Chapter 3rd

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Chapter No.3

GASES

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.

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 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 . 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 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.

Picture

Fig, Verification of Boyle,s Law 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, 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.

Picture

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 . 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 ( ).

Picture

Fig.(a) A plot between P and Fig (b) A plot between pressure and product pf P V

(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

=

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

b e V - 2 . I t c a n b e s h o w n t h a t t h e r a t i o .

=

Hence Charles’ s law is verified.

Picture.

Fig. Verification of Charles’s Law

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

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 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-273oC (more correctly-273.16 oC) which is called

zero Kelvin (0k) or zero absolute. The 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.

Table: volume –Temperature data for a given amount of a gas at constant pressure

Volume (cm3) Celsius

Temperature(oC)

Temperature (K) =k=cm3 k-1

1092 273 546 2

846 150 423 2

746 100 373 2

646 50 323 2

566 10 283 2

548 1 274 2

546 0 273 2

544 -1 272 2

526 -10 263 2

400 -73 200 2

346 -100 173 2

146 -200 73 2

0 -273 0 2

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

Picture

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 cm3 . According to

Charlie’s law.

==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=273.16oC

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=

of= (oC)+32

Picture

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)

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

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=

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 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 =

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 =

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

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.

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.

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

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

Picture.

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 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.

Picture

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.

Picture

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.”

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 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.

Picture

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

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.

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,

“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:

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.

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)

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=

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 =

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 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 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 31.1oC. 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.

Table : Critical Temperature and Critical Pressure of Some Gases Gas Critical Temperature Tc

(oC)

Critical Pressure Pc9 (atm)

Carbon dioxide , CO2 31.142(304.3k) 73.0

Freon -12 , CC12F2 111.54(384.7k) 39.6

Ammonia , NH3 132.44(405.6k) 111.5

Water Vapours, H2O 374.44(647.6k) 217

Oxygen , O2 -118.75(154.4k) 49.7

Argon , Ar -112.26(150.9k) 48

Nitrogen, N2 -147.06(126.1k) 33.5

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 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.

Picture

Fig. linde’s method for the liquefaction of air

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 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 100oc.

(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.

Picture

Fig Non-ideal behavior of gases at 0o C

(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 0oC, 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 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 0oC. 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.

Picture

Fig. Non-ideal behaviour of gases at 100oC

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 .

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.

Picture

Fig. A gas at low pressure when Fig . A gas at high pressure when

Actual volume is negligible Actual volume is negligible

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

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.

Picture

Fig : Force of attraction and Pressure correction

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? 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=

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 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.” 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 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 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.

EXERCISE

Q1: Select the correct answer out of the following alternative suggestions: (i) Pressure remaining constant, at which temperature the volume of a gas will become twice of what

it is at 0oC .

a. 546oC b.200oC c. 546 k d. 273k

Hind: V T.

(ii) Number of molecules in one dm3 of water is close to

a. b. c. d.

Hint: 1 dm3 of H2O =1000cm3 of H2O; 1000cm3 of H2O =1000g of H2O

No of moles of H2O = moles ( 1 ml of H2O =1 f of H2O)

(iii) Which of the following will have the same number of molecules at STP?

a. 280cm3 of CO2 and 280 cm3 of N2 O

b. 11.2dm3 of O2 and 32 g of O2

c. 44g of CO2 and 11.2 dm3 of CO

d. 28 g of N2 and 5.6 dm3 of oxygen

(iv) If absolute temperature of a gas is doubled and the pressure is reduced to one half , the volume of

the gas will

a. remain unchanged b. Increase four times

c. reduce to d. be doubled

Hint: PV = RT, V= : V

(v) How should the conditions be changed to prevent the volume of a given gas from expanding when

its mass is increased?

a. Temperature is lowered and pressure is increased.

b. Temperature is increased and pressure is lowers.

c. Temperature and pressure both are lowered

d. Temperature and pressure both are increased.

