.0 mol 5.0 L 5.0 L 298 K 298 K ?P ?P P1V1/n1T1 = P2V2/n2T2 This simplifies to: P1 = P2. Therefore, the partial pressures are the same.
21.
Chem 162-2004 Hourly exam II Chapter 5 - Gases Kinetic Molecular Theory/Effusion and Diffusion/Real gases Which of the following expressions properly calculates the average velocity, in m/s, of an O2 molecule at 25 C?
A.
B.
3(8.314)(298) 0.032 3(0.0821)(298) 0.032 3(0.0821)(298) 32 3(8.314)(25) 32 3(8.314)(298) 32
C.
D.
E.
RMS = (3RT/MW), where R is the energy constant (8.314 J/degmol), T is temperature in Kelvin, and MW is the MW in kg.
3.
CHEM 161- 2004 FINAL EXAM + ANSWERS CHAPTER 5 - GASES KINETIC MOLECULAR THEORY/EFFUSION AND DIFFUSION/REAL GASES The average kinetic energy for an ideal gas depends on
A.
B. C. D. E.
temperature only. pressure and volume only. pressure only. temperature, pressure and volume. temperature and pressure only.
KEavg = (3/2)RT KEavg is a function only of temperature.
36.
CHEM 161- 2004 FINAL EXAM + ANSWERS CHAPTER 5 - GASES KINETIC MOLECULAR THEORY/EFFUSION AND DIFFUSION/REAL GASES Which statement is not true concerning the kinetic-molecular model of gases? A. Under most conditions there is a lot of empty space between molecules.
B.
C. D. E.
At the same temperature and pressure, all gases have the same average molecular velocity. Pressure is caused by molecular collisions with the walls of the container. Molecules are in constant random motion. At the same temperature and pressure, one mole of any gas occupies the same volume.
A: True B: Not true. RMS = 3RT/MW Therefore, larger molecules have smaller velocities. C: True D: True E. True. P1V1/n1T1 = P2V2/n2T2 If the pressures, temperatures and moles are equal for the two gases, then their volumes are equal. V1 = V2
32.
CHEM 161- 2004 FINAL EXAM + ANSWERS CHAPTER 5 - GASES KINETIC MOLECULAR THEORY/EFFUSION AND DIFFUSION/REAL GASES
Gas X effuses through a small hole 4 times faster than gas Y under the same conditions of temperature and pressure. The molar mass of gas Y is:
A. B. C. D.
2 times that of X 4 times that of X 1/4 that of X 1/2 that of X 16 times that of X. Gas Y v2 = Y ?MW
E.
Gas X v1 = 4Y v1/v2 = M2/M1 4Y/Y = M2/M1 4 = M2/M1 16 = M2/M1
33.
CHEM 161- 2004 FINAL EXAM + ANSWERS CHAPTER 5 - GASES KINETIC MOLECULAR THEORY/EFFUSION AND DIFFUSION/REAL GASES Which one of the following statements is true? A. B. PV/nT is always smaller for a real gas than for an ideal gas. Under the same conditions of volume and temperature, real gases always have higher pressures than that calculated for ideal gases. Some real gases have lower pressures than that calculated for the ideal gas because of attraction between molecules. PV/nT is always larger for a real gas than for an ideal gas. Some real gases have lower pressures than that calculated for the ideal gas because of the finite volume occupied by the molecules.
C.
D. E.
A. False. Sometimes PV/nT is smaller and sometimes PV/nT is larger for a real gas than for an ideal gas. B. False. Sometimes real gases have higher pressures and sometimes lower pressures than for an ideal gas. C. True. That's the reason for the dip in the curve when PV/nT is plotted against pressure. D. False. Sometimes PV/nT is smaller and sometimes PV/nT is larger for a real gas than for an ideal gas. E. False. The finite volume will result in a larger volume for a real gas, not a lower pressure.
71 Calculate the average kinetic energy of the CH4 molecules in a sample of CH4 gas at 273 K and at 546 K. (KE)avg = (3/2)RT (KE)avg = (3/2) x (8.314 J/Kmol) x 273 K = 3405 J/mol (KE)avg = (3/2) x (8.314 J/Kmol) x 546 K = 6710 J/mol
