CHAPTER 5: Gases

•Chemistry of Gases•Pressure and Boyle’s Law•Temperature and Charles’ Law•The Ideal Gas Law•Chemical Calculations of Gases•Mixtures of Gases•Kinetic Theory of Gases•Real Gases

CHEM 1310 A/B Fall 2006

Gases

• The states of matter:– Gas: fluid, occupies all available volume– Liquid: fluid, fixed volume– Solid: fixed volume, fixed shape– Others? …

• Gases are the easiest to understand – can model them more precisely than liquids or solids using simple equations

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Ways to produce gases include…

• Decomposition2 HgO (s) → 2 Hg (l) + O2 (g)CaCO3 (s) + heat → CaO (s) + CO2 (g)etc.

• Acids reacting with carbonates or hydrogen carbonates to release CO2 (chapter 4)NaHCO3 (s) + HCl (aq) →NaCl (aq) + H2O (l) + CO2 (g)

• Acids react with metalZn (s) + 2 HCl (aq) → ZnCl2 (aq) + H2 (g)

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Some properties of gases

• Pressure (P) : how much force it exerts per unit area

• Temperature (T) : how hot or cold (kinetic energy of gas molecules)

• Volume (V) : space it takes up (the whole volume of the container it’s in)

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Pressure• Often measured by a

barometer• The height of a

column of mercury is related to the air pressure…Standard atmospheric pressure gives a column of mercury 76 cm high

• How does this work?

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Pressure by barometer• Atmosphere pushes down on

mercury in beaker, causes it to rise in column

• Push of atmosphere exactly balanced by force due to weight of mercury (F=ma)

• Mass of mercury is m = d A h, and a=g, so F = ma = dAhg

• P = F/A = dgh

CHEM 1310 A/B Fall 2006

Units of pressure

• P = F / A = N / m2 = kg m s-2 / m2 = kg / ms2. SI unit “pascal” (Pa). 105 Pa = 1 bar.

• Atmospheric pressure is 76 cm of Hg (or 760 mm Hg). Density is 13.596 g cm-1 (at 0o C) or 1.3596 x 104 kg m-3

• P = gdh = 9.80665 m/s2 x 1.3596 x 104 kg/m3 x 0.7600 m = 1.0133 x 105 kg / ms2 = 1.01325 x 105 Pa = “1 atmosphere” = 1 atm

CHEM 1310 A/B Fall 2006

Connection between P & V: Boyle’s Law

• Pressure and volume of a gas are related

• Use a “J tube” to figure out how

• In (a), pressure of atm exactly balances pressure of trapped gas Pgas = Patm

• In (b), pressure of gas is pressure balanced by Patm + that due to extra mercury added (=gdh), Pgas = Patm + gdh

• Gas is compressed (takes less V) and at higher pressure if more Hg is added

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Boyle’s Law• Boyle did experiments to show that if the

pressure doubled, the gas took up ½ as much room, etc. Pressure and volume are inversely relatedP1 V1 = P2 V2 (for fixed T and amt of gas)or more generally, PV = C (constant at fixed T and amount of gas), where C is independent of the particular gas chosen!

• C = 22.4 L atm at 0o C and 1 mol of gas; 0o C, 1atm are “standard temperature and pressure” (STP). At STP, one mol of gas occupies 22.4L

CHEM 1310 A/B Fall 2006

Relationship between V&T:Charles’ Law

• As T goes up (at constant pressure Patm), gas expandsV = V0 + α V0 Tcel, Tcel in Celcius

• All gases expand by the same relative amount when heated! (i.e., α is nearly the same for all gases!)

• Celcius temperature scale: water freezes at 0oC, water boils at 100oC by definition! Easy.

• On Celcius scale, α = 1/(273.15oC)… has a weird result when T=-273.15oC… “Absolute zero”. Can’t get below -273.15oC.

• “Absolute temperature scale”T (Kelvin) = 273.15 + Tcel (Celcius)

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Relationship between V and T

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Charles’ Law in Kelvin T scale

• If we substitute Tcel = T (Kelvin) – 273.15, we get V = a T, where a = V0 / 273.15

• So now (V1/V2) = (T1/T2) (for a fixed pressure and amount of gas)

• Lots of easy problems can be worked with this… for example, if T (in Kelvin!) is doubled, what happens to V?

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Ideal Gas Law

• Combines Charles’ Law and Boyle’s LawV α nT / P

Volume is proportional to amount of gas and temperature, and inversely proportional to pressure. Call the constant of proportionality R (“universal gas constant”)PV = nRT R = 0.082058 L atm mol-1 K-1

= 8.3145 J mol-1 K-1

• Can do lots of easy problems relating P, V, n, T of a gas using this simple equation

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Example problem• A gas cylinder weighs 1.5 kg empty and 2.0 kg when

filled with CO2 gas. If the cylinder has a volume of 1.0 L, what’s the pressure of the gas at 25oC?