Hint: PV = RT, V= : V ( M and R being constant)

(vi) The molar volume of CO2 is maximum at

a. STP b. 127oC and 1 atm

c. 0oC and 2 atm d. 273 oC 1 atm

Hint:For1 mole of CO2 PV = RT, V= : V or Vm

(vii) The order of the rate of diffusion of gases NH3, SO2 C12 and CO2 si :

a. NH3 > SO2 > C12 > CO2

b. NH3 > CO2 > SO2 > C12

c. C12 >SO2 > CO2 >NH3

Hint: r

(viii) Equal masses of methane and oxygen are mixed in an empty container at 25 oC . The fraction of total

pressure exerted by oxygen is

a. b.

c. d.

Hint: In an empty container, the partial pressure of gas is directly proportional to the mole-

fraction of the gas. Partial pressure mole fraction, at STP.

(viii) Gases deviate from ideal behavior at high pressure. Which of the following is correct for non-

ideality?

a. At high pressure, the ga molecules move in one direction only.

b. At high pressure, the collision between the gas molecules are increased manifold.

c. At high pressure, the intermolecular attractions become significant.

d. At high pressure, the intermolecular attractions become significant.

(ix) The deviation of a gas from ideal behavior is maximum at

a. -10oC and 5.0atm b. -10oC and 2.0atm

c. 100 oC and 2.0 atm d. 0.oC and 2.0atm

(xi) A real gas obeying van der Waals equation will resemble ideal gas if

a. both ‘a’ and ‘b’ are larger b. both ‘a’ and ‘b’ are small

c. ‘a’ is small and ‘b’ is larger d. ‘a’ is larger and ‘b’ is small

Ans: (i)c (ii)d (iii)a (iv)b (v)a (iv)b (vii)b (viii)a (ix)d (x)a (xi)b

Q2. Fill in the blanks. (i) The product of PV has the S.I. unit of _______.

(ii) Eight grams each of O2 and H2 at 27oC will have total K.E in the ratio of ________.

(iii) Smell of the cooking gas during leakage from a gas cylinder is due to of the property of _______of

gases.

(iv) Equal _________of ideal gases at the same temperature and pressure contain ________number

of molecules.

(v) The temperature above which a substance exists only as gas is called _____.

Ans. (i) atm dm3 (ii) :16 (iii) diffusion (iv) Volumes:

equal (v) Critical temperature

Q3. label the following sentences as true or false. (i) Kinetic energy of molecules is zero at 0oC.

(ii) A gas in a closed container will exert much higher pressure at the bottom due to gravity than at

the top.

(iii) Real gases show ideal gas behavior at low pressure and high temperature.

(iv) Liquefaction of gases involves decrease in intermolecular spaces.

(v) An ideal gas on expansion will show Joule-Thomson effect.

Ans. (i) False (ii) False (iii) True (iv) False (v) False

Q4: (a) What is Boyle’s law of gases? Give its experimental verification.

(b) What are isotherms? What happen to the positions of isotherms when they plotted at high

temperature for a particular gas.

(c) Why do we get a straight line when pressures exerted on a gas are plotted against inverse of

volumes. This straight line changes its position in the graph by varying the temperature. Justify it.

(d) How will you explain that the value of the constant k in equation PV=k depends upon

(i) The temperature of the gas (ii) the quantity of the gas.

Ans: (a) Isotherms: The P-V curves obtained at constant temperature are called isotherms. These

curves are obtained by plotting a graph between pressure on the x-axis and volume on the y-axis.

Similar curves are obtained at fixed temperatures.

When the isotherms (P-V curves are plotted at high temperature, they go away from the axes.

The reason is that, at higher temperature, the volume of the gas increase.

Ans.(c) A plot of P versus gives a straight line at constant temperature. This shows that p is directly

proportional to . This straight line will meet at the origin where both P and are zero. The P goes

down as the gas expands, falling away to zero as the volume approaches infinity (= =0)

This straight line changes its position because both pressure and volume varies on varying the

temperature. When temperature is increased both pressure and volume will increase. Keeping T

constant and plotting P versus another straight line is obtained. This straight line goes away

from x-axis. However , when temperature is decreased both the values of P and V will decrease.

