3 chapter-gases-long-questions

1st
year n0tes chemistry new
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Chapter No.3
GASES
LONG QUESTIONS
States of Matter
Matter exists in four states solid , liquid , gas and plasma.
The gases are the simplest form of matter.
Properties of Gases
The general properties of gases are the following:
1. Gases have no definite volume: They occupy all the available
space. The volume of the gas is the volume of the container.
2. Gasses have no definite shape. They take the shape of the
container.
3. Gasses have low densities. Gases have low densities as compared
with liquids and solids. The gases bubble through liquids and tend
to rise up.
4. Gasses con diffuse and effuse. They can diffuse and effuse rapidly
through each other in all directions. The odour spreads in this way.
The property of diffusions operates in liquids as well but is
negligible in solids.
5. Gases are compressible. They can easily be compressed by
applying pressure because there are large empty spaces between
the molecules.
6. Gases are expendables. They can expand on heating or increasing
the available volume . They expand to fill the entire volume of
their containers.

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Liquids and solids do not show an appreciable increase in volume
when they are heated. On sudden expansion of gases, cooling
effect is observed. this phenomenon is called Joule-Thomson
effect.
7. Gases exert pressure. They exert pressure on the walls of the
container.
8. Gases are miscible. They can be mixed in all proportions forming
a homogeneous mixture.
9. weak intermolecular forces. The intermolecular forces in gases are
very weak.
10. Gases are liquefiable. They may be converted to the liquid
state with sufficiently low temperature and high pressure.
Properties of Liquids
All liquids show the following general properties.
1. Liquids have definite volume but indefinite shape. They adopt
the shape of the container in which they are placed but their
volume does not change.
2. Liquids molecules are in constant random. The evaporation and
diffusion of liquid molecules are due to this motion.
3. Liquids are denser than gases. The densities of liquid are much
greater than those of gases but are close to those of solids.
4. Liquid molecules are very close together. Molecules of liquid lie
close together with very little space between them so they cannot
be compressed.
5. Intermolecular attractive forces. The intermolecular attractive
forces in liquids are intermediate between gases and solids. The
melting and boiling points of gases, liquid and solids depend upon
the strength of such forces. The strength of these forces is different
in different liquids.
6. Liquid molecules possess kinetic energy. Molecules of liquids
possess kinetic energy due to their motion. Liquid can be
converted into solids on cooling, e I , by decreasing their kinetic

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energy. Molecules of liquid collide among them-selves and
exchange energy but those of solid cannot do so.
Properties of Solids
1. Solid particles are very close together. The particles of
solid substances are very close to each other and they are tightly
packed. Due to tight packing of particles solids are non-
compressible and non-diffusible.
2. Strong attractive force .There are strong attractive forces in
solids which hold the particles together firmly. Due to this reason
solids have definite shape and volume.
3. The solid particles possess only vibration motion.
Units of Pressure
“The pressure of air (Earth‟s atmosphere) at sea level that will
support a column of mercury 760 mm in height is called one
atmosphere.”
It is the force exerted by 76 cm long column of mercury on an
area of 1 cm2
at O0
C .It is the average pressure of atmosphere at
sea level. A second common pressure unit is the torr. One torr is
the pressure exerted by a column of mercury 1 mm in height.
So,
1 atm=760 torr=760mm Hg
The SI unit for pressure is the Pascal (pa). Remember that
pressure is force per unit area, so Pascal can by expressed using the
SI units for force and area. The SI unit of force is the Newton (N),
and are area is measured in square (m2
).Thus, I pa =1Nm-2
, and
IN =1 kg ms-2
,
So,
1 pa =1 kg m-1
S-2
Hence, 1 atm =760 torr =760mm Hg=101325 pa =101.325 kpa
(kilopascal).
The most commonly unit of pressure used in engineering work is
pounds per square inch (psi).
1 atm -760=760torr =14.7 pounds inch 2 .

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The unit of pressure mill bar is commonly used by
meteorologists.
Gas Laws
“The relationships between the volume of a given amount of gas
and the prevailing conditions of temperature and pressure are
called the gas laws.”
When the external conditions of temperature and pressure are
changed, the volume of a given quantity of all gases in affected.
This effect is nearly the same irrespective of the nature of the gas.
Thus gases show a uniform behavior towards the external
conditions. The gas laws describe this uniform behavior of gases.
Boyle,s Law
“At constant temperature, the volume of a given mass of a gas is
inversely proportional to the pressure applied a the gas.”
Mathematically:
(T and n
constant)
PV=K ----------(1)
Where”K”is a proportionality constant. The value of k is
different for different amounts of the same gas .
According to Eq (1), boyle‟s law can be stated as follows:
“At constant temperature, the product of pressure and volume of
a fixed amount of a gas remains constant.”
So, P1V1 =k and P2V2=k
Hence, P1V1 =P2 V2
Where P1 and V1 are the initial pressure and volume while P2 and
V2 are the final pressure and volume.
Experimental Verification of Boyles,s Law
Consider a fixed amount of a gas in a cylinder at constant
temperature say at 25o
C. The cylinder is fitted with a moveable piston
and a manometer to read the pressure of the gas directly .Let the initial

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volume of the gas is 1 dm3
and its pressure is 2 atm when the piston has
one weight on it. When the piston is loaded with a weight two times
greater with the help of two equal weights, the pressure becomes 4 atm
and the volume is reduced to dm 3
. Similarly when the piston is loaded
with a weight three times greater.
Then the pressure becomes 6 atm and volume is reduced to dm3
.Thus
P1 V1 =2atmx1dm3
=2 atm dm3
=k
P2 V2 =4atm x dm3
= 2atm dm3
=k
P2 V2 =4atm x dm3
= 2atm dm3
=k
Hence the Bye‟s law is verified.
The value of k will remain the same for the same quantity of a
gas at the same temperature.
Example 1 : A gas having a volume of 10 dm3
is enclosed in a vessel
at 0o
C and the pressure is 2.5 atmospheres . This gas is allowed to
expand until the new pressure is 2 atmospheres. what will be the new
volume of this gas, if the temperature is maintained at 273 k.
Solution: Given: P1 =2.5atm ; P2
=2atm
V1 =10dm3
; V2 =?
Since temperature is constant ( 0o C =273k )
Formula Used: V2 =P1 V1
V2 =
V2 =
V2 =12.5 dm3
Answer
Graphical Explanation of Boyles,s Law
Take a given amount of gas and enclose it in a cylinder
having a piston in it. The Boyle,s law can be represented graphically in
three different ways as described below:
(i) Keeping the temperature constant at 0o C, when the
pressure of the gas is varied, its volume changes. On plotting a
graph between pressure on the x-axis and volume on the y-axis,

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a curve is obtained. Fig (a). This graph shows that volume is
inversely proportional to pressure. Increase in pressure
decreases the volume.
“The pressure –volume curve obtained at constant temperature is
called isotherm”
Now increase the temperature of the same gas to 25o C .keeping the
temperature constant graph between pressures of the gas is varied, its
volume changes. Again on plotting a obtained .this curve goes away
from both the axes, fig (b). The reason is that, at higher temperature, the
volume of the gas has increased. Again, this graph shows that volume is
inversely proportional to pressure. The increase in pressure decreases the
volume.
Similarly, if we increase the temperature further, keeping it
constant, and again plot another isotherm, It further goes away from the
axes and produces the same result.
Fig (a) . Isotherm of a gas at 0o
C Fig (b). Isotherm of a gas at
different temperatures
(ii) Keeping the temperature constant at T1, when the volume
of the gas is increased, the gas spreads over a larger region of
space. Consequently, it exerts less pressure on the container. It
means increase in column decreases the pressure of the gas.
On plotting a graph between the reciprocal of volume ( )n x-
axis and pressure (P) on the y-axis. It shows that p is directly
proportional ( )to . It indicates that the pressure goes down as
the gas expands. This straight line will meet at the origin where
both p and ( )are zero. Thus we can say that p goes to zero as
V goes to infinity ( =0). Now increase the temperature of the
same gas from T1 to T2 . Keeping the temperature constant at T2 .