ZUMDAHL 5TH EDITION CHEM 161-2002 RECITATION 6TH WEEK CHAPTER 5 - GASES KINETIC MOLECULAR THEORY/EFFUSION AND DIFFUSION/REAL GASES
73 Calculate the root mean square velocity of the CH4 molecules in a sample of CH4 gas at 273 K and at 546 K. RMS = (3RT/MW) RMS = (3x 8.314 J/Kmol x 273 K/0.01605 kg/mol) = 651 m/s* *units of RMS are in m/s RMS = (3x 8.314 J/Kmol x 546 K/0.01605 kg/mol) = 921 m/s*
ZUMDAHL 5TH EDITION CHEM 161-2002 RECITATION 6TH WEEK CHAPTER 5 - GASES KINETIC MOLECULAR THEORY/EFFUSION AND DIFFUSION/REAL GASES
81 The effusion rate of an unknown gas is measured and found to be 31.50 mL/min. Under identical experimental conditions, the effusion rate of O2 is found to be 30.50 mL/min. If the choices are CH4, CO, NO, CO2, and NO2, what is the identity of the unknown gas? (Effusion rate1/Effusion rate2) = (MW2/MW1) (31.50 mL/min)/(30.50 mL/min) = (32.00/X) ((31.50 mL/min)/(30.50 mL/min))2 = (32.00/X) X = 30.00 = NO
CHEM 161-1998 FINAL EXAM + ANSWERS CHAPTER 5 - GASES KINETIC MOLECULAR THEORY/EFFUSION AND DIFFUSION/REAL GASES 44. How fast does oxygen diffuse compared with hydrogen at the same temperature and pressure? A. 16 times as fast B. 4 times as fast C. 1/16 times as fast D. times as fast
E. times as fast k1/k2 = (M2/M1) Let "1" = O2 Let "2" = H2 k1/k2 = (2/32) = 1/4
3.
CHEM 161-2000 EXAM 2
CHAPTER 5 - GASES KINETIC MOLECULAR THEORY/EFFUSION AND DIFFUSION/REAL GASES
ET: Given MW1, MW2, rate1, find rate2.
A cylinder contains equal amounts (moles)of H2 and He. A small leak allows a small number of moles of H2 to escape in 5.0 minutes. Under these conditions, how long will it take for the same number of moles of He to escape? A. 10 minutes 7.0 minutes 2.5 minutes 5.0 minutes 12 minutes
B.
C. D. E.
(Effusion rate1/Effusion rate2) = (MW2/MW1) (moles/5 min)/(moles/X min) = (4/2) X = 7.1 minute
7. CHEM 161-2000 EXAM 2 CHAPTER 5 - GASES KINETIC MOLECULAR THEORY/EFFUSION AND DIFFUSION/REAL GASES 1.0 mol of H2 and 1.0 mol of CO2 are placed in a 5.0 liter container at 298 K. Which of the following will be the same for the H2 and the CO2? X. Y. Z. A. Frequency of collisions with the wall. Partial pressure. Average kinetic energy of each gas.
X, Y, and Z Y and Z only X only X and Z only Xand Y only
B.
C. D. E.
15. CHEM 161-2000 EXAM 2 CHAPTER 5 - GASES KINETIC MOLECULAR THEORY/EFFUION AND DIFFUSION/REAL GASES Which statements are part of the ideal gas model?
W X.
Molecules of an ideal gas are in constant random motion. Molecules of an ideal gas have extremely small volume compared to the volume of the container. Y. Pressure is caused by molecular collisions with the walls of the container. Z. Higher temperature increases the average kinetic energy of the molecules. A. W, Y, and Z only All the statements are part of the ideal gas model X and Z only W, X, and Y only X, Y and Z only
B.
C. D. E.
CHEM 161-2000 FINAL EXAM + ANSWERS CHAPTER 5 - GASES KINETIC MOLECULAR THEORY/EFFUSION AND DIFFUSION/REAL GASES 25. A sample of O2(g) effuses at a rate three times that of an unknown gas. What is the molar mass of the unknown gas?
A.
B. C. D. E.
288 g/mol 96.0 g/mol 55.0 g/mol 4.00 g/mol 10.7 g/mol
(Rate 2/Rate 1) = (M1/M2) (X/3X) = (M1/M2) Y = 288.58
CHEM 161-2000 FINAL EXAM + ANSWERS CHAPTER 5 - GASES KINETIC MOLECULAR THEORY/EFFUSION AND DIFFUSION/REAL GASES This problem does not belong in this chapter. 27. In which substance would the dispersion forces be the greatest? A. B. C. D. H2 F2 Cl2 Br2
