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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
CHEM 1310 A/B Fall 2006
Ways to produce gases include…
• Decomposition
2 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 metal
Zn (s) + 2 HCl (aq) → ZnCl2 (aq) + H2 (g)
CHEM 1310 A/B Fall 2006
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)
CHEM 1310 A/B Fall 2006
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?
CHEM 1310 A/B Fall 2006
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
CHEM 1310 A/B Fall 2006
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
related
P1 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 expands
V = 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)
CHEM 1310 A/B Fall 2006
Relationship between V and T
CHEM 1310 A/B Fall 2006
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?
CHEM 1310 A/B Fall 2006
Ideal Gas Law
• Combines Charles’ Law and Boyle’s Law
V 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
CHEM 1310 A/B Fall 2006
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?
CHEM 1310 A/B Fall 2006
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, T
CHEM 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?
CHEM 1310 A/B Fall 2006
Mixtures of gases
• Suppose 3 containers of equal volume V, each
of which contained 1 mol of gas at 1 atm of
pressure
H2
V, T, 1atm, 1mol
O2
V, T, 1atm, 1mol
N2
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
V)? What’s the total pressure?
CHEM 1310 A/B Fall 2006
Mixtures of gases, cont’d
H2, O2
N2
V, T
1 mol H2,
1 mol O2,
1 mol N2
• 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)
CHEM 1310 A/B Fall 2006
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)
CHEM 1310 A/B Fall 2006
Origin of pressure
• P = F/A
P α (impulse per collision) x (rate of
collisions with walls)
P α (m x u) x [(N/V) x u] = Nmu2 / V, where
u = speed of molecules
m = mass of molecules
(N/V) = number of molecules per unit
volume (more molecules = more collisions)
note faster molecules (u) = more collisions
CHEM 1310 A/B Fall 2006
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)
CHEM 1310 A/B Fall 2006
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!
CHEM 1310 A/B Fall 2006
Maxwell-Boltzmann distribution of
molecular speeds in N2 at 3 temps
Note: urms ≠ ump ≠ uav
CHEM 1310 A/B Fall 2006
Measurement of speed
distributions
CHEM 1310 A/B Fall 2006
Root mean square (RMS) speed
• Urms = sqrt(<u2>) = sqrt(3RT/M)
• Example: He at 298K has what RMS
speed?
CHEM 1310 A/B Fall 2006
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)
CHEM 1310 A/B Fall 2006
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
CHEM 1310 A/B Fall 2006
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) = nRT
is more accuate. a,b depend on gas, are tabulated.
n2/V2 is # of pairs per volume, and nb is excluded volume
CHEM 1310 A/B Fall 2006
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
CHEM 1310 A/B Fall 2006