3. What property and two conditions determine the phase of a sample of matter is? Property = Equilbirum Conditions = Temperature and Pressure

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1 9.A Calorimetry: Basics Instructions: Provide a response for each question that is well thought out, satisfies the prompt, is clearly explained, and LEGIBLE. 1. In a closed system what basic fact must always hold true for the energy content of the overall system? Energy Is neither lost nore gained, only transferred between components of the system 2. What is the difference between heat and temperature, including units and symbol? Heat (q) measured in J = total KE of particles (meaning dependent on amount of material) Temperature (T) measured in K = reflects the average KE of particcles (meaning independent of the amount of material as it is averaged out) 3. What property and two conditions determine the phase of a sample of matter is? Property = Equilbirum Conditions = Temperature and Pressure 9.B Calorimetry: Temperature Conversions and Calculations Remember: C = K K = C AND Sig Figs are by position of least precise decimal as it is +/- order of operation K, NOT K!!!!!!!!!!! Instructions: ON A SEPARATE SHEET, convert C to K or vice versa.. Remember, Sig Figs, NW = NC, Boxed Answers and N 3 (this includes labeling and in your set-up!!!) C = K K = C K = -30 C C = 1507 K C = K C = K K = C K = C Instructions: ON A SEPARATE SHEET, calculate the change in temperature ( ΔT ) from the following data. Remember, Sig Figs, NW = NC, Boxed Answers and N 3. Remember: ΔT = Tfinal - Tinitial C to 12.4 C C to C 12.4 C C = C C C = C C to 49.0 C 49.0 C ( 33.0 C) = 82.0 C C to C C 1234 C = C 9.C Calorimetry: Phase Diagrams Instructions: Answer the question 1 9 in relation to the following phase diagram. Remember N 3 1. Which letter denotes the solid phase? A _ 2. Which letter denotes the liquid phase? C _ 3. Which letter denotes the gas phase? B _ 4. Which letter denotes the triple point? d _ In your own words, what is the definition of a triple point? The condition of equlibrium between solid, liquid and gas due to pressure and temperature conditions 5. What is the melting point at 1 atm of pressure? 60 C _ 6. What is the boiling point at 1 atm of pressure? 100 C _ 7. Above what temperature is it impossible to liquefy this substance, no matter what the pressure? 110 C _ 8. At what temperature and pressure do all three phases Triple Point = 45 C 0.5 atm _ 9. At a constant temperature, what would you do to cause this substance to change from the liquid phase to the solid phase?? INCREASE in pressure

2 Instructions: Refer to the phase diagram below when answering the questions NOTE: Normal refers to 1.0 atm and remember N What are the values for temperature and pressure at STP? T= 0 C _P= 1.00 atm _ 11. What is the normal freezing point of this substance? 100 C _ normal boiling point of this substance? normal melting point of this substance? 340 C _ 100 C _ 12. What is the phase (s, l, g) of a substance at 2.0 atm and 100 C? s _ 13. What is the phase (s, l, g) of a substance at 0.75 atm and 100 C? l _ 14. What is the phase (s, l, g) of a substance at 0.5 atm and 100 C? g _ 15. What is the phase (s, l, g) of a substance at 1.5 atm and 50 C? s _ 16. What is the phase (s, l, g) of a substance at 1.5 atm and 200 C? l _ 17. What is the phase (s, l, g) of a substance at 1.5 atm and 800 C? g _ 18. What is the condition of the triple point of this substance? T= 180 C _P= 0.70 atm _ 19. A sample was at an initial pressure of 1.25 atm and a temperature of C and was lowered to a pressure of 0.25 atm, what phase transition(s) would occur? boiling _ l g 20. A sample was at an initial pressure of 0.5 atm and a temperature of C was lowered to a temperature of C, what phase transition(s) would occur? desublimation _ g s 21. If this substance was at a pressure of 2.0 atm, at what temperature would it melt? 180 C _ boil? Never _ 22. If this substance was at a pressure of 0.75 atm, at what temperature would it melt? 90 C _ boil? 180 C _ 23. At what temperature do the gas and liquid phases become indistinguishable from each other? 