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Amsat Chem IH

Gas Notes Topic#10

Table of Contents

Gases

1. Kinetic-Molecular Theory of Matter

2. Pressure and Dalton’s Law of Partial Pressure

3. Gas Laws

Molecular Composition of Gases

1. Volume-Mass Relationships of Gases

2. Ideal Gas Law

3. Effusion and Diffusion

4. Stoichiometry

Kinetic-Molecular Theory of matter Section#1

• Gases (Ideal gases) – Kinetic Molecular Theory of Gases

(1) Gases consist of large numbers of tiny particles - very far apart

(2) Gas collisions (between gas particles and container) are elastic

▪ No energy (KE) lost in collision (no speed/momentum lost)

(3) Gas particles in constant, random motion - high KE (motion)

(4) Gas particles not attracted/repulsed by other particles/container.

▪ Own entities with no interaction with other entities

(5) KEav = ½ mv2 depends on T (T in K for gases; K = 273 + °C)

▪ As T ↑’s, KEav (average KE) ↑’s

• The Kinetic-Molecular Theory and the Nature of Gases

o Expansion – No definite shape/volume (expand to fill container)

o Fluidity – Gases are fluids (they flow)

o Low Density – 1/1000 the density of liquids; gas density units, g/L

o Compressibility – Gases are compressible (at high compression (g) becomes (l)). Compression changes density.

o Diffusion/effusion

▪ Diffusion

• Spontaneous mixing of gases (gas moving through a gas)

▪ Effusion

• The speed at which a gas flows through a very small hole

• Balloon with a pin hole (how fast the balloon deflates)

• Deviations of Real Gases from Ideal Behavior

o The assumptions above are for an Ideal gas

o Real gases behave differently (deviate) from ideal gases

▪ The bigger the gaseous molecule - the further it deviates

• Cl2(g) deviates more than F2(g)

▪ Gases closest to ideal behavior

• He and H2 (very small and nonpolar)

▪ Ideal gas behavior

• Small, and nonpolar gas molecules at low P and high Tons

o P – pressure (atm, mmHg, kPa, psi, torr, Pa)

o V – volume (L or mL)

o n – number of moles of a gas (mol)

o R – Ideal gas constant

▪ 0.0821 when P is in atm (L•atm/mol•K)

▪ 8.31 when P is in kPa (L•kPa/mol•K)

▪ 62.40 when P is in mmHg (L•mmHg/mol•K)

o T – temperature (K)

o STP – Standard Temperature and Pressure is 0°C (273K) and 1.00atm

▪ a universal condition to standardize gas measurements

Pressure and Dalton’s Law of Partial Pressure Section#2

• Pressure and Force

o P = F/A = N/m2 = 1Pa (Pascal - SI pressure unit)

▪ F = m x a, A = l x l (area)

• N – Newton (unit for force)

• 1N = 1kg x 1m/s2 (m/s2 is an acceleration)

o Earth’s surface - 1kg mass exerts 9.8N of force

▪ F = 1kg x 9.8m/s2 = 9.8N (a = 9.8m/s2, gravity)

o Sample Problem 10.1 – Pressure

A ballet dancer exerts a 500N force on a wooden stage over an area equal to 325cm2.

(1) What is the pressure the dancer exerts on the floor? Ans: 1.5N/cm2

(2) On tip of toes with an area of 13cm2? Ans: 38.5N/cm2

(3) On the tip of one toe (A = 6.5cm2)?Ans: 77N/cm2

o Atmospheric Pressure

▪ Atmospheric pressure measured using a barometer

• Invented by Torricelli

• Reference point is sea level

▪ Individual gas pressure produced in a reaction is measured using a manometer

[pic][pic]

o Units for Pressure

▪ At sea level: 1atm = 760mmHg = 760torr = 101.3kPa = 1.013x105Pa = 14.7psi

o Sample Problem 10.2 – Pressure Unit Conversions

1) The average atmospheric pressure in Denver, Colorado, is 0.830atm. Express this pressure (a) in mmHg and (b) in kPa. Ans: 631mmHg, 84.1kPa

2) Convert a pressure of 570.torr to atms and kPa. Ans: 0.750atm, 76.0kPa

• Dalton’s Law of Partial Pressure

o A mixture of gases

o States the total pressure, PT, of a mixture of gases is equal to the sum of the partial pressures (Pa, Pb, etc.) of the component gases.

