chemistry. states of matter – session 2 session objectives
TRANSCRIPT
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 issteam volatile?
OH
Cl
(a) OH
OH
(b)
CH3
OH
(c) F
Cl
(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 compositionof the mixture
Solution:
mixture
ρRTM =
P
1.5 0.082 298 760
740 137.64gmol
2Let x mole of CO and (1 x) mole of CO
x 28 (1 x) 44 37.64x 0.3975
2
%CO 39.75
%CO 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:
2
2
moles of H 0.03
moles of O 0.0125
2 2 22H (g) O (g) 2H O
2
2
Since O is the limiting reagent
moles of H O vapour 0.025
2H O
0.025 0.082 373P
1.5
2H
0.005 0.082 373P
1.5
0.102 atm
0.51 atm
Animation for diffusion
Graham’s Law of Diffusion
(i) For same volume of two gases
1 1 2
2 1
2
vr t M
vr Mt
2 2
1 1
t M
t M
(ii) For two gases with different volumes and same diffusion time
1
1 2
22 1
vr Mt
vr Mt
1 2
2 1
v M
v M
Effusion
Rate of effusion
PA
r2 RTM
It happens under pressure through a Small aperture.
A=area of aperture
i oP P P (For same gas) Normally considered against vacuum
Question
Illustrative example 34:1 molar mixture of He and CH4 iseffusing through a pinhole at aconstant temperature.What is thecomposition of the mixture effusingout initially?
Solution:
He4
P P5
4CH
1P P
5 4
4 C
CHHe
CH H
4P Mr 5
1r MP5
4 16
8 : 11 4
Solution
Let x mole of CH4 is effusing in t sec.
Then, 8x mole of He is effusing in same time
He8x
Mole fraction of He, x 0.898x 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) ispricked with a sharp point & quicklydropped in a tank full of hydrogen at the same pressure. After a while theballoon will have
(a) Shrunk (b) Enlarged(c) Completely collapsed (d) Remain unchanged in size
Solution
1
rM
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,
21PV mn'c
3
Where,
m = mass of each gas molecule
n’ = number of gas molecules
c = velocity of gas molecule
KE of one molecule = 21mc
2
Kinetic Theory of Gases
21 2PV mc n'
2 3
23 1nRT mc n'
2 2(where n = number of moles,n’ = number of molecules)
2A
3 1RT mc . N
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?
Average kinetic energy of one molecule = A
3 R 3T kT
2 N 2
k = Boltzmann constant
= 1.38 × 10–16 ergs k–1 molecule–1
Questions
Illustrative example 5
Calculate the average kinetic energyper molecule and total kinetic energyof 2 moles of an ideal gas at 25oC.
Solution:
Average KE per molecule of the
gas = 23
A
3 R 3 8.314T 298
2 N 2 6.023 10
= 6.17 × 10–21 J
Solution
Average KE per mole of the gas = RT = x 8.314 x 298
= 3.72 kJ/mole
3
2
3
2
Total KE of 2 moles of the gas = RT x 2 = 7.44 kJ3
2
Molecular velocity
1 2 3 n
AVC C C ..... C
CN
AV
8 RT 8 PV 8PAlso C
M M
Average velocity
Most probable velocity (CMP)
MP
2 RT 2 PV 2PC
M M
Molecular velocity
RMS
3 RT 3 PV 3 PC
M M
Interrelation of molecular velocities
Root mean square velocity (CRMS)
2 2 21 2 n
RMS
C C ...... CC
N
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:
3RMS
3RTC 10 m/ s (1)
M
Avg
8RTC (2)
M
3Avg
8C 10 921.5 m/ s
3
From (1) and (2)
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
RMS
3RTC
M
For O2 at 27° C,
1/ 2
RMS 2(3) (R) (300 )
C O32
For SO2 at t° C,
1/ 2
RMS 2(3) (R) (273 t)
C SO64
Solution
Since both these velocities are equal,
1/ 2 1/ 2(3) (R) (300) (3) (R) (273 t)
32 64
or 600 = 273 + t
or t = 600 – 273 = 327° C
Maxwell-Boltzmann velocity distributionF
ract
ion
of
mo
lecu
les
Molecular speed ( cm / sec)0
0.1
0.2
0.3
2 × 104
4 × 104
6 × 10 8 × 1044
10 × 104
12 × 104
Maxwell Boltzm ann distribution
T = 500 K
T = 1500 K
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
2 21 2 1PV mn c mn'c
3 3 2
2A
2 1Mc where M n m and n N
3 2
21Mc Kinetic energy of the gas
2
2PV KE
3
KE kT
2PV kT
3
KE Absolute temperature(T)
Explanation of Boyle’s law on the basis of kinetic theory
2As is a cons tant quantity and
3k is also a cons tant,
therefore, if T is kept cons tant,
2kT 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
2PV kT
3
V 2 kT 3 P
Hence, if P is kept constant,V
T= constant which is Charles’ law.