Again a straight line is obtained. This straight line will be closer to the x-axis

Ans.(d) General equation: PV=nRT _______(1)

The general gas equation contains the Boyle’s law, for which nRT are constant at fixed T and n.

Bolyle’s law: PV=k _______(2)

Form Eq (1) and Eq (2), we get

K=nRT

K=

K=constant x mT (at fixed R and M )

K mT _______(3)

This relation indicates that

(i) k T; it means k depends upon the temperature of the gas

(ii) k m; it means k depends upon the quantity of the gas.

Q5. (a) What is the Charles’s law? Which scale of temperature is used to verify that = k (pressure and

number of moles are constant)

(b) A sample of carbon monoxide gas occupies 150.0mL at 25.0o C. it is then cooled at constant

pressure until it occupies 100.0mL. What is the new temperature?

(c) Do you think that the volume of any quantity of a gas becomes zero at -273oC. Is it not against

the law of conservation of mass? How do you deduce the idea of absolute zero from this

information?

Ans.(a) The relation, =k can be verified only when T is taken on the Kelvin scale.

Ans.(b) V1 =150 mL T1 =273+25=298K

V2 =100 mL T2 =?

Formula Used:

=

or

T2 =

T2 =

T2 =198.67k

T2 =273+toC

toC=T2 – 273

toC=198.67-273

=-74.33oC Answer

Ans.(c) We know that : V1 =Vo (1+)

At -273oC, V-273 = Vo (1-)= Vo (1-1) =Vo x0=0

The volume of the gas become zero at- 273oC. But it is impossible to imagine that a

Gas which is matter occupies no space. It goes against the law of conservation of mass.

Hence, it follows that -273oC is the lowest temperature which a body can ever have. So -273o C is

called the absolute zero of temperature. The scale of temperature on which -273oC is taken as zero

is called absolute scale of temperature.

Q6. (a) What is Kelvin scale of temperature ? Plot a graph for one mole of an ideal gas

to prove that a gas become liquid, carlier than -273o C .

(b) Throw some light on the factor in Charles’s law.

Ans.(b) In Charles’s law, the factor (0.00366 ) is the coefficient of expansion of given mass of gas at

constant pressure. It shows that a gas expands by parts of its volume at 0o C for a rise of

temperature of 1oC.

Statement of Charles’s Law: “At constant pressure, the volume of a given mass of gas increases or

decreases by of its volume at0oC for every 1oC rise or fall in temperature .”

Mathematically.

Vt -=Vo +

It means that if we have 273 cm3 of gas at 0oC, its volume will increase by 1 cm3 for every 1oC

rise in temperature if it is heated at constant pressure Thus,

At 1oC , the volume will become 274 cm3.

At 2oC , the volume will become 271 cm3.

At 273oC , the volume will become 0 cm3.

Similarly, if the gas is cooled by 1oC at constant pressure, its volume will decrease by 1 cm3 .

Thus,

At -1oC , the volume of the gas becomes 272 cm3.

At -2oC , the volume of the gas becomes 275 cm3.

At -273oC , the volume of the gas becomes 546 cm3.

Q7. (a) What is the general gas equation? Derive it in various forms.

(b) Can we determine the molecular mass of an unknown gas if we know the pressure, temperature

and volume along with the mass of that gas?

(c) How do you justify from general gas equation that increase in temperature or decrease of

pressure decreases the density of gas?

(d) Why do we feel comfortable in expressing the densities of gases in the units of g dm-3 rather than

g cm-3, a which is used to express the densities of liquids and solids.

Ans.(b) Yes, we can determine the molecular mass of an unknown gas if we know its, P,T,V and m by

applying the following formula:

M=

Ans.(c) We know form general gas equation,

d=

d=constant x (M and R being constant)

d=

Density is directly proportional to pressure and inversely proportional to temperature or

decrease of pressure, decreases the density of the gas.

Ans.(d) The densities of gases are very low. They are about 1000 times smaller than the densities of

liquids and solids. So, if gas densities are expressed in g cm-3, then the values will be very small.