▪ PT = Pa + Pb + Pc + …

▪ Each gas contributes a pressure to PT

o Gases collected over H2O

▪ H2O(g) is present over H2O

▪ The amount of H2O(g) over H2O(l) varies with temperature

• The warmer H2O(l) , the more H2O(g) above liquid

▪ The PT above the H2O(l) for a gas collected over H2O(l) is the sum of the PH2O (at the T of the H2O(l)) and the collected gas.

• PT = Pdry gas + PH2O (dry gas is absent water vapor)

▪ Pdry gas = PT – PH2O

o Sample Problem 10.3 – Dalton’s Law of Partial Pressure

1) O2(g) from the decomposition of KClO3, was collected by water displacement. The barometric pressure and the temperature during the experiment were 731.0torr and 20.0°C, respectively. What was the PO2 collected? (PH2O = 17.5torr) Ans: 713.5torr

2) Some H2 gas is collected over water at 20.0°C. The levels of water inside and outside the gas-collection bottle are the same level. The PH2 is 0.947atm. What is the barometric pressure at the time the gas was collected? (PH2O = 17.5torr) Ans: 0.970atm

Mole Fraction

▪ The partial pressure of a gas is directly related to the moles of gas in a mixture

• Xa = na/nT (written in decimal form)

o na is the moles of gas (a)

o nT is the total moles of gas in the mixture

• Pa = XaPT

o Multiplying the mole fraction by the total P (PT)will give the partial P of that gas

o Sample Problem 10.4 – Mole Fraction

(1) A system has a mixture of O2, N2, CO2, CH4, and Ar gases. The number of moles for each gas is 0.23, 0.82, 0.040, 0.0023, and 0.035 moles, respectively. What is the mole fraction for each gas? Ans: 0.204, 0.727, 0.035, 0.002, and 0.031

(b) If the total pressure for the system is 345.6kPa, what is the partial P for each gas?

Ans: 70.5kPa, 251kPa, 12.1kPa, 0.7kPa, and 10.7kPa

Gas Laws Section#3

• Boyle’s Law: Pressure-Volume Relationship

o Robert Boyle discovered that doubling the P halved the volume

▪ Gas is at constant T and moles (no Δ)

o Kinetic molecular explanation

▪ P is caused by moving particles hitting the container

▪ When the V is decreased, with no change in # of particles, there will be an increase in the particles per unit volume and the number of collisions with the walls of the container will increase, so P increases

o States a gas at constant T and amount, V varies inversely with P

▪ Inverse proportion: as V↑P↓, and V↓P↑

• PiVi = PfVf

o PiVi → initial P and V

o PfVf → final P and V

• P and V Δ, T and amount do not Δ

o Sample Problem 10.5 – Boyle’s Law

1) A sample of oxygen gas has a V of 150.mL when its P is 0.947atm. At constant T and n, what is the V of the gas at a P of 0.987atm? Ans: 144mL

2) A gas has a P of 1.26atm and occupies a V of 7.40L. If the gas is compressed to a V of 2.93L, what will its P be, assuming constant T? Ans: 3.18 atm

3) Divers know the P exerted by the water increases about 100kPa with every 10.2m of depth. This means that at 10.2m below the surface, the P is 201kPa; at 20.4m, the P is 301kPa; and so forth. Given the V of a balloon is 3.5L at STP and the T of the water remains constant, what is the V 51m below the water’s surface? Ans: 0.59 L

o Graphing Exercise for Boyles Law (V vs. P)

▪ Create a 3x7 table in your notebook

▪ Fill data table and graph (V vs. P) data on paper provided

|Volume-Pressure Data for a Gas Sample (at constant mass and T) |

|Volume (cm3) |Pressure (psi) |V x P |

| | | |

| | | |

| | | |

| | | |

| | | |

o Question

▪ What can be done to the P/V data to change an inverse proportion (curved) graph into a straight line?

▪ What happens to the volume if the pressure is tripled?

▪ If the pressure is halved, what happens to the volume?

• Charles’s Law: Volume-Temperature Relationship

o Gases expand when heated and contract when cooled

o V and T can change (n and P are held constant (do not Δ))

o Kinetic molecular explanation

▪ Under constant pressure, as a gas is heated, the gas particles have more KE (move faster) and the volume has to increase to maintain the constant pressure.

