Gaseous State A

8/10/2019 Gaseous State A

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Page # 1

Lecture # 1 & 2Introduction :

Matter as we know broadly exists in three states.

There are always two opposite tendencies between particles of matter which determine the state of matter

!nter molecular attractive forces.

The molecular motion / random motion.

In this chapter the properties and behaviours of the gases will be analysed and discussed in detail. These

properties are measured with the help of the gas laws as proposed Boyle,Charles,Gay lussac etc

Boyles law and measurement of pressure :Statement :

For a fixed amount of gas at constant temperature, the volume occupied by the gas is inversely proportional

to the pressure applied on the gas or pressure of the gas.

V "P

1

hence PV = const

this constant will be dependent on the amount of the gas and temperature of the gas.

P1V

1= P

2V

2

P

v

P2P1

V2 V1

BA

Applications of Boyles Law. For the two parts Aand B P1V

1= K & P

2V

2= K

hence it follows that P1V

1= P

2V

2.


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Page # 2

Units

Volume Pressure : Temperture

Volume of the gas is the Pressure = N/m2= Pa $S.I. unit Kelvin scale $Boiling point = 373 KVolume of the container C.G.S unit = dyne-cm2 ice point = 273 K

S.I. unit $m3 Convert 1N/m2into dyne/cm2 Farenheit scale $B.P. = 212 Fice point = 32 F

C.G.S. unit $cm3 24

5

2 cm10

dyne10

m1

N1% Celcius scale $B.P. = 100C

1 != 103m3 1N/m2= 10 dyne/cm2 ice point = 0C

1 != 103cm9

1dm3= 1 != 103m3 1 atm = 1.013 105N/m232212

32F

273373

273K

0100

0C

&&

%&

&%

&&

1ml = 103 != 1 cm3 =)0(R)100(R

)0(RR

&&

where R = Temp. on unknown scale.

Atmospheric pressure :

The pressure exerted by atomosphere on earths surface at sea level is called 1 atm.

1 atm = 1.013 bar

1 atm = 1.013 105N/m2= 1.013 bar = 760 torr

Solved ExamplesExample A rubber balloon contains some solid marbles each of volume 10 ml. A gas is filled in the balloon at

a pressure of 2 atm and the total volume of the balloon is 1 litre in this condition. If the external

pressure is increased to 4atm the volume of Balloon becomes 625 ml. Find the number of marbles

present in the balloon.

Sol. Let the no. of marbles be = n .volume of marble = 10 n ml.

volume of balloon earlier = 1000 ml.

later = 625 ml.

Now for the gas inside the balloon temperature and amount of the gas is constant, hence boyles law

can be applied

P1V

1= P

2V

2

4 (625 10n) = 2 (1000 10n)

625 4 = 2000 20n + 40n

625 4 2000 = 20n

20

20004625 '= n.

5

125= n n = 25

MEASUREMENT OF PRESSURE

Barometer :A barometer is an instrument that is used for the measurement of pressure.The construction of the barometer

is as follows

Hg is filled upto brim

Thinnairowglasstube P0

P0

P0

P = P0 atm

Mg

Perfect Vaccum

Cross sectional view of the capillary column


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Page # 3

A mercury barometer is used to measure atmospheric pressure by determining the height of a mercury

column supported in a selaled glass tube. The downward pressure of the mercury in the column is exactly

balanced by the outside atmospheric pressure that presses down on the mercury in the dish and pushes it

up the column.

A thin narrow callibrated capillary tube is filled to the brim, with a liquid such as mercury, and is inverted into

a trough filled with the same fluid.Now depending on the external atmospheric pressure, the level of the

mercury inside the tube will adjust itself, the reading of which can be monitored. When the mercury columninside the capillary comes to rest, then the net forces on the column should be balanced.

Applying force balance, we get,

Patm

A= mg (Ais the cross-sectional area of the capillary tube)

If (is the density of the fluid, then m = ( vhence, P

atm A = ( ( g h) A (v = A h)

(his the height to which mercury has risen in the capillary)

or, Patm

= (gh

! Normal atmospheric pressure which we call 1 atmosphere or 1 atm, is defined as the pressure

exerted by the atmosphere at mean sea level. It comes out to be 760 mm of Hg = 76 cm of Hg. (at mean sea

level the reading shown by the barometer is 76 cm of Hg)

1 atm = (13.6 103) 9.8 0.76 = 1.013 105Pas.

1 torr = 1 mm of Hg.

1 barr = 105N/m2(Pa)

Faulty Barometer :An ideal barometer will show a correct reading only if the space above the mercury

column is vacuum, but in case if some gas column is trapped in the space above the mercury column, then

the barometer is classified as a faulty barometer. The reading of such a barometer will be less than the true

pressure.

For such a faulty barometer

mg

P Aatm

P Agas

P0A = Mg + P

gasA

P0= (gh + P

gasor (gh = P

0P

gas

Solved ExamplesExample The reading of a faulty barometer is 700 mm of Hg. When actual pressure is 750 mm of Hg. The

length of the air column trapped in this case is 10 cm .Find the actual value of the atmosphericpressure when reading of this barometer is 750 mm of Hg. Assume that the length of the Barometer

tube above mercury surface in the container remains constant.


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Page # 4

Sol. P0= P

gas+ 700 (g

) Pgas

= 750 (g 700 (g = 50 (g

Now for the gas column in the capillary, amount and temp are constant hence P1V

2 = P

2V

2

(50 (g) (100 A) = gasP* (50 A)

) gasP* = 100 (g

Now, applying force balance in the new scenario.

atmP* = gasP* + 750 (g = 100 (g + 750 (g = 850 (g

Hence, the atmospheric pressure is now, 850 cm of Hg.

Example In each of the following examples, find the pressure of the trapped gas.

Hg 10 cm

gas column

P = 75 cm of Hg0

Sol. Total pressure of gas column= 75 + 10 = 85 cm of Hg.

Example

Sol. Pgas

= 65 cm of Hg.

Example

Pg= 75 + 10 sin +.


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Page # 5

Sol. From the above problem, it can be generalised that, applying force balance every single time is notnecessary. If we are moving up in a fluid, then substract the vertical length, and while moving downadd the vertical length.

Example The diameter of a bubble at the surface of a lake is 4 mm and at the bottom of the lake is 1 mm. If

atmospheric pressure is 1 atm and the temperature of the lake water and the atmosphere are equal.

what is the depth of the lake ?

