CHM 101 American University of Sharjah Chapters 5 to 8 Molar Ratio Midterm i need help in a midterm the mid term will be 13/july/2020 at 5 pm uae timingyou

CHM 101 American University of Sharjah Chapters 5 to 8 Molar Ratio Midterm i need help in a midterm the mid term will be 13/july/2020 at 5 pm uae timingyou will find the content below GENERAL CHEMISTRY
Chapter 5
Gases
Yehya Elsayed
Ebbing, Gammon – 9th Edition
Physical Characteristics of
Gases
• Gases are highly compressible
• They occupy the full volume of their containers and
take the shape of the container.
• When a gas is subjected to pressure, its volume
decreases.
• Gases always form homogeneous mixtures with other
gases.
Elements that exist as gases at 250C and 1 atmosphere
5.1
5.1
Pressure
• Pressure is defined as the force acting on an object per
unit area:
Newtons (N)
F m g
P 
A
A
m2
(force = mass x acceleration)
• The SI unit for pressure is newton per square meter also
called pascal (Pa). Sometimes kPa is used
1 pascal (Pa) = 1 N/m2
Pressure
•SI Units: 1 N = 1 kg.m/s2;
1 Pa = 1 N/m2.
•Gravity exerts a force on the
earth’s atmosphere
•A column of air 1 m2 in cross
section exerts a force of about
105 N.
•The pressure of a 1 m2
column of air is
P=105N/1 m2
=105 Pa
=100 kPa.
1 atm = 101,325 Pa
10 miles
4 miles
Sea level
0.2 atm
0.5 atm
1 atm
5.2
Gas Pressure
Measuring Atmosphere Pressure:
Barometers
• Atmospheric pressure is
measured with a barometer.
• If a tube is inserted into a
container of mercury open to
the atmosphere, the mercury
will rise 760 mm up the tube.
• Standard atmospheric
pressure is the pressure
required to support 760 mm
of Hg in a column.
• Units: 1 atm = 760 mmHg =
760 torr = 1.01325  105 Pa =
101.325 kPa.
Manometers Used to
Measure Gas Pressures
Measuring Other Gas
Pressures: Manometers
• The pressures of gases not open to
the atmosphere are measured in
manometers.
• A manometer consists of a bulb of
gas attached to a U-tube containing
Hg:
Δ
Δh
– If Pgas Patm then Pgas = Patm + Δh .
Hg
– If Pgas = Patm then Δh =0
Manometers Used to Measure Gas Pressures
5.2
The Gas Laws
• There are four variables to specify a sample of gas in
calculations: its amount in moles (n), its volume (V), its
temperature (T), and its pressure (P)
• These variables are related to each other through several
simple laws.
The Gas Laws
Boyle’s Law
The Pressure-Volume Relationship:
Boyle’s Law
• Boyle’s Law: the volume of a fixed quantity of gas at
constant temperature is inversely proportional to its
pressure.
Apparatus for Studying the Relationship Between
Pressure and Volume of a Gas
As P (h) increases
V decreases
5.3
The Gas Laws
•
•
•
•
•
The Pressure-Volume Relationship:
Boyle’s Law
Weather balloons are used as a practical consequence to
the relationship between pressure and volume of a gas.
As the weather balloon ascends, the volume decreases.
As the weather balloon gets further from the earth’s
surface, the atmospheric pressure decreases.
Boyle’s Law: the volume of a fixed quantity of gas is
inversely proportional to its pressure at constant
temperature.
Boyle used a manometer to carry out the experiment.
The Gas Laws
Boyle’s Law
The Pressure-Volume Relationship:
Constant temperature
Boyle’s Law
Constant amount of gas
(a) hyperbola
(b) straight line passing through
the origin.
The Gas Laws
Boyle’s Law
The Pressure-Volume Relationship:
Boyle’s Law
• Mathematically:
1
1
which
means
V  constant 
V
P
P
• Then
PV  constant
• Another way to state Boyle’s law is as follows:
for a fixed amount of gas at a constant temperature, the
product of the pressure and volume is a constant.
PiVi  constant  Pf V f
WORKED EXAMPLE 6.2 Boyle’s Law: Pressure–Volume Relationships
A cylinder of oxygen has a volume of 2.25 L. The pressure of the gas is 1470
pounds per square inch (psi) at 20 °C. What volume will the oxygen occupy
at standard atmospheric pressure (14.7 psi) assuming no temperature change?
Solution
We find it helpful to first separate the initial from the final condition.
Then use the equation P1V1 = P2V2 and solve for the desired volume or
pressure. In this case, we solve for V2.
Because the final pressure (14.7 psi) is less than the initial pressure
(1470 psi), we expect the final volume (225 L) to be larger than the
original volume (2.25 L), and we see that it is.
A sample of chlorine gas occupies a volume of 946 mL at a
pressure of 726 mmHg. What is the pressure of the gas (in
mmHg) if the volume is reduced at constant temperature to
154 mL?
P x V = constant
P1 x V1 = P2 x V2
P2 =
P1 = 726 mmHg
P2 = ?
V1 = 946 mL
V2 = 154 mL
P1 x V1
V2
726 mmHg x 946 mL
=
154 mL
= 4460 mmHg
5.3
The Gas Laws
The Temperature-Volume Relationship:
Charles’s Law
• Charles’s Law: the volume of a fixed quantity of gas at
constant pressure increases as the temperature in
Kelvins increases.
• Mathematically:
V is directly proportional to kelvin T at constant P
VT
⇒
V  constant  T
V
i
T
i