E.
I2
CHEM 161-2001 EXAM II + ANSWERS CHAPTER 5 - GASES KINETIC MOLECULAR THEORY/EFFUSION AND DIFFUSION/REAL GASES 15. Gas X effuses through a small hole 3.32 times faster than CO2 under the same conditions of temperature and pressure. Which of the following is gas X? A. H2 He NH3 O2 N2
B.
C. D. E.
(rate 1/rate 2) = (MW2/MW1) (3.32/1) = (44/X) X = 3.99; He = 4
CHEM 161-2001 FINAL EXAM WITH ANSWERS CHAPTER 5 - GASES KINETIC MOLECULAR THEORY/EFFUSION AND DIFFUSION/REAL GASES 5. Five identical balloons are inflated to the same extent with the indicated gases. Which balloon will collapse first? A.Ne B. C2H6 C. CO2
D. CH4
E. N2 Kinetic molecular theory The smallest molecular weight will have the highest rate of effusion. Ne = 20 C2H6 = 30 CO2 = 44 CH4 = 16 N2 = 28
CHEM 161-2001 SUMMER EXAM II + ANSWERS CHAPTER 5 - GASES KINETIC MOLECULAR THEORY/EFFUSION AND DIFFUSION/REAL GASES 13. If SF6 effuses through a hole at a rate of 1.25 mmol/s, which gas will effuse at a rate of 1.10 mmol/s under the same conditions? A. RhS2
B. TiCl4
C. BrF4 D. FeF3 E. BrF3 A. RhS2 B. TiCl4 C. BrF4 D. FeF3 E. BrF3 SF6 MW = 102.9 + 64 = 166.9 MW = 47.9 + 4x35.45 = 189.7 MW = 79.90 + 4x19 = 155.9 MW = 55.85 + 112 = 167.85 MW = 79.80 + 57 = 136.80 MW = 32.06 + 6x19 = 146.06
k1/k2 = (M2/M1) (1.25 mmols-1/1.10 mmols-1) = (X/146.06) X = 189 CHEM 161-2001 SUMMER EXAM II + ANSWERS CHAPTER 5 - GASES KINETIC MOLECULAR THEORY/EFFUSION AND DIFFUSION/REAL GASES 14. Which of the following is not a cause of gases deviating from the ideal gas law? A. Slower speeds at low temperature make the intermolecular interactions significant. B. Many particles at high pressure make the volume significant.
C. The molecular bonds become weaker at high pressure.
D. The collisions become elastic as the kinetic energy decreases. E. The frequency of collisions that "stick" goes up as the as density increases. CHEM 161-2001 SUMMER FINAL EXAM + ANSWERS CHAPTER 5 - GASES KINETIC MOLECULAR THEORY/EFFUSION AND DIFFUSION/REAL GASES 10. At what location and season will air behave most like an ideal gas? A. New Brunswick, NJ (alt. 300 ft) in Sept. B Death Valley, CA (alt. -300 ft) in Jan.
C
Denver, CO (alt 5000 ft) in August D. Death Valley, CA (alt. -300 ft) in March E. New Brunswick, NJ (alt. 300 ft) in March.
Gases are ideal at low pressures and high temperatures. 5000 feet is high altitude. Pressures are low at high altitude. Annual temperature is at or near its highest in August. Hence, the conditions in Denver, Colorado in August are the best to provide an ideal gas.
CHEM 161-2001 SUMMER FINAL EXAM + ANSWERS CHAPTER 5 - GASES KINETIC MOLECULAR THEORY/EFFUSION AND DIFFUSION/REAL GASES 20. A sample of CO2 effuses through a hole at a rate of 16.0 mole/min. What would the rate be for a sample of UF6 under the same conditions? A. 45.25 mole/min.
B. 5.66 mole/min.
C. 0.250 mole/min. D. 2.00 mole/min E. 32.00 mole/min.
CHEM 161-2002 EXAM II + ANSWERS CHAPTER 5 - GASES KINETIC MOLECULAR THEORY/EFFUSION AND DIFFUSION/REAL GASES When 1.00 mol of NH3(g) is placed in a 20.0 L container at 0o C, the pressure inside the container is measured to be 1.00 atm. Based on these measurements, we can conclude: A. NH3 molecules have more kinetic energy than if it were an ideal gas. PV = nRT PV/nRT = 1 (1 atm x 20.0 L)/(1.00 mol x 0.08205 Latmdeg-1 mol-1 x 273.15) = 0.89. Therefore the NH3 is showing a negative deviation from an ideal gas. This can only be explained by