(The density of the lake water and mercury are 1 g/ml and 13.6 g/ml respectively. Also neglect the

contribution of the pressure due to surface tension)Sol. P

1V

1= P

2V

2

) (760 mm 13.6 g)3

4,(4 mm/2)3= (760 mm 13.6 g + h 1 g)

3

4,(1 mm/2)3

760 13.6 64 = (760 13.6 + h)h = 64 760 13.6 760 13.6h = 63 760 13.6 mm

h =10001000

6.1376063

'

''km = 0.6511 km = 651.1 m Ans.

Example A gas is initially at 1 atm pressure. To compress it to 1/4 th of initial volume, what will be the

pressure required ?Sol. P1= 1 atm V

1= V

P2= ? V

2=

4

V

P1V

1= P

1V

2at const. T & n

P2=

2

11

V

VP=

4

V

Vatm1 '= 4 atm Ans.

Example A gas column is trapped between closed end of a tube and a mercury column of length (h) when this

tube is placed with its open end upwards the length of gas column is (!1) the length of gas column

becomes (!2) when open end of tube is held downwards. Find atmospheric pressure in terms ofheight of Hg column.

Hg h

P0case !

P1

!1

Hg h

P0

case !!

P2!2

Sol. For gas P1

= (PO

+ h) P2

= (PO

h)

V1= ,r2!

1V

2= ,r2!

2

at const T. and moles.P

1V

1= P

2V

2; (P

O+ h) ,r2!

1 = (P

Oh) ,r2!

2

PO

!2+ h!

1 = P

O!

2h!

2

PO

!2P

o!

1 = h!, + h!

2

P0=

)(

)(h

12

21

!!

!!

&

-cm of Hg column Ans.

Example If water is used in place of mercury than what should be minimum length of Barometer tube to

measure normal atmospheric pressure.

Sol. gHP = OH2P = Patm.

0.76 m 13.6 g = OH2h 1 g

OH2h = 0.76 13.6 = 10.336 m Ans.


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Example A tube of length 50 cm is initially in open atmosphere at a pressure 75 cm of Hg. This tube is dipped

in a Hg container upto half of its length. Find level of mercury column in side the tube.

Sol. If after dipping the tube, the

length of air column be x cm situation shown in the adjoining figure.

Then by using, PiV

i= P

fV

f

We have.

75 cm Hg !A = Pf x A ....... (1) (!= 50 cm)

& also, Pf = 75 cm Hg + (x

2! ) ....... (2)

(2) & (1) . [75 + (x 25)] x = 75 50 . x2+ 50 x 3750 = 0

) x = 41.14 or 91.14But, x can't be ve ) x = 41.14) mercury column inside the tube = (50 41.14) cm = 8.86 cm Ans.

Lecture # 3

Charles law :For a fixed amount of gas at constant pressure volume occupied by the gas is directly proportional to

temperature of the gas on absolute scale of temperature.

V "T or V = kT = k (t + 273)

ttanconsT

V% where kis a proportionality constant and is

dependent on amount of gas and pressure.

2

2

1

1

T

V

T

V% Temperature on absolute scale, kelv in scale or ideal gas scale.

V = a + bt Temperature on centigrade scale.

V

T

Relation : T = t + 273

Since volume is proportional to absolute temperature, T the volume of a gas should be

theoretical ly at absolute zero is zero.

Infact no substance exists as gas at a temperature near absolute zero, though the straight l ine

plots can be extra ploated to zero volume. Absolute zero can never be attained practically

though it can be approch only.

By considering 273.15C as the lowest approachable limit, Kelvin developed temperature

scale which is known as absolute scale.


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Page # 7

Solved ExamplesExample If the temp. of a particular amount of gas is increased from 27C to 57C, find final volume of the gas,

if inital volume = 1 lt and assume pressure is constant.

Sol.2

2

1

1

T

V

T

V%

)57273(

V

)27273(

1 2

-%- So V2= 1.1 lt.

Example An open container of volume 3 litre contains air at 1 atmospheric pressure. The container is heated

from initial temperature 27C or 300 K to tC or (T + 273)K the amount of the gas expelled from the

container measured 1.45 litre at 17C and 1 atm.Find temperature t.

Sol. )T01= 300 K

It can be assumed that the gas in the container was first heated to (T + 273), at which a volume /Vescaped from the container

hence applying charles law :

300

3

= 273T

V3

-

/-

Now, this volume /Vwhich escapes trans the container get cooled

)273T

V

-/

=290

45.1

Solve the two equations and get the value of /V an a Tdetermine /v & calculate t that will be the answer.

Example An open container of volume V contains air at temperature 27C or 300 K.The container is heated to

such a temperature so that amount of gas coming out is 2/3 of

(a)amount of gas initially present in the container.(b)amount of gas finally remaining in the container.

Find the temperature to which the container should be heated.

Sol. (a) Here , P & V are constant, n & T are changing. Let, initially the amount of gas present be n & temp

is 27C or 300K. Finally amount of gas present in container = n 3

2n = 0

1

234

5' n

3

1& final temp. be T.T.

Then using n1T

1= n

2T

2, we have,

n 300 =3

n T

2= T

2= 900K

i.e., final temp = 900K Ans.

(b) Let there be x mole of gas remaining in the container,3

2of x comes out

) (x +3

2x) = n =

3

x5= n )x =

5

n3

) Using n1T

1= n

2T

2n 300 K =

5

n3 T

2

) T2= 500K

final temperature = 500 K Ans.


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Page # 8

Calculation of pay load :Pay load is defined as the maximum weight that can be lifted by a gas fi lled ballon.

M

Buoyancy

balloon

For maximum weight that can be lif ted, applying force balance

Fbuoyancy

= Mballoon

g + Mpay load

g

. (air

v.g. = (gas

v.g + Mg + mg.

mass of balloon = m net force on

volume of balloon = v balloon = 0

density of air = (air

(at equilibrium / when balloon is incoming

density of gas inside the with constant speed)

balloon = (gas

Solved ExamplesExample A balloon of diameter 20 metre weights 100 kg. Calculate its pay-load, if it is filled with He at 1.0 atm

and 27C. Density of air is 1.2 kg m3. [R = 0.0082 dm3atm K1mol1]

Sol. Weight of balloon = 100 kg = 10 104g

Volume of balloon =

33 100

2

20

7

22

3

4r

3

401

234

5'''%,

= 4190 106cm3= 4190 103litre

Weight of gas (He) in balloon =RT

PVM 0123

45 % RT

MwPV"

=300082.0

41041901 3

''''

= 68.13 104g

) Total weight of gas and balloon= 68.13 104+ 10 104= 78.13 104g

Weight of air displaced = 3

6

10

1041902.1 ''= 502.8 104g

) Pay load = wt. of air displaced (wt. of balloon + wt. of gas)

) Pay load = 502.8 104

78.13 104

= 424.67 104

g

Example Find the lifting power of a 100 litre balloon filled with He at 730 mm and 25C. (Density of air = 1.25

g /L).