V
T
V
 constant
T
f
f
As T increases
V increases
5.3
The Gas Laws
The Temperature-Volume Relationship:
Charles’s Law
• A plot of V versus T is a straight line.
• When T is measured in C, the intercept on the
temperature axis is -273.15C.
• We define absolute zero:
0 K = -273.15C.
Temperature must be
in Kelvin
WORKED EXAMPLE 6.3 Charles’s Law: Temperature–Volume
Relationship
A balloon indoors, where the temperature is 27 °C, has a volume of 2.00 L.
What would its volume be in a hot room where the temperature is 47 °C.
Solution
First, and most important, convert all temperatures to the Kelvin scale:
T(K) = t (ºC) + 273
Solving the equation
for V2, we have
As expected, because the temperature increases, the volume must also
increase.
A sample of carbon monoxide gas occupies 3.20 L at 125
0C. At what temperature will the gas occupy a volume of
1.54 L if the pressure remains constant?
V1 /T1 = V2 /T2
V1 = 3.20 L
V2 = 1.54 L
T1 = 398.15 K
T2 = ?
T1 = 125 (0C) + 273.15 (K) = 398.15 K
V2 x T1
1.54 L x 398.15 K
3.20 L
V1
t2 =192- 273.15 = – 81.15 0C
T2 =
=
= 192 K
5.3
The Gas Laws
The Mole-Volume Relationship:
Avogadro’s Law
• Avogadro’s Hypothesis: equal volumes of gas at the same
temperature and pressure will contain the same number
of molecules (also moles).
• Avogadro’s Law: the volume of a gas at a given
(constant) temperature and pressure is directly
proportional to the number of moles of gas.
The Gas Laws
The Mole-Volume Relationship:
Avogadro’s Law
• Mathematically: V  number of moles (n)
V  constant  n
Constant temperature
Constant pressure
V1 / n1 = V2 / n2
• 22.4 L of any gas at 0C (273.15K) and 1 atm (760 torr)
contain 6.02  1023 gas molecules.
• These conditions of 0C (273.15K) and 1 atm (760 torr)
are known as the standard temperature and pressure
conditions (STP)
The Mole-Volume Relationship: Avogadro’s Law
• The molar volume of a gas is the volume occupied
by one mol of the gas
• At STP: I mole of
The Combined Gas Law
• From the three gas laws.
• Charles’s Law:
1
V  (constant n, T )
P
V  T (constant n, P)
• Avogadro’s Law:
V  n (constant P,T )
• Boyle’s Law:
• We can combine these into a general gas law:
PV
PV
PV
nT
1 1
2 2

constant

constant

V
nT
n1T1
n2T2
P
PV
1 1
n1T1

PV
2
2
n2T2
The Combined Gas Law
• All gas laws can then be derived from the combined gas
law:
• Charle’s Law (fixed n and P): P1V1  P2V2
n1T1
V

1
n2T2
T1
V
2
T2
• Other laws can also be derived similarily (fixed V and n):
PV
1 1
n1T1
• This law is called Amonton’s law

