Transcript PowerPoint 프레젠테이션

Chapter 4.
THE PROPERTIES OF GASES
THE NATURE OF GASES
4.1 Observing Gases
4.2 Pressure
4.3 Alternative Units of Pressure
THE GAS LAWS
4.4 The Experimental Observations
4.5 Applications of the Ideal Gas Law
4.6 Gas Density
4.7 The Stoichiometry of Reacting Gases
4.8 Mixtures of Gases
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Prelude to Chapters 4 and 5. The Three
States of Matter
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An Overview of the Physical States of Matter
The Distinction of Gases from Liquids and Solids
1. Gas volume changes greatly with pressure.
2. Gas volume changes greatly with temperature.
3. Gases have relatively low viscosity.
4. Most gases have relatively low densities under normal
conditions.
5. Gases are miscible.
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THE NATURE OF GASES (Sections 4.1-4.3)
4.1 Observing Gases
Many of physical properties of gases are very
similar, regardless of the identity of the gas.
Therefore, they can all be described simultaneously.
Samples of gases large enough to study are
examples of bulk matter – forms of matter that
consist of large numbers of molecules
Two major properties of gases:
Compressibility – the act of reducing the volume of
a sample of a gas
Expansivity - the ability of a gas to fill the space
available to it rapidly
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4.2 Pressure
– Pressure arises from the
collisions of gas molecules
on the walls of the container.
-
– SI unit of pressure is the pascal
(Pa)
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Measurement of Pressure
 Barometer – A glass tube, sealed
at one end, filled with liquid
mercury, and inverted into a
beaker also containing liquid
mercury (Torricelli)
where h = the height of a
column, d = density of
liquid, and g = acceleration
of gravity (9.80665 ms-2)
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 Manometer
This is a U-shaped tube filled
with liquid and connected to
an experimental system,
whose pressure is being
monitored.
-Two
types of Hg manometer:
(a) open-tube and (b) closed
tube system
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Self-Test 4.1B
The density of water at 20 o C is 0.998 g.cm -3. What height
would the column of liquid be in a water barometer at 20 oC
when the atmospheric pressure corresponds to 760. mm of
mercury?
Solution
P = dw hw g = dHghHgg
Hence
hw =
dHghHg
dw
(13.595 g cm-3)(760. mm)
=
(0.998 g cm-3)
= 1.04 x 104 mm (10.4 m)
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Self-Test 4.2A
What is the pressure in kilopascals in a system when the
mercury level in the system-side column in an open-tube
mercury manometer is 25 mm lower than the mercury level
in the atmosphere-side column and the atmospheric pressure
corresponds to 760. mm of mercury at 15 oC?
Solution
P = dhg = (13595 kg m-3)(0.785 m)(9.80665 m s-1)
= 104657 Pa = 105 kPa
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Example 4.3
A student attaches a glass bulb containing neon gas to an open-tube
manometer and calculates the pressure of the gas to be 0.890 atm.
(a) If the atmospheric pressure is 762 Torr, what height difference
between the two sides of the mercury in the manometer did the
student find?
(b) Which side is higher, the side of the manometer
attached to the bulb or the side open to the
atmosphere?
(c) If the student mistakenly switches the numbers
for the sides of the manometer when recording
the data in the laboratory notebook, what
would be the reported pressure in the gas bulb?
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Solution to Exercise 4.3
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4.3 Alternative Units of Pressure
- 1 bar = 105 Pa = 100 kPa
- 1 atm = 760 Torr = 1.01325×105 Pa (101.325 kPa)
- 1 Torr ~ 1 mmHg
Weather map
mbar
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Self-Test 4.3A
The US National Hurricane Center reported that the eye
of Hurricane Katrina (2005) fell as low as 902 mbar. What
is the pressure in atmospheres?
Solution
1 atm = 1.01325 bar, hence 0.920 bar =
= 0.890 atm
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0.920 bar x 1 atm
1.01325 bar
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THE GAS LAWS (Sections 4.4-4.6)
4.4 The Experimental Observations
 Boyle’s law: For a fixed amount of gas at
constant temperature, volume is inversely
proportional to pressure.
This applies to an isothermal system
(constant T) with a fixed amount of gas
(constant n).
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- For
isothermal changes between two states (1 and 2),
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Self-Test 4.4B
In a petroleum refinery a 750.-L container containing ethylene
gas at 1.00 bar was compressed isothermally to 5.00 bar. What
was the final volume of the container?
Solution
Isothermally means at constant temperature, hence Boyle's
law can be used.