Sol. Since, PV = nRT

PV =M

WRT ) W =

RT

PVM=

760

730

298082.0

4100

'

'g

i.e., Wt. of He = 15.72 g

Wt. of air displaced = 100 1.25 g/L = 125 g

) Lifting power of the balloon = 125 g 15.72 g = 109.28 g Ans.


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Page # 9

Gay-lussacs law :For a fixed amount of gas at constant volume, pressure of the gas is directly proportional to temperature of

the gas on absolute scale of temperature.

P "T

T

P= constant $dependent on amount and volume of gas

2

2

1

1

T

P

T

P% $temp on absolute scale

originally, the law was developed on the centigrade scale, where it was found that pressure is a linear

function of temperature P = P0+ bt where bis a constant and P

0is pressure at zero degree centigrade.

P

T O

P

P0

t(273C)

Example PV = K.V = K1/p

T

V= K

2.V = K

2T

P

K1 = K2T

PT =2

1

K

K= const .P "=

T

1. ?

## $## %&where are we wrong ?

This is wrong because we are varying temperature &

K1= f(1) thus K

1will change according to temperature

So2

1

K

Kwill be a function of temp & not constant.

Solved ExamplesExample The temperature of a certain mass of a gas is doubled. If the initially the gas is at 1 atm pressure.

Find the % increase in pressure ?

Sol.1

1

T

P=

2

2

T

P;

T

1=

T2

P2

% increase =1

1x 100 = 100%

Example The temperature of a certain mass of a gas was increased from 27C to 37C at constant volume.

What will be the pressure of the gas.

Sol.1

1

T

P=

2

2

T

P;

300

P=

310

P2; P

2=

30

31P


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Page # 10

Lecture # 4Avogadros Hypothesis :

For similar values of pressure & temperature equal number of molecules of different gases will occupy equal

volume.

N1 6$6 V (volume of N

1molecules at P & T of one gas)

N1 6$6 V (volume of N

1molecules at P & T of second gas)

. Molar volume & volume occupied by one mole of each and every gas under similar conditions will be equal.

One mole of any gas of a combination of gases occupies 22.413996 L of volume at STP.

The previous standarad is still often used, and applies to all chemistry data more than decade old, in this definition

Standard Temperature and Pressure STP denotes the same temperature of 0C (273.15K), but a slightly higher

pressure of 1 atm (101.325 kPa) .

Standard Ambient Temperature and Pressure (SATP), conditions are also used in some scientific works. SATP

c o n d i t i o n s m e a n s 2 9 8 . 1 5 k a n d 1 b a r ( i . e . e x a c t l y 1 05Pa) At SATP (1 bar and 298.15 K), the molar volume of an

ideal gas is 24.789 L mol1 (Ref. NCERT )

Equation of State :

Combining all the gas relations in a single expression which discribes relationship between pressure, volumeand temperature, of a given mass of gas we get an expression known as equation of state.

T

PV= constant (dependent on amount of the gas (n)).

1

11

T

VP=

2

22

T

VP

Ideal gas Equation :

nT

PV= constant [universal constant]

= R (ideal gas constant or universal gas constant)

experimentaly

R = 8.314 J/Kmole 7 25/3 = 1.987 cal/mole 7 2 = 0.08 Latm/mole 7 1/12

Solved ExamplesExample Four one litre flasks are separately filled with the gases, O

2, F

2, CH

4and CO

2under the same

conditions. The ratio of number of molecules in these gases :

(A) 2 : 2 : 4 : 3 (B) 1 : 1 : 1 : 1 (C) 1 : 2 : 3 : 4 (D) 2 : 2 : 3 : 4

Sol. According to avogadros hypothesis.All the flasks contains same no. of molecules

) Ratio of no. of molecules of O2, F

2, CH

4& CO

2

= 1 : 1 : 1 : 1 Ans (B)

Example Some spherical balloon each of volume 2 litre are to be filled with hydrogen gas at one atm & 27C

from a cylinder of volume V litres. The pressure of the H2gas inside the cylinder is 20 atm at 127C.

Find number of balloons which can be filled using this cylinder. Assume that temp of the cylinder is

27C.

Sol.1

11

T

VP=

2

22

T

VP

3002x1 '' = 2

60 = x

x = 30 balloons


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Page # 11

also cylinder is equivalent to 2 balloons gas will be transferred only till the time pressure in cylinder

is > outer pressure.

When pressure of cylinder = pressure of balloon.

hence only = 30 2 = 28 balloons will be filled.

Example A weather balloon filled with hydrogen at 1 atm and 300 K has volume equal to 12000 litres. On

ascending it reaches a place where temperature is 250 K and pressure is 0.5 atm. The volume of the

balloon is :

(A) 24000 litres (B) 20000 litres (C) 10000 litres (D) 12000 litres

Sol. Using1

11

T

VP=

2

22

T

VP;

K300

L12000atm1 '=

K250

Vatm0.5 2'

) V2= 20,000 L

Hence Ans. (B)

Daltons law of partial pressure :

Partial pressure :In a mixture of non reacting gases partial pressure of any component gas is defined as pressure exerted by

this component gas if whole of volume of mixture had been occupied by this component only.

Partial pressure of first component gas

v

RTnP 11%

v

RTnP 22%

v

RTnP 33%

Total pressure = P1+ P

2+ P

3.

Daltons law :

For a non reactinggaseous mixture total pressure of the mixture is summation of partial pressure of the

different components gases.

PTotal

= P1+ P

2+ P

3

v

RT)nnn( 321 --%

1T

1

T

1 xn

n

P

P%% (mole fraction of first component gas)

2T

2

T

2 xn

n

P

P%% (mole fraction of second component gas)

3T

3

T

3 xn

n

P

P%% (mole fraction of third component gas)


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Page # 12

Solved ExamplesExample The stop cock connecting the two bulbs of volume 5 litre and 10 lit re containing as ideal gas at

9 atm and 6 atm respectively, is opened. What is the f inal pressure if the temperature remains

same.

Sol. After the opening of the stop cock the pressure of the each bulb wil l remain same.

