Transcript Document 7600373

Vanessa Prasad-Permaul
Valencia Community College
CHM 1045
1
Elements that exist as gases at 250C and 1 atmosphere
2
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.
3
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
4
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.
Pressure =
Force
Area
• If you’ve ever inflated a tire, you’ve probably made a pressure
measurement in pounds (force) per square inch (area).
5
Units of Pressure
1 pascal (Pa) = 1 kg/m·s2
1 atm = 760 mmHg = 760 torr
1 atm = 101,325 Pa
Height is
proportional
to the
barometric
pressure
760mm
1 bar = 105 Pa
1 atm = 14.69 lb/in2
Hg is used instead of H2O :
• more dense
• better visibility
• accuracy
mercury
Barometer
6
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
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EMPIRICAL GAS LAWS
You can predict the behavior of gases based on the following
properties:
• Volume
• Amount (moles)
• Pressure
• Temperature
* If two of these physical properties are held constant,
it is possible to show a simple relationship between the other two…
8
Boyle’s experiment:
A manometer to study the relationship between
pressure (P) & Volume (V) of a gas
Hg
As P (h) increases
Hg
V decreases
9
BOYLE’S 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)
&
@ constant temperature (T) :
If pressure (P) increases, the volume (V) of the gas decreases
P1 * V1 = P2 * V2
10
• Pressure–Volume Law (Boyle’s Law):
11
Boyle’s Law
P a 1/V
P * V = constant
P1 * V1 = P2 * V2
Constant temperature
Any given amount of gas
12
EXERCISE 5.2
A volume of carbon dioxide gas equivalent to 20.0 L
was collected @ 23oC and 1.00atm pressure. What
would be the volume of gas at constant temperature
and 0.830atm?
P1 * V1 = P2 * V2
Application of Boyle’s law gives
V2 = V1 x P1 = 20.0 L x 1.00atm = 24.096 = 24.1 L
P2
0.830atm
13
Charles’ Law: relating volume and temperature
As T increases
V increases
14
Charles’ Law
• Temperature–Volume Law (Charles’ Law):
15
Charles’ Law
Variation of gas volume with temperature
VaT
Temperature must
be in Kelvin
V / T = constant
For a given amount of gas
at constant pressure
T (K) = t (0C) + 273.15
16
CHARLES’ LAW
the volume occupied by any sample of gas at a constant
pressure is directly proportional to the absolute temperature
So….
For a given amount of gas (n)
&
@ constant pressure (P) :
If temperature (T) increases, the volume (V) of the gas increases
V1
V2
T1 = T2
17
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
Vf = ?
Pf = 101 kPa
Ti = 292 K
Tf = 298 K
Apply Charles’s law to obtain
Vf = Vi x Tf = 4.38 dm3 x 298K = 4.470 = 4.47 dm3
Ti
292K
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COMBINED GAS LAW
Relating Volume, Temperature and Pressure
Taking Boyle’s 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
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Combined Gas Law
We can combine Boyle’s and charles’ law to come
up with the combined gas law
P1  V1
P2  V2

T1
T2
Use Kelvins for temp, any pressure, any volume
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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
Vf = ?
Pf = 102.8 kPa
Ti = 297 K
Tf = 308 K
Apply both Boyle’s law and Charles’s 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
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equal volumes of any two gases at the same
temperature & pressure contain the same number of molecules
V a number of moles (n)
V = constant x n
V1/n1 = V2/n2
Constant temperature
Constant pressure
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Avogadro’s Law
• The Volume–Amount Law (Avogadro’s
Law):
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Avogadro’s Law
• The Volume–Amount Law (Avogadro’s Law):
V n
• 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
 k1
n1
V1 = V2
n1
n2
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Avogadro’s 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).
Experiments show that at STP, 1 mole of an ideal gas
occupies 22.4 L.
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Boyle’s law: V
a
1
P
(at constant n and T)
Charles’ law: V a T (at constant n and P)
Avogadro’s law: V a n (at constant P and T)
V a
V = constant x
nT
P
nT
P
= R nT
P
R is the molar
gas constant
PV = nRT
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IDEAL GAS EQUATION
PV = nRT
R=
PV
nT
=
(1 atm)(22.414L)
(1 mol)(273.15 K)
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)
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
Ideal gases obey an equation incorporating the laws
of Charles, Boyle, and Avogadro.
P V  n  R  T

