Transcript Chapter 1 Gases Dr. Hisham E Abdellatef Professor of pharmaceutical analytical chemistry 2011 - 2012

Chapter 1
Gases
Dr. Hisham E Abdellatef
Professor of pharmaceutical
analytical chemistry
2011 - 2012
Characteristics of Gases
• Expand to fill and
assume the shape
of their container
• Compressible
Next
• Diffuse into one another and mix in all
proportions. homogeneous mixtures
• Particles move from an area of high concentration
to an area of low concentration.
•
Next
Properties
that determine physical behavior of a
gas
1. VOLUME
2. PRESSURE
3. TEMPERATURE
lxwxh
1. Volume
πr²h
capacity of the container enclosing it.
(m3), (dm3), or liter.
For smaller volumes (cm3), (ml).
2. Temperature
• Temperature: Three temperature
scales
•
Fahrenheit (ºF)
•
Celsius
(ºC)
•
Kelvin
K (no degree symbol)
• K performing calculations with the gas
law equations.
gas
• Expand when heated
Temperature
• K = (ºC) + 273.15
• Example: ºC = 20º
• K = 293.15
• Absolute or Kelvin Scale: =-273.15 (ºC
as its zero). (T when V= 0 )
• 1 K = 1 ºC
3. Pressure
The molecules in a gas are in
constant motion. gaseous
atoms that collide with
each other and the walls of
the container.
"Pressure" is a measure of
the collisions of the atoms
with the container.
Pressure
• Force per unit area
• Equation: P = F/A
F = force
A = area
Next
Barometer
- Atmospheric pressure
is measured with a
barometer.
• Height of mercury
varies with
atmospheric
conditions and with
altitude.
Measurement of Gas Pressure
Mercury
Barometer
- Standard atmospheric pressure is the
pressure required to support 760 mm
of Hg in a column.
- There are several units used for
pressure:
- Pascal (Pa), N/m2
- Millimeters of Mercury (mmHg)
- Atmospheres (atm)
Manometers
• Used to compare
the gas pressure
with the
barometric
pressure.
Next
Types of Manometers
• Closed-end manometer
The gas pressure is equal
to the difference in
height (Dh) of the
mercury column in the
two arms of the
manometer
Open-end Manometer
The difference in mercury levels (Dh)
between the two arms of the
manometer gives the difference
between barometric pressure and the
gas pressure
Three Possible Relationships
1. Heights of mercury in both columns are equal if
gas pressure and atmospheric pressure are
equal.
Pgas = Pbar
2. Gas pressure is
greater than the
barometric pressure.
∆P > 0
Pgas = Pbar + ∆P
3. Gas pressure is less
than the barometric
pressure.
∆P < 0
Pgas = Pbar + ∆P
The Simple Gas Laws
1. Boyle’s Law
2. Charles’ Law
3. Gay-Lussac’s Law
4. Combined Gas Law
variables required
to describe a gas
•
•
•
•
Amount of substance: moles
Volume of substance: volume
Pressures of substance: pressure
Temperature of substance: temperature
The Pressure-Volume Relationship:
Boyle’s Law
Boyle’s Law - The volume of a fixed
quantity of gas is inversely
proportional to its pressure.
1
P
(constantn and T )
V
P1V1  P2V2
25
The Pressure-Volume Relationship:
Boyle’s Law
Example
An ideal gas is enclosed in a Boyle's-law
apparatus. Its volume is 247 ml at a pressure
of 625 mmHg. If the pressure is increased
to 825 mmHg, what will be the new volume
occupied by the gas if the temperature is
held constant?
28
Solution
Method 1:
P1V1 = P2V2
or, solving for V2 the final volume
P V
V  1 1
2
p
2
625 mmHgX247 ml
V 
 187 ml
2
825 mmHg
Solution
Method 2:
The pressure of the gas increases by a
factor 825/625, the volume must decrease by a
factor of 625/825
V2= V1 X (ratio of pressures)
625 mmHg
V  247 x
 187 ml
2
825 mmHg
Charles’ Law
The volume of a fixed
amount of gas at constant
pressure is directly
proportional to the Kelvin
(absolute) temperature.
V1 = V2
T1
T2
or V1T2 = V2T1
Example . Charles’ Law
A 4.50-L sample of gas is
warmed at constant
pressure from 300 K to
350 K. What will its final
volume be?
Given:
V1 = 4.50 L
T1 = 300. K
T2 = 350. K
V2 = ?
Equation:
V1 = V 2
T1
T2
or V1T2 = V2T1
(4.50 L)(350. K) = V2 (300. K)
V2 = 5.25 L
Gay-Lussac’s Law
The pressure of a sample of gas is directly
proportional to the absolute temperature when
volume remains constant.
P1 = P2
T1
T2
or
P1T2 = P2T1
On the next slide
The amount of gas and its volume are the
same in either case, but if the gas in the
ice bath (0 ºC) exerts a pressure of 1 atm,
the gas in the boiling-water bath (100 ºC)
exerts a pressure of 1.37 atm. The
frequency and the force of the molecular
collisions with the container walls are
greater at the higher temperature.
Combined Gas Law
Pressure and volume are inversely proportional to
each other and directly proportional to
temperature.
P1V1
T1
or
= P2V2
T2
P1V1T2 = P2V2T1
Example. Combined Gas Law
A sample of gas is pumped
from a 12.0 L vessel at
27ºC and 760 Torr
pressure to a 3.5-L vessel
at 52ºC. What is the final
pressure?
Given:
P1 = 760 Torr
V1 = 12.0 L
T1 = 300 K
P2 = ?
V2 = 3.5 L
T2 = 325 K
Equation:
P1V1
T1
or
= P2V2
T2
P1V1T2 = P2V2T1
(760 Torr)(12.0 L)(325 K) =
( P2)(3.5 L)(300 K)
P2 = 2.8 x 10³ Torr
Avogadro’s Law
Volume & Moles
Avogadro’s Law
At a fixed temperature and pressure, the
volume of a gas is directly proportional to
the amount of gas.
V=c·n
V = volume c = constant
n= # of moles
Doubling the number of moles will cause the
volume to double if T and P are constant.
Ideal Gas Law
Equation
Includes all four gas variables:
• Volume
• Pressure
• Temperature
• Amount of gas
Next
PV = nRT
•
Gas that obeys this equation if said to be an ideal gas (or perfect gas).
•
No gas exactly follows the ideal gas law, although many gases come very close
at low pressure and/or high temperatures.
•
Ideal gas constant, R, is
R = PV
nT
= 1 atm x 22.4 L
1 mol x 273.15 K
R = 0.082058 L·atm/mol· K
Example
• Suppose 0.176 mol of an ideal gas
occupies 8.64 liters at a pressure of
0.432 atm. What is the temperature
of the gas in degrees Celsius?
Solution
PV = nRT
(0.432atm)(8.64liters)