At the beginning, the no. of moles of gas in A =RT

6x10

At the beginning, the no. of moles of gas in B =RT

9x5

)total no. of mole at the beginning =RT

105

Total no. of mole of gas before opening the stop cock

= total no. of moles of gas after opening stop cock =RT

105

)pressure after the opening of the stop cock

P =RT

105x

totalV

RT=

510

105

-= 7 atm

Example The partial pressure of hydrogen in a flask containing two grams of hydrogen and 32 gm of

sulphur dioxide is :

(A) 1/16th of the total pressure (B) 1/9th of the total pressure

(C) 2/3 of the total pressure (D) 1/8th of the total pressure

Sol.2H

n =mol/g2

g2= 1mol.

2SOn =

mol/g64

g32= 0.5 mol

) 2H

P 8 92

2

SO

H

n2nH

n

- P

T=

)5.01(

1

- P

T=

3

2P

T.

)(C)

Example Equal volumes of two gases which do not react together are enclosed in separate vessels.

Their pressures are 10 mm and 400 mm respectively. If the two vessels are joined together,

then what will be the pressure of the resulting mixture (temperature remaining constant) :

(A) 120 mm (B) 500 mm (C) 1000 mm (D) 205 mm

Sol. Let, vol of containers be V & temp be T

P1= 10mm P

2= 400mm

) n1= RT

VP1& n

2=

RT

VP2

)n1+ n2= RTV)PP( 21 '-

After joining two containers final vol = (V+V) = 2V(for gases)

) Pfinal=8 9

final

21

V

RTnn -=

RT

V)PP( 21 '-

V2

RT=

8 92

PP 21-

=

8 9

2

mm40010-

= 205 mm. Ans. (D)


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Page # 13

Example A mixture of NO2& CO having total volume of 100 ml contains 70 ml of NO

2at 1 atm, mixture is left

for some time and same NO2get dimerised to N

2O

4such that final volume of the mixture become

80 ml at 1 atm calculate mole fraction of NO2in final equilibrium mixture.

Sol. Initial volume of NO2= 70 ml

Initial volume of CO = 100 70 = 30 ml

Final volume of mixture = 80 ml

Let the volume of NO2in final mixture be x

Let vml NO2 be converted to N

2O

4

2NO2 6$6 N

2O

4

V V/2

Hence final volume

= volume of CO + volume of NO2left + volume of N

2O

4 formed

= 30 + 70 V + V/2 = 80

V = 40 ml

Hence volume of NO2left = 70 V = 30 ml

Now as volume :moles

) mole fraction = volume fraction =80

30=

8

3

Analysis of gaseous mixture :

Vapour density :Vapour density of any gas is defined as the density of any gas with respect to density of the H

2gas under

identical conditions of temperature T and pressure P.

vapour density =T&PsameunderHofdensity

P&Tatgasofdensity

2

P =M

RT.

V

m. P = (

M

RT. (=

RT

PM

vapour density =2H

gasPMRT

RTPM

=2H

gasM

M

=2

Mgas

Mgas

= 2 vapour density

Average molecular mass of gaseous mixture :total mass of the mixture div ided by total no. of moles in the mixture

Mmix

=mixtureinmolesof.noTotal

mixtureofmassTotal

If we have

n1, n

2and n

3 are moles of three different gases having of molar mass M

1, M

2 and M

3respectively.

Mmin

=321

332211

nnn

MnMnMn

--

--

Solved ExamplesExample Calculate the mean molar mass of a mixture of gases having 7 g of Nitrogen, 22 g of CO

2and 5.6

litres of CO at STP.

Sol. Moles of N2= 7/28 = 1/4

Moles of CO2= 22/44 = 1/2

Moles of CO = 5.6 / 22.4 = 1/4

mean molar mass = Mmin

=321

332211

nnnMnMnMn

---- = ( 7 + 7 + 22 ) / 1 = 36


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Page # 14

Lecture # 5

Grahams Law of Diffusion/Effusion :

Diffusion :

Net spontaneous flow of gaseous molecule from region of high concentration (higher partial pressure) to

region of lower concentration or lower partial pressure

flow will be from both sides, N2will try to equalise its partial

pressure in both the vessels, and so will O2.

Grahams Law :

Under similar condition of pressure (partial pressure) the rate of diffusion of different gases is inversely

proportional to square root of the density of different gases.

rate of diffusion r :d

1d = density of gas

2

1

r

r=

1

2

d

d=

1

2

M

M=

1

2

D.V

D.VV.D is vapour density

r = volume flow rate =dt

dVout

r = mole flow rate =dt

dnout

r = distance travelled by gaseous molecule per unit time =dt

dx

The general form of the grahams law of diffusion can be stated as follows, when one or all of the

parameters can be varied.

rate :TMP A

P Pressure, A area of hole, T Temp. , M mol. wt.

If partial pressure of gases are not equal.

Then rate of diffusion is found to be proportional to partial pressure & inversely proportional to square

root of molecular mass.

r :P

r :M

1

2

1

r

r=

2

1

P

P

1

2

M

M


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Page # 15

Selective diffusion :If one or more than one components of a mixture are allowed to diffuse and others are not allowed then it is

selective diffusion of those components.

O2H2 Diffusion

Pt membrane

although total pressure

may be different both sides.

Diffusion will happen till thetime partial pressure of Hbecomes same on both sides.

2

! Platinum allows only H2gas to pass through

Effusion : (forced diffusion) a gas is made to diffuse through a hole by application of external pressure.

Solved ExamplesExample In is a tube of length 5 m having 2 idendical holes at the opposite ends. H

2& O

2are made to effuse

into the tube from opposite ends under identical conditions. Find the point where gases will meet for

the first time.

Sol.2

1

r

r=

dt

ax

dx

dt=

1

2

m

m=

2

1

dx

dx=

2

32

2

1

dx

dx= 4 .

2

2

Obytravelledcetandis

Hbytravelledcetandis= 4

)x5(

x

& = 4

x = (5 x) 4

x = 20 4x

5x = 20

x = 4 from H2side

Example 5 ml of H2gas diffuses out in 1 sec from a hole. Find the volume of O

2that wil diffuse out from the

same hole under identical conditions in 2 sec.

Sol. Rate of diffusion of H2=

15

ml5= 5ml/s =

2Hr (say)

)2o

r =2H

r 4

1= 5ml/s

4

1

) Volume of O2diffused in 2.0 seconds

=45 2 ml = 2.5 ml Ans.


16/34

Page # 16

Example A vessel contains H2& O

2 in the molar ratio of 8 : 1 respectively. This mixture of gases is allowed to

diffuse through a hole, find composition of the mixture coming out of the hole.