They may go to fourth place of decimal for some gases. When we express the densities in g dm-3,

then the values of the densities become reasonable to be expressed. For example, the density of

CH4 gas is 0.7168 g dm-3 at STP, but if it is expressed in g cm-3 , then it is 0.0007168 g cm-3 at STP.

Therefore, we feel comfortable in expressing the densities of gases in the units of g dm-3 rather

than in g cm-3.

Q8. Derive the units for gas constant R in general gas equation:

(a) When the pressure is in atmosphere and volume in dm3.

(b) when the pressure is in Nm-2 and volume in m3 .

( c) when energy Is expressed in ergs.

Q9. (a) what is Avogadro’s law of gases?

(b) Do you think that 1 mole of H2 and 1 mole NH3 at 0oC and 1 atm pressure will have equal

Avogadro’s number of particles? If not, why?

(c) Justify that 1 cm3 of H2 and 1 cm3 of CH4 at STP will have the same number of molecules, when

one molecules of CH4 is 8 times heavier than that of hydrogen.

Ans (b) 1 mole of H2 and 1 mole of NH3 at 0oC and 1 atm pressure will have equal number of molecules

under the same conditions of temperature and pressure. Hence, 1 cm3 of H2 and 1 cm3 of CH4 at

STP will have the same number of molecules.

Q10. (a) Dalton’s law of partial pressure is only obeyed by those gases which do not have attractive

forces among their molecules. Explain it.

(c) Derive an equation to find out the partial pressure of a gas knowing the individual moles of

component gases and the total pressure of the mixture.

(d) Explain that the process of respiration obeys the Dalton’s law of partial pressure.

(e) How do you differentiate between diffusion and effusion? Explain Graham’s law of diffusion.

Ans. (a) For Dalton’s law of partial pressure to hold, there will be no attractive forces among the

molecules on the walls of the gases. The pressure of a gas is due to the collisions of the molecules

on the walls of the container. In the absence of attractive forces each molecules of gas mixture will

hit the walls of the container with the same number of times and with the same force. Thus the

partial pressure of a given gas is unaffected by the presence of other gases. In this case, the total

pressure. Hence the law will not hold in the presence of attractive forces among the molecules.

Q11. (a) What is critical temperature of a gas? What is its importance for

Liquefaction of gases? Discuss Linde’s method of liquefaction of gases.

(b) What is Joule-Thomson Effect? Explain its importance in linde’s method of liquefaction of gases.

Ans. (a) Importance of Critical temperature for liquefaction of gases.

The critical temperature of the gases provides us the information about the

Condition under which gases liquefy. For example,O2 ,has a critical temperature 154.4k(-118,75 oC). It must be cooled below this temperature before it can be liquefied by applying high pressure.

Q12. (a) What is Kinetic molecular theory of gases? Give its postulates.

(b) How dose Kinetic molecular theory of gases explain the following gas laws:

(i) Boyle’s law (ii) Charles’s law

(iii) Avogadro’s law (iv) Graham’s law of diffusion.

Q13. (a) Gases show non-ideal behavior at low temperature and high pressure.

Explain this with the help of a graph.

(b) Do you think that some of the postulates of Kinetic molecular theory of gases are faulty? Point

out these postulates.

(c) Hydrogen and helium are ideal at room temperature, but SO2 and C12 are non ideal. How will you

explain this?

Ans. (c) Hydrogen (b.p-253oC) and helium b.p-269oC)have a very low boiling points .They are far

away from their boiling points at room temperature. Also, they have smaller number of electrons

in their molecules and smaller molecular sizes, i.e., molecular weight. So, intermolecular forces

are negligible at room temperature. Hence, they behave as an ideal gases at room temperature.

On the other hand, SO2 (b.p-10oC) and C12 (b.p-34oC) have boiling points near to room

temperature. They are not far away from their boiling points at room temperature. Also, they have

larger number of electrons in their molecules and larger molecular sizes. So, sufficient

intermolecular attractive forces are present at room temperature. Hence, they behave as non-ideal

at room temperature.

