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Chapter 5: the Gaseous state

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Title: Chapter 5: the Gaseous state


1
Chapter 5 the Gaseous state
  • Vanessa Prasad-Permaul
  • Valencia Community College
  • CHM 1045

2
GAS LAWS PRESSURE AND MEASUREMENT
Elements that exist as gases at 250C and 1
atmosphere
3
GAS LAWS PRESSURE AND MEASUREMENT
Physical Characteristics of Gases
  • Gases assume the volume and shape of their
    containers.
  • Gases are the most compressible state of matter.
  • Gases will mix evenly and completely when
    confined to the same container.
  • Gases have much lower densities than liquids and
    solids.

4
GAS LAWS PRESSURE AND MEASUREMENT
a) Gas is a large collection of particles moving
at random throughout a volume b) Collisions of
randomly moving particles with the walls of the
container exert a force per unit area that we
perceive as gas pressure
5
GAS LAWS PRESSURE AND MEASUREMENT
HOW IS PRESSURE DEFINED? The force the gas
exerts on a given area of the container in which
it is contained. The SI unit for pressure is
the Pascal, Pa.
  • If youve ever inflated a tire, youve probably
    made a pressure measurement in pounds (force) per
    square inch (area).

6
GAS LAWS PRESSURE AND MEASUREMENT
Units of Pressure
1 pascal (Pa) 1 kg/ms2 1 atm 760 mmHg 760
torr 1 atm 101,325 Pa 1 bar 105 Pa 1 atm
14.69 lb/in2
  • Hg is used instead of H2O
  • more dense
  • better visibility
  • accuracy

7
GAS LAWS PRESSURE AND MEASUREMENT
  • EXERCISE 5.1
  • A GAS CONTAINER HAD A MEASURED PRESSURE OF
  • 57kPa. CALCULATE THE PRESSURE IN UNITS OF ATM
  • AND mmHg.
  • First, convert to atm (57 kPa 57 x 103 Pa).
  • 57 x 103 Pa x 1 atm
    0.562 0.56 atm
  • 1.01325 x 105 Pa
  • Next, convert to mmHg.
  • 57 x 103 Pa x 760 mmHg 427.5
    4.3 x 102 mmHg
  • 1.01325 x 105 Pa

8
GAS LAWS EMPIRICAL GAS LAWS
EMPIRICAL GAS LAWS
You can predict the behavior of gases based on
the following properties
If two of these physical properties are held
constant, it is possible to show a simple
relationship between the other two
9
GAS LAWS EMPIRICAL GAS LAWS
Boyles experiment A manometer to study the
relationship between pressure (P) Volume (V)
of a gas
As P (h) increases
V decreases
10
GAS LAWS EMPIRICAL GAS LAWS
BOYLES LAW the volume of a sample of gas at a
given temperature varies inversely with the
applied pressure So. For a given amount of gas
(n) _at_ constant temperature (T) If pressure
(P) increases, the volume (V) of the gas
decreases P1 V1 P2 V2
11
Boyles Law
  • PressureVolume Law (Boyles Law)

12
GAS LAWS EMPIRICAL GAS LAWS
Boyles Law
Constant temperature Any given amount of gas
13
GAS LAWS EMPIRICAL GAS LAWS
EXERCISE 5.2 A volume of carbon dioxide gas
equivalent to 20.0 L was collected _at_ 23oC and
1.00atm pressure. What would be the volume of gas
at constant temperature and 0.830atm? P1 V1
P2 V2 Application of Boyles law gives V2
V1 x P1 20.0 L x 1.00atm 24.096
24.1 L P2
0.830atm
14
GAS LAWS EMPIRICAL GAS LAWS
Charles Law relating volume and temperature
As T increases
V increases
15
Charles Law
  • TemperatureVolume Law (Charles Law)

16
GAS LAWS EMPIRICAL GAS LAWS
Charles Law
Variation of gas volume with temperature
V a T
For a given amount of gas at constant pressure
Temperature must be in Kelvin
T (K) t (0C) 273.15
V / T constant
17
GAS LAWS EMPIRICAL GAS LAWS
CHARLES LAW the volume occupied by any sample
of gas at a constant pressure is directly
proportional to the absolute temperature So. Fo
r a given amount of gas (n) _at_ constant
pressure (P) If temperature (T) increases, the
volume (V) of the gas increases
V1 V2 T1 T2
18
GAS LAWS EMPIRICAL GAS LAWS
EXERCISE 5.3 A chemical reaction is expected to
produce 4.3dm3 of oxygen at 19oC and 101kPa. What
will the volume be at constant pressure and
25oC? First, convert the temperatures to the
Kelvin scale. Ti (19 273) 292 K Tf
(25 273) 298 K Following is the data
table. Vi 4.38 dm3 Pi 101 kPa Ti 292
K Vf ? Pf 101 kPa Tf 298 K Apply
Charless law to obtain Vf Vi x Tf 4.38
dm3 x 298K 4.470 4.47 dm3
Ti 292K