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When volume is varied, its pressure changes. Again plot the
graph between ( )on x-axis, and P on y-axis, again a straight
line is obtained which meet at the origin . This straight line will
be away from the x-axis, Fig(a) . Again the straight line straight
line indicates that P is directly proportional to ( ).
(iii) Keeping the temperature constant at T1, when the
pressure of the gas is varied, its volume changes. On plotting a
graph between P on x-axis and the product PV on y-axis, a
straight line parallel to x-axis is obtained, Fig (b). This straight
line shows that PV remains constant even if we change pressure.
Now increase the temperature of the same gas form T1 to T2.
Keeping the temperature constant at T2, when the pressure of
the agars is varied, its volume change. On plotting graph
between P on x-axis and the product PV on y-axis, again a
straight line parallel to x-axis is obtained, Fig (b) . This
straight line shows that PV remains constant even if we
change pressure. However, the value of PV increases with
increase in temperature.
Charles’s Law
“At constant pressure, the volume of a given mass of a gas is
directly proportional to the absolute temperature.”
Mathematically: V = T (P and N are
constant)
V = kT
= k
Where k is a proportionality constant.
If the temperature is changed from T1 to T2 and the volume is
changed from V1 to V2 then we have
= k
and = k
so, = = k

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=
Experimental Verification of Charles, s Law
Consider a fixed amount of a gas enclosed in a cylinder fitted
with a moveable piston. Let the volume of the gas is V1 and its
temperature is T1. If the gas in the cylinder is heated, the
temperature and the volume of the gas will increase . The new
temperature be T2 and volume will be V-2 . It can be shown that
t h e r a t i o .
=
Hence Charles‟ s law is verified.
Example 2 : 250cm3
of hydrogen is cooled from 127o
C to -27o C by
maintaining the pressure constant. Calculate the new volume of the
gas at the low temperature.
Solution: Given: V1 =250cm3
; V2 =?
T1 =273+127=400k ; T2 =273-27=246k
Formula Used: =
V2 = xT2
V2 =
V2 = 153.75 cm3 Answer
Derivation of Absolute Zero
In order to derive absolute zero of temperature considers the
following quantitative statement of Charles‟s law.
Statement: “At constant pressure , the volume of a given mass of gas
increases or decreases by of its original volume at 0o C for
every 1 oC rise or fall in temperature respectively.”
In order to understand the above statement .suppose V o is the
volume of a given

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Mass of gas at 0 o
C and V1 is the volume at T o
C .Then from the
statement of Charles,s law.
For 1 o
C rise in temperature, the increase in volume =Vo x =
And, for to C rise in temperature ,the increase in volume =xt
=
The volume , Vt of the gas at to C becomes ,
Vt =vol of gas at 0o
C =increase in vol.at
to
C
Vt=Vo +
Vt=Vo (1+ )
This general equation can be used to know the volume of the gas at
various temperatures.
Suppose at 0o
C , the original volume , Vo of the gas is 546
cm 3
.Thus
Vo =546 cm3
At 10o C, V10 =546(1+ dm3
At 100 oC V100 =
cm3
Thus , the increase in temperature from 10o
C to 100 o
C, increases
the volume from 566 cm3
to 746cm3
.
Applying the Charles‟s Law,
=
The two ratios are not equal. So , Charles‟s law is not being
obeyed when temperature is measured on the Celsius scale. For this
reason a new temperature scale has been developed.
The new temperature scale starts from-273o
C (more correctly-
273.16 o
C) which is called zero Kelvin (0k) or zero absolute. The

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advantage of this scale is that all the temperatures on this scale are in
positive figures .In order to develop the new temperature scale, the best
way is to plot a graph between the variables of Charles,s law . That is, V
and T
Graphical Explanation of Charles,s Law
Keeping the pressure of a given mass (say 1 mole) of gas
constant, when the temperature of a gas is varied, its volume changes.
The volumes at different temperatures are measured, Now if a graph is
plotted between temperature (o
C) on x-axis and volume on y-axis , a
straight line is obtained ,this straight line cuts the temperature-axis and
(x-axis) at -273. 16o
Cif it is extrapolated. The volume of a gas becomes
zero at -273.16o
C. This temperature is the lowest temperature which is
attainable if the substance remains in the gaseous stat. Actually, all real
gases first converted into liquid and then into solids before reaching this
temperature
Fig. the graph between volume and temperature for a gas according
to table.
If many plots of this type are examined. It is found that a given gas
follows different straight lines for different masses of gas and for
different pressures. Greater the mass of gas taken, greater will be the
slope of the straight line. The reason is that greater the number of moles
greater the volume occupied .All these straight lines when extrapolated
meet at a common point of -273.16o
C (0k). It is clear that this
temperature of -273.16oC will be attained when the volume becomes
zero. It is true for an ideal gas. But for a real gas. Thus-273.16oC
represents the coldest temperature. This is the zero point (0K) for an
absolute scale of temperature.
Charles‟s law is obeyed when the temperature is taken on the
Kelvin scale. For example, at 283K (10o
C) the volume is 566 cm3
and at
373 k (100o
C) the volume is 746 cm3
. According to Charlie‟s law.

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==k
Hence , Charles‟s law is obeyed .
Scales of Thermometry
For measuring temperature, three scales of thermometry are
used.
(a) Celsius or Centigrade Scale
One this scale, the temperature of ice at 1 atmospheric pressure is
0o
C, and the temperature of boiling water at 1 atmospheric pressure is
100o
C. The space between these temperature marks is divided into 100
equal parts and each part is 1 o
C
(b) Fahrenheit Scale
On this scale, the temperature of ice at 1 atmospheric pressure is
32Fand the temperature of boiling water is 212o
F .The space between
these temperature marks is divided into 180equal parts and each part is
1o
F
(c) Absolute or Kelvin scale
One the absolute or Kelvin scale, the temperature of ice at 1
atmospheric pressure is 273 K and that of boiling water is 373 K. On the
Kelvin scale, absolute zero (0K) is the temperature at which the volume
of a gas becomes zero. It is the lowest possible temperature that can be
achieved. Thus, temperatures on the Kelvin scale are not divided into
degrees .Temperatures on this scale are reported in units of Kelvin , not
in degrees temperature on Kelvin Scale=273.16o
C or 0K=273.16o
C
Temperature on Kelvin scale=273+temperature o
C
Or
T (k)=273+ o
C
The following relationship helps us to understand the
introversions of various scales of temperatures.
T (k)=273+ o
C
o
C=