80 C _ 24. At what pressure would it be possible to find this substance in the gas, liquid, and solid phase? 0.70 atm_ 25. If I had a quantity of this substance at a pressure of 1.00 atm and a temperature of C, what phase change(s) would occur if I increased the temperature to C? At what temperature(s) would they occur? (NOTE: multiple answers needed for this question) 100 C s l 340 C _ l g Error = +/- 10 / OR +/ atm 9.D Calorimetry: Calometric Calculations 1. A sample of mercury is heated from 25.5 C to 52.5 C. In the process, 3050J of heat are absorbed. What mass of mercury was in the sample? The specific heat of mercury is J/g C. Ti = 25.5 C Tf = 52.5 C ΔT = 27.0 C q = 3050 J m = X g Cp = J/g C 3. A cube of gold with a mass of g is heated from 30.0 C to some higher temperature with the absorption of 921 J of heat. The specific heat of gold is J/g C. What is the final temperature of the gold? Ti = 30.0 C Tf = X C ΔT = Y C q = 921 J m = g Cp = J/g C 2. A block of aluminum with a mass of 140g is cooled from 98.4 C to 62.2 C with a release of 1137J of heat. From this data, calculate the specific heat of aluminum. Ti = 98.4 C Tf = 62.2 C ΔT = C q = J m = 140 g Cp = X J/g C 4. A total of 226 J of heat are absorbed as 58.3 g of lead is heated from 12.0 C to 42.0 C. From this data, what is the specific heat of lead? Ti = 12.0 C Tf = 42.0 C ΔT = 30.0 C q = 226 J m = 58.3 g Cp = X J/g C Instructions: ON A SEPARATE SHEET OF PAPER, preform the following calorimetric calculations. Remember, Sig Figs, NW = NC, Boxed Answers and N 3 (this includes labeling and in your set-up!!!) NOTE: I HAVE PULLED OUT VARIABLES WITH MY WORK TO TRY AND HELP YOU FIND THE ERRORS IN YOUR CALCUATIONS 5. A sample of mercury is heated from 25.5 C to 52.5 C. In the process, 3050J of heat are absorbed. What mass of mercury was in the sample? The specific heat of mercury is J/g C. Ti = 25.5 C Tf = 52.5 C ΔT = 27.0 C q = 3050 J m = X g Cp = J/g C Base Equation: q = m Cp ΔT Algebraic Rearrangement: m = _ q _ Cp ΔT Required Work to Show: m = 3050 J _ J/g C x 27.0 C Answer: 807 J 6. A block of aluminum with a mass of 140g is cooled from 98.4 C to 62.2 C with a release of 1137J of heat. From this data, calculate the specific heat of aluminum. Ti = 98.4 C Tf = 62.2 C ΔT = C q = J m = 140 g Cp = X J/g C Base Equation: q = m Cp ΔT Algebraic Rearrangement: Cp = _ q _ m ΔT Required Work to Show: Cp = J _ 140 g x C Answer: J/g C

3 7. A total of 226 J of heat are absorbed as 58.3 g of lead is heated from 12.0 C to 42.0 C. From this data, what is the specific heat of lead? Ti = 12.0 C Tf = 42.0 C ΔT = 30.0 C q = 226 J m = g Cp = X J/g C Base Equation: q = m Cp ΔT Algebraic Rearrangement: Cp = _ q _ m ΔT Required Work to Show: Cp = 226 J _ 58.3 g x 30.0 C Answer: J/g C 8. What is the change in heat energy when 132 g of copper at 32.2 C is raised to 45.0 C? The specific heat of copper is J/g C. Ti = 32.2 C Tf = 45.0 C ΔT = 12.8 C q = X J m = 132 g Cp = J/g C Required Work to Show: q = 132 g x J/g C x 12.8 C Answer: 650. J 9. What is the change in heat energy when 75.0 g of water at 9.8 C is raised to 22.4 C and the specific heat of liquid water is 4.18 J/g C? Ti = 9.8 C Tf = 22.4 C ΔT = 12.6 C q = X J m = 75.0 g Cp = 4.18 J/g C Required Work to Show: q = 75.0 g x 4.18 J/g C x 12.6 C Answer: 3950 J 10. What is the change in heat energy when 125 g of water at 10.0 C is raised to 90.0 C Ti = 10.0 C Tf = 90.0 C ΔT = 80.0 C q = X J m = 125 g Cp = 4.18 J/g C Required Work to Show: q = 125 g x 4.18 J/g C x 80.0 C Answer: J 11. What is the change in heat energy when g of aluminum metal at