an intermolecular attraction which results in lower frequency of collisions with the container walls, i.e., lower pressure than expected for an ideal gas. More kinetic energy means greater velocity. Greater velocity means greater pressure, the opposite of what is expected based on a negative deviation from an ideal gas. B. Some NH3 has decomposed into N2 and H2. If the moles of gas increase then the pressure will increase, offsetting each other, not converting an ideal gas into a real gas. C. The volume of the NH3 molecules is significant compared to the container's volume. If the volume is significant than V would be higher than that of an ideal gas, causing the result to be greater than 1, but it is less than 1.
D.
NH3 molecules are exhibiting intermolecular attractions. Intermolecular attractions result in the pressure being lower than that of an ideal gas. If the pressure decreases, the numerator decreases, and the result would be less than 1, which is the case here. E. NH3 is behaving as an ideal gas If NH3 was behaving as an ideal gas, then PV/nRT would equal to 1, which is not the case.
PV = nRT PV/nRT = 1 (1 atm x 20.0 L)/(1.00 mol x 0.08205 Latmdeg -1 mol-1 x 273.15) = 0.89 Since PV/nRT under these conditions is not equal to 1, then the NH 3 is not behaving as an ideal gas.
21.
CHEM 161-2002 EXAM II + ANSWERS CHAPTER 5 - GASES KINETIC MOLECULAR THEORY/EFFUSION AND DIFFUSION/REAL GASES Carbon dioxide gas escapes through a small hole at a rate of 2.00 mL/minute. At what rate would CH4 gas escape through the same hole under the same conditions of temperature and pressure. A. B. 1.21 mL/minute 2.75 mL/minute 3.31 mL/minute 0.73 mL/minute 5.50 mL/minute
C.
D. E.
(k1/k2) = (M2/M1) MW CO2 = 44; MW CH4 = 16 2/X = (16/44) X = 3.32 mL/minute
CHEM 161- 2002 FINAL EXAM + ANSWERS CHAPTER 5 - GASES KINETIC MOLECULAR THEORY/EFFUSION AND DIFFUSION/REAL GASES 26. What is the molar mass of a gas that diffuses 1.48 times faster than CO 2 ?
A.
B. C. D. E.
20.1 g/mol 96.4 g/mol 65.1 g/mol 29.7 g/mol 37.5 g/mol
Graham's Law of Effusion
____ k1/k2 = (M2/M1) ____ (1/1.48) = (X/44) X = 20.09
6.
CHEM 161-2003 EXAM II + ANSWERS CHAPTER 5 - GASES KINETIC MOLECULAR THEORY/EFFUSION AND DIFFUSION/REAL GASES What is the molecular weight of a gas that would effuse through a small hole approximately twice as fast as Cl2 does under the same conditions? Molecular weight of Cl2 = 71
A.
B. C. D. E.
18 30 64 32 44 Gas 2 (Cl2) Rate = X MW = 71
Gas 1 Rate = 2X ? MW
(Effusion rate1/Effusion rate2) = (MW2/MW1) (2X/X) = (71/X) 2 = (71/X) 4 = 71/X X = 17.8 gmol-1
10.
CHEM 161-2003 EXAM II + ANSWERS CHAPTER 5 - GASES KINETIC MOLECULAR THEORY/EFFUSION AND DIFFUSION/REAL GASES An amount of a certain gas has a volume of 20 mL at a pressure of 500 atm. At the same temperature, the pressure of this amount of the gas is doubled to 1000 atm with its volume decreasing to 15 mL (more than half of its original volume). Which one of the following statements is true? A. B. The gas is showing evidence of intermolecular attractions. The gas is behaving as an ideal gas.
C.
D. E.
The gas is showing evidence of significant molecular volume. The increased pressure causes a greater molar kinetic energy. The data is incorrect since Boyle's law is not followed.