Q14. (a) Derive van der Waals equation for real gases.

(b) What is the physical significance of van der Waals constant, ‘a’ and ‘b’.

Give their units.

Ans. (b) Physical Significance of van der Waals constant ‘a’ and ‘b’

(i) Significance of’ a’: The value of constant ‘a’ is a measure of the intermolecular attractive

forces and greater will be the ease of its liquefaction.

Units of ‘a’ :

The units of ‘a’ are related to the units of pressure, volume and number of moles.

P= or

a= =

a= atm dm6 mol-2

In SI units: a= ==Nm4 mole -2

(iii) Significance of ‘b’: The value of constant ‘b’ us related to the size of the molecule. Larger the

size of the molecule, lager is the value of ‘b’. It is effective volume of the gas molecules.

Units of ‘b’: ‘b’ is the compressible volume per mole of gas. So the units of ‘b’ are related to the units

of volume and moles.

V=nb or b =

b==dm3 mol-1

In SI units: b===dm3 mol-1

Q15. Explain the following facts:

(a) The plot of PV versus P is a straight line at constant temperature and with a fixed number of moles

of an ideal gas.

(b) The straight line in (a) is parallel to pressure-axis and goes away from the pressure axis at higher

pressure for many gases.

(c) The van der walls constant ‘b’ of a gas is four times the molar volume of that gas

(d) Pressure of NH3 gas at given conditions (say 1 atm pressure and room temperature) is less as

calculated by van der Waals equation than that calculated by general gas equation.

(e) Water vapors do not behave ideally at 273 k.

(f) SO2 is comparatively non-ideal at 273 k but behaves ideally at 327 K.

Ans. (a) At constant temperature and with a fixed number of moles of an ideal gas, when the pressure

of the gas is varied, its volume changes, but the product PV remains constant. Thus,

P1 V1 = P2 V2 = P3 V3 =

Hence, for any fixed temperature, the product PV when plotted against P.a straight line parallel to

P-axis is obtained. This straight line indicates that PV remains constant quantity.

Ans. (b) Now, increase the temperature of the same from T1 to T2 .At constant temperature T2 and

with the same fixed number of moles of an ideal gas, when the constant. However, the value of

PV increase with increase in temperature. On plotting graph between P on x-axis is obtained.

This straight line at T2 will be away from the x-axis. This straight line also shows that PV is a

constant quantity.

Ans. (c) Excluded volume, ‘b’ is four times the molar volume fo gas. The excluding with each

other as shown in Fig. The spheres are considered to be non-compressible. So the molecules cannot

approach each other more closely than the distance, 2r . Therefore, the space indicated by the

dotted sphere having radius, 2r will not be available to all other molecules of the gas. In other

words, the dotted spherical space is excluded volume per pair of molecules.

Let each molecules be a sphere with radius =r

Volume of one molecules (volume of sphere)=

The distance of the closest approach of 2 molecules =2 r

The excluded volume for 2 molecules=

The excluded volume for 1 molecule=

=

=4Vm =b

The excluded volume for ‘n’ molecules=n b

Where Vm is the actual volume of a molecule.

Hence, the excluded volume or co-volume or non-compressible volume is equal to 4 times the

actual volume of the molecules of the gas.

Ans. (d) The pressure of NH3 calculated by general gas equation is high because it is considered as

an ideal gas. In an ideal gas, the molecules do not exert any force of attraction on one another. On

the other hand, when the pressure of NH3 is considered as a real gas. Actually, NH3 is a real gas. It

consists of polar NH3 molecules approaches the walls of the container, it experiences an inward

pull. Clearly, the molecule strikes the wall with a lesser force than it would have done it these are

no attractive forces. As a result of this , the real gas pressure is less than the ideal pressure.

Ans. (e) Water vapors present at 273K do not behave ideally because polar water molecules exert

force of attraction on one another.