▪ Jacques Charles found volume changes by 1/273 of the original volume for every 1°C (1K)

• Constant P and initial T of 0°C

• Same for all gases

• If the T is increases by 273°C, the volume doubles

• Theoretically no volume for a gas at absolute zero (0K, -273°C)

o Not actually possible due to the fact a gas will become a (l)/(s) at a temperature greater than absolute zero

o Charles’s Law states the V of a fixed mass of gas at constant P varies directly with the Kelvin T.

▪ Direct proportion between V and T: As T↑V↑ and T↓V↓

• Vi/Ti = Vf/Tf

o Vi/Ti → initial V and T

o Vf/Tf → final V and T

o Sample Problem 10.6 – Charles’ Law

1) A sample of neon gas occupies a V of 752mL at 25°C. What V will the gas occupy at 50°C if the P remains constant? Ans: 815mL

2) A helium filled balloon has a volume of 2.75L at 20°C. The volume of the balloon decreases to 2.46L after it is placed outside on a cold day. What is the temperature in K and °C? Ans: 262K, -11°C

o Graphing Exercise for Charles Law (V vs T)

▪ Fill last column of data table. Graph (V vs. T) on graph paper

|Volume-Temperature Data for a Gas Sample (at Constant mass and P) |

|Volume (mL) |Temperature (K) |V/T |

|1092 |546 | |

|746 |373 | |

|566 |283 | |

|548 |274 | |

|546 |273 | |

|544 |272 | |

|400 |200 | |

|100 |50 | |

• Gay-Lussac’s Law: Pressure-Temperature Relationship

o States: P of moles of gas at constant V varies directly with the T(K)

▪ n and V are held constant (does not Δ)

▪ Pi/Ti = Pf/Tf - direct proportion (P↑T↑ the P↓T↓)

o Sample Problem 10.7 – Gay-Lussac’s Law

1) The gas in an aerosol can is at a pressure of 3.00atm at 25°C. Directions on the can warn the user not to keep the can in a place where the temperature exceeds 52°C. What would the gas pressure in the can be at 52°C? Ans: 3.27atm

2) Before a trip from NY to Boston, the pressure in an automobile tire 1.8atm @ 20°C. At the end of the trip, the pressure gauge reads 1.9atm. What is the new Celsius temperature of the air inside the tire? (Assume tires with constant volume.) Ans: 36°C

• Combined Gas Law – P, T, and V Relationship

o When a gas sample undergoes changes in V, P, and T.

▪ n is the only condition held constant

o Expresses the relationship between P, V, and T for constant n

▪ PiVi = PfVf

Ti Tf

o Sample Problem 10.8 – Combined Gas Law

1) A He filled balloon has a volume of 50.0L at 25°C and 1.08atm. What volume will it have at 0.855atm and 10.°C? Ans: 60.0L He

2) A gas was collected over water with a volume of a gas is 27.5mL at 22.0°C and 0.974atm. What will the volume be at 15.0°C and 0.993atm (VPH2O at 20oC is 2.6kPa)? Ans: 25.6mL

3) A 700.mL gas sample at STP is compressed to a volume of 200.mL, and the temperature is increased to 30.0°C. What is the new pressure of the gas in Pa? Ans: 3.94 x 105Pa or 394kPa

Review

o Kinetic Molecular Theory for Gases

▪ Not attracted/repulsed by walls or each other

▪ Fast, constant, and random moving

▪ Collisions are elastic

▪ KE directly proportional to temperature

▪ Particles take up no volume

o P = F/A measured in atm, mmHg, torr, kPa, Pa, or psi

o 1atm = 760mmHg = 760torr = 101.325kPa = 1.01325x105Pa = 14.7 psi

o The P underwater increases by 1atm for every 30 feet. Do not forget the initial 1atm of P at the surface.

o Variables – P, V, n, and T

o Gas Laws

|Gas Laws |

| |Boyle’s Law |Charles’s Law |Guy-Lussac’s Law |Combined Gas Law |

|Relationship |P and V |V and T |P and T |P, V, and T |

|Proportionality |inverse |direct |direct |N/A |

|k |k = PV |k = V/T |k = P/T |k = PV/T |

|Constants |T and n |P and n |V and n |n |

|Graph |curved |straight |straight |N/A |

▪ Dalton’s Partial Pressure

• PT = Pa + Pb + Pc +…

o Use the problem solving process I demonstrate

▪ Construct a table for initial and final values, solve equation for unknown variable, plug in values for knowns, and solve for the unknown quantity.