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Ideal gas law w/ molar mass

• Could have done the last problem more directly by re-casting the ideal gas law

• N = m (mass) / M (molar mass)• PV = (m/M) RT• Also, since density d=m/V, d= PM/RT• Can predict density from P, T, M• Can determine M (and not just empirical

formula!) from d, P, TCHEM 1310 A/B Fall 2006

Gases in chemical reactions• Can use ideal gas law to do stoichiometry problems now

using P, T, V… instead of just masses• Example: We want to make 15.0 kg of the rocket fuel

hydrazine, using the reaction 2 NH3 (g) + NaOCl (aq) →N2H4 (aq) + NaCl (aq) + H2O (l)

• If our ammonia is at 10oC and 3.63 atm, how much of it (in L) do we need?

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Mixtures of gases• Suppose 3 containers of equal volume V, each

of which contained 1 mol of gas at 1 atm of pressure

H2 O2 N2

V, T, 1atm, 1mol V, T, 1atm, 1mol V, T, 1atm, 1mol

• What happens if we take the O2 and N2 from the 2nd and 3rd containers and put them in the 1st

container at constant T (and of course constant

CHEM 1310 A/B Fall 2006V)? What’s the total pressure?

Mixtures of gases, cont’d

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• Partial pressure: pressure exerted by each gas in a gas mixture

• Total pressure = sum of partial pressures• Each gas obeys ideal gas law separately for the number

of moles of that gas and the partial pressure• Ideal gas law also holds for Ptot, ntot• PH2 = nH2 R T / V, etc.• Ptot = ntot R T / V• Therefore, PH2 / Ptot = nH2 / ntot ( = 1/3 for this example).

This is also called the “mole fraction”, XH2• PH2 = Ptot x XH2 ( = 1/3 Ptot for this example)

V, T1 mol H2, 1 mol O2,1 mol N2

H2, O2N2

Kinetic Theory of Gases• Tries to explain gas behavior using basic

physics• Molecules of a gas fly around in random

directions with a distribution of speeds• The molecules are pictured as hard objects

undergoing elastic collisions with each other and the walls of the container

• Pressure results from the many collisions of the gas molecules with the walls of the container. P = F/A, P α (impulse per collision) x (rate of collisions with walls)

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Origin of pressure

• P = F/AP α (impulse per collision) x (rate of collisions with walls)P α (m x u) x [(N/V) x u] = Nmu2 / V, whereu = speed of moleculesm = mass of molecules(N/V) = number of molecules per unit volume (more molecules = more collisions)note faster molecules (u) = more collisions

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Origin of pressure, cont’d

• So far we have PV α Nmu2

• Problem: not just one speed u, lots of them• Solution: ok to use an average over u2, the

mean square speed <u2>• Proportionality constant works out to be 1/3, so

PV = (1/3) Nm<u2>. But we also know PV = nRT, so nRT = (1/3) Nm<u2>, or RT = (1/3) N0 m<u2> (because N is just n times N0, Avogadro’s number)

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Origin of pressure, cont’d

• So we see that RT = (1/3) N0 m<u2> • Temperature is therefore due to the kinetic

energy of the gas molecules• KE = ½ m u2

• Average KE per mol is ½ (N0m) <u2> = 3/2 RT, or <u2> = 3RT / M (since molar mass M = N0m)

• So… molecules move faster at higher T, and slower if they are more massive!

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Maxwell-Boltzmann distribution of molecular speeds in N2 at 3 temps

Note: urms ≠ ump ≠ uav

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Measurement of speed distributions

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Root mean square (RMS) speed

• Urms = sqrt(<u2>) = sqrt(3RT/M)• Example: He at 298K has what RMS

speed?

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Diffusion• Result of previous example

seems to large… e.g., odors don’t seem to travel this fast!

• Reason: direction of molecule keeps changing due to collisions with other gas molecules

• Diffusion: random motion of a molecule due to collisions with other molecules. Typically about 1010 collisions per second! Typically goes about ~ 10-7 m between collisions (mean free path)

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Rates of effusion and gaseous diffusion

• Effusion: molecules escape through a very small hole into vacuum

• Gaseous diffusion: molecules pass through a (large) porous barrier into another container

• The rates of both of both processes are inversely proportional to the molar mass of the gas

• Gaseous diffusion used to separate 235U from 238U in World War II

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Real gases• PV=nRT is the ideal gas law• Gases are not “ideal”. Implicitly assumes

– Gas molecules are ideal “point particles” with no size– No attraction between molecules

• In reality, gas molecules do take up some volume, and there is some attraction between gas molecules which can cause clustering

• First effect reduces volume available to gas, and second reduces pressure exerted by gas

• Van der Waals equation of state(P + a n2/V2) (V - nb) = nRTis more accuate. a,b depend on gas, are tabulated. n2/V2 is # of pairs per volume, and nb is excluded volume

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Reliability of ideal gas law• Different molecules

deviate differently• Works best at low

pressures• Deviations at very large

pressures can be a factor of 2 or more

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