General Chemistry I 1
KINETIC MOLECULARDESCRIPTION OF THESTATES OF MATTER
CHAPTER 9The Gaseous State
General Chemistry I
U N I T III
CHAPTER 10Solids, Liquids, and Phase Transitions
CHAPTER 11Solutions
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THE GASEOUS STATE
9.1 The Chemistry of Gases
9.2 Pressure and Temperature of Gases
9.3 The Ideal Gas Law
9.4 Mixtures of Gases
9.5 The Kinetic Theory of Gases
9.6 Real Gases: Intermolecular Forces
9CHAPTER
General Chemistry I
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397
CaCO3(s) + 2HCl(g) →
CaCl2(s) + CO2(g) + H2O(l)
K2SO3(s) + 2HCl(g) →
2KCl(s) + SO2(g) + H2O(l)
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9.2 PRESSURE AND TEMPERATURE OF GASES
398
Pressure
Torricelli’s barometerEvangelista Torricelli (ITA, 1608-1647)
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Force exerted by the mercury column at its base F = mg
Pressure : /
F mg mg mP gh gh
A A V h V
= 13.5951 g cm–3 → density of Hg(l) at 0 oC
g = 9.80665 m s–2 → gravitational acceleration
h = 76 cm → height of mercury column
399
1 atmosphere pressure (1 atm) is
1.01325 x 105 kg/ms2 or Pa; the SI unit= 101.325 kPa
1.01325 bar (1 bar = 105 Pa760 mm Hg (at 273 K)760 torr (at any temperature)
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Pressure and Boyle’s Law
~ Experiments on the compression and expansion of air. “The spring of the Air and Its Effects” (1661).
Boyle’s J-tube experiment
Trapped air at the closed end of the J-
tube:
Add Hg and measure the volume of air (V):
1
(in mm)1 atm
760 mm atm
hP
1CP C
V V
Robert Boyle (UK, 1627-1691) PV Cor
C: a constant at constant T and fixed amount of gas
400
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Fig. 9.3 Boyle’s J-tube.
(a)Same Hg height on two sides
Pconfined = Pair
(b) Hg added
Difference in Hg heights, h
Vconfined compressed
400
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(a) P vs. V hyperbola
(b) P vs. 1/V straight line passing through the origin (slope: C)
(c) PV vs. P straight line independent of P
(parallel to P-axis, intercept C on PV-axis)
The value of C at 0oC and for 1 mol of gas,
~ good for all gases at very low pressure
C = PV = 22.414 L atm
401
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Temperature and Charles’s law
Jacques Charles (France,1746-1823)
constant (at constant and ) V T n P
o
0
273.15 C 1V
tV
0 0o1
273.15 C 273.15
t TV V V
(Kelvin) 273.15 (Celsius)T t
402
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Fig. 9.5 Volume of a gas confined at constant P increases as T increases.
403
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Fig. 9.6 The volume of a sample of a gas is a function of temperature at constant pressure.
404
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◈ Absolute Temperature Scale, K Kelvin, Lord William Thomson (UK, 1824-1907) 0 K : ‘absolute zero’ temperature 273.15 K : triple point of water 0 K = - 273.15 oC 0 oC = 273.15 K
Kelvin scale: absolute, thermodynamic temperature
scale - ‘absolute zero’ is the temperature at which all ther
mal motion ceases in the classical description of thermo
dynamics.
405
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9.3 THE IDEAL GAS LAW405
√ Equation of state
√ Limiting law for real gases
as P 0
√ Universal gas constant, R
R = 8.314 J·K–1·mol–1
= 8.206 x 10–2 L·atm·K–1
PV = nRT
Boyle’s law: V 1/P (at constant T and n)Charles’ law: V T (at constant P and n)Avogadro’s hypothesis: V n (at constant T and P)
nT
VP
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408
EXAMPLE 9.5 Concentrated nitric acid acts on copper
to give nitrogen dioxide and dissolved copper ions according
to the balanced chemical equation
Cu(s) + 4H+(aq) + 2NO3-(aq) → 2NO2(g) + Cu2+(aq) + 2H2O(l)
Suppose that 6.80 g copper is consumed in this reaction, and
that the NO2 is collected at a pressure of 0.970 atm and a
temperature of 45oC. Calculate the volume of NO2 produced.
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9.4 MIXTURES OF GASES408
▶ Partial pressure (Pi) of the ith gas in a mixture of gases →
pressure that the ith gas would exert if it occupied the container
alone
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◆ Dalton’s Law of Partial Pressures
The total pressure of a mixture of gases is the
sum of the partial pressures of its component.
A B ii
P P P P
▶ Mole fraction of the component A is xA
AA A B
A B
, 1n
x x xn n
AA ,
n RTP
V A B
nRT RTP n n
V V A
A AA B
n P
P x Pn n
409
PA = xAP
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409
EXAMPLE 9.6 When NO2 is cooled to room temperature, some of it reacts
to form a dimer, N2O4, through the reaction
2NO2(g) → N2O4(g)
Suppose 15.2 g of NO2 is placed in a 10.0 L flask at high temperature and the
Flask is cooled to 25oC. The total pressure is measured to be 0.500 atm.
What partial pressures and mole fractions of NO2 and N2O4 are present?