P1V1 = P2V2,
(1.00 bar)(750. L) = (5.00 bar)V2
V2 = 150. L
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 Charles’s law: For a fixed amount of gas
under constant pressure, the volume varies
linearly with the temperature.
This applies to
an isobaric
system
(constant P)
with a fixed
amount of gas
(constant n).
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The Kelvin Scale of Temperature
If a Charles’ plot of V versus T (at constant P
and n) is extrapolated to V = 0, the intercept
on the T axis is ~-273 oC.
- Kelvin temperature scale
T = 0 K = -273.15 oC,
when V → 0.
- Celsius temperature scale
t (oC) = T (K) - 273.15
0 oC = 273.15 K
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Another aspect of gas behavior (Gay-Lussac’s Law)
This applies to an
isochoric system
(constant V) with a
fixed amount of gas
(constant n).
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Self-Test 4.5A
A rigid oxygen tank stored outside a building has a pressure
of 20.00 atm at 6:00 am when the temperature is 10. oC. What
will be the pressure in the tank at 6:00 pm, when the temperature
is 30. o C?
Solution
Volume is constant, hence Gay-Lussac's Law can be used.
P1
T1
=
P2
(20.00 atm)
T2
(283.15 K)
=
P2
(303.15 K)
P2 = 21.41 atm
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Avogadro’s Principle
 Under the same conditions of temperature and
pressure, a given number of gas molecules occupy
the same volume regardless of their chemical
identity.
- This defines molar volume
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Self-Test 4.6A
A helium weather balloon was filled at -20. oC and a
certain pressure to a volume of 2.5 x 104 L with 1.2 x 103
mol He. What is the molar volume of helium under these
conditions?
Solution
Molar volume =
Volume
No. moles
=
(2.5 x 104 L)
(1.2 x 103 mol)
= 21 L mol -1
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The Ideal Gas Law
This is formed by combining the laws of Boyle,
Charles, Gay-Lussac and Avogadro.
 The ideal gas law:
Gas constant, R = PV/nT.
It is sometimes called a “universal constant” and
has the value 8.314 J K-1 mol-1 in SI units, although
other units are often used (Table 4.2).
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Table 4.2. The Gas Constant, R
-The ideal gas law, PV = nRT, is an equation of state that
summarizes the relations describing the response of an
ideal gas to changes in pressure, volume, temperature,
and amount of molecules; it is an example of a limiting law.
(it is strictly valid only in some limit: here, as P 0.)
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4.5 Applications of the Ideal Gas Law
- For conditions 1 and 2,
- Molar volume
- Standard ambient temperature and pressure (SATP)
298.15 K and 1 bar, molar volume at SATP = 24.79 L·mol-1
- Standard temperature and pressure (STP)
0 oC and 1 atm (273.15 K and 1.01325 bar)
- Molar volume at STP
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EXAMPLE 4.4
In an investigation of the properties of the coolant gas used in an
air-conditioning system, a sample of volume 500 mL at 28.0 oC was
found to exert a pressure of 92.0 kPa. What pressure will the sample
exert when it is compressed to 30 mL and cooled to -5.0 oC?
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Calculating the pressure of a given sample
Self-Test 4.7B
A idling, badly tuned automobile engine can release as
much as 1.00 mol of CO per minute into the atmosphere.
At 27 oC, what volume of CO, adjusted to 1.00 atm, is
emitted per minute?
Solution
This question requires the use of PV = nRT, where
P = 1.00 atm, n = 1.00 mol (min-1), and T = 300.15 K.
(R = 8.206 x 10-2 L atm K-1 mol-1)
V =
(1.00 mol (min-1)(0.08206 L atm K-1 mol-1)(300.15 K)
(1.00 atm)
= 24.6 L (min-1)
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Using the combined gas law when one variable is changed
Self-Test 4.8A
A sample of argon gas of volume 10.0 mL at 200. Torr is
allowed to expand isothermally into an evacuated tube
with a volume of 0.200 L. What is the final pressure of the
argon in the tube?
Solution
The volume is increased by a factor of 20, so we expect a
decrease in pressure by the same factor, under isothermal
conditions.
P1V1
n1 T1
=
P2V2
n2 T2
, where T1 = T2 and n1 = n2
(reduces to Boyle's Law)
(200. Torr)(10.0 mL) = P2(Torr)(200 mL)
P2 = 10.0 Torr
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Using the combined gas law when two variables are changed
Self-Test 4.9A
A parcel of air (the technical term in metereology for a
small region of the atmosphere) of volume 1.00 x 103 L
at 20. o C and 1.00 atm rises up the side of a mountain
range. At the summit, where the pressure is 0.750 atm,
the parcel of air has cooled to -10. oC. What is the volume
of the parcel of air at that point?