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o
f= (o
C)+32
Remember that the Fahrenheit and Celsius scales are both relative
temperature scales. They define two reference points (such as 0o
C and
100o
C). The Kelvin scale, however, is an absolute scale. Zero on this
scale is the lowest temperature that can achieved.
Standard Temperature and Pressure (STP)
The temperature 273.16 k (0o C) and pressure 1 atmosphere (760
mm Hg) have been chosen as standard temperature and standard
pressure.
STP 273.16K (0o C) and 1 atm (760 mm Hg)
General Gas Equation [Ideal Gas Equation]
“ A gas equation which is obtained by combining Boyle „s law ,
Charles ,s law and Avogadro ,s law is called general gas equation.”
It relates pressure, volume temperature and number of moles
of gas.
According to Boyle‟s Law V (at constant T and n)
According to Charles‟s Law V T (at constant P and n)
According to Avogadro‟s Law V T (at constant P
and T)
If the variables P,T and n are not Kept constant then all the above
three relationships can be combined together.
V
V=R
Where „R‟ is a general gas constant.
pV=nRT(First Form)

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The above equation is called ideal or general gas equation.
The general gas equation shows that if we have any quantity of
an ideal gas then the product of its pressure and volume is equal to the
product of number of moles,
Variation on the Gas Equation
During chemical and physical processes, any of the four
variables (P,V,n ,T) in the ideal gas equation ,PV =nRT may be fixed
and any of them may change.
For examples:
1. For a fixed amount of gas at constant temperature.
PV= nRT =constant [ n and T constant]
PV=k (Boyle‟s Law)
2. For a fixed amount of gas at constant pressure,
=[ n and P constant]
V=kT
3. For a fixed amount of gas at constant volume,
=[ n and V constant]
P=kT
4. For a fixed pressure and temperature,
[ P and T constant ]
V=kn (Avogadro‟s Law)
5. For a fixed volume and temperature,
[ T and V constant]
P=kn
General Gas Equation For one mole of gas

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For one mole of gas , the general gas equation is
PV==RT [ n=1]
For two different pressure, volumes and absolute temperatures, the
above equation can be written as,
Hence, we can write,
= (Second Form)
Numerical Value of Ideal Gas Constant , R
The numerical value of R depends upon the units chosen for P,V
and T.
(i) When pressure is expressed in atmosphere and volume in dm3
.
For 1 mole of a gas at STP, we have
N= 1 mole, P=1 atm, T =273.16k, V=22.414dm3
Putting these value in general gas equation,
R=
R=
R=0.0821dm3
atm k -1
mol-1
(ii) when pressure is expressed in mm of mercury or torr
and volume of gas in cm3
.
R= 0.0821dm3
atm k-1
mol-1
Since 1 dm3
=1000cm3
: 1 atm =760mm of Hg
R=0.0821x1000 cm3
x760mm Hg K-1
mol-1
R=62400 cm3
mm Hg K-1
mol-1
Since 1 mm=1 torr
R=62400 cm3
torr K-1
mol-1
(iii) When pressure is expressed in Nm-2
and volume in m3
(SI Units)
For 1 mole of gas at STP,
N= 1 mol, P=101325 Nm-2
, T=273.16 K,
V=0.022414m3
Substituting the values in general gas equation,
R=
R=

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R=8.3143 NmK-1
mol-1
Since 1 Nm=1 j
R=8.3143jk -1
mol-1
Remember that, whenever the pressure is given in Nm-2
and the
volume in m3
,then the value of used must be 8.3143 jk-1
mol-1
.
Remember that when the pressure is given in Nm-2
and the
volume in m3
, then the value of R used must be 8.3143JK-1
mol-
1
.joule (j) is unit of work and energy in the system.
(iv) When energy is expressed in ergs.
R=8.3143jk -1
mol-1
Since 1 J =107
erg
R=8.3143x107
erg K-1
mol -1
(v) When energyis expressed in calories.
R=8.3143jk -1 mol-1
Since 1 cal=4.184j
Or 1 j=
R=
R = 1.987 cal K-1
mol-1
(vi) R can be expressed in units of work or energy per Kelvin per
mole .
Form ideal gas equation, we can write
R =
If the pressure is written as force per units area and volume as
area x length,
R====
Hence R can be expressed in units of work or energy per Kelvin per
mole.
Physical Meaning of Value of R
The physical meaning of the value 0.0821 dm3 atm K-1
mol-1
of
R is that if we have one mole of an ideal gas at 273.16 K and one
atmospheric pressure and its temperature is increased by 1 K then it will
absorb 0.0821 dm3
atm of energy , dm3 atm being the unit of energy
.Hence the value of R is a universal parameter for all the gases . It tells

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us that the Avogadro‟s number of molecules of all the ideal gases have
the same demand of energy.
Density of an Ideal Gas
We know that, n=
Where a is the number of moles of gas .m is the mass of the
substance in grams and M is the molar mass of the substance.
On substituting the above expression into the ideal gas equation,
We obtained
PV=RT (Third From)-------(6)
PM==RT
PM=dRT (d=)
d= (Fourth From)---------(7)
Hence the density of an ideal gas is directly proportional to its molar
mass.
Greater the pressure on the gas , the closer will be the molecules
and greater the density temperature of the gas ,the lower will be the
density of the gas.
With the help of equation (7) ,we can calculate the relative molar
mass, M of an ideal gas if its pressure ,temperature and density are
know.
Example3:
A sample of nitrogen gas is enclosed in a vessel of volume 380
cm3
at 120o
C and pressure of 101325 Nm-2
.This gas is transferred to a
10dm3
.
Flask and cooled to 27o
C.Calculate the pressure in Nm-2
exerted
by the gas at27o
C.
Solution:
V1 =380cm3
=0.38 dm3
; V2 =10dm3
P1 =101325 Nm-2
; P2 =?
T1 =273+120=393k ; T2 =273+27=300k
Formula used:
=
P2 =

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P2 =
P2 =0.028atm
P2 =0.029x101325=2938.4 Nm-2
Answer
Example 4:
Calculate the density of CH4 at 0o
C and 1 atmospheric pressure,
what will happen to the density if
(a) temperature is increase to 27o
C.
(b) the pressure is increased to 2 atmospheres at 0o
C.
Solution:
MCH4=12+4=16g mol-1
P=1 atm
R= 0.0821dm3
atm K-1
mol-1
T = 273+0=273K
Formula Used:
d =
d =
d=0.7138 g dm3 Answer
(a) Density at 27o
C
MCH4=16g mol -1
P= 1 atm
R=0.0821 dm3
atm K-1
mol-1
T=273+27=300K
Formula Used:d=
d =
d = 0.6496g dm3
Answer
Thus by increasing the temperature from 0o
C , the density of gas has
decreased form 0.7143 to 0.6496 g dm3
.
(b) Density at 2 atm pressure and 0o
C.
MCH4=16g mol-1
P = 2 atm
R =0.0821 dm3 atm K-1 mol-1
T = 273+0=273k
Formula Used: d =