Ans. (f) At low temperature, the molecules of SO2 possess low kinetic energy. They come close to

each other. The e intermolecular attractive forces become very high. So, it behave non-ideally at

273K. At high temperature, the molecules of SO2 have high kinetic energy. The molecules are at

larger distances from one other another. The intermolecular attractive forces become very weak.

So, it behaves ideally at 327K.

Q16. Helium gas in a 100 cm3 container at a pressure of 500 torr is transferred to a container with a

volume of 250cm3.What will be the new pressure

(a) if no change in temperature occurs

(b) if its temperature changes from 20 oC to 15 oC?

Solution: (a) Given: P1=500 torr P2 =?

V1 =100cm3 V2 =250cm3

Formula used: P2 V2 = P1V1

P2 =

P2 =

=250torr Answer

(b) Given: P1 =500torr ; P2 =?

V1 =100cm3 ; V2 =250 cm3

T1 =273 +20=293K : T2=273+15=288K

Formula Used: =

P2 = x

=

=196.58 torr Answer

Q17. (a) What are the densities in kg/m3 of the following gases at STP

(i) Methance, (ii) oxygen (iii) hydrogen

(P=101325Nm-2, T=273k, molecular masses are in kg mol-1 )

(b) Compare the values of densities in proportion to their mole masses.

(c) How do you justify that increase of volume upto 100 dm3 at 27 oC of 2 moles of NH3 will

allow the gas behave ideally.

Solution: (a) (i)Given P=101325Nm-2

T=273K

Molar mass of CH4 =12+4=16g mol-1 =16x10-3 kg mol-1 R=8.3143NmK-1 mol-1

d=?

Formula Used: d=

d=

d= 0.714kgm-3

P=101325Nm-2

T=273k

Molar mass of O2 =32g mol-1

d= ?

Formula used: d=

d=

d=1.428kgm-3

(iii) P=101325Nm-3 ; T=273k

Molar mass of H2 =2x10-3kg mol-1 :R=8.3143 NmK-1 mol-1

d=?

Formula Used: d=

d=

d=0.089kg m-3

Q18. A sample of krypton with a volume of 6.25 dm3 , a pressure of 765 torr and a temperature

of 20oC is expanded to a volume of 9.55 dm3 and a pressure of 375 torr. What will be its final

temperature in oC?

Solution: P1 = 765torr ; P2 =375torr

V1 =6.25dm3 ; V2 =9.55 dm3

T1 =273+20 =273k ;T2 =?

Formula used: x

T2 =

T2 =

T2 =

T1 =273+o C

oC =T-273=219.46-273

=-53.54 oC Answer Q19. Working at a vacuum line, a chemist isolated a gas in a weighing bulb with a volume of 255 cm3 ,

at a temperature of 25 oC and under a pressure in the bulb of 10.0torr. The gas weight 12.1 mg.

what is the molecular mass of this gas?

Solution: V=255 cm3 =0.255 dm3

P=10.0torr =

T1 =273+25=298K

m=12.1mg=0.0121g

R=0.0821dm3 atom K-1 mol-1

Formula Used: PV= RT

M=

M=

M=87.93 g Mol-1 Answer

Q20. What pressure is exerted by a mixture of 2.00g of H2 and 8.00g of N2 at 273 K in a dm3

vessel?

Solution: Given: V=10dm3 ; T=273k

R=0.0821 dm3 atm K-1 mol-1

P H2 =? PN2 =?

Mass of H2 =2.00g Mass of N2 =8.00g

Molar mass of H2 =2g Molar mass of N2 =28g mol-1

nH2 = =1 mole ;n N2 = =0.286 mole

n=mH2 +nN2 =1+0.286=1.286 moles

PV=nRT

P x10 dm3 =1.286 molx 0.0821dm3 atm k-1 mol-1 x 273K

P=

P=2.88 atm Answer

Q21. (a) The relative densities of two gases A and B are 1:1.5 find out the volume of B which will

diffused in the same time in which 150 dm3 of A will diffuse?