PV
2
2
n2T2
P
1
T1

P
2
T2
A certain light bulb containing argon at 1.20 atm and 18 0C
is heated to 85 0C at constant volume. What is the final
pressure of argon in the light bulb (in atm)?
n, V are constant
P1
P2
=
T1
T2
P1 = 1.20 atm
T1 = 291 K
P2 = ?
T2 = 358 K
T2
358
K
=
1.20
atm
x
= 1.48 atm
P2 = P1 x
291 K
T1
5.4
The Ideal Gas Law and its Application
• If R is the constant of proportionality (called the gas
constant), then
PV
 constant
nT
PV
R
nT
• The ideal gas equation is:
PV  nRT
• The ideal gas is the gas that obeys all the simple gas laws
and has a molar volume at STP of 22.4L
• R = 0.08206 L·atm/mol·K = 8.314 J/mol·K
The Ideal Gas Law and its
Application
• Since R is constant, we can use the data associated with
the molar volume of a gas at STP:
1 atm X 22.414 L
L atm
PV
RR
 R  0.08206
nT
1 mol X 273.15 K
mol K
The Ideal Gas Equation
Relating the Ideal-Gas Equation and the
Gas Laws
• In general, if we have a gas under two sets of conditions,
then
P1V1

n1T1
P2V2
n2T2
/
Kg.m2/(s2.K.mol)
/
/
/
/
The Gas Laws
• Volume of 1 mol of gas at STP is:
(note: 4 known and one unknown. The units should be
identical to those shown in the gas constant)
PV  nRT
nRT 1 mol 0.08206 L·atm/mol·K 273.15 K 
V

 22.41 L
P
1.000 atm
What is the volume (in liters) occupied by 49.8 g of HCl at
STP?
T = 0 0C = 273.15 K
P = 1 atm
PV = nRT
nRT
V=
P
n = 49.8 g x
1.37 mol x 0.0821
V=
L•atm
x
mol•K
1 mol HCl
= 1.37 mol
36.45 g HCl
273.15 K
1 atm
V = 30.6 L
5.4
Applications of the Ideal-Gas
Equation
Molar (or molecular) Mass and Gas Densities
• Density has units of mass over volume.
• Rearranging the ideal-gas equation with M as molar mass
(n= m/M) we get
mRT
mRT
PV  nRT  PV 
 MPV  mRT  M 
M
PV
mRT
m MP
MP
PV  nRT  PV 
 MPV  mRT  
d 
M
V
RT
RT
• Also we can write:
mRT
dRT
M
M 
PV
P
A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and 27.0
0C. What is the molar mass of the gas?
mRT
mRT
PV  nRT  PV 
 MPV  mRT  M 
M
PV
M=
L•atm
mol•K
4.65 g x 0.0821
2.10 L
1 atm
x 300.15 K
M = 54.6 g/mol
OR
dRT
M=
P
M=
4.65 g
m
=
= 2.21
d=
V 2.10 L
g
2.21 L
L•atm
mol•K
x 0.0821
1 atm
x 300.15 K
g
L
Gases in Reaction Stochiometry
The Law of Combining Volumes: “Gay-Lussac’s
Law”
• In a reaction involving gases, the volumes of gaseous
reactants and products are in small whole-number ratios
as long as they are measured at the same temperature and
pressure.(2H2 + O2 →2H2).
(2mol H2+ 1mol O2→2 mol H2O)
(2volumes H2 reacts with 1volume O2 at STP to
produce 2volumes of H2O)
• This is because n and V are directly proportional to each
other at constant T and P. Either n or V can then be used
in stoichiometric calculations.
Gases in Reaction Stoichiometry
The Law of Combining Volumes
• Gay-Lussac’s Law of combining volumes: In a reaction
involving gases, at a given temperature and pressure, the
volumes of gaseous reactants and products are ratios of
small whole numbers.
The combining ratios 2:1:2
Ammonia burns in oxygen to form nitrogen oxide (NO) and
water vapor. How many volumes of NO are obtained from one
volume of ammonia at the same temperature and pressure?
4NH3 + 5O2
4NO + 6H2O
4 mole NH3
4 mole NO
1 mole NH3
1 mole NO
At constant T and P
1 volume NH3
1 volume NO
5.3
Gas Stoichiometry
What is the volume of CO2 produced at 37 0C and 1.00 atm
when 5.60 g of glucose are used up in the reaction:
C6H12O6 (s) + 6O2 (g)
g C6H12O6
5.60 g C6H12O6
V=
x
nRT
=
P
mol C6H12O6
6CO2 (g) + 6H2O (l)
mol CO2
V CO2
6 mol CO2
1 mol C6H12O6
x
= 0.187 mol CO2
180 g C6H12O6
1 mol C6H12O6
L•atm
x 310.15 K
mol•K
1.00 atm
0.187 mol x 0.0821
= 4.76 L
Mixtures of Gases: Dalton’s Law of Partial
Pressures
• Since gas molecules are so far apart, we can assume they
behave independently.
• Dalton’s Law: the total pressure exerted by a mixture of
gases is equal to the sum of the partial pressures exerted
by each gas:
P
total
 P  P  P  P
1
2
3
• Each gas obeys the ideal gas equation:
RT 