C is cooled to 82.0 C? The specific heat of aluminum is J /g C. Ti = C Tf = 82.0 C ΔT = 18.0 C q = X J m = g Cp = J/g C Required Work to Show: q = g x J/g C x 18.0 C Answer: 1050 J 12. What is the mass of a sample of iron if that sample lost 2300J of heat energy when it cooled from 80 C to 30 C? The specific heat of iron is J /g C. Ti = 80 C Tf = 30 C ΔT = - 50 C q = J m = X g Cp = J/g C Base Equation: q = m Cp ΔT Algebraic Rearrangement: m = _ q _ Cp ΔT Required Work to Show: m = J _ J/g C x - 50 C Answer: 100 g without sig figs = FYI 13. What is the specific heat of metal if its mass is g and it requires J of heat energy to raise its temperature from 27.4 C to 67.3 C? Ti = 27.4 C Tf = 67.3 C ΔT = 39.9 C q = J m = g Cp = X J/g C Base Equation: q = m Cp ΔT Algebraic Rearrangement: Cp = _ q _ m ΔT Required Work to Show: Cp = J _ g x 39.9 C Answer: J/g C 14. If a 38 g sample of water releases 621 J of heat energy and cools to 4.0 C. What was the initial temperature of the water? Ti = X C Tf = 4.0 C ΔT = Y C q = 621 J m = 38 g Cp = 4.18 J/g C Base Equation: q = m Cp ΔT Algebraic Rearrangement: ΔT = _ q _ m Cp ΔT = Tf - Ti Ti = Tf + ΔT Required Work to Show: ΔT = 621 J _ = 3.9 C 38 g x 4.18 J/g C Ti = 4.0 C C Answer: 7.9 C 15. If 41 grams of water at 24 C absorbs 2208 J of heat energy, what will be the final temperature of the water? Ti = 24 C Tf = X C ΔT = Y C q = 2208 J m = 41 g Cp = 4.18 J/g C Base Equation: q = m Cp ΔT Algebraic Rearrangement: ΔT = _ q _ m Cp ΔT = Tf - Ti Tf = ΔT + Ti Required Work to Show: ΔT = 2208 J _ = 13 C 41 g x 4.18 J/g C Tf = 24 C + 13 C Answer: 37 C 16. How much heat energy is absorbed by 10 g of silver if it increases in temperature from 10 C to 310 C? The specific heat of silver is J/g C. Ti = 10 C Tf = 310 C ΔT = 3.0 x 10 2 C q = X J m = 10 g Cp = J/g C Required Work to Show: q = 10 g x J/g C x 3.0 x 10 2 C Answer: 700 J 17. What is the change in temperature in a 128 g sample of titanium if it absorbs 2808 J of heat energy at a temperature of 2.2 C? The specific heat of titanium is J/ g C. Ti = 2.2 C Tf = Not Needed ΔT = Y C q = 2808 J m = 128 g Cp = J/g C Base Equation: q = m Cp ΔT Algebraic Rearrangement: ΔT = _ q _ m Cp Required Work to Show: ΔT = 2808 J _ = 128 g x J/g C Answer: 41.9 C 20. What is the change in temperature in a 128 g sample of water if it absorbs 2808J of heat energy at a temperature of 3.21 C? Ti = 3.21 C Tf = Not Needed ΔT = Y C q = 2808 J m = 128 g Cp = 4.18 J/g C Base Equation: q = m Cp ΔT Algebraic Rearrangement: ΔT = _ q _ m Cp Required Work to Show: ΔT = 2808 J _ = 128 g x 4.18 J/g C Answer: 5.25 C

4 9.E Calorimetry: Heating and Cooling Curves Instructions: Answer the question 1 2 using this heating curve for water. 1. Identify the sections where the following phases are found: a. D E _gas b. A B _solid c. B D E _liquid d. B _solid and liquid e. D _liquid and gas 2. Identify by letter (A-E) in which section the following are found: a. B _ Freezing (if cooling) b E _ Particles farthest apart c. D _ Boiling d. A _ Particle motion is most restricted e. B _ Heat of fusion f. B D _ All areas where energy change is potential only g. D _ Heat of vaporization h. B C D E _ All areas where particles move past each other. i. A _ Least kinetic energy j. A C E _ All areas where kinetic energy is changing k A _ most potential energy l. B D _ All areas where phase changes occur m. A C E _ All areas in which the heat is making the particles move faster n. B D _ All areas in which the heat is breaking the attractions or bonds between particles o. B D _ All areas in which the particles are not changing their speed Heat Removed Instructions: Given the following BP and MP information, sketch and label heating and cooling curves for the substances under the following changes in temperature AND phase. BPAg = 2162 C MPAg = 962 C BPPb = 1749 C MPPb = 328 C BPCu = 2927 C MPCu = 1085 C BPN = -196 C MPN = C 3. Water going from 24 C to 123 C. 6. Water going from 10 C to steam at100 C 9. Water going from 130 C to -30 C. 4. Water going from -24 C to 110 C. 7. Pb going from 900 C to -90 C. 10. Ice going from 0 C to water at 10 C. 5. Silver going from 1300 C to 50 C. 8. Cu going from 150 C to 1200 C. 11. Water at 0 C going to -50 C. 9.F Calorimetry: The Full Picture Instructions: ON A SEPARATE SHEET OF PAPER, (1) Sketch the heating or cooling curves (BE SURE TO LABEL!!!) associated with each of the following temperature changes and: (2) Calculate the amount of heat energy lost or gained during each temperature change given the information below right. Remember, Sig Figs, NW = NC, Boxed Answers and N 3 (this includes labeling and in your set-up!!!) g of H2O changing in temperature from C to 85.0 C q1 = m Cl ΔTl = 45.0 g 4.18 J/g C C = J q2 = m HVAP = 45.0 g J/g = J q3 = m Cg ΔTg = 45.0 g 2.08 J/g C C = J qt = J

5 g of H2O changing in temperature from C to C. q1 = m Cs ΔTs = 65.2 g 2.11 J/g C 35.0 C = 4820 J q2 = m HFUS = 65.2 g 334 J/g = J q3 = m Cl ΔTl = 65.2 g 4.18 J/g C C = J q4 = m HVAP = 65.2 g 2260 J/g = J q5 = m Cg ΔTg = 65.2 g 2.08 J/g C 5.0 C = J qt = J g of H2O(g) changing in temperature from C to C q1 = m HVAP = 15 g J/g = J q2 = m Cl ΔTl = 15 g 4.18 J/g C C = J q3 = m HFUS = 15 g J/g = -5.0 x 10 2 J q4 = m Cs ΔTs = 15 g 2.11 J/g C C = J qt = -48,000 J g of Au changing in temperature from 100. C to 1234 C q1 = m HFUS = g 64.4 J/g = J q2 = m C ΔT = g J/g C 1134 C = J qt = J g of O2 changing in temperature from C to C q1 = m HVAP = 57.2 g 13.8 J/g = 789 J q2 = m HFUS = 57.2 g 213 J/g = J q3 = m C ΔT = 57.2 g J/g C C = J qt = J g of H2O changing in temperature from C to C q1 = m Cs ΔTs = 8934 g 2.11 J/g C 90.0 C = 1.70 x 10 6 J ( ) q2 = m HFUS = 8934 g 334 J/g = J q3 = m Cl ΔTl = 8934 g 4.18 J/g C C = J q4 = m HVAP = 8934 g 2260 J/g = J q5 = m Cg ΔTg = 8934 g 2.08 J/g C 40.0 C = J qt = J g of H2O changing in temperature from C to H2O(s) at 0.00 C q1 = m Cg ΔTg = 41.5 g 2.08 J/g C C = J q2 = m HVAP = 41.5 g J/g C = J q3 = m Cl ΔTl = 41.5 g 4.18 J/g C C = J q4 = m HFUS = 41.5 g -334 J/g = J qt = J

6 g of He changing in temperature from C to C q1 = m HVAP = 57.2 g 5.23 J/g = 299 J q2 = m HFUS = 57.2 g 20.9 J/g = 1.20 x 10 3 J (1200) q3 = m C ΔT = 57.2 g 5.19 J/g C C = J qt = J 9.G Gases: KMT Instructions: Provide a response for each question that is well thought out, satisfies the prompt, is clearly explained, and LEGIBLE. 1. What theory explains the behavior of gases with respect to conditions such as temperature and pressure? Kinetic Molecular Theory (KMT) _ 2. Complete the followign statements in relation to the KMT. a. Gases consist of large numbers of tiny particles that are far apart relative to their size. This means that indiviual particle volume of the gas I negliabe_ (i.e. a gas is considered as a whole group rather than as individual interacting particles) b.collisions between gas particles and between particles and container walls are elastic collisions. This means that they collided with (No loss of KE/conversion to PE due to collisions) the will leave with the same amount of energy c. Gas particles are in constant, rapid, random motion. This can be inferred because that molecules travel in straight l ine pather between the collision at the wall of their container and collisions with each other d. There are no forces of attraction or repulsion between gas particles. This means that that no energy is required to separate the particles of a gas as there is no mutual attraction (No IMF s) e.the average kinetic energy of gas particles depends on the temperature of the gas. This means that temperature is proportional to the KE of the particles of a gas (But only when temperature is expressed in Kelvins) 3. Real _ gasses conforms to all postulates of the KMT. 4. In order