19
GAS LAWS EMPIRICAL GAS LAWS
COMBINED GAS LAW Relating Volume, Temperature
and Pressure Taking Boyles Law and Charles
Law The volume occupied by a given amount of
gas is proportional to the absolute temperature
divided by the pressure V constant x T or
PV constant (for a given amount of gas)
P T P1V1 P2V2 T1 T2
20
Combined Gas Law
We can combine Boyles and charles law to come
up with the combined gas law Use Kelvins
for temp, any pressure, any volume
21
GAS LAWS EMPIRICAL GAS LAWS
EXERCISE 5.4 A balloon contains 5.41dm3 of helium
at 24oC and 10.5kPa. Suppose the gas in the
balloon is heated to 35oC and the pressure is
now 102.8kPa, what is the volume of the
gas? First, convert the temperatures to
kelvins. Ti (24 273) 297 K Tf (35
273) 308 K Following is the data
table. Vi 5.41 dm3 Pi 101.5 kPa Ti
297 K Vf ? Pf 102.8 kPa Tf 308
K Apply both Boyles law and Charless law
combined to get Vf Vi x Pi x Tf 5.41 dm3
x 101.5kPa x 308K 5.539 5.54 dm3
Pf Ti
102.8kPa 297K
22
AVOGADROS LAW relating volume and amount
GAS LAWS EMPIRICAL GAS LAWS
equal volumes of any two gases at the same
temperature pressure contain the same number
of molecules
Constant temperature Constant pressure
23
Avogadros Law
  • The VolumeAmount Law (Avogadros Law)

24
Avogadros Law
  • The VolumeAmount Law (Avogadros Law)
  • At constant pressure and temperature, the volume
    of a gas is directly proportional to the number
    of moles of the gas present.
  • Use any volume and moles
  • V1 V2
  • n1 n2

25
GAS LAWS EMPIRICAL GAS LAWS
Avogadros Number one mole of any gas contains
the same number of molecules 6.023 x 1023 Must
occupy the same volume at a given temperature and
pressure
The conditions 0 0C and 1 atm are called standard
temperature and pressure (STP).
26
IDEAL GAS LAW
GAS LAWS THE IDEAL GAS LAW
Charles law V a T (at constant n and P)
Avogadros law V a n (at constant P and T)
R is the molar gas constant
PV nRT
27
GAS LAWS THE IDEAL GAS LAW
IDEAL GAS EQUATION
PV nRT
R 0.082057 L atm / (mol K) R 8.3145
J / (mol K) R 1.9872 cal / (mol K)
The units of pressure times volume are the units
of energy joules (J) or calories (cal)
28
The Ideal Gas Law
  • Ideal gases obey an equation incorporating the
    laws of Charles, Boyle, and Avogadro.
  • 1 mole of an ideal gas occupies 22.414 L at STP
  • STP conditions are 273.15 K and 1 atm pressure
  • The gas constant R 0.08206 LatmK1mol1
  • P has to be in atm
  • V has to be in L
  • T has to be in K

29
GAS LAWS THE IDEAL GAS LAW

EXERCISE 5.5 Show that the moles of gas are
proportional to the pressure for constant volume
and temperature
Use the ideal gas law, PV nRT, and solve for
n n PV RT n V x P
RT
Note that everything in parentheses is constant.
Therefore, you can write n constant x P Or,
expressing this as a proportion, n ? P
30
GAS LAWS THE IDEAL GAS LAW

EXERCISE 5.6 What is the pressure in a 50.0L gas
cylinder that contains 3.03kg of oxygen at
23oC? T 23oC 273K 296K V 50.0L R
0.08206 L . atm/K . mol n 3.03kg x 1000g x
1mol 94.688 mol O2
1 kg 31.998g P ? PV nRT P nRT
0.0347mol x 0.08206 L . atm x 296K V
K . mol
46.0 atm
50.0L
31
The Ideal Gas Law
  • Density and Molar Mass Calculations
  • You can calculate the density or molar mass (MM)
    of a gas. The density of a gas is usually very
    low under atmospheric conditions.

