Chapter 6: Gases 6.1 Measurements on Gases MH5, Chapter 5.1 Let’s look at a certain amount of gas, i.e. trapped inside a balloon. To completely describe the state of this gas one has to specify the following quantities: Volume Symbol: V Units: m3 1 m3 = 1000 L = 1000 dm3 1 L = 10-3 m3 Temperature Symbol: T Units: K (“Kelvin”). The Kelvin scale is the “absolute temperature” scale. To convert ° C to K we simply add 273 (actually it is 273.15, but in most cases “273” will do) Example: 25° C = (25 + 273) K = 298 K Pressure Symbol: p Units: force per area 1Pa (Pascal) = 1 N/m2 1000 Pa = 1 kPa Other units are still used for pressure:

• psi = pound per square inch (we are not going to use that) • 105 Pa = 1 bar • 101300 Pa = 1.013 bar = 1 atm = 760 mm Hg (= 760 Torr)

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A mercury barometer can be used for measuring gas pressure “Normal pressure” corresponds to 101300 Pa = 760 mm Hg ( =1 atm). The Hg level will be lower when you move this device from the basement to the third floor, or when a rainstorm is moving in (why?) MH5, Figure 5.1

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Number of Moles With gases, like with any other substances we can use the relationship

m = MM × n

to convert mass (m) to number of moles (n), if we know the molar mass (MM). Example: How many moles of N2 molecules are there in 2.50 g? Calculations involving gases can be confusing because of the different units that are sometimes used for the same quantity. That means: before you do any calculations

• Convert the volume to L • Convert the pressure to kPa • Convert the temperature to K • Convert the “amount” of material to moles

… and you’ll be OK ☺

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6.2 The Ideal Gas Law MH5, Chapter 5.2 Pressure, temperature, volume, and number of moles are related by the ideal gas law:

p × V = n × R × T where R = 8.31 L kPa mol-1 K-1 is a constant If you use this value of R in calculations, V has to be in L, p has to be in kPa !!!! (In SI units: R = 8.31 J mol-1 K-1) Although this law only applies to “ideal” gases (meaning that the molecules do not interact with each other, individual gas molecules have zero volume) it describes the behavior of real gases (He, O2, N2, … remarkably well). Remember: n is the number of gas “particles” (not gas

atoms !!!!), i.e. for 1 mol He n = 1 for 1 mol N2 n = 1

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What does the ideal gas law tell us? Let’s try this for V • The volume is proportional to the number of

moles (for fixed T, fixed p): V = • The volume is proportional to the temperature

(for fixed n, fixed p): V = • The volume is inversely proportional to the

pressure (for fixed n, fixed T) V =

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Example: How much volume is occupied by one mole of gas at standard temperature and pressure (“STP”), i.e. 0° C and 101.3 kPa (1 atm)?

This result does not depend on the nature of the gas! 1 mol He occupies the same volume as 1 mol Cl2 … Figure MH5 p.107

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A question from the Dec. 2002 exam: … and another one

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Gases are mostly empty space, and the molecules of an “ideal gas” are assumed to have zero volume. Q: So why do gases occupy volume, and where does gas pressure come from? A: Gas molecules are in constant “chaotic motion”, they frequently collide with one another and with the container walls Figure MH5 5.7 What we “feel” as gas temperature is related to the speed of the gas particles. When the molecules are fast the gas feels hot.

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6.3. More Gas Law Calculations MH5, chapter 5.3 Final and Initial State Problems A gas often undergoes a change from an initial to a final state. Example: Consider a sample of gas at 25° C and 101.3 kPa (1.000 atm) [initial state, state1]. What is the pressure if the gas is heated to 100° C at constant volume [final state, state2]? Using the ideal gas law, and remembering that n = const, V = const (and, of course R = const), we find that Initial state: p1V = nRT1 Final state: p2V = nRT2 Dividing one equation by the other cancels V, n, R and leaves p2/p1 = T2/T1 or p2 = T2/T1 × p1

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Example MH5, Example 5.2 A 250 mL flask, open to the atmosphere, contains 0.0110 mol of air at 0° C, on heating, part of the air escapes; how much remains in the flask at 100° C ?

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Molar Mass and Density The ideal gas law offers a simple approach to the experimental determination of the molar mass of a gas. This method can also be applied to volatile liquids (i.e. liquids that evaporate easily, e.g. “nail polish remover” [= acetone]). Why are most volatile liquids smelly? Example MH5, Example 5.4 A sample of liquid acetone is placed in a 3.00 L flask and vaporized by heating to 95° C at 103.3 kPa (1.02 atm). The vapor filling the flask at this temperature and pressure weighs 5.87 g (you can assume that there is no air left in the flask, i.e. it contains only acetone vapor). a) What is the density of acetone vapor under these conditions? b) Calculate the molar mass of acetone. How does this compare to the expected molar mass of acetone [molecular formula CH3-CO-CH3 ] ?

