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Gases

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Published in: Chemistry
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Gases

Sachin S / Pune

14 years of teaching experience

Qualification: M.Sc (Pune University - 2008), B.Sc (Yashwant College - 2006)

Teaches: Chemistry, AIEEE, BITSAT, CET, IIT JEE Advanced, IIT JEE Mains, AIPMT, Medical Entrance Exams, NEET

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  1. Chapter 1 1 Gases An Introduction to Chemistry by Mark Bishop
  2. Chapter Map Particle nature of gases (Section 3. l) •eal ga so Relationship between velocity, temperature, and kinetic encr9' (Section 4. I) •ses and their provk•rt• Relationship between lume and press Ideal gas equation &nbincd gas law eq Combined gas law equation calculations Relationship between ure and tempera cure Ideal gas equation calculations Gases and equation Stoich calculations Relationship between moles and volume Relationship between e and temper Relationship between les and pressu Ternsxrature conversions (Section 2.6) Molar volume at stan temperature and pressure Equation stoichiometry calculations (Sections 10.1 and 10.3) rccnt yield calculations (Section 10.2) Dalton's law Of Dalton's Partial Pressures Partial Pressu calculations
  3. Gas 'Ille particles move rapidly and collide constantly. Particles occupy a small part of the total volume. Little mutual attraction or repulsion between particles Collisions causc changes in direction and velocity.
  4. Gas Model • Gases are composed of tiny, widely-spaced particles. — For a typical gas, the average distance between particles is about ten times their diameter.
  5. Gas Model (cont.) • Because of the large distance between the particles, the volume occupied by the particles themselves is negligible (approximately zero). — For a typical gas at room temperature and pressure, the gas particles themselves occupy about 0.1% of the total volume. The other 99.9% of the total volume is empty space. This is very different than for a liquid for which about 70% of the volume is occupied by particles.
  6. Gas Model (cont.) • The particles have rapid and continuous motion. — For example, the average velocity of a helium atom, He, at room temperature is over 1000 m/s (or over 2000 mi/hr). The average velocity of the more massive nitrogen molecules, N2, at room temperature is about 500 m/s. — Increased temperature means increased average velocity of the particles.
  7. Gas Model (cont.) • The particles are constantly colliding with the walls of the container and with each other. — In a typical situation, a gas particle moves a very short distance between collisions. Oxygen, 02, molecules at normal temperatures and pressures move an average of 10-7 m between collisions.
  8. Gas Model (cont.) • There is no net loss of energy in the collisions. A collision between two particles may lead to each particle changing its velocity and thus its energy, but the increase in energy by one particle is balanced by an equal decrease in energy by the other particle.
  9. Ideal Gas Assumptions • The particles are assumed to be point-masses, that is, particles that have a mass but occupy no volume. • There are no attractive or repulsive forces at all between the particles.
  10. Gas Pressure Pressure (P) = Force/Area Gas particles are constantly colliding with the walls of their container, and each time a particle hits a wall, it exerts a force against the wall. This force is proportional to its momentum, which is mass times velocity. — A more massive particle moving at a given velocity will exert more force against the wall. — A faster moving particle exerts more force against the wall than a slower moving particle of the same mass. Force due to particle collisions with the walls Gas pressure = Area of the walls
  11. Gas and their Units Pressure (P) = Force/Area — units • 1 atm = 101.325 kPa = 760 mmHg = 760 torr • 1 bar = 100 kPa = 0.9869 atm = 750.1 mmHg • Volume (V) — unit usually liters (L) • Temperature (T) c + 273.15 Number of gas particles expressed in moles (n)
  12. Gas Law Objectives For each of the following pairs of gas properties, (1 ) describe the relationship between them, (2) describe a simple system that could be used to demonstrate the relationship, and (3) explain the reason for the relationship. • V and P when n and T are constant • P and T when n and V are constant • V and T when n and P are constant • n and P when V and T are constant • n and V when P and T are constant
  13. Apparatus for Demonstrating Relationships Between Properties of Gases Valve to add and remove gas Movable piston Thermometer Pressure gauge
  14. Decreased Volume Leads to Increased Pressure Volume decreased p IN if n and T are constant https://preparatorychemistry.com/Boyles Law A constant number of gas particles Constant temperature Increased pressure Canvas.html
  15. Relationship between P and V Decreased volume number of gas particles Increased volume of container Increased number of particles close to any area of wall number of collisions per second Increased area of wall force due to collisions Increased area of wall Increased gas pressure