to fully describe a gas, 4 _ measurable quantities must be stated. a. Define PRESSURE: amount of force exerted per unit of area _ units: 1 atm ( atmosphere ) = 760 _ mm Hg ( millimeters mercury ) = 760 _ torr = _ kpa ( kilopascals ) measured with a barometer _ b. Define TEMPERATURE: measure of the average KE of the particles of a substance _ units: degrees Celsius ( C _) or Kelvin ( K _) how to convert from C to K? K = C K _ c. Define VOLUME: measure of 3-D space occupied by a sample of matter_ units: 1 Liter (L) = 1000 _ ml = 1000 _ cm 3 = 100 _ dm 3 d. Define QUANTITY: amount of substance_ units: mole_ Convert from grams to moles using molar mass _ abbreviated MM _ 5. STP stands for Standard Temperature and Pressure _. The conditions at STP are exactly 1_ atm of pressure and a temperature of exactly K _ and any gas at STP will occupy a volume of 22.4 _ L. 9.H Gases: Variables of Gas Laws Instructions: ON A SEPARATE SHEET OF PAPER, preform the following temperature and pressure conversions. Remember, Sig Figs, NW = NC, Boxed Answers and N 3 (this includes labeling and in your set-up!!!) atm to mm Hg kpa to atm x 10 4 torr to mm Hg 2.00 atm 760 mm Hg = 1520 mmhg 115 kpa 1 atm = 1.13 atm 3.50x10 4 torr 760 mm Hg = 3.50 x 10 4 mmhg 1 atm kpa 760 torr mm Hg to kpa 7. 93,500 Pa to atm atm to kpa mm Hg kpa = mmhg 93,500 Pa 1 kpa 1 atm = atm atm kpa = 49.6 kpa 760 mm Hg 1000 Pa kpa 1 atm mm Hg to atm torr to atm C to Kelvin 500. mm Hg 1 atm = atm 950. torr 1 atm = 1.25 atm TK = TC = 120 C = 390 K 760 mm Hg 760 torr C to Kelvin K to C C to Kelvin TK = TC = C = K TC = TK = K = C TK = TC = C = K K to C C to Kelvin Kelvin to C TC = TK = 100. K = 173 C TK = TC = C = K TC = TK = 5 K = 268 C

7 Instructions: ON A SEPARATE SHEET OF PAPER, identify the values and symbols of all variables present within the following data. Be sure to denote the unknown variable, what you need to solve for, as = x. Variables to denote in include but are not limited to include n1, V1, P1, T1, n2, V2, P2, and T2. Remember, temperature for Gas Laws must be reported in K. No Work must be shown. 16. If 2.00 mol of gas occupies 4.50L at STP. How much of the same gas will occupy 3.00L at STP? n1 = 2.00 mol V1 = 4.50 L P1 = 1 atm T1 = K n2 = X mol V2 = 3.00 L P2 = 1 atm T2 = K 17. A gas has an initial volume of 15 L. If the temperature increases from 330 K to 450 K, find the new volume. n1 = constant V1 = 15 L P1 = constant T1 = 330 K n2 = constant V2 = X L P2 = constant T2 = 450 K 18. A gas exerts 1.2 atm of pressure. If the temperature is raised from 225 K to 325 K, find the new pressure. n1 = constant V1 = constant P1 = 1.2 atm T1 = 225 K n2 = constant V2 = constant P2 = X atm T2 = 325 K 19. Suppose 5.00 L of a gas is known to contain mol. If the amount of gas is increased to 1.80 mol, what new volume will result? n1 = mol V1 = 5.00 L P1 = constant T1 = constant n2 = 1.80 mol V2 = X L P2 = constant T2 = constant 20. A sample of oxygen takes up 34 dm 3 of space when it is under 500 kpa of pressure. When the pressure is changed to 340 kpa, find the new volume. n1 = constant V1 = 34 dm 3 P1 = 500 kpa T1 = constant n2 = constant V2 = X dm 3 P2 = 340 kpa T2 = constant 21. The pressure of some N2 drops from 315 kpa to 220 kpa. If the initial volume is 1.4 L, find the new volume. n1 = constant V1 = 1.4 L P1 = 315 kpa T1 = constant n2 = constant V2 = X L P2 = 220 kpa T2 = constant 22. The pressure of neon changes from 786 mm Hg to 1811 mm Hg. If the initial temperature 87 o C, what is the new temperature (in K)? n1 = constant V1 = constant P1 = 786 mm Hg T1 = 360. K n2 = constant V2 = constant P2 = 1811 mm Hg T2 = X K 23. When the temperature of a gas changes, its volume decreases from L to 7.92 L. If the final temperature is measured to be K, what was the initial temperature (in K)? n1 = constant V1 = L P1 = constant T1 = X K n2 = constant V2 = 7.92 L