Transcript Document

Chemistry
States of matter – Session 2
Session Objectives
Session Objectives
1. Graham’s law of diffusion/effusion
2. Postulates of kinetic theory of gases
3. Kinetic gas equation and kinetic energy of
gases
4. Velocity of gas molecules
5. Maxwell-Boltzmann velocity distribution
6. Explanation of gas laws on the basis of
kinetic theory of gases
Questions
Illustrative example 7
Which of the following compounds is
steam volatile?
OH
Cl
OH
OH
(b)
(a)
Cl
OH
CH3
(c)
F
(d)
Solution
The compound which forms
intramolecular H-bonding has lower
boiling point and hence, steam volatile.
Hence, the answer is (b).
O
O
H
H
Illustrative example 1
A mixture CO and CO2 is found to
have a density of 2gL–1 at 250C
and 740 torr. Find the composition
of the mixture
Solution:
Mmixture
ρRT
=
P
1.5  0.082  298  760

740
 37.64 gmol1
Let x mole of CO and (1  x) mole of CO2
 x  28  (1  x)  44  37.64
x  0.3975
%CO  39.75
%CO2  60.25
Illustrative example 2
A 1500 ml flask contains 400 mg of O2
and 60 mg H2 at 1000C and they are
allowed to react to form water vapour.
What will be the partial pressure of the
substances present at that temperature?
Solution:
2H2(g)  O2(g) 
 2H2O
moles of H2  0.03
moles of O2  0.0125
Since O2 is the limiting reagent
moles of H2O vapour  0.025
 PH2O
0.025  0.082  373

1.5
 0.51 atm
0.005  0.082  373
PH2 
1.5
 0.102 atm
Animation for diffusion
Graham’s Law of Diffusion
(i) For same volume of two gases
v
r1
t
 1 
v
r2
t2
M2
M1
t2

t1
M2
M1
(ii) For two gases with different volumes and
same diffusion time
v1
r1
 t 
v2
r2
t
M2
M1
v1

v2
M2
M1
Effusion
It happens under pressure through a
Small aperture.
Rate of effusion
r
PA
2RTM
A=area of aperture
P  Pi  Po (For same gas)
Normally considered against vacuum
Question
Illustrative example 3
4:1 molar mixture of He and CH4 is
effusing through a pinhole at a
constant temperature.What is the
composition of the mixture effusing
out initially?
Solution:
 PHe 
PCH4
4
P
5
1
 P
5
4
P M
rHe
CH4

 5
1
rCH4
MHC
P
5

4 16
 8 :1
1 4
Solution
Let x mole of CH4 is effusing in t sec.
Then, 8x mole of He is effusing in
same time
 Mole fraction of He, xHe
8x

 0.89
8x  x
Composition of He in the mixture = 89%
Composition of CH4 in the mixture = 11%
Question
Illustrative example 4
A balloon filled with ethylene (C2H4) is
pricked with a sharp point & quickly
dropped in a tank full of hydrogen at
the same pressure. After a while the
balloon will have
(a) Shrunk
(b) Enlarged
(c) Completely collapsed (d) Remain unchanged in size
Solution
1
r
M
Since, molar mass of H2 is much less than C2H4.
H2 will diffuse into the balloon.
Hence, answer is (b).
Kinetic molecular theory
• All gases are made up of very large
number of extremely small particles
called molecules.
 The actual volume of the molecules is
negligible as compared to the total volume
of the gas.
 The distances of separation between the
molecules are so large that the forces of
attraction or repulsion between them are
negligible.
Kinetic molecular theory
• The molecules are in a constant
state of motion in all directions.
During their motion, they collide
with one another and also the walls
of the container.
• The molecular collisions are perfectly
elastic.
• The pressure exerted by the gas is due
to bombardment of the gas molecules
on the walls of the container.
 Average kinetic energy of molecule

absolute temperature.
Kinetic Theory of Gases
The kinetic gas equation,
PV 
1
mn'c 2
3
Where,
m = mass of each gas molecule
n’ = number of gas molecules
c = velocity of gas molecule
KE of one molecule =
1 2
mc
2
Kinetic Theory of Gases
PV 
1
2
mc 2  n'
2
3
3
1 2 (where n = number of moles,
 nRT  mc n'
2
2
n’ = number of molecules)
3
1
RT  mc 2 . NA
2
2
(for one mole of a gas, n =1 and n’ = NA )
= Average kinetic energy per mole
Ask yourself
What will be the
average kinetic
energy of one
molecule?
3 R
3


T

kT
Average kinetic energy of one molecule =
2 NA
2
k = Boltzmann constant
= 1.38 × 10–16 ergs k–1 molecule–1
Questions
Illustrative example 5
Calculate the average kinetic energy
per molecule and total kinetic energy
of 2 moles of an ideal gas at 25oC.
Solution:
Average KE per molecule of the
3 R
3
8.314