Explanation of Dalton’s law on the basis of kinetic theoryLet us consider only two gases. According to kinetic gas equation,
21PV mn'c
3
21 mn'cP
3 V
Now, if only the first gas is enclosed in the vessel of volume V, the pressure exerted would be
' 21 2 1
1m n c1
P3 V
Again, if only the second gas is enclosed in the same vessel (so that V is constant), then the pressure exerted would be
' 22 2 2
2m n c1
P3 V
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 ' 21 1 1 2 2 2m n c m n c1 1
P3 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 theoryLet us assume equal volume of two gases at the same temperature and pressure, then from kinetic gas equation.
' 21 1 1
1PV m n c (for first gas)
3
' 22 2 2
1PV m n c (For second gas)
3
' 2 ' 21 1 1 2 2 2m n c m n c (i)
Since average kinetic energy per molecule depends on temperature,
2 21 1 2 2
1 1m c m c (ii)
2 2
Explanation of Avogadro’s hypothesis on the basis of kinetic theory
Dividing equation (i) by (ii), we have
This is Avogadro’s hypothesis.
' '1 2n n
Explanation of Graham’s law on the basis of kinetic theory
From kinetic gas equation,
21PV mn'c
3
21P c
3
mn'where density of the gas
V
3Pc
1
i.e., c
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
21
657 1.6
M 28 252g mol (XeF )38 0.8
1 1 2 2 1
2 1 2 1 2
r n t M P
r t n M P
1 57 M 0.838 1 28 1.6
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:
For diffusion,
1
1 1 2
22 1
2
vr t M
vr Mt
1 for O2, 2 for SO2
2
2
SO
O
vM 6430
20 M 3260
310 2 14.14 dm
Ans. 14.14 dm3
Class exercise 2180 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?
1 2
2 1
r Mr M
2SO180
M15120 M20
180 20 64120 15 M
64M 16 g/ mol
4
Solution:
Class exercise 3The 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:
A B
B A
r M 1r M 4
B
A
M 1M 16
Let WA and WB are the weights of two gases in the mixture
WA : WB = 2 : 3
Solution
A BA B
A B
W WSince, n , n
M M
AA
A B
nMole fraction of A, x
n n
A
A
A B
A B
WM
W WM M
B
B BB
A BA B
A B
Wn M
xW Wn nM M
Similarly, mole fraction of B,
Solution
A A B
B A B
x W Mx M W
2 1 13 16 24
xA : xB = 1 : 24
Ans. 1 : 24
Class exercise 4Calculate 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
RMS3RT 3PV 3P
(i) CM M
P = 1 × 76 × 13.6 × 981 dyne cm–2
4 1
RMS3 76 13.6 981
C 1.94 10 cm sec0.0081
Solution:
-3= 0.0081 gm cm
Solution
7
RMS3RT 3 8.314 10 300
(ii) CM 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?
2 2RMS
H N
3RT 3R 300C
M M
3RT 3R 3002 28
Ans. 21.43 K
T = 21.43 K = -251.57°C
Solution:
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
Since K E T, the kinetic energy of molecules remains same
Hence, the answer is (a).
Solution:
Class exercise 7The relation between PV and kinetic energy of an ideal gas isPV = ____ KE
3(a)
23
(b) k2
2(c)
32
(d) k3
From kinetic gas equation,
21PV mn'c
3 21 2
mc n'2 3
2PV KE
3 (for one molecule)
Solution:
Class exercise 8Can we use vapour densities in place of densities in the formula?
1 2
2 1
rGiven
r
M
Vapour density,V.D2
Now, if we replace density with vapour density of two gases, then
21 2 2
12 1 1
Mr V.D. M2
Mr V.D. M2
Which is also valid.
So, we can replace density with vapour density.
Ans. Yes
Solution:
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
RMS3RT
CM
2 RMS3R 298
For SO , C64
RMS3R 298
For He, C4
Solution:
Solution
RMS 2
RMS
3R 298C SO 64C He 3R 298
4
41: 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
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.
Solution:
Hence, the answer is (c).
Thank you