32
GAS LAWS THE IDEAL GAS LAW
EXERCISE 5.7 Calculate the density of helium
(g/L) at 21oC and 752mmHg. The density of air
under these conditions is 1.188g/L. What is the
difference in mass between 1 liter of air and 1
liter of helium?
Variable Value
P 752 mmHg x 1 atm 0.98947 atm 760mmHg
V 1 L (exact number)
T (21 273) 294 K
n ?
32
33
GAS LAWS THE IDEAL GAS LAW
Using the ideal gas law, solve for n, the moles
of helium. n PV 0.98947 atm x
1L RT
0.08206 L . atm/ K . mol x 294K
0.4101 mol
Now convert mol He to grams. 0.04101 mol He x
4.00g He 0.16404g He 0.164g/L
1.00 mol He
1 L
Therefore, the density of He at 21C and 752 mmHg
is 0.164 g/L. The difference in mass between one
liter of air and one liter of He mass air -
mass He 1.188 g - 0.16404 g 1.02396
1.024 g difference
34
GAS LAWS THE IDEAL GAS LAW
EXERCISE 5.8 A sample of a gaseous substance at
25oC and 0.862 atm has a density of 2.26 g/L.
What is the molecular mass of the
substance? Variable Value
P
0.862 atm
V 1 L (exact
number)
T (25 273) 298 K
n
? From the ideal gas law, PV
nRT, you obtain n PV
0.862 atm x 1 L 0.03525
mol RT 0.08206 L . atm/K . Mol x
298K
34
34
35
GAS LAWS THE IDEAL GAS LAW
Dividing the mass of the gas by moles gives you
the mass per mole (the molar mass). Molar mass
grams of gas 2.26g
64.114 g/mol moles
of gas 0.03525mol Therefore, the
molecular mass is 64.1 amu.
36
GAS LAWS GAS MIXTURES
Daltons Law of Partial Pressures
V and T are constant
P1
P2
Ptotal P1 P2
37
Daltons Law of Partial Pressures
  • For a two-component system, the moles of
    components A and B can be represented by the mole
    fractions (XA and XB).
  • Mole fraction is related to the total pressure
    by

38
GAS LAWS GAS MIXTURES
EXERCISE 5.10 A 10.0L flask contains 1.031g O2
and 0.572g CO2 at 18oC. What are the partial
pressures of each gas? What is the total
pressure? What is the mole fraction of oxygen in
this mixture? Each gas obeys the ideal gas law.
1.031 g O2 x 1 mol 0.0322188 mol O2
32.00g P nRT 0.0322mol x
0.0821L.atm/K.mol x 291K 0.076936 atm
V
10.0L 0.572 g CO2 x 1 mol 0.012997 mol
CO2 44.01g P nRT
0.0130mol x 0.0821L.atm/K.mol x 291K
0.031036 atm V
10.0L
39
GAS LAWS GAS MIXTURES
The total pressure is equal to the sum of the
partial pressures PT PO2 PCO2
0.076936atm 0.031036atm 0.10797

0.1080
atm The mole fraction of oxygen in the mixture
is Mole fraction O2 PO2 0.076936atm
0.7122 0.712
PT 0.1080atm 0.712 x 100 71.2
mole of O2 in this gas mixture
40
Kinetic Molecular Theory
  • This theory presents physical properties of gases
    in terms of the motion of individual molecules.
  • Average Kinetic Energy ? Kelvin Temperature
  • Gas molecules are points separated by a great
    distance
  • Particle volume is negligible compared to gas
    volume
  • Gas molecules are in constant random motion
  • Gas collisions are perfectly elastic
  • Gas molecules experience no attraction or
    repulsion

41
Kinetic Molecular Theory
42
  • Average Kinetic Energy (KE) is given by

U average speed of a gas particle R 8.314 J/K
mol m mass in kg MM molar mass, in kg/mol NA
6.022 x 1023
43
  • The RootMeanSquare Speed is a measure of the
    average molecular speed.

Taking square root of both sides gives the
equation
44
Example 17
  • Calculate the rootmeansquare speeds of helium
    atoms and nitrogen molecules in m/s at 25C.

45
Kinetic Molecular Theory
  • Maxwell speed distribution curves.

46
Grahams Law
  • Diffusion is the mixing of different gases by
    random molecular motion and collision.

47
Grahams Law
  • Effusion is when gas molecules escape without
    collision, through a tiny hole into a vacuum.

48
Grahams Law
  • Grahams Law Rate of effusion is proportional to
    its rms speed, urms.
  • For two gases at same temperature and pressure

49
Behavior of Real Gases
  • At higher pressures, particles are much closer
    together and attractive forces become more
    important than at lower pressures.

50
Behavior of Real Gases
  • The volume taken up by gas particles is actually
    less important at lower pressures than at higher
    pressure. As a result, the volume at high
    pressure will be greater than the ideal value.