T


 298
gas =
23
2 NA
2 6.023  10
= 6.17 × 10–21 J
Solution
3
3
Average KE per mole of the gas =
RT = x 8.314 x 298
2
2
= 3.72 kJ/mole
3
Total KE of 2 moles of the gas = RT x 2 = 7.44 kJ
2
Molecular velocity
Average velocity
C1  C2  C3 .....  Cn
CAV 
N
Also CAV 
8 RT

M
8 PV

M
Most probable velocity (CMP)
CMP 
2 RT

M
2 PV

M
2P

8P

Molecular velocity
Root mean square velocity (CRMS)
CRMS 
CRMS 
C12  C22  ......  Cn2
N
3 RT

M
3 PV

M
3 P

Interrelation of molecular velocities
CAvg : CRMS = 0.9213
CMP : CRMS = 0.8165
CAvg: : CMP = 1.1286
Questions
Illustrative example 6
The gas molecules have root mean
square velocity of 1000 m/s.
What is its average velocity?
(a) 1000 m/s
(b) 921.58 m/s
(c) 546 m/s
(d) 960 m/s
Solution:
CRMS 
C Avg 
3RT
 103 m / s    (1)
M
8RT
     (2)
M
From (1) and (2)
CAvg 
8
 103  921.5 m / s
3
Illustrative example 7
Calculate the temperature at which root
mean square velocity of SO2 molecules
is same as that of O2 molecules at 27° C.
Solution:
1/ 2
CRMS
 3RT 


 M 
1/ 2
 (3) (R) (300 ) 
C
O

For O2 at 27° C,


RMS 2
32


For SO2 at t° C,
1/ 2
 (3) (R) (273  t) 
CRMSSO2  

64


Solution
Since both these velocities are equal,
1/ 2
 (3) (R) (300) 


32


1/ 2
 (3) (R) (273  t) 
 

64


or 600 = 273 + t
or t = 600 – 273 = 327° C
Fraction of molecules
Maxwell-Boltzmann velocity distribution
0.3
T = 500 K
0.2
T = 1500 K
0.1
0 2 × 10
4
4
4
4
4
4
4 × 10 6 × 10 8 × 10 10 × 10 12 × 10
Molecular speed (cm/sec)
Maxwell Boltzmann distribution
Characteristic features of
Maxwell’s distribution curve
• A very small fraction of molecules has
very low or very high speeds.
• The fraction of molecules possessing higher
and higher speed goes on increasing till it
reaches a peak. The fraction with still higher
speed then goes on decreasing.
• The peak represents maximum fraction of
molecules at that speed. This speed,
corresponding to the peak in the curve,is
known as the most probable speed.
• On increasing temperature, the value of the
most probable speed also increases.
Explanation of Boyle’s law on
the basis of kinetic theory
According to kinetic gas equation
PV 
1
2 1
mn`c2   mn'c2
3
3 2
2 1 2
 Mc where M  n`m and n` NA
3 2
1 2
Mc  Kinetic energy of the gas
2
2
PV  KE
3
KE  Absolute temperature(T)
KE  kT
PV 
2
kT
3
Explanation of Boyle’s law on
the basis of kinetic theory
As
2
is a cons tan t quantity and
3
k is also a cons tan t,
therefore, if T is kept cons tan t,
2
kT will be constant
3
Hence PV  cons tant, which is Boyle's law.
Explanation of charl’s law on
the basis of kinetic theory
As deduced from the kinetic
gas equation, we have
2
PV  kT
3
V 2k

T 3P
Hence, if P is kept constant,
V
= constant which is Charles’ law.
T
Explanation of Dalton’s law on
the basis of kinetic theory
Let us consider only two gases.
According to kinetic gas
equation,
1 mn' c2
P
3
V
1
mn' c2
3
Now, if only the first gas is enclosed in the vessel of
volume V, the pressure exerted would be
PV 
' 2
1 m1n2c1
P1 
3
V
' 2
Again, if only the second gas is enclosed in
c2
1 m2n2
P2 
the same vessel (so that V is constant),
3
V
then the pressure exerted would be
Explanation of Dalton’s law on
the basis of kinetic theory
Lastly, if both the gases are enclosed
together in the same vessel then since
the gases do not react with each other,
their molecules behave independent of
each other. Hence, the total pressure
exerted would be
' 2
' 2
1 m1n1c1 1 m2n2c2
P

3
V
3
V
= P1 + P2
Similarly, if more than two gases are present, then it can
be proved that P = P1 + P2 + P3 + ...
Explanation of Avogadro’s hypothesis
on the basis of kinetic theory
Let us assume equal volume of two
gases at the same temperature and
pressure, then from kinetic gas
equation.
1
PV  m1n1' c12    (for first gas)
3
1
' 2
PV  m2n2
c2    (For sec ond gas)
3
' 2
 m1n1' c12  m2n2
c2    (i)
Since average kinetic energy per molecule depends
on temperature,
1
1
2
2
m1c1  m2c2
   (ii)
2
2
Explanation of Avogadro’s
hypothesis on the basis of
kinetic theory
Dividing equation (i) by (ii), we have
'
n1
 n'2
This is Avogadro’s hypothesis.
Explanation of Graham’s
law on the basis of kinetic theory
From kinetic gas equation,
1
PV  mn' c2
3
1 2
P  c
3
where
c
mn'
   density of the gas
V
3P

i.e., c 
1

at constant pressure
This is in accordance with Graham’s law.
Illustrative example 8
One mole of N2 at 0.8 atm takes 38
seconds to diffuse through a pinhole,
whereas one mole of an unknown
compound of xenon with fluorine at
1.6 atm takes 57 seconds to diffuse
through the same hole. Calculate the
molecular mass of the compound.
(Xe = 131, F = 19)
Solution
r1 n1 t2



r2
t1 n2
1 57



38 1
M2 P1

M1 P2
M 0.8

28 1.6
2
 57 1.6 
1
M  


28

252
g
mol
(XeF6 )