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d =
d=1.428 g dm-3 Answer
Thus the increasing of pressure has increased the density . the
density becomes double by doubling the pressure .
Example 5:
Calculate the mass of 1 dm3 of NH3 gas at 30o
C and 1000 mm
Hg pressure, considering that NH4 is behaving ideally.
Solution: Given: P=1000 mm Hg = 1.316 atm ; V=1 dm3
MNH4 =14+3=17g mol-1 : R=0.0821dm3 atm K-1
mol-1
T=273+30-303l ; m=?
Formula Used: PV =
M=
M =
M =0.0908 g Answer
Avogadro ‘s Law
The law may be stated in a number of ways as follows;
1. “Equal volumes of ideal gases at the same temperature
and pressure contain equal number of molecules.”
2. “Equal number of molecules of ideal gases at the same
temperature and pressure occupy equal volumes.”
3. “At constant temperature and pressure, the volume of an
ideal gas is directly proportional to the number of moles or
molecules of gas.”
Mathematically: V n (at P and T constant)
V=kn
Explanation:
Since one mole of an ideal gas at STP has a volume of 22.414
dm3
,So 22.414dm3
of an ideal gas at STP will have Avogadro‟s number
of molecules , i.e, 6.02x1023
.
Mathematically , it can be written as,
22.414 dm3
of an ideal gas at STP has number of
molecules =6.02x1023

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1 dm3
of an ideal gas at STP has number of molecules
=molecules
=2.68x1022
molecules
In other words if we have one dm3
of each of H2
,O2
,N2
and CH4
in
separate vessels at STP, then the number of molecules in each will be
2.68x 1022
.
Similarly when the temperature or pressure are equally changed for
these four gases, then the new equal volumes will have the same number
of molecules , 2.68x1022
.
One dm3
of H2 at STP weight approximately 0.0899 grams and
one dm3
of O2 at STP weight 1.4384 grams, but their number of
molecules are the same .Although , oxygen molecule is 16 times heavier
than hydrogen but this does not disturb the volume occupied by the
molecules because molecules , of the gases are widely separated from
each other at STP .One molecule of gas is approximately at a distance of
300 times its own diameter from its neighbors at room temperature .
Daltan’s Law of Partial Pressure
“The total pressure exerted by a mixture of non-reacting gases is
equal to the sum of their individual partial pressure.”
Mathematically: Pt =P1 +P2 +P3
Where Pt is the total pressure of the mixture of gases and P1 ,P2
and P3 are the partial pressure of gas1 . gas2 and gas 3 respectively in the
mixture.
The partial pressure of a gas in a mixture of gases in the pressure
a gas would exert if it were the only gas in the container. Partial
pressures are commonly represented by small p„s.
Explanation: Suppose we have four 10 dm3
cylinders. The first one
contains H2 gas at a pressure of 400 torr; the second one contains CH4
gas at a pressure of 100 torr at the same temperature. Now transfer these
gases to the fourth cylinder of capacity 10dm3
at the same temperature.

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According to Dalton„s law, the pressure in the fourth cylinder is found to
be equal to the sum of the pressures that each gas exerted by itself.
Pt =PH2 +PCH 4 +Po2
Pt=400+500+100=1000 torr
There three non-reacting gases are behaving independently under
the normal conditions. The rapidly moving molecules of each gas in a
mixture have equal opportunities to collide with the walls of the
container. Hence, each gas in the container exerts a pressure independent
of the pressure of other gases in the container. The total pressure is the
result of total number of collisions per unit area in a given time. Since
the molecules of each gas move independently, so the general gas
equation can be applied to the individual gases in the gaseous mixture.
(i)PH2 V=n H2 RT PH2 =nH2 PH2 nH2
(ii)PH2 V=n H4RTPH2 =nH2 Pch2 nCh4
(iii)PO2 V=no2 RT Po2 =no2 Po2 no2
is a constant factor
H2 , CH4 and O2 have their own partial pressures . Since volume and
temperature are the same so their number of moles will be different and
will be directly proportional to their partial pressures .
Adding Equations (i),(ii) and (iii),we get ,
(PH2 +PCH4 +Po2 )V= (nH2 + nCH4 +no2 )RT
Pt = nt RT
Where Pt =PH2 +PCH4 +Po2 , nt =nH2 +nCH4 + no2
Hence, the total pressure of the mixture of gases depends upon the total
number of .moles of the gases.
Calculation of Partial Pressure of a Gas
The partial pressure of any gas in a mixture of gases can be
calculated provided we know the mass or number of moles of the gas,
the total pressure and the total number of moles present in the mixture of
gases.

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21
Suppose we have a mixture of gas A and gas B. The mixture is
enclosed in a container having volume (V). The total pressure of the
mixture is Pt . The number of moles of the gases A and B are NA and n B
respectively .If the temperature of the mixture is T, then we can write
general gas equations
For the mixture of gases: Pt V=n t RT --------(i)
For gas A: PAV=nART --------(ii)
For gas B: PBV=nBRT ---------(iii)
Divide Eq (ii) by Eq (i) ,we get,
=
=
PA=
Similarly, by dividing Eq (iii) by Eq (i), we get
PA =
In mole –fraction form, the above equations can be written as,
PA=XA Pt [ XA = ]
PB = XB Pt [ XB = ]
Example 6: There is a mixture of H2, He and CH4 occupying a
vessel of volume 13 dm3 at 37 o
C and pressure of 1
atmosphere. The masses of H2 and He are 0.8g and 0.12g
respectively. Calculate the partial pressure in torr Hg of each
gas in the mixture.
Solution: Given:
P=1 atm : V=13dm3
n=? ; T=273+37=310K
R=0.0821 dm3
atm K-1
mol-1
Formula Used:
PV=nRT

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22
n =
n=
nH2 =
nHe =
nCH4 = 0.51-(0.396+0.03)=0.51-0.426=0.084mol
PH2 =
PHe =
=0.776x760=589.76torr Answer
PHe =
PH2 =
=0.058x760=44.08 torr Answer
PCH4 =
PCH4 =
=0.164x760=124.64 torr Answer
Applications of Dalton’s Law of Partial pressure
The four important applications of Dalton‟s law of partial
pressures are the following:
1. Determination of the pressure of dry gas: The Daltons law of
partial pressure is used to find the pressure of dry gas collected over
water. When a gas is collected over water it becomes moist. The
pressure exerted by a moist gas is , therefore , me sum of the partial
pressure of the dry gas and the pressure of the water vapours which are
mixed with the gas .The partial pressure exerted by the water vapour is
called aqueous tension .Thus

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23
Remember that :
While solving the numerical, the aqueous tension is
subtracted from the total pressure (P moist gas )
2. Process of Respiration: In living beings, the process of
respiration depends upon differences in partial pressures. When animals
inhale air then oxygen moves into lungs as the partial pressure of oxygen
in the air is 159 torr while the partial pressure of oxygen in the lungs is
116 torr . On the other hand, CO2 produced during respiration moves out
in the opposite direction as its partial pressure is low in the air relative to
its partial pressure in the lungs .
3 Pilots feel uncomfortable breathing at higher altitudes: At
higher altitudes ,the pilot feel uncomfortable breathing because the
partial pressure of oxygen in the unperssurized cabin is low as compared
to torr in air where one feels comfortable breathing.
4 Deep sea divers feel uncomfortable breathing: deep Sea divers
take oxygen mixed with an inert gas (He) and adjust the partial pressure
of oxygen according to the requirement. In sea after every 100feet depth,
the diver experiences about 3 atm pressure .Therefore ,normal air cannot
be breathed in depth of sea . Moreover, the pressure of N2 increases in
depth of sea and it diffuses in the blood.
Diffusion and Effusion
Diffusion
“The spontaneous intermingling (intermixing) of molecules of
one gas with another at a given temperature and pressure is called
diffusion.”
Explanation of Diffusion from Kinetic Theory
According to the Kinetic molecular theory of gases, the
molecules of the gases move haphazardly. They collide among
themselves, collide with the walls of the container and change their