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Another approach to calculating gas densities is to use the ideal gas law and derive the following equation: pV = nRT

p/(nRT) = 1/V | × m mp/(nRT) = m/V

Left hand side: m = Mn, and therefore m/n = M Right hand side: m/V = density Therefore: Mp/(RT) = density The density depends on … • pressure: Increasing p increases the density • Temperature: Increasing T decreases density,

that’s why hot air balloons rise in air • Molar mass: Decreasing M decreases density,

that’s why He-filled balloons rise in air Figure: MH5, p. 110

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A question from the Dec 2002 exam:

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… and another one

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6.4 Stoichiometry of Gaseous Reactions MH5, Chapter 5.4 We already know how to relate moles of substance to grams of substance (using m = Mn). For gas reactions we can now also include volume and pressure. Example MH5, example 5.6 Hydrogen peroxide, H2O2 is the active ingredient in commercial preparations for bleaching hair. It decomposes into water and oxygen. What mass of hydrogen peroxide must be used to produce 1.00 L of oxygen at 25° C and 1 atm (101.3 kPa)?

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Example MH5, Example 5.7 Octane, C8H18 is one of the hydrocarbons in gasoline. How many liters of oxygen, measured at 98.7 kPa and 24° C are required to burn 1.00 g of octane?

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Gay-Lussac discovered the law of combining volumes: The volume ratio of any two gases in a reaction at constant temperature and pressure is the same as the reacting mole ratio. Example: Synthesis of ammonia N2 (g) + 3 H2 (g) 2 NH3 (g) 1 molecule of N2 and 3 molecules of H2 will produce 2 molecules of NH3 1 mole of N2 and 3 moles of H2 will produce 2 moles of NH3 and … 1 L of N2 and 3 L of H2 will produce 2 L of NH3 1 m3 of N2 and 3 m3 of H2 will produce 2 m3 of NH3 10 m3 of N2 and 30 m3 of H2 will produce 20 m3 of NH3 This works, because V = RT/p × n The volume is proportional to the number of moles!

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A question from the Dec. 2002 exam:

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6.5. Gas Mixtures: Partial Pressures and Mole Fractions MH5, chapter 5.5 The ideal gas law applies to all gases, therefore it should also apply to gas mixtures (e.g. air). Consider a mixture of two gases at constant volume and temperature (moles gas A = nA, moles gas B = nB), the total pressure is given by ptot = ntot × RT/V = (nA + nB) × RT/V = nA × RT/V + nB × RT/V = pA + pB pA and pB are the same pressures that gas A and gas B would exert if they were alone. They are referred to as partial pressures. The total pressure is the sum of the partial pressures of all gases In this case:

ptot = pA + pB The partial pressures are simply calculated from the ideal gas law:

pA = nART/V pB = nBRT/V

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If there are more than two gases: ptot = pA + pB + pC + … This is known as Daltons Law of Partial Pressures: “The total pressure of a gas mixture is the sum of the partial pressures of the components of the mixture.” Example: A H2/He mixture has pH2 = 100 kPa, pHe = 200 kPa. What is the total pressure?

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Wet Gases, Partial Pressure of Water Consider the following arrangement: A closed container contains some liquid water and a gas (say N2).

Mixture of N2 and water vapor

Liquid water

Q. What is the total pressure of the gas? A. It is the partial pressure of the N2, plus that of the water vapor.

ptot = pN2 + pH2O The partial pressure of the water vapor is equal to the so-called vapor pressure of liquid water. It has a fixed value at a given temperature. The vapor pressure of water is independent of the amount of liquid water (… as long as there is any liquid water)!

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Vapor pressure of water as a function of temperature. (taken from lab manual, page C5)

An example of an experiment where this is important is the following: Zn + 2H+ Zn2+ + H2. The H2 gas is collected by displacing water … MH5, Figure 5.6

The total pressure of the collected gas is pH2 + pH2O.

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Example: Electrolysis of water MH5, Example 5.8 Hydrogen gas is prepared by electrolyzing water at 25° C. 152 mL of hydrogen are collected at a total pressure of 101 kPa. What is the partial pressure of hydrogen under these conditions? How many number of moles of hydrogen were collected?

MH5, Figure 5.5

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Partial Pressure and Mole Fraction As pointed out on p. 147, the following equations apply to a mixture of two gases A and B. pA = nA RT/V and ptot = ntot RT/V Dividing these equation by one another gives pA/ptot = nA/ntot We call XA = nA/ntot

the mole fraction of A in the gas mixture It follows that

pA = XA × ptot The partial pressure of a gas in a mixture is equal to its mole fraction multiplied by the total pressure. Example: A gas mixture contains equal numbers of molecules A and B. Calculate the partial pressures of A and B.

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Example: Dry air at 25° C consists of 78.1 volume % nitrogen, 21.0% oxygen, and 0.9% Argon. • Calculate the mole fractions and the partial

pressures of these gases when the total pressure is 101.3 kPa.

• How many moles of nitrogen, oxygen, and argon

atoms are there in 0.5 L under these conditions?

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… and a question from the Feb. 2002 exam

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