Molecular Composition of Gases

Volume –Mass Relationships of Gases Section#1

• Volume-Mass Relationships of Gases

o Gay-Lussac

▪ Studied gas V relationships

o Constant T & P

• Observed 2L H2(g) reacts with 1L O2(g) to produce 2L H2O(g)

2H2(g) + O2(g) → 2H2O(g) 2:1:2

2L 1L 2L

o Mole ratios can be used as volume ratios

▪ Gay-Lussac’s law of combining volumes

• Constant T & P, volumes of gaseous reactants and products can be expressed as small whole number ratios

o Avogadro’s Law

▪ Equal volumes of gases at same T & P contain equal numbers of molecules

• Also, gas V directly proportional to amount of gas

2H2(g) + O2(g) → 2H2O(g) 2:1:2

2 molecules 1 molecules 2molecules

2 mol 1 mol 2 mol

2L 1L 2L

o Sample Problem 10.9 – Avogadro’s Law

1) How many moles of H2(g) are needed to completely react with two moles of N2(g)?

3H2(g) + N2(g) → 2NH3(g) Ans: 6mol H2

2) What volume of H2(g) is necessary to react with 5L of N2(g) to produce ammonia? Ans: 15L

o Molar Volume of Gases

▪ Standard molar volume of a gas

• The volume of 1 mol of a gas at standard conditions (STP)

o 1 mol of any gas has a volume of 22.4L (@ STP)

▪ Conversion: 1mol = 22.4L (@ STP)

▪ Use to find V or mol of a gas (@ STP)

o Sample Problem 10.10 – Molar Volume

1) A chemical reaction produces 0.0680 mol of O2(g). What V in L is occupied by this gas at STP? Ans: 1.52L

2) At STP, a sample of Ne(g) occupies 550.cm3. How many moles of Ne(g) does this represent? Ans: 0.0246 mol

Ideal Gas Law Section#2

• Derivation of the Ideal Gas Law

o The mathematical relationship between P, T, n, and V.

o PV = nRT (Pivnert)

▪ Volume always in liters (convert mL to L)

▪ Temperature always in Kelvin (convert °C to Kelvin)

▪ Universal gas constant, R

• What is the volume for 1 mole of any gas at STP?

o 22.4L = V (Solve for V in notes)

• Use the Ideal gas Equation to solve for R

o In atm? (Solve for R in notes)

▪ R = (PV)/(nT) = 0.0821 atm•L/mol•K

o In kPa? (Solve for R in notes)

▪ R = (PV)/(nT) = 8.31 kPa•L/mol•K

o In mmHg (or torr)? (Solve for R in notes)

▪ R = (PV)/(nT) = 62.4 mmHg•L/mol•K

o Sample Problem 10.11 – Ideal Gas Law I

What is the P in atmospheres exerted by 0.500mol sample of nitrogen gas in a 10.0L container at 298K? Ans: 1.22atm

o Sample Problem 10.12 – Ideal Gas Law II

What is the volume, in liters, of 0.250mol of oxygen gas at 20.0°C and 0.974atm of pressure? Ans: 6.17L

o Sample Problem 10.13 – Ideal Gas Law III

What mass of chlorine gas in grams, is contained in a 10.0L tank at 27°C and 3.50atm of pressure? Ans: 101g Cl2

o Gas Density

▪ Derived from Ideal Gas

• d = g/V

o Set n = g/MM

o Substitute g/MM for n in the Ideal Gas Equation

o PV=nRT

o PV=(g/MM)RT

o Solve for g/V

▪ g/V = MMxP = dgas

RxT

▪ Label is g/L

o Sample Problem 10.14 – Gas Density

1) At 28°C and 0.974atm, 1.00L of gas has a mass of 5.15g. What is the molar mass of this gas?

Ans: 131g/mol

2) What is the temperature of a gas with a molar mass of 44.01g/mol, a pressure of 0.980atm, and a density of 2.70g/L? Ans: 195K

3) What is the d of a sample of ammonia gas, NH3, if the P is 0.928atm and the T is 63.0°C?

Ans: 0.572g/L NH3

● Real gases

o Do not follow ideal gas law

o Most like ideal under low pressure and high temperature

o The smaller the gas particle, the more ideal it is.

Effusion and Diffusion Section#3

o Effusion

▪ The process by which a confined gas in a container randomly pass through a tiny opening in the container

▪ The faster the gas leaves the opening, the smaller the gas molecule (by mass)

▪ The slower the gas leaves the opening, the larger the gas molecule (by mass)

o Graham’s Law

▪ Relates the mass and velocity of two gases under the same condition.