(0.176mol)(0.082
1 liter atm K -1mol-1 )
 258 K
To degrees Celsius we need only
subtract 273 from the above result:
=258 - 273 = -15OC
Example
• Suppose 5.00 g of oxygen gas, O2, at
35 °C is enclosed in a container
having a capacity of 6.00 liters.
Assuming ideal-gas behavior,
calculate the pressure of the oxygen
in millimeters of mercury. (Atomic
weight: 0 = 16.0)
Solution
•
One mole of O2 weighs 2(16.0) =
32.0 g. 5.00 g of O2 is, therefore,
5.00 g/32.0 g mol-1, or 0.156 mol. 35
°C is 35 + 273 = 308 K
• PV = nRT
p
(0.156 mol)(0.082 1 liter atm K - I mol - 1)( 308 K)
 0.659 atm
6.00 liters
 0.659 atm
760 mmHg
 500 mm Hg
1 atm
Molar
volume of an
gas at STP
ideal
• The volume occupied by one mole, or
molar volume, of an ideal gas at STP
is
nRT
V
P
(1.0000mol)(0.082
057 lit er at m K -I mol-1 )( 273.15K)

1.0000at m
 22.414lit ers
Applications of the
Ideal Gas Law
Dalton’s Law of
Partial Pressure
Mixture of Gases
Total Pressure
The total pressure of a mixture of
gases is the sum of the partial
pressures of the components of the
mixture.
Ptotal = PA + PB + ……
Total Pressure: Mixture of
Gases
Example: Gas Mixtures & Partial
Pressure
A gaseous mixture made from 6.00 g O2 and 9.00 g
CH4 is placed in a 15.0 L vessel at 0ºC. What is
the partial pressure of each gas, and what is the
total pressure in the vessel?
Step 1:
nO2 = 6.00 g O2 x 1 mol O2
32 g O2
= 0.188 mol O2
Next ---- >
1 mol CH4
16.0 g CH4
= 0.563 mol CH4
nCH4 = 9.00 g CH4 x
Step 2: Calculate pressure exerted by each
PO2 = nRT
V
= (0.188 mol O2)(0.0821 L-atm/mol-K)(273 K)
15.0 L
= 0.281 atm
PCH4 =
(0.563 mol)(0.0821 L-atm/mol-K)(273 K)
15.0 L
= 0.841 atm
Step 3: Add pressures
Ptotal = PO2 + PCH4
= 0.281 atm + 0.841 atm
Ptotal = 1.122 atm
Try ?
• Q1, Q2, Q3 page 42(lecturer note )
Graham’s Law
Molecular Effusion and Diffusion
Molecular Effusion and Diffusion
Graham’s Law of Effusion
Effusion – The escape of gas through a
small opening.
Diffusion – The spreading of one
substance through another.
Molecular Effusion and Diffusion
Graham’s Law of Effusion
Graham’s Law of Effusion – The rate
of effusion of a gas is inversely
proportional to the square root of its
molecular mass.
Molecular Effusion and Diffusion
Graham’s Law of Effusion
Graham’s Law of Effusion – The rate
of effusion of a gas is inversely
proportional to the square root of its
molecular mass.
r1
M2

r2
M1
Molecular Effusion and Diffusion
Graham’s Law of Effusion
Graham’s Law of Effusion – The rate
of effusion of a gas is inversely
proportional to the square root of its
molecular mass.
• Gas escaping from a balloon is a
good example.
r1
M2
d2


r2
M1
d1
Molecular Effusion and Diffusion
Graham’s Law of Effusion
r1
M2
d2


r2
M1
d1
Chapter 10
63
Example 1. Rate of Effusion
• The rate of effusion of an unknown gas (X) through
a pinhole is found to be only 0.279 times the rate of
effusion of hydrogen (H2) gas through the same
pinhole, if both gases are at STP. What is the
molecular weight of the unknown gas?
• (Atomic weight; H = 1.01.)
Next
• Solution
Mx= 26.0
• Pv = nRt
n=m
M
mass
d
Pv= m RT
PM = m RT
M
V
PM = dRT
or d = PM
RT
Example2 . Rate of Effusion
Calculate the ratio of the
effusion rates of hydrogen
gas (H2) and uranium
hexafluoride (UF6), a gas used
in the enrichment process
fuel for nuclear reactors.
Known:
Molar Masses
H2 = 2.016 g/mol
UF6 = 352.02 g/mol
(Rate of effusion)² =
MU compound
MH gas
= 352.02
2.016
Rate of effusion = 13.21
Quize
• 1. what is the relative rates of
diffusion of H2 and CO2 under the
same condition ?
• 2. What is the density of gas which
it’s diffusion is 1.414 times of the
rate of diffusion of CO2 at STP ???.
Problem
Calcium hydride, CaH2, reacts with water to
form hydrogen gas:
CaH2(s) + 2 H2O(l)  Ca(OH)2(aq) + 2
H2(g)
How many grams of CaH2 are needed to
generate 10.0L of H2 gas if the pressure of
H2 is 740 torr at 23oC?
Problem
CaH2(s) + 2 H2O(l)  Ca(OH)2(aq) + 2 H2(g)
-Calculate moles of H2 formed
-Calculate moles of CaH2 needed
-Convert moles CaH2 to grams
Problem
CaH2(s) + 2 H2O(l)  Ca(OH)2(aq) + 2 H2(g)
P  740torr / 760  0.974
V  10.0 L
T  23o C273296K
r  0.08206L(atm) / m ol( K )
Problem
CaH2(s) + 2 H2O(l)  Ca(OH)2(aq) + 2 H2(g)
P  740torr / 760  0.974
V  10.0 L
T  23o C273296K
r  0.08206L ( atm) / m ol( K )
0.974atm(10.0 L )
nH 2 
0.08206( 296K )
Problem
CaH2(s) + 2 H2O(l)  Ca(OH)2(aq) + 2 H2(g)
P  740torr / 760  0.974
V  10.0 L
T  23o C273296K
r  0.08206L(atm) / m ol( K )
0.974atm(10.0 L)
nH 2 
0.08206( 296K )
nH 2  0.401m ol
Problem
CaH2(s) + 2 H2O(l)  Ca(OH)2(aq) + 2 H2(g)
P  740torr / 760  0.974
V  10.0 L
T  23o C273296K
r  0.08206L(atm) / m ol( K )
nH 2
nH 2
0.974atm(10.0 L)