  16. Boyle' s Law The pressure of an ideal gas is inversely proportional to the volume it occupies if the moles of gas and the temperature are constant. 1 if n and T are constant
  17. Increased Temperature. Leads to Increased Pressure Piston locked In position Heat added Constant number of gas particles Constant volume Increased temperature Dlncreasedpressure P T if n and V are constant https://preparatorychemistry.com/Gay_Lussacs Law Canvas.html
  18. between P and T„,—-— Increased temperature Increased average velocity of the gas particles Increased number of collisions with the wa s Increased force per collision Increased total force of collisions force due to collisions Increased area of wall Increased gas pressure
  19. Law The pressure of an ideal gas is directly proportional to the Kelvin temperature of the gas if the volume and moles of gas are constant. PTT if V and n are constant
  20. Increased TemperäfüFé Leads to Increased-Volume Piston free to move Heat added Constant number of gas particles Increased vo ume Increased temperature Constant pressure if n and P are constant https://preparatorychemistry.com/Charles Law Canvas.html
  21. between T and Increased temperature Increased average velocity of the gas particles Increased number of Increased force per collision collisions with the walls Initial increase in force per area—that is, in pressure Inside pressure is greater than external pressure Container expands —Y Increased volume Decreased pressure until the inside pressure equals the external pressure
  22. Charles' Law For an ideal gas, volume and temperature described in kelvins are directly proportional if moles of gas and pressure are constant. Voc T if n and P are constant
  23. Increased Moles of Gas Leads to Increased Pressure Ell* Piston locked in position Gas added Constant volume Constant temperature Increased pressure poc n if T and V are constant https://preparatorychemistry.com/Moles Pressure Law Canvas.html
  24. between n and Increased number of gas particles Increased number of collisions with the walls Increased total force of collisions Increased gas pressure
  25. Relationship Between—. Moles of Gas and Pressure • If the temperature and the volume of an ideal gas are held constant, the moles of gas in a container and the gas pressure are directly proportional. Pun if T and V are constant
  26. Increased Moles of Gas Leads to Increased Volume--•-— Piston free to move Gas added VT n if T and P are constant https://preparatorychemistry.com/Avogadros At constant pressure... Increased volume Constant temperature Law Canvas.html
  27. between n and Increased number of gas particles Increased number of collisions with the walls Increased total force of collisions Initial increased in force per area - that is, in pressure Inside pressure is greater than external pressure Container expands Increased volume Decreased pressure until the inside pressure equals the external pressure
  28. Avogadro' s Law For an ideal gas, the volume and moles of gas are directly proportional if the temperature and pressure are constant. VT n if T and P are constant
  29. Engines and Pressure Elhe combustion of the gasoline leads to an increase in moles of gas, which also causes the gas pressure to increase. Decreased volume leads to increased gas pressure inside the cylinder. Gaseous gasoline mixed with air moves into the cylinder. 2 3 "Ihe reaction is exothermic, so the temperature of the product gases increases, contributing to the increased gas pressure. The increased pressure pushes the piston down, turning the crankshaft and ultimately the car's wheels. Cylinder Piston Crankshaft Sample reaction: 2C8H18(g) + 2502(g)— 16C02(g) + 18H20(g) + energy 27 moles of gas -+ 34 moles of gas
  30. Breathing — 101.3kPa External Pressure Diaphragm The diaphragm contracts, and the chest expands, causing the lungs to expand. The increased volume decreases the pressure in the lungs to be ow the external pressure, causing air to move into the lungs faster than it moves Out. — 101.3 kPa The diaphragm relaxes and the chest returns to its original volume, causing the lung volume to decrease. This increases the pressure in the lungs, causing air to move out of the lungs faster than it moves in.
  31. Gas Problems • Ideal gas equation problems. • Combined gas equation problems • Gas stoichiometry problems Dalton's Law of partial pressures problems
  32. Algebraic Approach to Problem Solving Write down the values you are given, assigning variables to each. Write the appropriate equation. Solve for the variable of your unknown. Plug in values given, including their units. Cancel units and if necessary, do unit conversions to get the units to cancel. Do the calculation and report your answer with the correct significant figures and unit.
  33. Unit Check • If your units cancel to yield the correct unit or units, you know that you... — used the correct equation, — did the algebra correctly, — and did all the necessary unit conversions.