Gas (condition 1) Gas (condition 2) V = 0.020 L V = 0.015 L P = 500 atm P = 1000 atm T1 = T2 n1 - n2 P1V1/n1T1 = P2V2/n2T2 P1V1 = P2V2 500 atm x 0.020 L 1000 atm x 0.015 L If the ideal gas law was followed, then P1V1 would be equal to P2V2. But P2V2 is too large, i.e., this gas is not following the ideal gas law. Either P 2 is too large or V2 is too large, or the combination is too large. Deviations from the ideal gas law can be explained only by the volume being too large or the pressure being too small, so let's focus on the volume. The volume under condition 2 should be 0.010 L. Since the actual volume is greater than the theoretical volume then this would suggest that the gas has significant volume. This conclusion is consistent with what might be expected at a pressure of 1000 atm. (If the cause of non-ideality was that the gas is showing evidence of intermolecular attractions, then the pressure would be lower than it is (the product of the pressure and volume for condition 2 would be lower than the product of the pressure and volume for condition 1, but it is higher).)
15.
CHEM 161-2003 EXAM II + ANSWERS CHAPTER 5 - GASES KINETIC MOLECULAR THEORY/EFFUSION AND DIFFUSION/REAL GASES Which of the following is not part of the ideal (kinetic) gas model? A. B. C. The volume of the molecules is much less than the volume of the container. The molecules are in constant random motion Attractive forces between molecules are negligible Collisions with the walls of the container are negligible compared with collisions with other molecules. Collisions are elastic with no loss of kinetic energy.
D.
E.
In an ideal gas, there are numerous collisions with the walls of the container, but the size of the molecules is considered to be so small that the molecules essentially don't collide with each other. This is the opposite of "D".
10.
Chem 161-2003 Final exam Chapter 5 - Gases Kinetic molecular theory The average speed of an O2 molecule at 298 K is 482 m/s. At what temperature will its average speed be 964 m/s? A. B. C. D. 421 K 682 K 596 K 791 K 1192 K
E.
The average speed of O2 at 298 K is irrevant to this problem. I don't know why this information was provided. RMS = (3RT/MW) 964 m/s = (3x8.314xT/0.032 kg) T = 1192 K
13.
Chem 161-2003 Final exam Chapter 5 - Gases Chem 161-2003 Final exam Chapter 5 - Gases Kinetic molecular theory/Effusion and Diffusion/Real gases One of the assumptions of the kinetic theory of gases is that molecules in an ideal gas exert no forces on each other. If molecules actually attracted each other, then which of the following consequences would be observed? X. Liquefaction of the gas would occur at sufficiently low temperatures Y. The molar volume of the gas at STP would be less than 22.4 Liters Z. A 22.4 Liter sample containing 1.00 mol of the gas at 273 K would have a pressure greater than 1.00 atm.
A. B. C.
none of them X, Y, and Z Y and Z only X and Y only
D.
E.
X and Z only
X. Liquefaction occurs because the gas molecules attract each other. So this would be observed when the gas molecules attract each other. Y. If the molecules of gas are ideal, then one mole at STP would require 22.4 Liters. But if the molecules attract each other then they would require less than 22.4 Liters. (Note that under very high pressures another factor, the volume of the molecules, comes into play and so at very high pressures the volume of the molecules is greater than expected for an ideal gas. Z. If the molecules of gas are ideal, then they would collide with the walls of the container providing a pressure of 1.00 atm. But if they interact with each other, then the collisions on the wall would be slightly less, providing a pressure of slightly under (not greater than) 1.00 atm.