Solution
P1V1
n1T1
=
P2V2
n2T2
where n1 = n2
(1.00 atm)(1.00 x 103 L)
(293.15 K)
=
(0.750 atm)V2(L)
(263.15 K)
V2 = 1.20 x 103 L
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Self-Test 4.10A
Calculate the volume occupied by 1.0 kg of hydrogen at 25 oC
and 1.0 atm.
Solution
We can use the ideal gas equation PV = nRT, after first finding
the number of moles of H2 in 1.0 kg.
n=
Mass
=
Molar mass
1.0 x 103 g
= 496 mol
2.016 g/mol
(1.0 atm)V(L) = (496 mol)(8.206 x 10-2 L atm K-1mol -1)(298.15 K)
V = 1.2 x 104 L
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Example 4.25
A sample of methane gas, CH4, was slowly heated at a constant pressure
of 0.90 bar. The volume of the gas was measured at a series of different
Temperatures and a plot of volume vs. temperature was constructed.
The slope of the line was 2.88×10-4 L K-1. What was the mass of the
sample of methane?
Solution
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4.6 Gas Density
Molar concentration of a gas is the number moles divided by the volume
occupied by the gas.
Molar concentration of a gas at STP (where molar volume is 22.4141 L):
This value is the same for all gases, assuming ideal behavior.
Density, however, does depend on the identity of the gas.
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Gas Density Relationships
• For a given P and T, the greater the molar mass, the greater its density.
• At constant T, the density increases with P. In this case, P is increased
either by adding more material or by compression (reduction of V).
• Raising T allows a gas to expand at constant P, increases V and
therefore reduces its density.
Density at STP
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Self-Test 4.11A
The oil produced from eucalyptus leaves contains the volatile
organic compound eucalyptol. At 190. oC and 60 Torr, a sample
of eucalyptol vapor had a density of 0.320 g L-1. Calculate the
molar mass of eucalyptol.
Solution
Density =
0.320 g L
-1
MP
where M is molar mass
RT
=
M(g mol-1)(60 Torr)
(62.324 L Torr K-1 mol -1)(463.15 K)
M = 154 g mol-1
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4.7 The Stoichiometry of Reacting Gases
-Molar volumes of gases are generally > 1000
times those of liquids and solids.
e.g. Vm (gases) = ~ 25 L mol-1; Vm (liquid water)
= 18 mL mol-1
-Reactions that produce gases from condensed
phases can be explosive.
e.g. sodium azide
(NaN3) for air bags
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EXAMPLE 4.6
The carbon dioxide generated by the personnel in
the artificial atmosphere of submarines and
spacecraft must be removed form the air and the
oxygen recovered. Submarine design teams have
investigated the use of potassium superoxide, KO2,
as an air purifier because this compound reacts with
carbon dioxide and releases oxygen:
4 KO2 (s) + 2 CO2(g) → 2 K2CO3(s) + 3 O2(g)
Calculate the mass of KO2 needed to react with 50 L
of CO2 at 25 oC and 1.0 atm.
Vm = 24.47 Lmol-1; 1 mol CO2 -> 2 mol KO2; MKO2 = 71.10 gmol-1
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Self-Test 4.12A
Calculate the volume of carbon dioxide, adjusted to 25 o C and 1.0 atm,
that plants need to make 1.00 g of glucose, C6H12O 6, by photosynthesis
in the reaction
6CO 2(g) + 6H 2O(l)
C6H12O6(s) + 6O 2(g)
Solution
From the equation, the stoichiometry of CO2:glucose is 6:1.
The molar mass of glucose is 180 g/mol.
The molar volume of CO 2 at 25 o C and 1 atm is 24.47 L mol -1
Volume of CO 2 = 1.00 g glucose x
1 mol glucose x 6 mol CO 2 x 24.47 L
180 g glucose
1 mol glucose 1 mol CO 2
= 0.82 L
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Example 4.59
A 15.0-mL sample of ammonia gas at 1.00×102 Torr and 30 oC is mixed
with 25.0 mL of hydrogen chloride gas at 1.50×102 Torr and 25 oC, and
the following reaction takes place:
NH3(g) + HCl(g)
NH4Cl(s)
(a) Calculate the mass of NH4Cl that forms.
