Chapter No.3 GASES LONG QUESTIONS · Chapter No.3 GASES LONG QUESTIONS States of Matter Matter...

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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 O0C .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 25oC. 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 0oC 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 0oC 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 127

o 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 oC and V1 is the volume at T

oC .Then from the

statement of Charles,s law.

For 1 oC 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 0oC =increase in vol.at

toC

Vt=Vo +

Vt=Vo (1+ )

This general equation can be used to know the volume of the gas at

various temperatures.

Suppose at 0oC , 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

oC, increases

the volume from 566 cm3 to 746cm

3 .

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-273oC (more correctly-

273.16 oC) 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 (oC) 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. 16oCif it is extrapolated. The volume of a gas becomes

zero at -273.16oC. 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 -273.16oC (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 cm

3 . 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

0oC, and the temperature of boiling water at 1 atmospheric pressure is

100oC. The space between these temperature marks is divided into 100

equal parts and each part is 1 oC

(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

1oF

(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=273.16oC or 0K=273.16

oC

Temperature on Kelvin scale=273+temperature oC

Or

T (k)=273+ oC

The following relationship helps us to understand the

introversions of various scales of temperatures.

T (k)=273+ oC

oC=

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of= (

oC)+32

Remember that the Fahrenheit and Celsius scales are both relative

temperature scales. They define two reference points (such as 0oC and

100oC). 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 =1000cm

3 : 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 dm

3 ; V2 =10dm

3

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 0

o 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 30

o 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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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 oC 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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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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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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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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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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“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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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

According to kinetic gas equation,

PV=

Multiplying and dividing by 2 on right hand side.

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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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N1 = N2

Thus ,number of molecules of gas1 =number of molecules of gas2

4. Graham’s law of Diffusion According to kinetic gas equation,

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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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.

According to kinetic gas equation,

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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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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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 31.1

oC. It must

be cooled below this temperature before it can be liquefied by

applying high pressure. However, if temperature is maintained

below 31.1oC. Then lower pressure than critical pressure is

required to liquefy it.

(2) NH3 is a polar gas. Its critical temperature is 132.44 oC. 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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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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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 0oC and then at 100

oc.

(a) Effect of Pressure Variation On Deviations at 0oC

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 0oC.For both he and H2 at 0

oC, 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 0oC, 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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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 100oC

In the fig, the curves between compressibility factor and pressure

for the four real gases, He, H2, and CO2at 100oC 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 100oC than at 0

oC. 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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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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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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Pies 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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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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mol) of H2 is taken, then it will occupy 0.0266 dm

3 mol

-1 . It means that

if 2.0266 dm3 or 266 cm

3 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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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 oC. 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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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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Chapter No.3 GASES LONG QUESTIONS · Chapter No.3 GASES LONG QUESTIONS States of Matter Matter...

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