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SMonday, March 17, 2008Florida wants to have mail in vote, to re-do the primary. The Florida delegation says, "no" to senator. State law doesn't allow for mail in elections. New York Governor Elliot Spietzer resigned from office because of a prostit

UCSD - BIBC - 110

ASU - POS - 110

A More Perfect Union Barack Obama is one of the most influential speakers I have ever personally seen speak to a group of people. When he was campaigning in my hometown of Cedar Rapids, IA, I had the opportunity to listen to his message about, "Chang

ASU - POS - 110

Gettysburg Address The Gettysburg Address is very important part of history and for that matter was a very important speech given by President Lincoln. At this time in history the U.S. Civil war had just ended and many American soldiers had lost thei

ASU - POS - 110

2/13/2008February 5, 2008 Op-Ed ColumnistThe Cooper ConcernsBy DAVID BROOKS Im not a Hillary-hater. Shes been an outstanding senator. She hung tough on Iraq through the dark days of 2005. In this campaign, she has soldiered on bravely even thoug

UCSD - BIMM - 110

Name_Quiz 4. BIMM 110 1. Provide a rough diagram of the structure of a T-cell receptor, including the position of the hypervariable binding regions in the receptor, and the position of receptors in the cell. (3 pts) A dimeric protein diagram, no mat

UCSD - BIMM - 110

Name_Quiz #3. BIMM 110 1. The epidermal growth factor receptor activates downstream kinases that are not directly bound to the the transmembrane receptor. Name one of these kinases and the class of macromolecules the downstream kinases modify. (2 pt

UCSD - BIMM - 110

Quiz #2 BIMM 110 25 points totalName_1.The insulin receptor has an enzymatic function, what is it? Which domain of the receptor contains this enzymatic function? (2 pts). tyrosine kinase or adds phosphate groups to tyrosines, Intracellular domain

UCSD - BICD - 130

BICD 130 - Embryos, Genes, and DevelopmentSummer 2007Final Exam (Five parts, 170 points total)ANSWER KEY _ Please PRINT your name herePlease write your answers in ink. Requests for regrades will not be considered for exams done in pencil. EXAM

UCSD - MATH - 10C

Math 10B Final Exam Review OutlineBasic Information for the Final Exam:The final will consist of approximately 8-9 questions. The final exam will be cumulative; it will not focus too much on any particular topic. Most likely, the recent material wi

UCSD - MATH - 10C

Math 10C Final Exam Review OutlineBasic Information for the Final Exam:The exam will consist of approximately 8 questions, with multiple parts. You will not be allowed a calculator on the exam, so please do not bring one. You should bring a number

UCSD - MATH - 10C

Math 10A Final Review SessionTime: 6:30pm-8:00pm Date: Mar. 19th, 2008 Location: Center 216The exam will consist of approximately 10 questions. Many questions are quite similar to the previous midterms. You should look back over those exams. Also,

UCSD - MATH - 10C

Homework 5 Solutions14.1.16 We have ft (18, 6) P 90 - 93 = = -1.5 percent/month. t 20 - 18Eighteen months after rates are exposed to a formaldehyde concentration of 6 ppm, the percent of rats surviving is decreasing at a rate of about 1.5 per mon

UCSD - MATH - 10C

Name: TA: Math 10C Quiz 4 March 7, 2008 Sec. No:PID: Sec. Time:1. Given z = f (x, y), x = x(u, v), and y = y(u, v), suppose that x(2, 3) = 4 and y(2, 3) = 5. Find zv (2, 3) in terms of some of the numbers a, b, c, d, e, r, s, t, where fx (2, 3) =

UCSD - MATH - 10C

Name: TA: Math 10C Quiz 3 Feb 15, 2008 1. Let P = (-2, 1, 0), Q = (1, 2, -1), and R = (-3, 2, 1). - - a. Find the vectors P Q and P R. Sec. No:PID: Sec. Time:- - Solution: P Q = 3i + j - k, P R = -i + j + k. - - b. Compute the cross product

UCSD - MATH - 10C

Name: TA: Math 10C Quiz 1 Jan 18, 2008 Sec. No:PID: Sec. Time:1. Consider a group of new lightbulbs. For a given bulb, let t be the number of months the bulb lasts before burning out. The density function for the distribution of t is given by p(t

UCSD - MATH - 10C

Name: TA: Math 10C Quiz 2 Feb 8, 2008 Sec. No:PID: Sec. Time:1. Let f (x, y) = xy. Draw a contour diagram with contours at c = -1, c = 0, and c = 1. Describe in words what shape these contours are (ie, circles, ellipses, hyperbolas, lines, or som