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24
directions. In other words the molecules of gases are scattered or
intermixed due to collisions and random motion to from a homogeneous
mixture.
Example: (1) The spreading of fragrance of a rose or a scent is
due to diffusion.
(2) Suppose NO2 (a brown gas an d O2 ( a colorless
gas )are separated from each other by a partition .
When the partition is removed, both diffused into each
other due to collisions and random motion. They
generate a homogeneous mixture. The partial pressure
of
Both are uniform throughout the mixture.
Effusion:
“The escape of gas molecules through an extremely small
opening into a region of low pressure is called effusion.”
The spreading of molecules in effusion is not due to collisions
but to their escape one by one .Actually the molecules of a gas are
habitual in colliding with the walls of the vessel. When a molecule
approaches just in front of the opening it enters the other portion of the
vessel. This type of escape of gas molecules through a small hole into a
region of low pressure or vacuum is called effusion.
Graham’s Law of Diffusion
“The rate of diffusion or effusion of a gas is inversely
proportional to the square root of it’ s density at constant temperature
and pressure.”

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Mathematically. Rate of diffusion (at constant T
and P)
Rate of diffusion =
R =
Rx = =k
The constant k is same for all gases, when they are all studied at the
same temperature and pressure .
Suppose we have two gases 1 and 2 whose rates of diffusion are
r1 and r2 and densities are D1 and d2 respectively. According to
Graham‟s law,
r1 x 1=k
r2 x 2=k
Divide the two equations and rearrange
Since the density of a given gas is directly proportional to its molecular
mass.
d M
Therefore, Graham‟s law of diffusion can also be written as,
Where M 1 and M2 are the molecular masses of gases.
Experimental Verification of Graham‟s Law of Diffusion
This law can be verified in the laboratory by noting the rates of
diffusion of two gases in a glass tube when they are allowed to move
from opposite ends. Two cotton plugs soaked in NH4 OH and HCI
solution are introduced in the open ends of 100cm long tube at the same
time .NH3 molecules travel a distance of 59.5 cm when HCI molecules

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26
cover 40.5 cm in the same time. Dense white fumes of NH4CI are
produced at the junction of two gases.
=
=
Form Graham‟s law the rates of diffusion of NH3 and HCI can be
calculated by using their molecular masses as follows:
=
This result is the same as obtained form the experiment.
Hence the law is verified
Remember that light gases diffuse more rapidly than heavy gases.
Fig: Verification of Graham‟s Law
Example 7: 250 cm3
of the sample of hydrogen effuses four times as
rapidly as 250 cm3
of an unknown gas . Calculate the molar mass of
unknown gas .
Solution: Let the symbol for the unknown gas is x.
Rate of effusion of unknown gas rx =1
Rate of effusion of hydrogen gas rH2=1x4=4
Molar of effusion of hydrogen gas MH2 =2g mol-1
Molar mass of unknown gas, Mx =?
Formula used:
Taking square of both the sides,
Mx =16x2g mol-1
Mx=32g mol -1
Answer

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Kinetic Molecular Theory of Gases
“A gas model which explains the physical behavior of gases
and the gas laws is called the kinetic molecular theory of gases.
The kinetic molecular theory of gases was first suggested by
Bernoulli (1738).R.J .Clausisus (1857) derived the Kinetic gas equation
and deduced all the gas laws from it .Maxwell (1859) gave the law of
distribution of velocities .Bolt Mann (1870) gave the concept of
distribution of energies among the gas molecules. Vander walls (1873)
modify the Kinetic gas equation in order to apply for real gases.
Postulates of Kinetic Molecular Theory of Gases
The basic assumptions of the Kinetic molecular theory of an
ideal gas are the following:
1. All gases consist of a large number of very small particles called
molecules. Noble gases (He,Ne, Ar,etc )have mono=atomic
molecules .
2. The molecules of a gas mve randomly in straight lines in all
directions with various velocities. They collide with one another
and with the walls of the container and change their direction.
3. The pressure exerted by a gas is due to the collections of the gas
molecules with the walls fo the container. The collisions among the
molecules and with the walls of the container are perfectly elastic.
There is no loss energy during collision or mutual friction.
4. The molecules of a gas are widely separated from one another
.There is large empty spaces between the gas molecules.
5. There are no attractive forces between the gas molecules .There is
no force of gravity on gas molecules.
6. The actual volume of the gas molecules is negligible as compared
to the volume of the gas.
7. The motion of the gas molecules due to gravity is negligible as
compared to the effect of continued collisions between them.
8. The average kinetic energy of the gas molecules is directly
proportional to absolute temperature of the gas.

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Average K.E T
Average K.E=
Note : The average kinetic energy is considered because the gas
molecules are moving with different velocities . The velocity of
each molecules change with each collision .Some of the molecules
has very low velocities, while others move very rapidly.
Kinetic Gas Equation
Keeping in view the basic assumptions of kinetic molecular
theory of gases, Clausius derived an expression for the pressure of
an ideal gas . This equation relates the pressure of the gas with
mass of a molecule of the gas, the number of molecules of the gas
in the volume of the gas and the mean square velocities of gas
molecules .The kinetic gas equation has the following form.
PV=
Where P=pressure of the gas
V=Volume of the gas
M=mass of one molecule of gas
N=number of gas molecules in volume
=mean square velocity (average of the
square of the
Velocities of the gas
molecules)
The idea of the mean square velocity is important because all the
molecules of a gas at a particular temperature have different velocities
.These velocities are distributed among the molecules.
Mean velocity ,Mean Square Velocity ,Root Mean Square Velocity
Mean(average ) velocity,

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29
“The average value of the different velocities of all the molecules
in a sample of gas at a particular temperature is called the mean velocit
.”
If there are n1 molecules with velocity c1 , n2 molecules with
velocity c2 and n3 molecules with velocity c3,then ,
Where is known as the mean of average velocity .The bar (-) over the
velocity (c ) indicates average or mean value.
In this reference n1 + n2 +n3 =N
mean square velocity ,c2
“The average value of the square of the different velocities of
all the molecules in a sample of gas at a particular temperatures is
called the square velocity.”
Where is known as the mean square velocity.
Root mean square Velocity ,crms
“The square root of the mean of the squares of the different
velocities of all the molecules in a sample of gas at a particular
temperature is called the root mean square velocity.”
Crms =
“The square root of the mean square velocity.
The expression for the root mean square velocity deduced from the
kinetic gas equation is written as:

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30
Where , Cmns root mean square velocity
M =molecular mass of the gas
T = absolute temperature
This equation is a quantitative relationship between the absolute
temperature and the velocities of the gas molecules .According to this
equation ,the higher the temperature of a gas ,the greater the velocities .
Explanation of Gas Law From kinetic theory of Gases
1. Bo lye’s Law
According to kinetic gas equation,
PV=
Multiplying and dividing by 2 on right hand side.
PV=
But =KE
PV= (K.E)
Now, according to kinetic theory of gases, kinetic is directly
proportional to absolute temperature
K.E T
K.E =kT
PV =
If temperature is constant, then is constant
PV=constant (which is boyle‟s law)
Thus boyle‟s law is derived.
2. Charles,s Law
PV=
Multiplying and dividing by 2 on right hand side.