 38 0.8 
Class Exercise
Class exercise 1
20 dm3 of SO2 diffuses through a
porous partition in 60 s. What
volume of O2 will diffuse under
similar conditions in 30 s?
Solution:
v1
r1
t1
M2
For diffusion,


v2
r2
M1
t2
v
MSO
30
2 
1 for O2, 2 for SO2  20 
MO
2
60
 10 2  14.14 dm3
Ans. 14.14 dm3
64
32
Class exercise 2
180 cm3 of an organic compound
diffuses through a pinhole in vacuum
in 15 minutes, while 120 cm3 of SO2
under identical condition diffuses in
20 minutes. What is the molecular
weight of the organic compound?
Solution:
r1

r2
M2
M1
180  20


120  15
M 
180
 15 
120
20
64
M
64
 16 g / mol
4
MSO
M
2
Class exercise 3
The ratio of rates of diffusion of
gases A and B is 1 : 4, if the ratio of
their masses present in the mixture
is 2 : 3, calculate the ratio of their
mole fractions.
Solution:
rA

rB

MB
1

MA
4
MB
1

MA 16
Let WA and WB are the weights of
two gases in the mixture
WA : WB = 2 : 3
Solution
nA
Mole fraction of A, xA 
nA  nB

WA
WB 
Since,
n

,
n



A
B
M
M

A
B
WA
MA

WA WB

MA MB
Similarly, mole fraction of B, xB 
WB
MB
nB

WA WB
nA  nB

MA MB
Solution
xA WA  MB


xB MA  WB

2 1
1


3 16 24
 xA : xB = 1 : 24
Ans. 1 : 24
Class exercise 4
Calculate the root mean
square velocity of
(i) O2 if its density is
0.0081 g ml–1 at 1 atm.
(ii) ethane at 27° C and 720 mm of Hg
Solution:
(i) CRMS 
3RT

M
3PV

M
3P

P = 1 × 76 × 13.6 × 981 dyne cm–2
= 0.0081 gm cm-3
3  76  13.6  981
CRMS 
 1.94  104 cm sec1
0.0081
Solution
(ii) CRMS 
3RT

M
3  8.314  107  300
30
= 4.99 × 104 cm sec–1
Ans. (i) 1.94 × 104 cm sec–1, (ii) 4.99 × 104 cm sec–1
Class exercise 5
At what temperature will H2 molecules
have the same root mean square
velocity as N2 gas molecules at 27° C?
Solution:
CRMS 
3RT

MH
2

3R  300
MN
2
3RT 3R  300

2
28
 T = 21.43 K = -251.57°C
Ans. 21.43 K
Class exercise 6
If a gas is expanded at constant
temperature, the kinetic energy of the
molecules
(a) remains same (b) will increase
(c) will decrease
(d) None of these
Solution:
Since K  E T, the kinetic energy of molecules remains same
Hence, the answer is (a).
Class exercise 7
The relation between PV and
kinetic energy of an ideal gas is
PV = ____ KE
(a)
3
2
3
(b) k
2
2
(c)
3
(d)
2
k
3
Solution:
From kinetic gas equation,
1
PV  mn' c2
3
PV 
2
KE
3
1
2 2
 mc  n'
2
3
(for one molecule)
Class exercise 8
Can we use vapour densities in
place of densities in the formula?

Given

r1

r2
2 

1 
Solution:
M
Vapour density,V.D 
2
Now, if we replace density with vapour density of two gases, then
M2
r1
V.D.2
2 


M1
r2
V.D.1
2
Which is also valid.
M2
M1
So, we can replace density with vapour
density.
Ans. Yes
Class exercise 9
The ratio of root mean square
velocities of SO2 to He at 25° C is
(a) 1 : 2
(b) 1 : 4
(c) 4 : 1
(d) 2 : 1
Solution:
CRMS 
3RT
M
For SO2 , CRMS 
For He, CRMS 
3R  298
64
3R  298
4
Solution
CRMS  SO2 


CRMS He 

3R  298
64
3R  298
4
4
 1: 4
64
Hence, the answer is (b).
Class exercise 10
There is no effect of ____ on the motion
of gas molecules.
(a) temperature
(b) pressure
(c) gravity
(d) density
Solution:
According to kinetic theory of gases, velocity of
gas molecules will not be affected by the
gravity as these are considered to be point
masses.
Hence, the answer is (c).
Thank you