UCSD - MATH - 10C

Name: TA: Math 10C. Midterm Exam 2 February 27, 2008 Sec. No:PID: Sec. Time:Turn off and put away your cell phone. You may one page of notes, but no calculators, books or other assistance. Read each question carefully, and answer each question co

UCSD - MATH - 10C

Name: TA: Math 10C . Midterm Exam 1 January 30, 2008 Sec. No:PID: Sec. Time:Turn off and put away your cell phone. You may use one page of notes, but no calculators, books or other assistance. Read each question carefully, and answer each questio

UCSD - MATH - 10C

Homework 6 Solutions14.7.2 Calculating the partial derivatives: f = 2(x + y), x Therefore, we get f = 2(x + y), y 2f = 2, y 2 2f = 2, yx 2f = 2. xy 2f = 2. x214.7.4 Since f (x, y) = xey , the partial derivatives are fx = ey , fy = xey , fxx = 0, f

UCSD - MATH - 10C

Homework 3 Solutions12.1.2 The distance of a point P = (x, y, z) from the yz-plane is |x|, from the xz-plane is |y|, and from the xy-plane is |z|. So A is closest to the yz-plane, since it has the smallest x-coordinate in absolute value. B lies on t

UCSD - MATH - 10C

Homework 4 Solutions13.1.2 i + 2j - 6i - 3j = -5i - j. 13.1.6 13.1.12 13.1.14 2a + 7b - 5z = 2(2j + k) + 7(-3i + 5j + 4k) - 5(i - 3j - k) = (4j + 2k) + (-21i + 35j + 28k) - (5i - 15j - 5k) = (-21 - 5)i + (4 + 35 + 15)j + (2 + 28 + 5)k = -26i + 54j +

UCSD - MATH - 10C

Homework 2 Solutions9.2.6 No. Ratio between successive terms is not constant: 9.2.8 Yes, a = y 2 , ratio of successive terms is y. 9.2.16 The series has 10 terms. The first term is a = 0.2 and the constant ratio is x = 0.1, so Sum = 0.2(1 - (0.1)10

UCSD - MATH - 10C

8.7.4 Since the function takes on the value of 4, it cannot be a cdf (whose maximum value is 1). Also, the function decreases, which means that it is not a cdf because a cdf is increasing. Thus, this function is a pdf. Since the area under a pdf is 1

UCSD - BIBC - 110

UCSD - BIBC - 110

UCSD - BIBC - 100

BIBC 100 FINAL EXAM WINTER 2008BIBC 100 FINAL EXAMWINTER 2008SECTION I CHOOSE ANY 4 OUT OF THE NEXT 5 QUESTIONS TO ANSWER. 5 points each. WRITE "OMIT" in LARGE LETTERS ON THE PAGE YOU DON'T WISH TO ANSWER. 41) Draw a diagram of a three-stranded

Georgetown - MATH - 200

Quiz Schedule for General Chemistry II Chem 010 Spring 2008 Quiz Meeting times: Tuesdays 3:15 - 4:05 Thursdays 9:15-10:05 10:15 - 11:05 11:15 - 12:05 2:15 - 3:05 7:15 - 9:05 Fridays 2:15 - 3:05 Date: Jan 15, 16, 17 T Th F Jan 22, 24, 25 Topic No Quiz

Georgetown - MATH - 200

PROGRAMA DE CLASE - SPANISH INTERMEDIATE 1 (021) Primavera 2003 Textos: Video: Pasajes: Lengua (2002) Pasajes: Cultura (2002) Pasajes: Cuaderno de prctica (CP) (2002) ValentinaImportant Notes: 1) All pages listed must be prepared before coming to c

Washington - TC - tc 333

Assignment: Short ReportHere is your scenario for the short report assignment: You have an internship with a local engineering firm, and your manager has sent you an e-mail requesting that you write up a memo providing information on demographics on

University of Louisville - CHEM - 202

This syllabus is tentative and subject to change. Changes will be announced in lecture. It is the student's responsibility to be aware of any and all changes.Chemistry 202 General Chemistry II Spring 2008 Professor: Meeting: Recitation: Office Hrs:

Virginia Tech - GEOG - 1004

THINKING ABOUT POPULATION: PART TWO12 11 10 9 Old Stone 7 Age 8 6 5 4 3 1950 2 1 Black Death -The Plague 1900 1800 1975 Bronze Age Iron Age Modern Age Middle Ages 2000 Future 2100New Stone AgeBillions1+ million 7000 6000 5000 4000 3000 2000 1

Virginia Tech - AAEC - 1005

Virginia Tech - AAEC - 1005

Virginia Tech - AAEC - 1005

Virginia Tech - AAEC - 1005

Georgetown - MATH - 200

FIRST LEVEL MODERN STANDARD ARABIC 011Fall 04 Instructor: Phone: E-mail: Office hours: Hana Zabarah ICC 306 202-687-8006 (O) 703-448-8248 (H) zabarahh@georgetown.edu TBATeaching Assistant: E-mail: Office hours:TBACourse Description: This cours

Georgetown - MATH - 200

GEORGETOWN UNIVERSITY DEPARTMENT OF SPANISH AND PORTUGUESEINTENSIVE SPANISH PROGRAMPROGRAM INTENSIVE INTERMEDIATE SPANISH (SPAN 032)FALL 2003CRISTINA SANZ, DIRECTOR E-MAIL: sanzc@georgetown.edu MAILBOX: ICC 403 OFFICE: ICC 412 PHONE #: 7-7213

Mines - PHGN - 320

Georgetown - MATH - 200

MATH 040: Probability and Statistics Course Policies and SyllabusInstructor: Dr. Kimberly Sellers Room 340, St. Mary's Hall 202-687-8829 kfs7@georgetown.edu DeVeaux RD, Velleman PF, and Bock DE Stats: Data and Models, 2nd edition, Pearson Addison-We

UMiami - MTH - 112

1n0v.w, .(\dIY,'4x, 2_lntUol.rd.ltL:lJ =' Sra/= Uz tV,I Lolr, =+_d{ tl l;'I7II, 1e- |l-llrrIr-'fIt,/frpnf'a-,tr IV - gx- ol*vCoiltfE<-* + 2tln hK-\-f,^ 4r +6o( lr y\J

Syracuse - MFE - 331

4/8/2008Intellectual PropertyIntellectual Property Legal mechanisms to protect the rights of intellectual property owners for: ideas concepts names designs processes The legal mechanisms are intended To provide an incentive and reward to

Syracuse - MFE - 331

Realizing the productsProduct Realization ProcessesStanley Tools Jobmaster ScrewdriverRollerblade InLink SkateHewlett-Packard DeskJet PrinterVolkswagen New Beetle AutomobileBoeing 777 AirplaneAnnual production volume Sales lifetime Sale

Syracuse - MFE - 331

New Product Development Project You are asked to develop a new product for the University student market (both undergraduate and graduate students.) The constraints on your project are: The product is highly likely to require fewer than 10 custom c

Tiffin - PSY - 101

Psychology Movie REACTION TO MOVIE AWAKENINGS1Psychological Reaction to Awakenings Movie Nic Hetzel Tiffin University Psychology 101 Professor Heck January 14, 2008Psychology Movie The film Awakenings was made from a true story about a research

RIT - EN - 362

11.1Arguments, Validity and SoundnessArguments An argument is a two-part structure composed of a set of sentences P1 , P2 , . . . , Pn called the premises of the argument and a sentence C called the conclusion of the argument. P1 . . . . P2 . .

Tiffin - POL - 151

Military Sub-Saharan The sub-Saharan is mainly known for its devastating numbers in the issue of AIDS but there is also a good deal of military issues throughout the sub-Saharan. The Burundi cease-fire was a big topic in 2001. One other event that yo

Tiffin - ENG - 142

Nic Hetzel Prof. Galindo English 142-02 February 22, 2008 Barn Burning 5. The exposition of the story show that Abner already has enemies and that tells a lot about the character as you read on. Abner's behavior in the story tells a lot about his cha

Tiffin - PSY - 101

Psychology Movie Running head: REACTION TO MOVIE MEMENTO1Psychological Reaction to Memento Movie Nic Hetzel Tiffin University Psychology 101 Professor Heck March 14, 2008Psychology Movie2Abstract Anterograde amnesia is a disorder that deals