Sol. Here,2H

n :2O

n = 8 : 1 &2

2

O

H

r

r=

2

2

O

H

n

n

2

2

H

O

M

M

. 2

2

O

H

r

r=

1

8

2

32=

1

32

.8 98 9 t/outgmincoOofmolesof.no

t/outgmincoHofmolesof.no

2

2

/

/=

1

32

) Required composition of H2: O2coming out = 32 : 1 Ans.

Example The rate of diffusion of a sample of a ozonized oxygen is 0.98 times than that of oxygen. Find the %

(by volume of ozone in the ozonized sample.

Sol. Let, rate of dif fusion of ozonized oxygen be rg

& Let, reat of diffusion of oxygen be2O

r

) 2O

g

rr = 0.98 (given) ......... (1)

but2O

g

r

r=

2/1

g

O

M

M2

00

1

2

33

4

5(mean molar mass of ozonized oxygen = M

g)

Form (1).0.98 =gM

32

.(0.98)2=gM

32) M

g

=

8 92

98.0

32= 33.32

Let % of O3be x )% O2= (100 x) in ozonized oxygen.

) 8 9

100

32x100x48 '-= 33.32

.3200 + 16x = 33.32 ) x =16

132= 8.25 %

i.e. % of O3= 8.25 % Ans.

Example Assume that you have a sample of hydrogen gas containing H2, HD and D

2that you want to separate

into pure components (H = 1H and D = 2H). What are the relative rates of diffusion of the three

molecules according to Grahams law ?

Sol. Since D2is the heaviest of the three molecules, it will diffuse most slowly, and well call its relative

rate 1.00. We can then compare HD and H2with D

2.

Comparing HD with D2, we have

diffusionDofRate

diffusionHDofRate

2= HDofMass

DofMass 2 = amu0.3

amu0.4= 1.15

Comparing H2with D

2we have

diffusionDofRate

diffusionHofRate

2

2=

2

2

HofMass

DofMass=

amu0.2

amu0.4= 1.41

Thus, the relative rates of diffusion are H2(1.41) > HD (1.15) > D2(1.00).


17/34

Page # 17

Lecture # 6Kinetic Theory of Gases :

Postulates / assumptions of KTG :

A gas consists of tiny spherical particles called molecules of the gas which are identical in shape &

size (mass)

The volume occupied by the molecules is negligible in comparision to total volume of the gas.

For an ideal gas, volume of the ideal gas molecule ~ 0. Gaseous molecules are always in random motion and collide with other gaseous molecule & with

walls of the container.

Pressure of the gas is due to these molecular collisions with walls of the container

These collisions are elastic in nature

Molecular attraction forces are negligible. Infact, for an ideal gas attraction or repulsion forces are

equal to zero.

Newtons laws of motion are applicable on the motion of the gaseous molecule.

Effect of gravity is negligible on molecular motion.

The average K.E. of gaseous molecules is proportional to absolute temp of the gas.

2

1M )u( 2 "T (bar is for average)

Kinetic equation of gaseous state (expression for pressure of gas)

Derivationm = mass of one molecule

U'

= Ux i + Uyj + Uz k

Consider collision with ABCD

initaliP

'= mU

x i ; final fP

'= mU

x i

change in momentum due to collision = 2 Uxm

time taken between two succussive collision with force ABCD = t =xU

2!

frequency of collision =t

1=

!2

Ux

change in momentum in one sec. = force = 2 m!2

UU xx'=

!

2xUm

force due to all the molecules

=!

m }Ux........UU{

2N

2x

2x 21

---

average value of 2NU =2NU = N

U...........UUx 2x2x

21 N2

--

Fx=

!

M }UN{ 2x

all the three direction are equal as the motion is totally random in all directions, hence

2xU =

2yU =

2zU

2U =2z

2y

2x UUU --

= 3 2xU

Fx=

!

M. N

3

1 2U


18/34

Page # 18

Pressure = 2xF

!=

3

1 3

N

!2U The volume of the container v= !

3

) PV =3

1mN 2U Kinetic equation of gases

where 2U is mean square speed

root mean square speed = Urms

= 2U = 001

2334

5 ---N

U......UUU 2N23

22

21

Verification of Gaseous Laws Using Kinetic Equation :

From postulates ; PV =3

1mN 2U

2

1m 2U :T = ;T Where ;is a proportionality constant

PV =32

0123

45 2Um

21

N ; PV =32

;NT

Boyles Law : N : constantT : constantPV = constant

Charles law : N : constantP : constantv :T

Kinetic energy of gaseous molecule (translation K.E.)To calculate ;we have to use ideal gas equation (experimental equation)

PV = nRT

kinetic equation PV =32 ;nRT = 32 ;(nNA) T

on comparing ;=2

3

AN

R

;=2

3K where K =

AN

R= Boltzmann constant

Average K.E. of molecules =2

1m

2U = ; T

Average K.E. =2

3K T (only dependent on temperature not on nature of the gas.)

Average K.E. for one mole = NA0

1

234

5 2Um2

1

= 2

3

K NAT = 2

3

RT

Root mean square speed

Urms

= 2U = m

kT3 =

ANm

TR3Where m-mass of one molecule

" Dependent on nature of gas i.e mass of the gas

Urms

=M

TR3molar mass must be in k/mole.

Average speedU

av= U

1+ U

2+ U

3+ ............ U

N

Uav

=M

RT8

,=

m

KT8

,K is Boltzmman constant


19/34

Page # 19

Most probable speed : The speed possesed by maximum number of molecules at the given temperature

UMPS

=M

RT2=

m

KT2

Solved ExamplesExample In a container of capacity 1 litre there are 1023molecules each of mass 1022gms. If root mean

square speed is 105cm/sec then calculate pressure of the gas.

Sol. PV =3

1MN 2U

P = ?V = 103m3

m = 1025kg

N = 1023

2U = 105cm/sec = 103m/sec

2U = 106m2/sec2

P 103 =3

1 1025 1023 0 106

P =3

1 102 106 103

P =3

1 107pascals

Example (i)Calculate the pressure exerted by 1023gas molecules, each mass 1022g in a container of volume

one litre. The rms velocity is 105cm/sec.

(ii)What is the total kinetic energy (in cal) of these particles ?

(iii)What must be the temperature ?

Sol. (i) Here N = 1023molecules m = 1022g , V = 1L = 1000 cm3

rms velocity = 105cm/second

) PV =3

1Nm 2U

) P =31 Nm 2U

= 3cm10003

1

' 1023 1022g (105)2cm2/s2

= 3.33 107dyne/cm2 Ans.