P2 = constant T2 = K 24. If 22.5 L of nitrogen at 748 mm Hg are compressed to 725 mm Hg at constant temperature. What is the new volume? n1 = constant V1 = 22.5 L P1 = 748 mm Hg T1 = constant n2 = constant V2 = X L P2 = 725 mm Hg T2 = constant 25. A gas with a volume of 4.0L at a pressure of 205kPa is allowed to expand to a volume of 12.0L. What is the pressure in the container if the temperature remains constant? n1 = constant V1 = 4.0 L P1 = 205 kpa T1 = constant n2 = constant V2 = 12.0 L P2 = X kpa T2 = constant 26. What pressure is required to compress liters of air at 1.00 atmosphere into a cylinder whose volume is 26.0 liters? n1 = constant V1 = L P1 = 1.00 atm T1 = constant n2 = constant V2 = 26.0 L P2 = X atm T2 = constant 27. A 40.0 L tank of ammonia has a pressure of 12.7 kpa. Calculate the volume of the ammonia if its pressure is changed to 8.4 kpa while its temperature remains constant. n1 = constant V1 = 40.0 L P1 = 12.7 kpa T1 = constant n2 = constant V2 = X L P2 = 8.4 kpa T2 = constant 9.I Gases: A,B,C and D of Gas Laws Instructions: ON A SEPARATE SHEET OF PAPER, preform the following basic gas law calculations. Remember, Sig Figs, NW = NC, Boxed Answers and N 3 (this includes labeling and in your set-up!!!) 1. If 2.00 mol of gas occupies 4.50L at STP. How much of the same gas will occupy 3.00L at STP? Base Equation: V1 = V2 Algebraic Rearrangement: n2 = _ n1 V2 _ n1 n2 V1 Required Work to Show: n2 = 2.00 mol 3.00 L 4.50 L Answer: 1.33 mol gas 4. Suppose 5.00 L of a gas is known to contain mol. If the amount of gas is increased to 1.80 mol, what new volume will result (at an unchanged temperature and pressure)? Base Equation: V1 = V2 Algebraic Rearrangement: V2 = _ V1 n2 _ n1 n2 n1 Required Work to Show: V2 = 5.00 L 1.80 mol mol Answer: 9.33 L gas 2. A gas has an initial volume of 15 L. If the temperature increases from 330 K to 450 K, find the new volume. Pressure is Constant, so cancel it out Base Equation: P1V1 = P1V2 Algebraic Rearrangement: V2 = _ V1 T2 _ T1 T2 T1 Required Work to Show: V2 = 15 L 330 K 450 K Answer: 20. L gas 5. A canister contains 425 kpa of carbon dioxide, 750 kpa of nitrogen, and 525 kpa of oxygen. What is the total pressure of the container? Base Equation: PT = P1 + P2 + P3 Algebraic Rearrangement: PT = PCO2 + PN2 + PO2 Required Work to Show: PT = 425 kpa kpa kpa Answer: 1.70 x 10 3 kpa 3. A gas exerts 1.2 atm of pressure. If the temperature is raised from 225 K to 325 K, find the new pressure. Volume is Constant, so cancel it out Base Equation: P1V1 = P1V2 Algebraic Rearrangement: P2 = _ P1 T2 _ T1 T2 T1 Required Work to Show: P2 = 1.2 atm 325 K 225 K Answer: 1.7 atm gas 6. A sample of oxygen takes up 34 dm 3 of space when it is under 500 kpa of pressure. When the pressure is changed to 340 kpa, find the new volume. Temperature is Constant, so cancel it out Base Equation: P1V1 = P1V2 Algebraic Rearrangement: V2 = _ P1 V1 _ T1 T2 P2 Required Work to Show: V2 = 34 dm kpa 340 kpa Answer: 50 dm 3 O2

8 7. The pressure of some N2 drops from 315 kpa to 220 kpa. If the initial volume is 1.4 L, find the new volume. Temperature is Constant, so cancel it out Base Equation: P1V1 = P1V2 Algebraic Rearrangement: V2 = _ P1 V1 _ T1 T2 P2 Required Work to Show: V2 = 1.4 L 315 kpa 220 kpa Answer: 2.0 L N2 8. The pressure of neon changes from 786 mm Hg to 1811 mm Hg. If the initial temperature 87 o C, what is the new temperature (in K)? Volume is Constant, so cancel it out Base Equation: P1V1 = P1V2 Algebraic Rearrangement: T2 = _ T1 P2 _ T1 T2 P1 Required Work to Show: T2 = 360. K 1811 mm Hg 786 mm Hg Answer: 829 K 9. When the temperature of a gas changes, its volume decreases from L to 7.92 L. If the final temperature is measured to be K, what was the initial temperature (in K)? Pressure is Constant, so cancel it out Base Equation: P1V1 = P1V2 Algebraic Rearrangement: T2 = _ T1 V2 _ T1 T2 V1 Required Work to Show: T2 = K 7.92 L L Answer: 202 K 10. If 22.5 L of nitrogen at 748 mm Hg are compressed to 725 mm Hg at constant temperature. What is the new volume? Temperature is Constant, so cancel it out Base Equation: P1V1 = P1V2 Algebraic Rearrangement: V2 = _ P1 V1 _ T1 T2 P2 Required Work to Show: V2 = 22.5 L 748 mm Hg 725 mm Hg Answer: 23.2 L N2 11. A tank containing ammonia and argon has a total pressure equal to 1.8 atm. The pressure of the ammonia is 1.2 atm. What is the pressure of the argon gas? Base Equation: PT = P1 + P2 + P3 Algebraic Rearrangement: PAr = PT PNH3 Required Work to Show: PAr = 1.8 atm 1.2 atm Answer: 0.6 atm Ar 12. A gas with a volume of 4.0L at a pressure of 205kPa is allowed to expand to a volume of 12.0L. What is the pressure in the container if the temperature remains constant? Temperature is Constant, so cancel it out Base Equation: P1V1 = P1V2 Algebraic Rearrangement: P2 = _ P1 V1 _ T1 T2 V2 Required Work to Show: P2 = 205 kpa 4.0 L 12.0 L Answer: 68.3 L N2 13. What pressure is required to compress liters of air at 1.00 atmosphere into a cylinder whose volume is 26.0 liters? Temperature is Constant, so cancel it out Base Equation: P1V1 = P1V2 Algebraic Rearrangement: P2 = _ P1 V1 _ T1 T2 V2 Required Work to Show: P2 = 1.00 atm L L Answer: 7.54 atm air 14. A 40.0 L tank of ammonia has a pressure of 12.7 kpa. Calculate the volume of the ammonia if its pressure is changed to 8.4 kpa while its temperature remains constant. Temperature is Constant, so cancel it out Base Equation: P1V1 = P1V2 Algebraic Rearrangement: V2 = _ P1 V1 _ T1 T2 P2 Required Work to Show: V2 = 12.7 kpa 40.0 L 8.4 kpa Answer: 60. L NH3 15. A container containing 5.00 L of a gas is collected at K and then allowed to expand to 20.0 L. What must the new temperature be in order to maintain the same pressure? Pressure is Constant, so cancel it out Base Equation: P1V1 = P1V2 Algebraic Rearrangement: T2 = _ T1 V2 _ T1 T2 V1 Required Work to Show: V2 = K 5.00 L 20.0 L Answer: 25.0 K 16. A gas occupies ml at a temperature of 27.0 C. What is the volume at K? Pressure is Constant, so cancel it out Base Equation: P1V1 = P1V2 Algebraic Rearrangement: V2 = _ V1 T2 _ T1 T2 T1 Required Work to Show: V2 = ml K K Answer: ml gas 17. If 15.0 liters of neon at 25.0 C is allowed to expand to 45.0 liters, what must the new temperature be to maintain constant pressure? Pressure is Constant, so cancel it out Base Equation: P1V1 = P1V2 Algebraic Rearrangement: T2 = _ T1 V2 _ T1 T2 V1 Required Work to Show: V2 = K 45.0 L 15.0 L Answer: 895 K 18. The pressure of a gas changes from kpa to 50.0 kpa. The volume changes from 45 L to 40 L. If the initial temperature is K, what is the final temperature in K? Base Equation: P1V1 = P1V2 Algebraic Rearrangement: T2 = _ T1 P2 V2 _ T1 T2 P1 V1 Required Work to Show: T2 = K 50.0 kpa 40 L kpa 45 L Answer: 100 K 19. A sample of nitrogen goes from 21 L to 14 L and its pressure increases from 100. kpa to 150. kpa. The final temperature is 300. K. What was the initial temperature in Kelvins? Base Equation: P1V1 = P1V2 Algebraic Rearrangement: T1 = _ T2 P1 V1 _ T1 T2 P2 V2 Required Work to Show: T1 = 300. K 150. kpa 14 L 100. kpa 21 L Answer: 300. K 20. A sample of argon goes from 500 K to 350 K and its pressure changes from 280 kpa to 380 kpa. If the initial volume is 18 dm 3, what is the final volume? Base Equation: P1V1 = P1V2 Algebraic Rearrangement: V2 = _V1 P1 T2 _ T1 T2 T1 P2 Required Work to Show: V2 = 18 dm kpa 350 K 500 K 380 kpa Answer: 9 dm 3 Ar 21. A sample of neon experiences a pressure drop from 75 kpa to 53 kpa. The temperature increases from K to K. If the initial volume is 12 L, what is the final volume? Base Equation: P1V1 = P1V2 Algebraic Rearrangement: V2 = _V1 P1 T2 _ T1 T2 T1 P2 Required Work to Show: V2 = 12 L 75 kpa K K 53 kpa Answer: 27 L Ne