  34. Ideal Gas Equation Derivation Poc n PTT 1 if Tand Vare constant if n and Vare constant if n and Tare constant P oc P = (a constant) — PV- constant) T 0.082058 L 8.3145 L. kl)a PV=nRT R = or K • mol K • mol
  35. Expanded Gas Equation nRT R- n = moles = -g 0.082058 L • atm 8.3145 L • kl)a or K • mol grams grams mole K. mol 1 mol mass in grams molar mass ART g = mass M = molar mass
  36. Ideal Gas Problems • Tip-off — The usual tip-off that you can use the ideal gas equation to answer a question is that you are given three properties of a sample of gas and asked to calculate the fourth. A more general tip-off is that only one gas is mentioned, there's not chemical reaction mentioned, and there are no changing properties.
  37. Ideal Gas Problem Step 1 • Step 1: Assign variables to the values given and the value that is unknown. Use P for pressure, V for volume, n for moles, T for temperature, g for mass, and M for molar mass.
  38. Ideal Gas Problem Step 2 Step 2: Write the appropriate form of the Ideal Gas Equation. — If the number of particles is given or desired in moles, use the most common form of the ideal gas equation. 0.082058 L atm 8.3145 L. 1<Pa PV- nRT or K • mol K • mol — If mass or molar mass is given or desired, use the expanded form of the ideal gas equation. P V = -CRT g = mass M = molar mass
  39. Ideal Gas Problem Steps 3-6.,.--— Step 3: Rearrange the equation to isolate the unknown. Step 4: Plug in the known values, including units. Be sure to use kelvin temperatures. Step 5: Make any necessary unit conversions and cancel your units. Step 6: Calculate your answer and report it to the correct significant figures and with the correct unit.
  40. Example Most incandescent light bulbs contain argon, but krypton gas does a better job than argon of slowing the evaporation of the tungsten filament in the light bulb. Because of its higher cost, however, krypton is only used when longer life is considered to be worth the extra expense. How many moles of krypton gas must be added to a 175-mL incandescent light bulb to yield a gas pressure of 117 kPa at 21.6 oc?
  41. Example 1 How many moles of krypton gas must be added to a 175-mL incandescent light bulb to yield a gas pressure of 117 kPa at 21.6 oc? n = ? V: 175mL P = 117 kPa PV = nRT T = 21.6 oc + 273.15 = 294.8 K 117kPä (175 n,Æ) 103 8.3145-E.kPx 294.8k k • mol = mol
  42. Example 2 What is the volume of an incandescent light bulb that contains 1.196 g Kr at a pressure of 1.70 atm and a temperature of 97 oc? 1.196 g 1.70 atm T = 97 oc + 273.15 = 370 K P V = g RT g = mass M = molar mass 1.196* gRT PM 0.082058 k • mol 370 k 0.255 L Kr (1.70 aem) 83.80
  43. Example 3 What is the density of krypton gas at 18.2 oc and 762 mmHg? = ? T = 18.2 oc +273.15= 291.4K 762 mmHg P V = g RT g = mass M = molar mass g PM 83.80 g 762 mmHg I-mol 0.082058 L•atm 291.4k = 3.51 g/L 1 -atm- 760 mmHg
  44. Combined Gas Law Equation Derivation nRT 1b 16 - nlRTl 1)2V2 so 711 T 1 16 so
  45. Combined Gastäw Equation Problems—m Tip-off — The problem requires calculating a value for a gas property that has changed. In other words, you are asked to calculate a new pressure, temperature, moles, or volume of gas given sufficient information about the initial and final of the other properties. A more general tip-off is that only one gas is mentioned, there's no chemical reaction mentioned, and there are changing properties.
  46. Combined Gas Equation Problem Steps 1 and 2 Step 1: Assign the variables P, T, n, and V to the values you are given and to the unknown value. Use the subscripts 1 and 2 to show initial or final conditions. Step 2: Write out the combined gas law equation, and eliminate the variables for any constant properties. (You can assume that the properties not mentioned in the problem remain constant.)
  47. Combined Gas Law—.-. Equation Problem steps 3-6 • Step 3: Rearrange the equation to isolate the unknown property. • Step 4: Plug in the values for the given properties. • Step 5: Make any necessary unit conversions and cancel your units. • Step 6: Calculate your answer and report it with the correct units and significant figures.