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31
PV=
=KE
PV= (K.E)
Now ,according to kinetic theory of gases , kinetic energy is directly
proportional to absolute temperature.
K.E T
K.E =kT
PV =
V =
If pressure is constant, then is constant
V=constant x T
V T (which is Charles‟s law)
3 Avogadro’s Law
Consider two gases 1 and 2 at the same pressure P and the same volume
V.Suppose their number of molecules are N1 and N2 ,masses are m1 and
m2 and mean square velocities are respectively.
For gas 1: PV- …………..(1)
For gas 2: PV- ………….(2)
From Eq(1) and Eq (2) ,we get,
= ……..(3)
If the temperature of the gas 1 and gas 2 are the same, their kinetic
energy will also be the same.
Kinetic energy of gas1 =kinetic energy of gas2
=
Dividing eq(3) by eq(4)

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32
N1 = N2
Thus ,number of molecules of gas1 =number of molecules of gas2
4. Graham’s law of Diffusion
PV =
For 1 mole gas , N=NA
PV=
PV = (mNA=M)
Where M is the molecular mass of gas.
P=
P= [ =d(density of gas)]
=
Taking the square root of both the sides
=
cnms =
Since the root mean square velocity of the gas is equal to the rate of
diffusion of the gas,
Cnms =r
So, r=

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33
r=
At constant pressure,
r=
r=constant x
r
Thus , Graham „s law of diffusion of gases is derived
Kinetic Interpretation of Temperature
Consider one mole of the gas containing N molecules. Suppose
m is the mass of one molecule of the gas and is the mean square
velocity of the molecule.
PV= …………..(1)
The Eq(1) can be rewritten as,
PV= ……………(2)
Now , the kinetic energy of one molecule of a gas due to its
translational motion is given by the following equation ,
Ek= …………….(3)
Where Ek is the average translational kinetic energy of a gas.
On putting eq(3) in Eq (2) , we get,
PV =
Since the number of gas molecules in one mole of gas in NA
(Avogadro‟s-number)
So,
N=NA
PV =

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For I mole of the ideal gas,
PV=RT
Therefore,
=RT
Ek=
Ek
We can draw the following conclusions from the above equation:
1. Concept of Absolute Zero of Temperature
According to the equation, T= , absolute temperature of a gas is
directly
Proportional to the average translational kinetic energy of the gas
molecules. It means that a change in temperature brings about a change
in the intensity of molecular motion .Thus, when T=0k (-273.16o
C), the
translational kinetic theory, absolute zero of temperature, both velocity
and kinetic energy of gas molecules become zero. Also at this
temperature, the gas exerts no pressure on the walls of the container.
The absolute zero temperature cannot be attained .However, a
temperature as low as 10-5 K is obtained.
2. Concept of Average Kinetic Energy of a gas
The average translational kinetic energy of a gas depends only on its
absolute temperature and is independent of the pressure, volume, or type
of gas. The mean kinetic energy of each gas molecule is the same at the
same temperature. The mean kinetic energy of gas molecule does not
depend upon its mass. Therefore, a small molecule such as H2 will have ,
at the same temperature , the same average kinetic energy as a much
heavier molecule such as CO2 . Hence, for different gases average
kinetic energy of molecules at the same temperature is the same. Thus at
the same temperature.
K.E of gas 1 = K.E of gas2
3. Concept of Heat Flow

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35
When heat flows from one body to another, the molecules in the
hotter body give up some of their kinetic energy through collisions to
molecules in the colder body. This process of flow of heat continues
until the average translational kinetic energies of all the molecules
become equal. Consequently, the temperature of both the bodies
becomes equal.
4. Concept of Temperature of Gases, Liquids and solids
Temperature is the measure of average translational kinetic
energies of molecules. In gases and liquids, temperature is the measure
of average translational of kinetic energies of their molecules , In solids ,
temperature is the measure of vibrational kinetic energy of molecules
because there is only vibrational motion about mean position .
Liquefaction of Gases
General Principle of Liquefaction of Gases
Under suitable conditions of temperature and pressure all gases
may be liquefied. Liquefaction occurs only when the attractive forces
between the molecules are greater than the kinetic energy of the
molecules.
Two conditions favour this situation.
1. High pressure. High pressure is applied on gases. This brings the
molecules of a gas close to each other. Thus the forces between the
molecules increase.
2. Low temperature. Low temperature is created in gases .This
decreases the kinetic energy of the molecules. Thus the attractive
forces between the molecules increase.
In general, the liquefaction of a gas requires high pressure and
low temperature.
Liquefaction means the conversion of a gas into a liquid.
For each gas, there is a certain temperature above which the gas
cannot be liquefied even if a very high pressure is applied. This
temperature is known as critical temperature. The critical
temperature of a gas may be defined as, “the highest temperature
which a gas con be liquefied by increasing the pressure is called

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36
critical temperature.‟ The critical temperature is represented by Tc.
Below the critical temperature a gas can be liquefied however great
the pressure may be. The value of the critical temperature of a gas
depends upon its size, shape and intermolecular forces present in it.
“The pressure which is required to liquefy the gas at its critical
temperature is called critical pressure.”
The critical pressure is represented by Pc.
The critical temperature and the critical pressure of the gases are
very important because they provide us the information about the
condition under which gases liquefy.
Examples:
(1) CO2 is a nonpolar gas its critical temperature is 31.1o
C. It must
be cooled below this temperature before it can be liquefied by
applying high pressure. However, if temperature is maintained
below 31.1o
C. Then lower pressure than critical pressure is
required to liquefy it.
(2) NH3 is a polar gas. Its critical temperature is 132.44 o
C. It must
be cooled below this temperature before it can be liquefy by
applying sufficient high pressure.
Joule-Thomson Effect (Joule-Kelvin Effect)
“The cooling effect produced when a compressed gas is allowed
to undergo sudden expansion through the nozzle of a jet is called joule-
Thomson effect.”
The effect is due to energy used in overcoming the
intermolecular attractive forces of the gas. Joule-Thomson effect is used
in the liquefaction of gases by cooling.
Explanation: In a compressed gas, the molecules are very close to each
other. Therefore, the intermolecular attraction is fairly large. When the
compressed gas is allowed to undergo sudden expansion in passing
through a small orifice (hole) into a region of low pressure, the
molecules move apart. In doing so, the intermolecular attractive forces