(b) Hydrogen (H2 ) diffuses through a porpous plate at a rate of 500cm3 per minute at 0oC. What is

the rate of diffusion of oxygen through the same porpous plate 0oC?

(c) The rate of effusion of an unknown gas A through a pinhole is found to be 0.279 times the rate of

effusion of H2 gas through the same pinhole.

Calculate the molecular mass of the unknown gas at STP.

Solution: Given:

dA =1 rA=150dm3

rB=

(b) rH 2=500 cm3 per minute MH2 = 2 g mol-1

ro2=? Mo2 =32 g mol-1

=

=4

ro2==125cm3 Answer

(c) Given: rH2=1 rA=0.279

MA =? MH2=g mol-1

=

=

=

0.078=

MA =

MA =25.64 g mol-1 Answer

Q22. Calculate the number of molecules and the number of atoms in the given amounts of each gas

(a) 20cm3 of cH4 at 0oC and pressure of 700 mm of mercury

(b) 1 cm3 of NH4 at 100oC and pressure of 1.5 atm

Solution: (a) Given: P=700 mm =0921atm

V=20cm3 =0.02dm3

R=0.0821dm3 atm K-1 mol-1

T=273+0=273K

N=?

Formula Used: PV=nRT

n=

n=

No of molecules =No of moles x NA

=8.22x1020 molecules Answer

Since 1 molecule of CH4 contains =5 atoms

Therefore, No of atoms =5x4.948x1020

=24.74x1020 atoms Answer

(b) Given P=105 atm ; R=0.0821dm3 atm K-1 mol-1

V=1 cm3 =0.001dm3 ; T=273+100=373K

n=?

Formula Used: n=

n=

No of molecules =4.89x10-5 x6.02x1023

=29.44x1018 =2.944x1019 molecules Answer

No of atoms =2.944x1019 x4=1.18x1020 atoms Answer

Q23. Calculate the masses of 1020 molecules of each of H2 ,O2 and CO2 at STP. What will happen to the

masses of these gases, when the temperature of these gases are increased by 100oC and pressure is

decreased by 100 torr.

Solution: (a) Given: Molecules of H2 =1020

Now , 6.02x1023 molecules of H2 at STP =1mole =2g

1020 molecules of H2 at STP =

=0.332x10-5 g

=3.32x10-4 g Answer

(b) Given: Molecules of O2 =1020

Now, 6.02x1023 molecules of O2 at STP =32g

1020 molecules of CO2 STP =

=5.32x 10-3 g Answer

(c) Given: Molecules of CO2 =1020

Now, 6.02x1023 molecules of CO2 at STP =44g

1020 molecules of CO2 at STP =

=7.30x10-3 g Answer

Q24. (a) Two moles of NH3 are enclosed in a 5dm3 flask at 27oC. calculate the pressure exerted by

the gas assuming that

(i) it behaves like an ideal gas

(ii) it behaves like a real gas

a=1.17 atm dm6 mol-2

b=0.0371 dm3 mol-1

(b) Also calculate the amount of pressure lessened due to forces of attractions at these conditions of

volume and temperature.

(c) Do you expect the same decreased in the pressure of 2 moles of NH3 having a volume of 40dm3

and at temperature of 27oC.

Solution: (a.i) Given: V=5dm3 ; T=273+27=300K

n=2mole ; R=0.821dm3 atm K-1 mol-1

P=?

Formula Used: PV=nRT

P=

P==

P=9.852 atm Answer

(a.ii) Given: n=2mol

a=4.17atm dm3 mol-1

b=0.0371dm3 mol-1

V=5dm3

R=0.0821atm dm3 K-1 mol-1

T=273+27=300K

P=?

Formula Used: (V-nb)=nRT

On rearranging the equation

P=-

On substituting the values

P=-

P=-

P=-

P=9.99-0.667

P=9.32atm Answer

(b) Difference of pressure , P=9.852-9.32=053atm Answer

(c) Given: n = 2 mol ; V=40dm3;T=273+27=300K

p =?

Formula Used: PV=nRT

P=

P=

P=1.232 atm

The decrease in pressure is not the same