(a)
(b)
(Dalton’s Law)
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9.5 THE KINETIC THEORY OF GASES410
1. A gas consists of a collection of molecules in
continuous random motion.
2. Gas molecules are infinitesimally small (mass) points.
3. The molecules move in straight lines until they collide.
4. The molecules do not influence one another except
during collisions.
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- Collision with walls: consider molecules traveling only in one dimension, x, with a velocity of vx.
All the molecules within a distance vxt of the walland traveling toward it will strike the wall during theinterval t.
If the wall has area A, all the particles in a volumeAvxt will reach the wall if they are traveling towardit.
411
The change in momentum (final – initial)of one molecule: -2mvx = 2mvx momentumchange for the wall
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The number of molecules in the volume Avxt is that fraction of the total volume V, multiplied by the total number of molecules:
The average number of collisions with the wall during the interval t is half the number in the volume Avxt:
The total momentum change = number of collisions × individual wall momentum change
411
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Force = rate of change of momentum = (total momentum change)/t
mean-square speed
412
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412
- Kinetic energy of NA molecules,
- average kinetic energy per molecule, kB = R/NA
- root-mean-square speed
M = molar mass = NAm
Root mean square speeds of some gases at 25 oC
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Maxwell-Boltzmann distribution of speed
23/2
2 /2 ( ) with ( ) 4 2
M RTMf f e
RT
vN
= v v v vN
James Clerk Maxwell Ludwig Eduard Boltzmann (Scotland, 1831-1879) (Austria, 1844-1906)
2B
3/2
/22
B
or ( ) 4 2
m k Tmf e
k T
vv v
23 1A/ 1.38066 10 J KBk R N Boltzmann constant:
414
Fraction ofmolecules withspeeds between v and v + v
NN
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415
Fig. 9.14 Maxwell-Boltzmann distribution of molecularspeeds in N2 at three different temperatures.
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mpvrmsv
v
416Different speeds associated with a collectionof gas molecules or atoms
1. Most probable speed (vmp)
df(v)dv
= 0v = vmp
vmp = 2kBTm
or 2RTM
2. Average speed (v)_
v = vf(v)dv0
oo_ = 8kBTm
or 8RTM
3. Mean square speed (v2)_
v2_
v2f(v)dv0
oo
= = 3kBTm
or 3RTM
4. Root mean square speed (vrms)
vrms = 3kBTm
or 3RT
Mv2_
=
vmp < v < vrms
_= 1.000:1.128:1.225
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9.6 REAL GASES: INTERMOLECULAR FORCES
417
▶ Compression (or Compressibility) factor, Z z is a measure of deviation from ideality
- For an ideal gas, z = 1
- For real gases, z deviates from
1 as P increases:
z < 1 when attractive forces dominate
z > 1 for repulsive forces dominate
z =Vm
Vmideal =
Vm
RT/P=
PVm
RT=
nRT
Vm = molar volume = V/n
PV
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◈ The Van der Waals Equation of State
▶ Corrections to the ideal equation of state
- Attraction at long distance:
Reduction in collision frequency
Reduction in intensity of collision
/n V/n V
2
ideal 2
nP P a
V
- Repulsion at short distance:
No overlap of molecules → excluded volume effect
Reduction in free volume n
idealV V bn
418
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2
2
nP a V nb nRT
V
▶ Van der Waals equation:
a: atm L2 mol-2
b: L mol-1
R: L atm mol-1 K-1
419
Rearranging this equation to solve for P gives the form of the equation that is most often used in calculations:
See Slide 37
P =nRT
V nb
_ an2
V2_
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Repulsive forces (through b) increase z above 1.
Attractive forces (through a) reduce z.
The Boyle temperature TB
Influence of van der Waals parameters a and b onthe compressibility factor z
Substituting the approximation1
1 nb/V~ 1 +
nb
V_
into the equation above gives,
z ~ 1 + baRT
n
V_
+...
+....
(at low n/V)
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-Note that the Constant b is the volume excluded by 1 mol of molecules
and should be close to Vm, the volume per mole in the liquid state.
The temperature at which the coefficient in brackets is 0 (so z = 1)is the Boyle temperature (TB)
TB =a
Rb
For T > TB repulsive forces dominate and z > 1
For T < TB attractive forces dominate and z < 1
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Example of the use of the van der Waalsequation for the calculation of pressure
Solution
A 10.0-L tank containing 25 mol of O2 is stored in a diving supplyshop at 25 oC. Use the data in table 4.5 and the van der Waalsequation to estimate the pressure in the tank.
P =(25 mol) x (0.08206 L atm K-1 mol-1) x (298 K)
10.0 L (25 mol) x (3.19 x 10-2 L mol-1)_
_ (1.364 L2 atm mol-2) x(25 mol)2
(10.0 L)2
= 58 atm
Rewriting the van der Waals equation so as to solve for P,
P + a n2
V2(V nb) = nRT becomes P = nRT
(V nb)a
n2
V2__ _
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Intermolecular Forces
12 6
LJ ( ) 4 V RR R
Lennard-Jones Potential:
where is the depth and is the distance at which V(R) passesthrough zero.
421
(Ar)
(He) (Ar)
(He)