  48. Example 4 A helium weather balloon is filled in Monterey, California, on a day when the atmospheric pressure is 102 kPa and the temperature is 18 oc. Its volume under these conditions is 1.6 x 104 L. Upon being released, it rises to an altitude where the temperature is -8.6 oc, and its volume increases to 4.7 x 104 L. Assuming that the internal pressure of the balloon equals the atmospheric pressure, what is the pressure at this altitude?
  49. Example 4 A helium weather balloon is filled in Monterey, California, on a day when the atmospheric pressure is 102 kPa and the temperature is 18 oc. Its volume under these conditions is 1.6 x 104 L. Upon being released, it rises to an altitude where the temperature is -8.6 oc, and its volume increases to 4.7 x 104 L. Assuming that the internal pressure of the balloon equals the atmospheric pressure, what is the pressure at this altitude? PI = 102 kPa +273.15=291 K VI = L = -8.6 +273.15=264.6 K VI =4.7x104L 1b 16 264.6 k 103 kPa 291 K = 32 kPa
  50. Measurable Property and Moles Measurable property Mass for (s) and (l) Volume of solution for (aq) Volume of gas for (g) Molar mass Molarity 3 ways Moles Moles of pure substance Moles of solute Moles of gas
  51. Conversion between moles and volume of gas Using molar volume at STP (only for STP, which is rare) Using the Ideal Gas Equation nRT R as a conversion factor K•mol 8.3145 L•kPa K' mol 0.082058 L•atm or or 8.3145 L•kPa K•mol 0.082058 L•atm K•mol
  52. Standard Temperature and Pressure (STP) and Molar Volume • Common: 0 oc (273.15 K) and 1 atm 8.3145 L•kPx (273.15 K) k •mol 101.325kPc • Correct: O oc (273.15 K) and 1 bar (100 kPa) 8.3145 L•kPx (273.15 K) k •mol IOOkPa I mol STP 22.711 L I mol STP
  53. Eq u ation Stoichiometry Start here when Mass substance I mass IS given. Start here when volume of solution is given. Volume of solution I Molarity I R as a conversion factor Volume of gas I Mass substance 2 Molar mass 2 Molar mass I (coeffflcient 2) mol 2 (coefficient 1) mol I moles I •lhis is the core Of any equation stoichiometry moles 2 problem. or = Start here when volume of gas is given. Ras a conversion factor Volume Of solution 2 Molarity 2 Can be converted into mass, into volume of solution, or into volume of nRT or V = Volume Of gas 2 atP2&T2
  54. Example 1 Iron is combined with carbon in a series of reactions to form pig iron, which is about 4.3% carbon. The first step in this process is the reaction of carbon with oxygen to form carbon monoxide. For this reaction, what is the maximum volume of carbon monoxide at 105 kPa and 35 oc that could form from the conversion of 8.74 x 105 L of oxygen at 99.4 kPa and 27 oc? Start here when Mass substance I mass grvcn. Start hem when u•lume Mass substance 2 Molar mass 2 of solution is given, Volume of solution I Molarity I R as a conversion fact or Volume Of gas I mass I 2) mol (coefficient I) mol I moles 2 moles I Ibis is the core of any equation stoichiomerry problem. conversion Start here when volume of gas is given. Volume of solution 2 Can bc converted into mass, into 4_ volume of solution, or into volume of Volume of gas 2 at P2&T2
  55. Example 1 The first step in this process is the reaction of carbon with oxygen to form carbon monoxide. For this reaction, what is the maximum volume of carbon monoxide at 105 kPa and 35 oc that could form from the conversion of 8.74 x 105 L of oxygen at 99.4 kPa and 27 oc? Conversion from units of one substance to units of another substance, both involved in a chemical equation, so it's equation stoichiometry. Write a balanced equation. 2C + 02 2C0
  56. Example 1 For this reaction, what is the maximum volume of carbon monoxide at 105 kPa and 35 oc that could form from the conversion of 8.74 x 105 L of oxygen at 99.4 kPa and 27 oc? 2C + 02 -i 2C0 Convert from the units of substance 1 you have to moles substance 1. n 02 99.4kPt(8.74 x 105 E) 8.3145E.kPa (300 K) k• mol = 3.48 x 10 mol 02