9 9.J Gases: Ideal Gas Law 1.What are the differnces between an ideal gas and a real gas? Ideal = no individual particle volume or mass = elastic collisions = no IMF s = ALL moleules KE proprtional to temperature_ Real = individual particles with inelastic collision influnced by IMF s possessing nonuniform KE _ 2. REAL GASES BEHAVE NEARLY IDEALLY UNDER CONDITIONS of high _ temperature, low_ pressure, & low_ molar mass. Instructions: In P V = n R T: Complete the following statements to remind yourself of the requirments of the Ideal Gas law before completing the calualtions that follow. "P" stands for pressure_, must be in units of mmhg / torr / atm / kpa_ "V" stands for volume_, must be in units of liters_ "n" stands for mole_, must be in units of moles_ "T" stands for temperature_, must be in units of Kelvins_ "R" stands for the Ideal Gas Constant, has a value that varies_ dependent on unit of pressure used_ Instructions: ON A SEPARATE SHEET OF PAPER, preform the following basic gas law calculations. Remember, Sig Figs, NW = NC, Boxed Answers and N 3 (this includes labeling and in your set-up!!!) 3. If 3.7 moles of propane are at a temperature of 28 C and are under kpa of 8. Given 4 moles of a gas at a pressure of 5.6 atm and a volume of 12 liters, what is pressure, what volume does the sample occupy? the temperature? Base Equation: PV = nrt Algebraic Rearrangement: V = _ n RkPa T _ L kpa P Required Work to Show: V = 3.7 mol K kpa Answer: 60. L Propane Base Equation: PV = nrt Algebraic Rearrangement: T = _ P V _ n RmmHg Required Work to Show: T = 5.6 atm 12 L _ 4 mol A sample of carbon monoxide at 57 C and under 0.67 atm of pressure takes up 85.3 L of space. What mass of carbon monoxide is present in the sample? Base Equation: PV = nrt Algebraic Rearrangement: n = _ P V _ Ratm T Required Work to Show: n = 0.67 atm 85.3 L = 2.1 mol CO K Answer: 59 g CO 2.1 mol CO g CO _ 1 mol CO 5. At 45 C, 71 g of fluorine gas take up 6843 ml of space. What is the pressure of the gas, in kpa? Base Equation: PV = nrt Algebraic Rearrangement: P = _ n RkPa T _ V Required Work to Show: 71 g F2 1 mol F2 _ = 1.9 mol F g F2 Answer: 530 kpa F2 P = 1.9 mol K L 6. At 971 mm Hg, 145 g of carbon dioxide have a volume of L. What is the temperature of the sample, in C? Base Equation: PV = nrt Algebraic Rearrangement: T = _ P V _ n RmmHg Required Work to Show: 145 g CO 1 mol CO _ = 5.18 mol CO g CO Answer: 103 K T = 971 mm Hg L 5.18 mol At 137 o C and under a pressure of 3.11 atm, a 67.3 g sample of an unknown noble gas occupies L of space. What is the gas? Base Equation: PV = nrt Algebraic Rearrangement: n = _ Ratm T _ P V Required Work to Show: n = K = mol X 3.11 atm L Answer: Kr L kpa 67.3 g X _ = 83.7 g/mol X mol X L mmhg Answer: 200 K 9. An unknown quantity of gas at a pressure of 1.2 atm, a volume of 31 liters, and a temperature of 87 0 C, how many moles of gas are present? Base Equation: PV = nrt Algebraic Rearrangement: n = _ Ratm T _ P V Required Work to Show: n = K 1.2 atm 31 L Answer: 0.79 mol Gas 10. A vessel contains 3.21 moles of gas with a volume of 60.9 liters and at a temperature of K, what is the pressure inside the container? Base Equation: PV = nrt Algebraic Rearrangement: P = _ n RkPa T _ V L kpa Required Work to Show: P = 3.21 mol K 60.9 L Answer: 175 kpa or 1310 torr/mmhg or 1.73 atm 11. A vessel contains 7.7 moles of gas at a pressure of 0.09 atm and at a temperature of 56 0 C, what is the volume of the container that the gas is in ml? Base Equation: PV = nrt Algebraic Rearrangement: V = _ n RkPa T _ P Required Work to Show: V = 7.7 mol K = 2310 L 0.09 atm Answer: ml or 2.31 x 10 6 ml 12. A vessel contains 1.37 moles of gas at a temperature of C, and a volume of liters, what is the pressure of the gas in atm? Base Equation: PV = nrt Algebraic Rearrangement: P = _ n RkPa T _ V Required Work to Show: P = 1.37 mol K 88.9 L Answer: atm