• Equation

o Derivation

▪ Since the gases are under the same conditions they have the same KE.

▪ KE = ½ Mv2

• We use M for m because we use the molar mass of each gas

▪ KEA = KEB

▪ Substitute the equation in for KE

▪ ½ MAvA2 = ½ MBvB2

▪ Solving for the ratio of vA/vB

• vA/ vB = √MB/√MA

▪ States rates of effusion for two gases at the same conditions are inversely proportional to the square roots of their molar masses.

o Applications of Graham’s Law

▪ vA/ vB = √MB/√MA = √dB/√dA

o Sample Problem 10.15 – Effusion Rates

1) Compare the rates of effusion of hydrogen and oxygen. Ans: 3.98

2) A sample of hydrogen effuses through a porous container 9 times faster than an unknown gas. Estimate the MM of the unknown gas. Ans: 160g/mol

3) Compare the rate of effusion of carbon dioxide with that of hydrogen chloride at the same T and P. Ans: CO2 0.9 times as fast as HCl

4) If a molecule of neon gas travels at an average of 400m/s at a given T, estimate the average speed of a molecule of butane gas, C4H10, at the same T. Ans: 235 m/s

▪ All gases at the same T have the same KE

• Compare: H2 and Rn @ Same T

o KEH2 = ½ MH2v2H2 = ½ MRnv2Rn = KERn

▪ Why?

• What can be said about the KE of each particle in a mix of gases?

• Is each particle’s velocity the same?

• Are their masses the same?

Stoichiometry of Gases Section#4

• If conditions on the two gases are the same

o Use volume to volume calculations

▪ Use mole ratios from a balanced equation

o Sample Problem 10.16 – Molar Volume and Mass

1) A chemical reaction produced 98.0mL of SO2(g) at STP. What was the mass (in grams) of the gas produced? Ans: 0.280g SO2

2) What is the volume of 77.0g of NO2(g) at STP Ans:37.5L NO2

o Sample Problem – 10.17 – Gas Stoichiometry I

1) Propane, C3H8, is a gas that is sometimes used as a fuel for cooking (lets have some BBQ!) and heating. The complete combustion of propane occurs according to the following equation: C3H8(g) + O2(g) →CO2(g) + H2O(g). Assume all volume measurements are made at STP.

(a) Balance the equation.

(b) What will be the volume, in L, of oxygen required for the complete combustion of 0.350L of propane? Ans: 1.75L O2

(c) What will be the volume of carbon dioxide produced in the reaction? Ans: 1.05L CO2

2) Assuming all volume measurements are made at the same temperature and pressure, what V of hydrogen gas is needed to react completely with 4.55L of oxygen gas to produce water vapor? Ans: 9.10L H2

3) What volume of oxygen gas is needed to react completely with 0.626L of carbon monoxide gas, CO, to form gaseous carbon dioxide? Assume all volume measurements are made at the same temperature and pressure. Ans: 0.313L O2

• If the V of gas A is given at standard conditions (STP)

o Convert gas A to moles of A

▪ Use 1mol = 22.4L

o Use stoich (mole ratios) to find moles of B

o Find mass of B using mole ratios and MMB

▪ Gas A V (at STP) → moles of A → moles of B → grams of B

• If the V of gas A has conditions other than STP

o Convert V of gas A into moles of A using Ideal Gas Law, PV=nRT

o Use stoich (mole ratios) to find moles of B

o Find mass of B using MMB

▪ Gas A V (n=PV/RT)→ moles of A → moles of B → grams of B

• If the unknown is for the V of gas B

o STP conditions

▪ Mass A → moles A → moles B → V of gas B (use 22.4L = 1mol)

o Other conditions

▪ Mass A → moles A → moles B → V of gas B (V = nRT/P)

o Sample Problem 10.18 - Gas Stoichiometry II

1) Tungsten, W, a metal used in light-bulb filaments, is produced industrially by the reaction of tungsten (VI) oxide with hydrogen. WO3(s) + H2(g) → W(s) + H2O(l) (unbalanced). How many liters of hydrogen gas at 35°C and 0.980atm are needed to react completely with 875g of tungsten (VI) oxide? Ans: 292L H2

2) What V of chlorine gas at 38°C and 1.63atm is needed to react completely with 10.4g of sodium to from NaCl? Ans: 3.54L Cl2

3) How many liters of gaseous carbon monoxide at 27°C and 0.247atm can be produced from the burning of 65.5g of carbon according to the following equation? Ans: 544L CO

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