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37
must be overcome. For this purpose energy is required. This energy is
taken from the gas itself which is thereby cooled.
Method from the liquefaction of Gases
Lind’s method of liquefaction of gases
The basic principle of the Linde‟s method of liquefaction of
gases is the joule-Thosmson effect. The apparatus used for this purpose
is shown in the fig.
For liquefaction of air, pure dry air is compression .The
compressed air is then passed through a spiral pipe having a jet at the
end. The air expands at the jet into the chamber where the pressure of
the air falls down from 200 atm to 1atm. As a result of joule-Thomson
effect a considerable drop of temperature occurs .The cooled air goes up
and cools the incoming compressed air of the spiral tube .the cooled air
goes again to the compression pump .The process of compression and
expansion is repeated again and again till the air is cooled to such an
extent that it liquefies. The liquefied air collects at the bottom of the
expansion chamber from where it is drawn off. All gases except H2 and
He can be liquefied by this method.
Non-Ideal Behavior of gases
An ideal gas obeys the gas laws and general gas equation for all
valyes of temperature and pressure. All real gases, however, show
marked deviations from ideal behavior. These deviations from ideal
behavior depends on:
(i) Pressure, and (ii) temperature
A convenient way of showing the deviations of a real gas from
ideal behavior is to plot a graph between on x-axis and on y-axis. The
ratio is called the compressibility factor. Its value is equal to 1 for 1
mole of an ideal gas for all values of temperature and pressure. For a real
gas, the compressibility factor deviate from 1.The extent of deviation

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38
from 1 depends upon the pressure and temperature. it may be either
greater than 1 or lesser than 1 . A value greater than one shows that thee
gas is less compressible than an ideal gas. On the other hand, a value
less than one indicate that the gas is more compressible than an ideal
gas.
For an ideal gas, since the increase of pressure decreases the
volume in such a way that remains constant at constant temperature, so
a straight line is obtained parallel to the pressure axis.
In order to understand the deviations of a real gas from ideal
behavior, let us study the plot of against p of a few real gases like He,
H2, N2 and CO2 first at 0o
C and then at 100o
c.
(a) Effect of Pressure Variation On Deviations at 0o
C
In the fig, the curves between compressibility factor and pressure
for the four real gases, He, H2, N2 and CO2 along with the behavior of an
ideal gas, are shown.
(i) At low pressures, all the curves for these four gases
approach to unity as the pressure approaches zero. Thus all real
gases behave as ideal gases at low pressures (upto10 atm).
(ii) At high pressure, the behavior of He and H2 differs from
that of N2 and CO2 at 0o
C.For both he and H2 at 0o
C, the
compressibility factor is always greater than one. This curves
show a continuous increase for both He. It means that these
gases are less compressible than expected from ideal behavior.
For both N2 and CO2 at 0o
C, the curve first decreases and then
increases again as the pressure increases. This decrease in the
compressibility factor below unity is more pronounced in CO2
than in N2. For CO2, the dip in the curve is greatest.
The extent of deviation of these four gases show that these
gases have their own limitations for obeying general gas

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39
equation. It depends upon the nature of the gas that at which
value of pressure, it will start disobeying .
(b) Effect of Pressure Variation on Deviation at 100o
C
In the fig, the curves between compressibility factor and pressure
for the four real gases, He, H2, and CO2at 100o
C along with the
behavior of an ideal gas, are shown. It is clear from the shape of
the curves, of these four gases that the deviations from the ideal
behavior become less at 100o
C than at 0o
C. The graphs come
closer to the expected straight line and the deviations are shifted
towards higher pressure. It means that the increase in temperature
makes the gases ideal.
Conclusions:
On the basis of experimental observations, we draw the following
two conclusions.
1. Gases are ideal at low pressure and non-ideal at high
pressure.
2. Gases are ideal at high temperature and non-ideal at low
temperature.
Causes for Deviation From Ideality
In the derivation of ideal gas equation, PV=nRT, it was assumed
that all gases behaved exactly alike under all conditions of pressures and
temperatures. In 1873, van der Waals pointed out that all real gases
deviate from ideal behavior at high pressures and low temperatures. It
was found that the following two incorrect postulates in the kinetic
theory of gases are the cause of deviation.
1. The molecules exert no forces of attraction on each other in a
gas.
When the pressure on a gas is high and the temperature is low
then the intermolecular attractive forces become significant.
Therefore, the ideal gas equation does not hold good .Actually,
under these conditions, the gas does not remain ideal .

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40
2. The actual volume of gas molecules is negligible as compared
to the volume of the container .
The actual volume of gas molecules is very small as compared to
the volume of the container, so it can be neglected. This volume,
however, does not remain negligible when there is high pressure
on the gas. This can be understood from the following figures.
Van der Waals equation For Real Gases
Keeping in view the molecular volume and the intermolecular
attractive forces,
Van derWallls pointed out that the ideal gas equation does not hold good
for real gases
He suggested that both volume and pressure factors in ideal gas equation
needed
Correction in order to make it applicable to the real gases.
Volume Correction
The volume of a gas is the free space in the container available
for the movement
of molecules . For an ideal gas, the actual volume of the molecules is
considered zero. So
the volume of an ideal gas is the same as the volume of the container . It
also represents the volume in which the molecules are free to move. For
a real gas, the molecules are rigid spherical particles which have a
definite volume. Consequently, they take up part of the space in the
container. Thus the volume in which the molecules of a real gas are
actually free to move is less than the volume of the container. Thus, van
der Walls suggested that a suitable volume correction term may be
subtracted from the volume of the vessel containing the gas . The
volume correction term for one mole of the gas is „b‟. If v is the volume
of the vessel containing the gas, then the volume available to gas
molecules for their free movement is written as.
V free = V vessel =b

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41
Where b is constant depending upon the nature of the gas . b is also
called the excluded volume or co-volume . It has been found that b is not
the actual volume of molecules but it is four times the actual volume of
the molecules. B =4 Vm .
Where Vm is the actual volume of one mole of gas molecules.
Pressure Correction
Consider a molecule in the interior of a gas. Since it is
completely surrounded by other molecules equally in all directions so
the resultant attractive force acting on the molecule is zero .However,
when the molecule approaches the wall of the container, the equal
distribution of molecules about it is upset. There is now a greater
number of molecules away from the wall with the result that these
molecules exert net inward pull on the molecule. Thus, when a molecule
is about to strike the wall of the container, it experiences a force of
attraction towards the other molecules in the gas .So it will strike the
wall with a lower velocity and will exert a lower pressure. Hence the
observed pressure, P Exerted by the molecules is less than the ideal
pressure, Pi by an amount Pi
.It is, therefore, necessary to add a pressure
correction term in order to obtain the ideal pressure.
Pi =P+ Pi
Pi is the true kinetic pressure, if the forces of attraction would have
been absent. Pi is the amount of pressure lessened due to attractive
forces.
It is suggested that a part of the pressure P for one mole of a gas
used up against intermolecular attractions should decrease as volume
increase. Consequently, the value of pi
in terms of a constant „a‟ which
accounts for the attractive forces and the volume V of vessel can be
written as
Pi
=
How to prove it?