(ii) Total K .E of the molecules

=2

1Nm 2U

=2

1 1023 1025kg (103m/s)2

=2

J10000 = 5000 J 7 1195.0 Cal Ans.


20/34

Page # 20

(iii) Avg K .E of one molecule =2

1 2U = 2

3kT

) T =3

1

k

Um 2

=3

1

8 9314.8

10023.6s/m10Kg10 232325 '''

K 01

234

5%

NA

Rk"

= 2414.8 K Ans.

Example If for two gases of molecular weights MAand M

Bat temperature T

Aand T

B; T

AM

B= T

BM

A, then which

property has the same magnitude for both the gases.

(A) Density (B) Pressure (C) KE per mol (D) RMS speed

Sol. Given that TAM

A= T

BM

A.

A

A

M

T=

B

B

M

T

But, r.m.s. =M

RT3

) r.m.sA= AA

M

RT3

& r.m.sB=B

B

M

RT3

) r.m.sA= r.m.sB ) Ans. (D)

Example It has been considered that during the formation of earth H2gas was available at the earth. But due

to the excessive heat of the earth this had been escaped. What was the temperature of earth during

its formation?

(The escape velocity is 1.1 x 106cm/s)

Sol. Escape velocity of H2should be equal to avg velocity of H

2.

) Avg velocity of H2= 1.1 106cm/s = 1.1 104m/s

But, avg. velocity = M

RT8

,

. 1.1 104= 3102T314.88

'',

'' 8 9kg102g2M 3H2 '%%

) T =8 9

314.88

102101.1 324

''',''

K = 11430.5 K = 11157.5oC Ans.


21/34

Page # 21

Lecture # 7Maxwells distributions of molecular speeds :

Postulates/Assumptions of speed distributions

It is based upon theory of probability.

It gives the statistical averages of the speed of the whole collection of gas molecules.

Speed of gaseous molecules of may vary from 0 to


22/34

Page # 22

Z=1 idealHe/H2

N2

NH /CO3 2

PVariation of Z with pressure at different temperature (for a gas ) :

Conclusions :

Z = 1 for ideal gasZ > 1 at all pressures for He/H

2

Z < 1 at low pressure (for all l other gases)Z > 1 at high pressur (for alll other gases)

Vander Waal Equation of real gases :The ideal gas equation does not considetr the effect of nattractive forces and molecular volume.

vander Waal corrected the the ideal gas equation by taking the effect of

(a)Molecular volume (b)Molecular attraction

Volume correction :Ideal gas equation :

PiV

i= nRT ; In the equation V

istands for the volume which is available for free movement of the molecules.

Videal

= volume available for free movement of gasesous moleculehence, V

i= V {volume not available for free movement} For an ideal gas

Vi= V {V = volume of container}

but for a real gasV

i=V , as all the volume is not available for free movement

Molecules have finite volume :

Excluded volume per molecule =2

1

>?@

ABC

, 3)r2(3

4= Co-volume per molecule.

The volume that is not available for free movement is called excluded volume.let us see, how this excluded volume is calculated.

r r

1 2

Excluded volume(not available for free momement)

For example 2, the entire shaded region is excluded, as its centre of mass cannot enter this region.If both molecules were ideal, then they would not have experienced any excluded volume but not in the case,of real gas as the centre of mass of 2cannot go further.Hence for this pair of real gas molecules.

Excluded volume per molecule =2

1

>?@

ABC

, 3)r2(3

4= 4

>?@

ABC

, 3r3

4

% excluded volume per mole of gas b = NA4

>?@

ABC

, 3r3

4

= 4 x NAx Volume of individual molecule

for n moles excluded volume = nb

Vi= V nb volume correction


23/34

Page # 23

Lecture # 8

Pressure correction or effect of molecular attraction forces :

Molecule in middle of container

Due to these attraction speed during coll ision will be reduced

Momentum will be less

Force applied will be less

Pressure will be less.

Pideal

= P + {correction term}

Correction term "no. of molecule attracting the colliding molecule. (n/v)Correction term "density of molecules (n/v).

no. of collision "density of molecules 01

234

5v

n

net correction term : 01

234

5v

n0

1

234

5v

n=

2

2

v

an

ais constant of proportionality

and this is dependent on force of attraction

Stronger the force of attraction greater will be a(Constant)

Pi= P +

2

2

v

an

Vander waals equation is

00

1

233

4

5-

2

2

v

anP (v nb) = nRT

VERIFICATION OF VANDER WAALS EQUATIONS

Variation of Z with P for vander wall equation at any temp.

Vander waal equation for 1mole

001

2334

5-

2mV

aP (V b) = RT

AT LOW PRESSURE(at separate temp.)

At low pressure Vmwill be high.

Hence b can be neglected in comparision to Vm. but 2

mV

acant be neglected as pressure is low

Thus equation would be

001

2334

5-

2mV

aP V

m= RT

PVm+

mV

a= RT

RT

PVm +RTV

a

m= 1

Z = 1

RTV

a

m Z < 1

Real gas is easily compressible in comparision to an ideal gas.


24/34

Page # 24

AT HIGH PRESSURE(moderate temp.)

VMwill be low

so b cant be neglected in comparision to Vm

but 2mV

acan be neglected in comparision to much high values of P..

Then vander Wall equation will be

P(Vmb) = RT

PVm

Pb = RT

RT

PVm =RT

Pb+ 1

Z =RT

Pb+ 1 Z > 1

If Z > 1 then gas is more diff icult to compress in comparision to an ideal gas.

At low pressure and very high temperature.

Vmwill be very large

hence bcan be neglected and 2mV

acan also be neglected as V

mis very large

PVm= RT (ideal gas condition)

For H2or He a ~ 0 because molecules are smaller in size or vander wall forces will be very weak,

these are non polar so no dipole-dipole interactions are present in the actions.

P(Vmb) = RT

so Z = 1 +RT

Pb

afactor depends on inter molecular attraction forces.

" afactor for polar molecule > afactor for non polar molecule.

Example Arrange following in decreasing afactor (H2O, CO

2, Ar)

H2O > CO

2> Ar

polar"For non polar molecules :

Greater the size surface area greater will be vander wall forces so greater will be aconstant.

Example Arrange following gases according to a

He, Ar, Ne, Kr.

aKr

> aAr> a

Ne> a

He

"More afactor means high boiling point.