  57. Example 1 For this reaction, what is the maximum volume of carbon monoxide at 105 kPa and 35 oc that could form from the conversion of 8.74 x 105 L of oxygen at 99.4 kPa and 27 oc? 2C + 02 -i 2C0 n 02 99.4kPt(8.74 x 105 E) 8.3145E.kPa (300 K) k• mol = 3.48 x 104 mol 02 Convert from moles of substance 1 to moles substance 2. 2 mol CO ? mol CO = 3.48 x I xuol--Oi = 6.96 x 104 mol CO
  58. Example 1 For this reaction, what is the maximum volume of carbon monoxide at 105 kPa and 35 oc that could form from the conversion of 8.74 x 105 L of oxygen at 99.4 kPa and 27 oc? 2C + 02-9 2C0 2 mol CO ? mol CO = 3.48 x I xuol--Oi = 6.96 x 104 mol CO Convert from the moles of substance 2 to units of substance 2 that you want. 6.96 x nRT 308 k k • mot - 1.70 x 106 L CO 105 4epa-
  59. Example 1 For this reaction, what is the maximum volume of carbon monoxide at 105 kPa and 35 oc that could form from the conversion of 8.74 x 105 L of oxygen at 99.4 kPa and 27 oc? 2C + 02 2C0 Or unit analysis, using R as a conversion factor. ? L CO ? L CO ? L CO ? L CO - 8.74 x 105 L 02 K • mol = 8.74 x 105% 02 8.3145 kPa K• mol = 8.74 x 105k 02 8.3145E•kPa = 8.74 x 105±-03 8.3145E•kPx 99.4kPa- 300k 99.4 300 k 2 mol CO 1 mol 02
  60. Example 1 For this reaction, what is the maximum volume of carbon monoxide at 105 kPa and 35 oc that could form from the conversion of 8.74 x 105 L of oxygen at 99.4 kPa and 27 oc? 2C + 02 2C0 Or unit analysis, using R as a conversion factor. ? L CO ? L CO ? L CO 8.3145E•kPa = 8.74 x 8.3145E•kPT K • mol - 8.74 x 8.3145-E•k4Xx = 1.70 x 106 L CO 99.4 kPa- 300 k 99 4kPa- 300 k 300 k 2 mol CO 1 mod 2 mol CO I mol 2 moi CO I mol 8.3145 L.kPa K • mot 8.3143 L.kPT K • mot 8.3145 1..kPa 308 105kPa 308 k 105kPa
  61. Example 2 In the reaction of carbon with oxygen to form carbon monoxide, what minimum volume of oxygen at STP (273.15 K and 1 bar) is necessary to convert 125 Mg of carbon to carbon monoxide? 2C + 02-9 2C0 Start here when Mass substance I mass us given. Start here when Of solution is given. Volume of solution I NIOlarity I Ras a conversion factor Volume of gas I at PI & Tl Mass substance 2 Molar mass 2 Molar mass I (coeffficicnt 2) mol 2 (coefffcietlt l) mol I moles 2 moles 1 lhis is the core Of any equation stoic-biometry problem. or u = Start here when volume Of gas is given. Ras a conversion factor Volume of solution 2 Molarity 2 Can be converted Into mass, Into volume of solution, or into volume of nRT or V = Volume of gas 2
  62. Example 2 • In the reaction of carbon with oxygen to form carbon monoxide, what minimum volume of oxygen at STP (273.15 K and 1 bar) is necessary to convert 125 Mg of carbon to carbon monoxide? ? L 02 2C + 02 2C0 106 & I mol-e = 125 Mgr Mg 12.011 - L 02 106* 2 - 123 Mge I Mg 1.18x105 L 02 I 02 8.3145 L.kPa 12.011 22.711 L 02 1 STP 273.15k 100 kPa
  63. Example 3 Sodium hypochlorite, NaOCl, found in household bleaches, can be made from a reaction using chlorine gas and aqueous sodium hydroxide: C12(g) + 2NaOH(aq) -9 NaOCl(aq) + NaCl(aq) + H20(/) What minimum volume of chlorine gas at 101.4 kPa and 18.0 oc must be used to react with all the sodium hydroxide in 3525 L of 12.5 M NaOH? Start when •lass substance Start when volume of is given Volume of solution I Mass substance 2 Volar mass 2 factor Volume Of I moles 2 moles I is the any equation stoichiometry problem. here when of gas is given. Volume of solution 2 Molarity 2 Can or into volume Of Volume of gas 2
  64. Example 3 C12(g) + 2NaOH(aq) NaOCl(aq) + NaCl(aq) + H20(/) What minimum volume of chlorine gas at 101.4 kPa and 18.0 oc must be used to react with all the sodium hydroxide in 3525 L of 12.5 M NaOH? ? L C12 = 35254*LaOH-sohr 1 12.5 mol NaOH ? L C12 — 3525 -L-NaOH-soh-r 12.5 me-V-NaO-H ? L C12 - 35254s-NaOH.ohv 1 mol C12