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42
Pi
es determined by the forces of attraction between molecules of
type A which are striking the wall of the container and molecules of type
B which are pulling inward. The net force of attraction is proportional to
the concentrations of A type and B type molecules.
Let „n‟ is the number of moles of A and B separately and total
volume of both gases is „v‟.Then concentration, C= in moles dm-3 of A
and B separately. Therefore,
Pi
x
Pi
x
Pi
=
Where‟ a‟ is a constant of proportionality.
If n=1 , then Pi
=
Greater the attractive forces among the gas molecules, smaller the
volume of container, greater the value of lessened pressure Pi
.
This „a‟ is called coefficient of attraction or attraction per unit
volume . It has a constant value for a particular real gas .Thus effective
kinetics pressure of a gas is given by Pi , which is the pressure if the gas
would have been ideal .
Pi
+ P=
The kinetic gas equation for one mole of a gas may be written as,
(P+ )(V-b) =RT
For „n‟ moles of a gas, the kinetic gas equation may be written as
(P+) (V-b) =RT
This is called van der Waals equation. „a‟ and „b‟ are known as van
der waals constants.
Units of ‘a’
Since Pi
=
a=

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43
a=
a=atm dm6
mol-2
In sI units, pressure is in Nm-2
and volume in m3.
So, a=
a= Nm4
mol-2
Units ‘b’: „b‟ is excluded or incompressible volume /mole of gas .So ,
its units should be dm3 mol-1 or m3 mol-1 .
The values of „a‟ and „b‟ can be determined by knowing the
values of P,V and T of a gaseous system under two different conditions .
The values of „a‟ and ‟b‟ for some common gases are given in the
following table.
Gas ‘a’ (atm dm6 mol-2)
‘b’(dm3 mol-1)
Hydrogen
0.245 0.0266
Oxygen 1.360
0.0318
Nitrogen
1.390
0.0391
Carbon dioxide
3.590 0.0428
Ammonia
4.170 0.0371
Sulphur dioxide 6.170
0.0564
Chlorine
6.493 0.0562
The presence of intermolecular forces in gases like CI2 and SO2
increase their „a‟ factor. The least valur of „a‟ for H2
is taken, then its
small size and non-palar character. The „b‟ value of H2
is 0.0266 g (1

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44
mol) of H2
is taken, then it will occupy 0.0266 dm3
mol-1
. It means that
if 2.0266 dm3
or 266 cm3
of volume at closet approach in the gaseous
stat.
Example 8: One mole of methane gas is maintained at 300 k. Its
volume is 250 cm3
.calculate the pressure exerted by the gas under the
following conditions.
(i) When the gas is ideal (ii) When the gas is non-ideal
a=2.253 atm dm6 mol-1 b=0.0428dm3 mol-1
Solution:
(i) when methane gas is behaving as ideal, general gas
equation is applied , that is , PV=nRT
v=250 cm =0.25 dm3
n=1 mol
T=300k R=0.0821atm dm3
k-1
mol-1
P=?
P=
P=
P=98.52 atm Answer
(ii) When methane gas is behaving as non-ideal, van der Waals
equation is applied,
(P)(V-nb)=nRT
One rearranging the above equation for P,
P=-
P=
P=
P=117.286-36.048
P=81.238atm Answer
In the non-ideal situation the pressure has lessened up to
=98.52-81.238=17.282atm Answer
Plasma Stat (or Plasma)
What is Plasma?
“A high temperature ionized gas mixture consisting of free
electrons, positive ions and natural atom sis called plasma.”

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It is the fourth state of matter, in addition to solid, liquid and gas.
It was first identified by William Crookes in 1879. It is a gaseous state.
It is a gaseous state. It is electrically neutral because it consists of almost
equal numbers of free electrons and positive ions. It conducts electricity
because it contains free electrons and positive ions. It constitutes more
than 99% of the visible universe. It is everywhere in our space
environment. Naturally occurring plasma is rare on earth. Plasma is used
to conduct electricity in neon signs and fluorescent bulbs. It occurs only
in lightning discharges and in artificial devices like fluorescent lights,
neon signs, etc. Scientists have constructed special chambers to
experiment with plasma in laboratories.
How is Plasma formed?
Plasma is formed by the interaction of energy with atoms or
molecules. Atoms or molecules may be ionized on heating. An atom on
losing an electron changes to a positive ion. In a sufficiently heated gas,
ionization occurs many times, creating clouds of free electrons and
positive ions. However, all the atoms are not necessarily ionized and
some of them may remain neutral atoms. “A high temperature ionized
gas mixture consisting of free electrons, positive ions and natural atom
sis called plasma.”
Plasma contains a significant number of electrically charged particles
which are free electrons and positive ions. It is due to this reason plasma
shows electrical properties and behavior.
What plasma can potentially do?
Plasma can generate explosions, carry electrical currents and
support magnetic fields within themselves. Space plasmas can contain
enough heat to melt the earth thousands of times over. Crystal plasmas
can freeze the earth at least a hundred times, one after the other.
Artificial and Natural plasma:
Artificial Plasma can be created by using charges on a gas, e.g,.
in neon signs .Plasma at low temperature is not exist. This is because
outside a vacuum, low temperature plasma reacts rapidly with any

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46
molecule it meets. This aspect makes it both very useful and not easy to
use.
Natural plasma exists only at very high temperatures or low
temperature vacuums. It does not break down or react rapidly. It is
extremely hot having a temperature over 20.000 o
C. Plasmas possess
very high energy. They vaporize any material they touch.
Characteristics of Plasma:
1. Plasma shows a collective response to electric and magnetic
fields. This is because it has sufficient numbers of charged
particles. The motions of the particles in the plasma generate fields
and electric currents from within plasma density. It refers to the
density of the charged particles.
2. Although plasma contains free electrons and positive ions and
conducts electricity but as whole it is electrically neutral. It
contains equal number of free electrons and positive ions.
Where is plasma found?
Plasma is the most abundant from of matter. The entire universe
is almost of plasma. It existed before any other forms of matter came
into being .Plasmas are found from the sun to qarks. Quarks are the
smallest particles in the universe. Most of the matter in inner-stellar
space is plasma. The sun is a 1.5 million kilometer ball of plasma which
is heated by nuclear fusion.
On the earth, plasma only occurs in a few limited places such as
lightning bolts, flames, auroras and fluorescent lights. On passing
electric current through neon gas, both plasma and light are produced.
Applications of plasma:
The important applications of plasma are the following.
(i) A fluorescent light bulbs and tube is not like an ordinary
light bulb. Inside the tube is a gas. When light is turned on,
electricity flows through the tube. This electricity acts as a
source of energy and chares up the gas. This charging and

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47
exciting of the atoms creates a glowing plasma inside the tube
or bulb.
(ii) Neon signs are glass tubes filled with gas. When they are
turned on, electricity flows through the tube. The electricity
charges the gas and creates plasma inside the tube. The plasma
glows a special colour depending on the nature of the gas inside
the tube.
(iii) They are used for processing go semiconductors,
sterilization of some medical products, lamps ,lasers, diamond
coated films , high power micro wave sources, and pulsed
power switches.
(iv) They provide foundation for the generation of electrical
energy from fusion pollution control, and removal of hazardous
chemicals.
(v) Plasmas are used to light up offices and homes, make
computers and electronic equipment to work.
(vi) They drive lasers and particle accelerators, help to clean
up the environment, pasteurize foods, and make tools corrosion-
resistant.
Future Horizons:
Scientists are working on putting plasma to effective use. Low energy
plasma should by able to survive without instantly reacting and
degeneration. The application of magnetic fields involves the use of
plasma. The magnetic fields create low energy plasma which relate
molecules which are in a metastable state. The metastable molecules do
not react until they collide with another molecule with just the right
energy. This enables the met stable molecules to survive long enough to
react with a designated molecule. The met stable molecules are
selectivity. It makes a potentially unique solution to problems like
radioactive contamination. Scientists are experimenting with mixtures of
gases to work as met stable agents on plutonium and uranium.

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