Pi  ni 

V 
i
Dalton’s Law of Partial Pressures
V and T
are
constant
P1
P2
Ptotal = P1 + P2
5.6
Mixtures of Gases: Dalton’s Law of
Partial Pressures
• Combing the equations
RT 

Ptotal  n1  n2  n3  

V 
Partial Pressures and Mole Fractions
• Let ni be the number of moles of gas i exerting a partial
pressure Pi, then
RT
)
Pi
V

RT
Pt
nt (
)
V
ni (
Pi
Pt

ni
nt
where i is the mole fraction (ni/nt).
Pi  i Ptotal
nA
XA =
nA + nB
A sample of natural gas contains 8.24 moles of CH4, 0.421
moles of C2H6, and 0.116 moles of C3H8. If the total
pressure of the gases is 1.37 atm, what is the partial pressure
of propane (C3H8)?
Pi = Xi PT
PT = 1.37 atm
Xpropane =
0.116
8.24 + 0.421 + 0.116
Ppropane = 0.0132 x 1.37 atm
= 0.0132
= 0.0181 atm
5.6
Gas Mixtures and Partial Pressures
Collecting Gases over Water
PT = Pg + PH O
2
2KClO3 (s)
2KCl (s) + 3O2 (g)
PT = PO2 + PH2 O
Mixtures of Gases: Dalton’s Law of Partial
Pressures
Collecting Gases over Water
• It is common to synthesize gases and collect them by
displacing a volume of water.
• To calculate the amount of gas produced, we need to
correct for the partial pressure of the water:
Ptotal  Pgas  Pwater
• The partial pressure of the water vapor is a quantity
called the vapor pressure (V.P.) of water. It depends only
on the temperature. The warmer the water is, the higher
its V.P.
Copyright © by Houghton Mifflin Company. All rights reserved.
• Suppose a 156 mL sample of H2 gas was collected over water at
19 oC and 769 mm Hg. What is the mass of H2 collected?
• The vapor pressure of water at 19 oC is 16.5 mm Hg (Table).
• From the ideal gas law, PV = nRT, you have
2.02 g H 2
0.00644 mol H 2 
 0.0130 g H 2
1 mol H 2
An 18.6 mL volume of hydrochloric acid reacts completely with a
solid sample of MgCO3. The reaction is:
2HCl + MgCO3—-> CO2 + H2O + MgCl2
The volume of CO2 formed is 159 mL at 23 oC and 731 mmHg. What
is the molarity of the HCl solution?
Find the number of moles of CO2 first. Then convert this to moles
of HCl and molarity of HCl.
n=
731/760 atm x 0.159 L
= 0.006296mol


0.08206 L at m/ K mol x 296 K
2 mol HCl
= 0.01259 mol
mol HCl = 0.006296 mol CO2 x
1 mol CO2
0.01259 mol
Molarity =
= 0.6770 = 0.677 M HCl
0.0186 L
Kinetic Molecular Theory Of Gases
1. A gas is composed of molecules that are separated from each
other by distances far greater than their own dimensions. The
molecules can be considered to be points; that is, they possess
mass but have negligible volume.
2. Gas molecules are in constant motion in random directions, and
they frequently collide with one another. Collisions among
molecules are perfectly elastic.
3. Gas molecules exert neither attractive nor repulsive forces on one
another.
4. The average kinetic energy of the molecules is proportional to the
temperature of the gas in kelvins. Any two gases at the same
temperature will have the same average kinetic energy
KE = ½ mu2
Maxwell’s distribution of molecular speeds.
Kinetic Molecular Theory
• As kinetic energy increases, the velocity of the gas
molecules increases.
• Average kinetic energy, , is related to root mean square
speed:
  1 mu 2
2
• Root mean square speed, u, is the speed of a gas molecule
having average kinetic energy.
Kinetic Molecular Theory
• Average kinetic energy of a gas is related to its mass:
2
1
  2 mu
• Consider two gases at the same temperature: the lighter
gas has a higher rms than the heavier gas.
• Mathematically:
3RT
u
R= 8.314 J/mol.K
M
Kinetic Molecular Theory
• Kinetic molecular theory explains the origin of a gas
pressure.
• Pressure of a gas results from the number of collisions
per unit time on the walls of container.
• In other words, the magnitude of pressure is given by
how often and how hard the molecules strike the walls of
the container.
Application of Kinetic Molecular
Theory to Explain Boyle’s Law
• However, as the volume increases, the area of the
container walls increase. As a result the gas molecules
will strike the container walls less often. Therefore,
pressure decreases.
• Note: As volume increases at constant temperature, the average
kinetic energy of the gas remains constant. Therefore, u is
constant.
Application of Kinetic Molecular
Theory to Explain Charle’s Law
• As we heat a gas, we supply the gas with increased
energy, and they begin to move faster and strikes the
container walls harder and more often.
• To keep the pressure constant, the volume of the
container must increase so that the surface area is also
increased.
Same logic can be used to explain avogadro’s law using kinetic molecular theory
Kinetic Molecular Theory: Effusion and
Diffusion
Graham’s Law of Effusion
• Effusion is the process in
which a gas escapes from its
container through a tiny hole
(a balloon will deflate over
time due to effusion).
• The rate of effusion can be
quantified.
Kinetic Molecular Theory
Effusion
• Consider two gases with molar masses M1 and M2, the
relative rate of effusion is given by:
r1
r2