  65. Example 3 C12(g) + 2NaOH(aq) NaOCl(aq) + NaCl(aq) + H20(/) What minimum volume of chlorine gas at 101.4 kPa and 18.0 oc must be used to react with all the sodium hydroxide in 3525 L of 12.5 M NaOH? ? L 12.5 ? L 12.5 ? L 12.5 I L--Na044-sol-n = 5.26x105 L C12 I -H:tcyl C,12 2 mol-NaOH- I C,12 2 mol—NaOH- I C,12 2 mol-NaOH- 8.3145 L.kPa 8.3145 L.kPx 8.3145 291.0 k 101.4 291.0k 1 01.4
  66. Gas Pressure • Gas pressure is the total force due to particle collisions with the walls of the container at an instant in time divided by the area of the walls. • The total force of collisions is determined by • The number of collisions • The average force per collision
  67. Force per Collision • The force per collision is proportional to the momentum of the colliding particle. • Momentum is mass times velocity. • At a constant temperature, the mass and average velocity are constant, so the average momentum is constant, so the average force per collision is constant.
  68. Ideal Gas Assumptions • The particles are assumed to be point-masses, that is, particles that have a mass but occupy no volume. • There are no attractive or repulsive forces at all between the particles.
  69. M Of Gases P total - PA + PB • Partial pressure can be defined as the portion of the total pressure that one gas in a mixture of gases contributes. • For ideal gases, this is the pressure that a gas would have if it were alone in the container.
  70. Dalton' s Law of Partial Pressures Dalton's Law of Partial Pressures states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of all the gases. Ptotal = PA + PB n BRTB PB VB total VA + VB = V Ptotal VB TA+TB (TIA + 11B) —
  71. Dalton' s Partial Pressures— tiff. Ptotal = EPpartial or P total - PA + PB Ptotal each gas)
  72. Dalton' s Law-ofPäifiäf Pressures Problems— • Tip-off — The problem involves a mixture of gases and no chemical reaction. You are asked to calculate a value for one of the variables in the equations below, and you are given (directly or indirectly) values for the other variables.
  73. Dalton' s Law of_Partial Pressures Problem steps 1 & 2 Step 1: Assign variables to the values that are given and the value that is unknown. Step 2: From the following equations, choose the one that best fits the variables assigned in Step 1. Ptotal = EPpartial or Ptotal = each gas) —
  74. Dalton' s Law Partial Pressures Problem steps Step 3: Rearrange the equation to solve for your unknown. Step 4: Plug in the values for the given properties. Step 5: Make sure that the equation yields the correct units. Make any necessary unit conversions. Step 6: Calculate your answer and report it with the correct units and significant figures.
  75. Example 1 A typical "neon light" contains neon gas mixed with argon gas. If the total pressure of the mixture of gases is 1.30 kPa and the partial pressure of neon gas is 0.27 kPa, what is the partial pressure of the argon gas? p 1.30 kPa p Ne = 0.27 kPa P Ar = ? Ar T Ne- 1.30 kPa -0.27 kPa = 1.03 kPa
  76. Example 2 If 6.3 mg of Ar and 1.2 mg Ne are added to a 375-mL tube at 291 K, what is the total pressure of the gases in kilopascals? 375mL T = 291K 11Ar total ? mol Ar = 6.3 mg-Ar 103 mg — ? mol Ne = 1.2 mg-Ne 103 mg I mol Ar 39.948g-Ar I mol Ne 20.1797 0.00016 mol Ar 0.000059 mol Ne P total = each gas Ptotal = (TIA r + n N) 8.31/15k.kPa (291 K) 103 mE (0.00016 mol-AT + 0.000059 mol-Ne)
  77. Tip-offs for Problems • Ideal Gas Equation: 1 gas, no chemical reaction, and no changing properties. Combined Gas Law Equation: 1 gas, no chemical reaction, changing properties. Gas Stoichiometry: converting from one substance to another, both in a chemical reaction, and one or both are gases. Dalton's Law of Partial Pressures: 2 or more gases, no changing properties, no chemical reaction.

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