" liquification pressure :

Is the pressure required to convert gas into liquid.

for easy liquefaction aDand PE

When Z < 1, Vm< Vm, ideal . easily liquifiableZ < 1, V

m< V

m, ideal . more difficult to compress.

Example Arrange the following according to liquification pressure.

n-pentane ; iiso

pentane , neo pentane.

an-pentene

> aiso pentane

> aneo pentane

liquif ication pressure = LP

etanpennPL

& < etanpenisoPL < etanpenneoPL

bis roughly related with size of the molecule. (Thumb rule)

b = NA4 >?

@

AB

C

,3

r3

4


25/34

Page # 25

Solved ExamplesExample Two vander waals gases have same value of b but different a values. Which of these would occupy

greater volume under identical conditions ?

Sol. If two gases have same value of b but different values of a, then the gas having a larger value of a will

occupy lesser volume. This is because the gas with a larger value of a will have larger force of

attraction and hence lesser distance between its molecules.


26/34

Page # 26

Lecture # 9Virial Equation of state :

It is a generalised equation of gaseous state all other equation can be written in the form of virial equation of

state.

Z is expressed in power series expansion of P or 001

2334

5

mV

1

Z = 1 +mV

B+ 2

mV

C+ 3

mV

D+ .....................

B second virial coefficient , C third virial coefficient , D fourth virial coefficient

Vander wall equation in virial form :

001

2334

5-

2mV

aP (V

mb) = RT

P =)bV(

RT

m& 2

mV

a

Z =RT

PVm=

)bV(

V

m

m

& RTVa

m=

)V/b1(

1

m&

RTV

a

m

x1

1

&= 1 + x + x2+ x3+ ..........

Z = 001

2334

5---- .........

V

b

V

b

V

b1

3m

3

2m

2

m

RTV

a

m

= 1 +mV

1 0

1

234

5&

RT

ab + 2

m

2

V

b+ 3

m

3

V

b+ .................

comparing with virial equation

according to vander walls equation

B = b

RT

a

, C = b2

, P = b3

at low pressure : Vmwill be larger

hence 2mV

1, 3

mV

1................ can be neglected

Z = 1 +mV

1 0

1

234

5&

RT

ab

If 01

234

5&

RT

ab = 0 . at T =

Rb

a

Z = 1 (ideal gas)

so at Z =Rba gas will behave an ideal gas (or follows boylesley)

But at constant temperature ideal gas equation is obeying Boyes law as T =Rb

a, so the temp is called

Boyles temp.

TB=

Rb

a

Z = 1 RTV

a

mfor a single gas if we have two graphs as above we must conclude T

2< T

1at Boyles temperature afactor

is compensated by factor so Z = 1.


27/34

Page # 27

Critical constant of a gas :When pressure is incerases at constant temp volume of gas decreases

AB $gasesBC $vapour + liquidCD $liquidcritical point : At this point all the phydical properties of liquid phase will be equal to phsical properties in

vapour such as

density of liquid = density of vapour

TCor critical temp :

Temperature above which a gas can not be liquidied

PCor critical pressure :

minimum pressure which must be applied at critical temp to convert the gas into liquid.

VCor critical volume :

volume occupied by one mole of gas at TC& P

C

Critical constant using vander wall equations :

001

2334

5-

2mV

aP (V

mb) = RT

( aPV2m- ) (Vmb) = RT Vm2

PVm

3+ aVmPbV

m2ab RTV

m2= 0

Vm

3+ Vm

2 01

234

5-

P

RTb +

P

aV

m

P

ab= 0

cubic can hence there will be three roots of equation at any temperature of pressure.

At critical point all three roots will coincide and will give singles dx = VC

at critical point Vander Waal equation will be

Vm

3Vm

2 001

2334

5-

C

C

P

RTb +

CP

aV

m

CP

ab= 0 ...(1)

But at critical point all three roots of the equation should be equal, hence equation should be :

Vm33Vm2VC+ 3Vm VC2VC3= 0 ...(2)

comparing with equation (1)

b +C

C

P

RT= 3V

C....(i)

CP

a= 3 V

C2 ...(ii)

CP

ab= V

C3 ...(iii)

PC= 2

CV3a

substituting PC= 2)b3(3

a= 2b27

a


28/34

Page # 28

by (i)C

C

P

RT= 3 V

Cb = 9b b = 8b

TC=

Rb27

a8

At critical point, the slope of PV curve (slope of isotherm) will be zero

CTmV

P

0

0

1

2

3

3

4

5

F

F

= 0 ...(i)

at all other point slope will be negative

O is the maximum value of slope.

mVFF

CTmV

P001

2334

5

FF

= 0 ....(ii)

{Mathematically such points an known as point of inf lection (where first two duivation becomes zero)}

using the two TCP

Cand V

Ccan be calculate

by

By any two a can be calculated but a by VCand T

Cand a by T

Cand P

Cmay differ as these values are

practical values and VCcant be accuratly calculated so whwn we have V

CT

C& P

Cgiven use P

C& T

Cto

deduce aas thet are more reliable.

Reduced Equation of state :Reduced Temp : Temperature in any state of gas with respect to critical temp of the gas

Tr=

CT

T

Reduced pressure : Pr=

CP

P

Reduced volume : Vr=

C

m

V

V

Vander wall equation

00

1

2334

5-

2mV

aP (V

mb) = RT

Substitute values :

001

2334

5- 2

C2r

CrVV

aPP (VrV

Cb) = R T

rT

C

Substiture the value of PCT

Cand V

C

002

334

52r b27

aP 0

01

2334

5-

22r

2r )b3(V

a

b27

aP (3b V

rb) = RTr

Rb27

a8

002

334

5-

r

r

V

1

3

P(3 V

r1) =

3

TR8 r

001

2334

5- 2

r

rV3P (3V

r1) = 8 T

rReduced equation of state

Equation is independent from a, b and R so will be followed by each and every gas independent of its nature.


29/34

Page # 29

Solved ExamplesExample The vander waals constant for HCI are a = 371.843 KPa and b = 40.8 cm 3mol1find the critical

constant of this substance.

Sol. The critical pressure, PC= 2b27

a= 62

3

10)8.40(27

10843.371&''

'= 2

9

)8.40(27

10843.371

'

'= 8.273 x 106

Pa = 8.273 MPa

The critical pressure, TC= Rb27

a8

R = 8.314 KPa dm3K1mol1

TC= Rb27

a8= 3108.4027314.8

843.3718&'''

'= 324.79 = 324.8 K

The critical volume, VC= 3b = 3 x 40.8 = 122.4 cm3

Example The vander waals constant for gases A, B and C are as follows :

Gas a/dm6KPa mol2 b/dm3mol1

A 405.3 0.027B 1215.9 0.030

C 607.95 0.032

Which gas has

(i) Highest critical temperature

(ii) The largest molecular volume

(iii) Most ideal behaviour around STP ?