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(b) Identify the gas in excess and determine the pressure of the excess
gas at 27 oC after the reaction is complete (in the combined volume
of the original two flasks).
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4.8 Mixtures of Gases
– A mixture of gases that do not react with one another behaves
like a single pure gas.
 Partial pressure: The total pressure of a mixture of gases is the sum
of the partial pressures of its components (John Dalton).
P = PA + PB + … for the mixture containing A, B, …
- Humid gas: P = Pdry air + Pwater vapor (Pwater vapor = 47 Torr at 37 oC)
 mole fraction: the number of moles of molecules of the gas
expressed as a fraction of the total number of moles of molecules
in the sample.
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EXAMPLE 4.7
Air is a source of reactants for many chemical processes. To determine
how much air is needed for these reactions, it is useful to know the
partial pressures of the components. A certain sample of dry air of
total mass 1.00 g consists almost entirely of 0.76 g of nitrogen and
0.24 g of oxygen. Calculate the partial pressures of these gases when
the total pressure is 0.87 atm.
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Self-Test 4.14A
A baby with a severe bronchial infection is in respiratory distress.
The anesthetist administers heliox, a mixture of helium and oxygen
with 92.3% by mass O 2. What is the partial pressure of oxygen being
administered to the baby if the atmospheric pressure is 730 Torr?
Solution
n(He) =
0.077 g
= 0.0193 mol
4.00 g mol-1
x(He) =
0.0193 mol
0.0481 mol
n(O 2) =
0.923 g
= 0.0288 mol
32.0 g mol-1
= 0.401
x(O 2) =
0.0288 mol
0.0481 mol
= 0.599
[= 1.00 - 0.401 (only two components)]
P(O 2) = x(O2)P = 0.599 x 730 Torr = 437 Torr
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Chapter 4.
THE PROPERTIES OF GASES
MOLECULAR MOTION
4.9 Diffusion and Effusion
4.10 The Kinetic Model of Gases
4.11 The Maxwell Distribution of Speeds
REAL GASES
4.12 Deviations from Ideality
4.13 The Liquefaction of Gases
4.14 Equations of State of Real Gases
2012
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MOLECULAR MOTION (Sections 4.9-4.11)
4.9 Diffusion and Effusion
 Diffusion: gradual dispersal
of one substance through
another substance
 Effusion: escape of a gas
through a small hole into a
vacuum
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 Graham’s law: At constant T, the rate of
effusion of a gas is inversely proportional to
the square root of its molar mass:
Strictly,Graham’s law relates to effusion, but
it can also be used for diffusion.
For two gases A and B with molar masses MA and MB,
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Rate of effusion and average speed increase as the
square root of the temperature:
 Combined relationship: The average speed of
molecules in a gas is directly proportional to the
square root of the temperature and inversely
proportional to the square root of the molar mass.
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Self-Test 4.15A
It takes 30. mL of argon 40. s to effuse through a porous
barrier. The same volume of vapor of a volatile compound
extracted from Caribbean sponges takes 120. s to effuse
through the same barrier under the same conditions. What
is the molar mass of the compound?
Solution
Time for Ar to effuse
=
M(unknown)
Time for unknown to effuse
40 (s)
=
120 (s)
M(unknown) =
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M(Ar)
39.95 (g mol-1)
M(unknown)
360 g mol -1
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152s
Example 4.73
A sample of argon gas effuses through a porous plug in 147 s.
Calculate the time required for the same amount of (a) CO2, (b) C2H4,
(c) H2, and (d) SO2 to effuse under the same conditions of pressure
and temperature.
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4.10 The Kinetic Model of Gases
 Kinetic molecular theory (KMT) of gases makes four assumptions:
1. A gas consists of a collection of molecules in
continuous random motion.
2. Gas molecules are infinitesimally small points.
3. The molecules move in straight lines until
they collide.
4. The molecules do not influence one another
except during collisions.
- Collision with walls: consider molecules
traveling only in one dimensional x with a
velocity of vx.
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The change in momentum (final – initial)
of one molecule: -2mvx = 2mvx momentum
change for the wall
All the molecules within a distance vxDt of the wall
and traveling toward it will strike the wall during the
Interval Dt.
If the wall has area A, all the particles in a volume
AvxDt will reach the wall if they are traveling toward it.
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The number of molecules in the volume AvxDt 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 Dt is half the number in the volume AvxDt:
The total momentum change = number of collisions × individual
molecule momentum change
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Force = rate of change of momentum =
(total momentum change)/Dt
for the average value of <vx2>
Mean square speed:
Pressure on wall:
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where vrms is the root mean square speed,
or
- The temperature is proportional to the mean square speed of the
molecules in a gas.