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 L·atm·K–1·mol–1

P has to be in atm

V has to be in L

T has to be in K
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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
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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
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The Ideal Gas Law
• Density and Molar Mass Calculations:
mass
n  MM P  MM
d


volume
V
R T
• You can calculate the density or molar mass
(MM) of a gas. The density of a gas is
usually very low under atmospheric
conditions.
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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
P
V
T
n
Value
752 mmHg x 1 atm = 0.98947 atm
760mmHg
1 L (exact number)
(21 + 273) = 294 K
?
32
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
Now convert mol He to grams.
0.04101 mol He x 4.00g He
1.00 mol He
1L
=
0.4101 mol
= 0.16404g He = 0.164g/L
Therefore, the density of He at 21°C 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
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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
P
V
T
n
Value
0.862 atm
1 L (exact number)
(25 + 273) = 298 K
?
From the ideal gas law, PV = nRT, you obtain
n =
PV =
0.862 atm x 1 L
RT
0.08206 L . atm/K . Mol x 298K
= 0.03525 mol
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Dividing the mass of the gas by moles gives you the mass per
mole (the molar mass).
Molar mass = grams of gas
moles of gas
=
2.26g
0.03525mol
= 64.114 g/mol
Therefore, the molecular mass is 64.1 amu.
35
Dalton’s Law of Partial Pressures
V and T are
constant
P1
P2
Ptotal = P1 + P2
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Dalton’s 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:
Pi  X i Ptot
nA
XA 
nA  nB
nB
XB 
nA  nB
X A  X B 1
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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
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V
10.0L
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
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
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
40
41

Average Kinetic Energy (KE) is given by:
1
2
KE  mu
2
U = average speed of a
gas particle
R = 8.314 J/K mol
m = mass in kg
3RT
3RT
u

mNA
MM
MM = molar mass, in
kg/mol
NA = 6.022 x 1023
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
The Root–Mean–Square Speed: is a measure
of the average molecular speed.
3RT
u 
MM
2
3RT
urms 
MM
Taking square root of both
sides gives the equation
43
Calculate the root–mean–square speeds of
helium atoms and nitrogen molecules in
m/s at 25°C.
44
Kinetic Molecular Theory
• Maxwell speed distribution curves.
45
Graham’s Law
• Diffusion is the
mixing of different
gases by random
molecular motion
and collision.
46
Graham’s Law
• Effusion is when
gas molecules
escape without
collision, through a
tiny hole into a
vacuum.
47
Graham’s Law
• Graham’s Law: Rate of effusion is
proportional to its rms speed, urms.
3RT
Ratea u rms 
MM
• For two gases at same temperature and
pressure:
Rate1

Rate2
MM 2
MM 1

MM 2
MM 1
48
Behavior of Real Gases
• At higher pressures, particles are much
closer together and attractive forces become
more important than at lower pressures.
49
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.
50
Behavior of Real Gases
• Corrections for non-ideality require van der
Waals equation.

n 
 P  a 2   V – n  b   nRT
V 

2
Intermolecular
Attractions
Excluded
Volume
51
Example 1: Boyle’s 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)
52
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)
53
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)
54
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?
55
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?
56
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
69.5°C.
57
What is the volume (in liters) occupied by 7.40
g of CO2 at STP?
58
What is the molar mass of a gas with a density
of 1.342 g/L at STP?
59
What is the density of uranium hexafluoride, UF6,
(MM = 352 g/mol) under conditions of STP?
60
The density of a gaseous compound is 3.38 g/L at
40°C and 1.97 atm. What is its molar mass?
61
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?
62
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?
63
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?
64
On a humid day in summer, the mole fraction
of gaseous H2O (water vapor) in the air at
25°C 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?
65

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)
66
All of the mole fractions of elements in a given
compound must add up to?
1.
2.
3.
4.
100
1
50
2
67
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)
68
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?
69

What are the relative rates of diffusion of the
three naturally occurring isotopes of neon:
20Ne, 21Ne, and 22Ne?
70
• 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.
71
Given that 3.50 moles of NH3 occupy 5.20 L
at 47°C, calculate the pressure of the gas
(in atm) using
(a)
the ideal gas equation
(b) the van der Waals equation. (a = 4.17, b =
0.0371)
72
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 L2·atm mol–2, and b = 0.0387
L/mol.
73