0.08206( 296K )
 0.401m ol
1CaH 2
x

2 H2
0.401m ol
Problem
CaH2(s) + 2 H2O(l)  Ca(OH)2(aq) + 2 H2(g)
P  740torr / 760  0.974
V  10.0 L
T  23o C273296K
r  0.08206L(atm) / m ol( K )
0.974atm(10.0 L)
0.08206(296K )
 0.401m ol
nH 2 
nH 2
1CaH 2
x

2 H2
0.401m ol
x  0.2005m ol
Problem
CaH2(s) + 2 H2O(l)  Ca(OH)2(aq) + 2 H2(g)
P  740torr / 760  0.974
V  10.0 L
T  23o C273296K
r  0.08206L(atm) / m ol( K )
0.974atm(10.0 L)
nH 2 
0.08206(296K )
nH 2  0.401m ol
1CaH 2
x

2 H2
0.401m ol
x  0.2005m ol
gCaH2  0.2005m ol(42.10g / m ol)
Problem
CaH2(s) + 2 H2O(l)  Ca(OH)2(aq) + 2 H2(g)
P  740torr / 760  0.974
V  10.0 L
T  23o C273296K
r  0.08206L(atm) / m ol( K )
nH 2
nH 2
0.974atm(10.0 L)

0.08206(296K )
 0.401m ol
1CaH 2
x

2 H2
0.401m ol
x  0.2005m ol
gCaH2  0.2005m ol(42.10g / m ol)
 8.44g
Kinetic-Molecular
Theory of Gases
Kinetic-Molecular Theory
-
Theory developed to explain gas behavior
To describe the behavior of a gas, we must first
describe what a gas is:
–
–
–
–
–
Gases consist of a large number of molecules in constant random
motion.
Volume of individual molecules negligible compared to volume of
container.
Intermolecular forces (forces between gas molecules)
negligible.
Energy can be transferred between molecules, but total kinetic
energy is constant at constant temperature.
Average kinetic energy of molecules is proportional to
temperature.
Nonideal (Real) Gases
Gases may be
1.
Ideal gas on obeys the gas law PV =
constant (at constant T)
2. Real gas Deviate (not obey to gas
law PV ≠ Constant)
• Real gas deviate from ideal gases
when (T is very low and P is very
high).
• two factors :
• Real gases posse’s attractive forces
between molecules.
• Every molecule in a real has a real
volume.
True volume (Real) =
• V container – non copressiable
volume (b)
= V- b for 1 mole
Actual volume = V – nb ( for n mole)
force of attraction between
the molecules of gases:
(pressure)
a
 p 2
V
1 mole
2
an
p 2
V
n moles
van der Waals Equation
van der Waals Equation
Equation corrects for volume and intermolecular forces
(P + n²a/V²)(V-nb) = nRT
•
n²a/V² = related to intermolecular forces of attraction
•
n²a/V² is added to P = measured pressure is lower than expected
•
a & b have specific values for particular gases
•
V - nb = free volume within the gas
EXAMPLE
Calculate the pressure exerted by 10.0 g of methane, CH4,
when enclosed in a 1.00-liter container at 25 °C by using (a) the
ideal-gas law and (b) the van der Waals equation.
Gas
a (L2.atm/mol2)
b (L/mol)
CO2
3.658
0.04286
Ethane C2H6
5.570
0.06499
Methane CH4
2.25
0.0428
Helium He
0.0346
0.0238
Hydrogen H2
0.2453
0.02651
Oxygen O2
1.382
0.03186
Sulfur dioxide SO2
6.865
0.05679
The molecular weight of CH4 is 16.0; so n, the number of moles
of methane, is 10.0 g/16.0 g mol-1, or 0.625 mol.
(a) Considering the gas to be ideal and solving for P, we obtain
nRT (0.625 mol)(0.0821 liter atm K -1 mol 1 )(298K)
P

 15.3 atm
V
1.00 liter
(b) Treating the gas as a Van der Waals gas and solving for P,
we have
nRT n 2 a
P

V  nb V 2
(0.625 mol)(0.082 1 liter atm K -1 mol 1 )(298K)
P
1.00 liter - (0.625 mol) ( 0.248 liter mol -1 )
(0.625 mol) 2 (2.25 liters 2 atm mol  2 )

(1.00 liter) 2
 14.8atm
Try ?
• Q1, Q2, Q3 page 50 (lecturer note )