u1
u2
3RT

3RT
M1
M2

M2
M1
r1 u1 t 2
M2

 
r2 u2 t1
M1
• Only those molecules that hit the small hole will escape
through it.
• Therefore, the higher the rms the more likelihood of a gas
molecule hitting the hole.
Kinetic Molecular Theory
Molecular Effusion
• The lower the molar mass, M, the higher the rms.
3RT
u
M
• Graham’s law of effusion states that the rate of effusion of gas
molecules from a particular hole is inversely proportional to the
square root of the molecular mass of the gas at constant T & P
Kinetic Molecular Theory
Diffusion and Mean Free Path
• Diffusion is the process by which the molecules of one gas mixes
with the molecules of one or more other gases as a result of
translation motion of molecules
• Diffusion is faster for light gas molecules.
r1
r2
=

M2
M1
• consider someone opening a perfume bottle: it takes a while to
detect the odor but rms speed at 25C is about 1150 mi/hr (511m/s)
• Diffusion is slowed by gas molecules colliding with each other.
Obtain the ratio of the rates of effusion of H2 and H2Se under
the same conditions
The rates of effusion are directly related to the rms speed. Assume the temperature is
the same for both, and write a ratio of two Maxwell distributions.
1/ 2
uH
uH
2
2
Se
=
 3RT 


M
(H
)
 m 2 
1/ 2
1/ 2
 3RT 


M
(H
Se)
 m 2 
 M m (H 2Se) 
 80.976 g/mol 
= 
=



2.016
g/mol
 M m (H 2 ) 


1/ 2
= 6.3377 = 6.338
Thus the ratio of the rates of effusion of H2 and H2Se is 6.338 to 1
Nickel forms a gaseous compound of the formula
Ni(CO)x.What is the value of x given that under the same
conditions methane (CH4) effuses 3.3 times faster than
the compound?
Let methane be gas 1 and Ni(CO)x be gas 2
r1 = 3.3 x r2
M1 = 16 g/mol
M2 =
=
r1 2
(r )
2
(3.3)2
r1

r2
M2
M1
x M1
x 16
= 174.2
58.7 + 28 x = 174.2
x = 4.1 ~ 4
Ni(CO)4
Real Gases: Deviations from Ideal
Behavior
• From the ideal gas equation, we have PV  n
RT
PV
• Ideal behavior, For 1 mol of gas,
1
RT
for all
pressures and temperatures.
• In a real gas, PV/RT varies from 1 significantly.
• Intermolecular forces
• Relative volume is not negligible
Repulsive Forces
Attractive Forces
Real Gases: Deviations
from Ideal Behavior
Real Gases: Deviations
from Ideal Behavior
• As the pressure on a gas increases, the molecules are
forced closer together.
• As the molecules get closer together, the volume of the
container gets smaller.
• The smaller the container, the more space the gas
molecules begin to occupy.
• Therefore, the higher the pressure, the less the gas
resembles an ideal gas.
Real Gases: Deviations
from Ideal Behavior
• The smaller the distance between gas molecules, the
more likely attractive forces will develop between the
molecules.
• Therefore, the less the gas resembles an ideal gas.
• As temperature increases, the gas molecules move faster
and further apart.
• Also, higher temperatures mean more energy available to
break intermo…
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