51
Behavior of Real Gases
  • Corrections for non-ideality require van der
    Waals equation.

Excluded Volume
IntermolecularAttractions
52
Example 1 Boyles Law
  • A sample of argon gas has a volume of 14.5 L at
  • 1.56 atm of pressure. What would the pressure be
  • if the gas was compressed to 10.5 L? (at constant
  • temperature and moles of gas)

53
Example 2 Charles Law
  • A sample of CO2(g) at 35?C has a volume of 8.56
  • x10-4 L. What would the resulting volume be if
  • we increased the temperature to 85?C? (at
  • constant moles and pressure)

54
Example 3 Avogadros Law
  • 6.53 moles of O2(g) has a volume of 146 L. If we
  • decreased the number of moles of oxygen to 3.94
  • moles what would be the resulting volume?
  • (constant pressure and temperature)

55
Example 4Combined Gas Law
  • Oxygen gas is normally sold in 49.0 L steel
    containers at a pressure of 150.0 atm. What
    volume would the gas occupy if the pressure was
    reduced to 1.02 atm and the temperature raised
    from 20oC to 35oC?

56
Example 5 Gas Laws
  • An inflated balloon with a volume of 0.55 L at
    sea level, where the pressure is 1.0 atm, is
    allowed to rise to a height of 6.5 km, where the
    pressure is about 0.40 atm. Assuming that the
    temperature remains constant, what is the final
    volume of the balloon?

57
Example 6 Ideal Gas Law
  • Sulfur hexafluoride (SF6) is a colorless,
    odorless, very unreactive gas. Calculate the
    pressure (in atm) exerted by 1.82 moles of the
    gas in a steel vessel of volume 5.43 L at 69.5C.

58
Example 7 Ideal Gas Law
  • What is the volume (in liters) occupied by 7.40 g
    of CO2 at STP?

59
Example 8 Density MM
  • What is the molar mass of a gas with a density of
    1.342 g/L at STP?

60
Example 9 Density MM
  • What is the density of uranium hexafluoride, UF6,
    (MM 352 g/mol) under conditions of STP?

61
Example 10 Density MM
  • The density of a gaseous compound is 3.38 g/L at
    40C and 1.97 atm. What is its molar mass?

62
Example 11 Dalton
  • Exactly 2.0 moles of Ne and 3.0 moles of Ar were
    placed in a 40.0 L container at 25oC. What are
    the partial pressures of each gas and the total
    pressure?

63
Example 12 Dalton
  • A sample of natural gas contains 6.25 moles of
    methane (CH4), 0.500 moles of ethane (C2H6), and
    0.100 moles of propane (C3H8). If the total
    pressure of the gas is 1.50 atm, what are the
    partial pressures of the gases?

64
Example 13 Mole Fraction
  • What is the mole fraction of each component in a
    mixture of 12.45 g of H2, 60.67 g of N2, and 2.38
    g of NH3?

65
Example 14 Partial Pressure
  • On a humid day in summer, the mole fraction of
    gaseous H2O (water vapor) in the air at 25C can
    be as high as 0.0287. Assuming a total pressure
    of 0.977 atm, what is the partial pressure (in
    atm) of H2O in the air?

66
Gas Stoichiometry Example
  • In gas stoichiometry, for a constant temperature
    and pressure, volume is proportional to moles.
  • Example Assuming no change in temperature and
    pressure, calculate the volume of O2 (in liters)
    required for the complete combustion of 14.9 L of
    butane (C4H10)
  • 2 C4H10(g) 13 O2(g) ? 8 CO2(g) 10 H2O(l)

67
Example 15
  • All of the mole fractions of elements in a given
    compound must add up to?
  • 100
  • 1
  • 50
  • 2

68
Example 16
  • Hydrogen gas, H2, can be prepared by letting zinc
    metal react with aqueous HCl. How many liters of
    H2 can be prepared at 742 mm Hg and 15oC if 25.5
    g of zinc (MM 65.4 g/mol) was allowed to react?
  • Zn(s) 2 HCl(aq) ? H2(g) ZnCl2(aq)

69
Example 18
  • Under the same conditions, an unknown gas
    diffuses 0.644 times as fast as sulfur
    hexafluoride, SF6 (MM 146 g/mol). What is the
    identity of the unknown gas if it is also a
    hexafluoride?

70
Example 19 Diffusion
  • What are the relative rates of diffusion of the
    three naturally occurring isotopes of neon 20Ne,
    21Ne, and 22Ne?

71
Behavior of Real Gases
  • Deviations result from assumptions about ideal
    gases.
  • 1. Molecules in gaseous state do not exert
  • any force, either attractive or repulsive,
    on
  • one another.
  • 2. Volume of the molecules is negligibly small
  • compared with that of the container.

72
Example 20 Ideal Vs. Van Der Waals
  • Given that 3.50 moles of NH3 occupy 5.20 L at
    47C, calculate the pressure of the gas (in atm)
    using
  • the ideal gas equation
  • (b) the van der Waals equation. (a 4.17, b
    0.0371)

73
Example 21 Ideal Vs. Van Der Waals
  • Assume that you have 0.500 mol of N2 in a volume
    of 0.600 L at 300 K. Calculate the pressure in
    atmospheres using both the ideal gas law and the
    van der Waals equation.
  • For N2, a 1.35 L2atm mol2, and b 0.0387
    L/mol.
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