- This was the first acceptable physical interpretation of temperature:
a measure of molecular motion.
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EXAMPLE 4.7
What is the root mean square speed of nitrogen
Molecules in air at 20 oC?
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Self-Test 4.16A
Estimate the root mean square speed of water molecules
in the vapor above boiling water at 100. o C.
Solution
Molar mass of water is 18.01 g mol-1 or 0.01801 kg mol-1.
From vrms = (3RT/M)1/2,
3 x (8.3145 J K-1 mol-1) x (373 K)
vrms
=
=
0.01801 kg mol -1
719 m s-1
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A molecular description of Boyle’s Law
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A molecular description of Charles’s Law
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A molecular description of Avogadro’s Law
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A molecular description of Dalton’s law of partial pressures
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4.11 The Maxwell Distribution of Speeds
Maxwell derived equation 22, for calculating the fraction of gas molecules
having the speed v at any instant, from the kinetic model.
v = a particle’s speed
DN = the number of molecules with speeds in the range
between v + Dv
N = total number of molecules; M = molar mass
f(v) = Maxwell distribution of speeds
For an infinitesimal range,
average speed
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-
Molar mass (M) dependence:
as M increases, the
fraction of molecules with
speeds greater than a
specific speed decreases.
- Temperature dependence:
as T increases, the
fraction of molecules
with speeds greater
than a specific speed
increases.
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REAL GASES (Sections 4.12-4.14)
- Deviations
from the ideal gas law are significant at
high pressures and low temperatures (where
significant intermolecular interactions exist).
4.12 Deviations from Ideality
- Gases condense to
liquids when cooled or
compressed (attraction).
- Liquids are difficult to
compress (repulsion).
Deviation from ideal gases
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 Compression factor (Z): the ratio of the actual
molar volume of the gas to the molar volume of an
ideal gas under the same conditions.
For an ideal gas, Z = 1
Long range attractions; smaller Z,
condensation of gases
Short range repulsions; larger Z, low
compressibility of liquids and solids,
finite molecular volume
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- For
many gases,
attractions
dominate at low
pressure (Z < 1),
while repulsive
interactions
dominate at high
pressure (Z > 1).
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4.13 The Liquefaction of Gases
 Joule-Thomson effect: when attractive forces
dominate, a real gas cools as it expands.
– In this case expansion requires energy, which
comes from the kinetic energy of the gas,
lowering the temperature.
– The effect is used in some refrigerators and
to effect the condensation of gases such as
oxygen, nitrogen, and argon.
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The Linde refrigerator for the liquefaction
of gases
i.e. Adiabatic
cooling;
temperature
decrease under
isentropic
expansion
of any gas (w
0)
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4.14 Equations of State of Real Gases
 Virial equation:
 van der Waals equation:
or
–nb volume excluded since molecules cannot overlap (repulsions)
b volume excluded by 1 mol ~ molar volume in the liquid state
pressure reduced due to attractions between pairs of
molecules
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The effect of intermolecular attractions on
measured gas pressure
Pideal = P + a(n/V)2
(P: actual pressure)
actual pressure is smaller than the ideal pressure
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The effect of molecular volume on measured gas volume
Videal = V - nb
actual volume is greater than the ideal volume
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Virial expansion of the van der Waals equation
At low particle densities
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Table 4.5 Van der Waals Parameters for
some Common Gases
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 Model of gas
1. A large number of gas molecules in ceaseless,
random, and straight motion.
2. The average speed and the spread of speeds
increase with T and decrease with m.
3. Molecules travel in straight lines until they collide
with other molecules or the container wall.
4. Widely separated. Intermolecular forces have only
a weak effect on the properties.
5. Repulsions increase the molar volume, whereas
attractions decrease the molar volume.
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EXAMPLE 4.9
Refrigerant gas (a = 16.2 L2 atm mol–2, b
= 0.084 L/mol), 1.50 mol in 5.00 L at 0 oC;
Estimate the pressure.
Use expanded form of the van der Waals equation:
P = nRT/(V – nb) - an2/V2
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Self-Test 4.17A
A 10.0-L tank containing 25 mol of O 2 is stored in a diving supply
shop at 25 o C. Use the data in table 4.5 and the van der Waals
equation to estimate the pressure in the tank.
Solution
From P = nRT/(V - nb) -an2/V2,
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
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