Sol. TC= Rb27

a8Since, R is constant, higher the value of a/b higher will be critical temperature.

VC= 3b and VC: Vm(for a particular gas) therefore higher the value of VChigher will be molarvolume of the gas.

If the critical temperature close to 273 K, gas will behave ideally around the STP. Let us

illustrate the result in a tabular form.

Gas a/dm6KPa mol2 b/dm3mol1 TC VC a/b

A 405.3 0.027 534.97 K 0.081 1.501 x 104

B 1215.9 0.030 1.444.42 K 0.09 4.053 x 104

C 607.95 0.032 677.07 K 0.096 1.89 x 104

(i) B gas has the largest critical temperature.

(ii) C gas has the largest molecular volume.

(iii) A gas has the most ideal behaviour around STP.


30/34

Page # 30

Lecture # 10Vapour pressure of a liquid (aqueous Tension of water) :

Vapour pressure depends on

(a) Temperature (T D. VPD )

(b) Nature of the liquid

Vapour pressure is independent of amount of liquid & surface area of liquid.

Vapour pressure of the liquid is independent of pressure of any gas in the container,

Ptotal

= Pgas

+ Pwater vapour

Solved ExamplesExample In a container of capacity 1 litre, air and some liquid water is present in equilibrium at total pressure of

200 mm of Hg this container is connected to another one litre evacuated container find total pressure

inside the container when equilibrium is again stablised (aqueous tension or vapour pressure at this

temp. is 96 mm Hg).

Sol. Total pressure = 100 mm of Hg = Pgas

+ Pvapour water

. Pgas

+ 96 = 200

Pgas

= 107 mm of Hg Initial ly

when second coataitive connected

P1= 107 mm of Hg P2= ?V

1= 1 V

2= 2 litre

P1V

1= P

2V

2

107 1 = P2 2

53.5 = P2

After equilibrium is established

Ptotal

= 53.5 + 93 = Pgas

+ Pwater

= 146.5 mm of Hg at equilibrium

Eudiometry : The analysis of gaseous mixtures is called eudiometry. The gases are identified by absorbing them

in specified and specific reagents.

Some Common Facts :

Liquids and solutions can absorb gases.

If a hydrocarbon is burnt gases liberated will be CO2& H

2O. [H

2O is seperated out by cooling the

mixture & CO2by absorption by aqueous KOH]

If organic compound contains S or P then these are converted into SO2& P

4O

10by burning the

organic compound.

If nitrogen is present then it is converted into N2.

[The only exception : if organic compound contains NO2group then NO

2is liberated]

If mixture contains N2gas & this is exploded with O

2gas, do not assume any oxide formation unless specified.

Ozone is absorbed in turpentine oil and oxygen in alkaline pyragallol.


31/34

Page # 31

Solved ExamplesExample Carbon dioxide gas (CO

2) measuring 1 litre is passed over heated coke the total volume of the gases

coming out becomes 1.6 litre. Find % conversion of CO2into carbon monoxide.

Sol. CO2+ C 6$6 2CO CO

2CO

1 0 at time t 1 x 2xInitial volume = 1 litre

final volume = 1.6 litre no of litres = (1 + x) litres1 + x = 1.6 x = 0.6 x = 0.6

1

6.0 10 = 60% of CO

2will be converted into CO

Example 100 ml of hydrocarbon is mixed with excess of oxygen and exploded on cooling the mixture wasreported to have a contraction of 250 ml the remaining gas when passed through a solution ofaqueous KOH the mixture shows a further contraction of 300 ml. Find molecular formula of thehydrocarbon.

Sol. CxH

y+ 0

1

23

4

5-

4

Yx O

2 6$6 xCO

2 + OH

2

y2

100 ml x100 "2

y100.

Example. 100 ml of an hydrocarbon is burnt in excess of oxygen in conditions so that water formed getscondensed out the total contraction in volume of reaction mixture was found to be 250 ml when thereaction mixture is further exposed to aqueous KOH a further contraction of 300 ml is observed findmolecular formula of hydrocarbon.

Sol. CxH

y + O

2 6$6 CO2 + H2O

100 ml excess 300 mlBy POAC on Catoms

x 100 = 300

x = 3POAC on Hatomsy 100 = 2 moles of H

2O

POAC on O atoms2 v = 2 300 + 1 H

2O {v = volume of O

2consumed}

2 v = 600 + 50 y

v =2

y50600 -volume of O

2consumed

The total volume contraction is 250 ml.Hence, 100 + v 300 = 250

200 + v = 250 = 450

2 450 600 = 50 y

50300 = y = 6

Hydro carbon will be C3H

6

Alternative :Using balanced chemical equation

CxH

y + 0

1

234

5 -4

yxO

2 6$6 xCO

2 +

2

yH

2O

t = 0 100 ml v ml 0 0

t 0 v 100 0

1

23

4

5 -

4

yx100 x ml

2

y100

volume remained


32/34

Page # 32

250 = (volume of reactants) (volume products volume of unused reactant)

= (100 + v) {100 x +2

y100+ v 100 {x + y/4}} ...(1)

volume of CO2

100 x = 300 . x = 3The above two equations can be solved to get the required answer.

Example. A sample of water gas has a composition by volume of 50% H2

, 45% CO and 5% CO2

. Calculate the

volume in litre at STP at water gas which on treatment with excess of steam will produce 5 litre H2.

The equation for the reaction is :

CO + H2O 6$6 CO

2+ H

2

Sol. If x L CO in needed then

volume of H2in water gas = L%50

45.0

x01

234

5' = L

9.0

xL

2

1

45.0

x%0

1

234

5'

But, from equation : CO + H2O 6$6 CO

2+ H

2

& Gay-Lussacs law, we get, that the volume of H2

produced = volume of CO taken.

) Volume of H2due to reaction = x L

) Total volume of H2= 0

1

234

5- x

9.0

xL = 5 L

.9.0

x9.1= 5 L

) x =9.1

59.0 '

) Volume of water gas = L45.09.1

59.0L

45.0

x

'

'% = 5.263 L Ans.

Example. One litre of oxygen gas is passed through a ozonizer & the final volume of the mixture becomes 820ml. If this mixture is passed through oil of turpentine. Find final volume of gas remaining.

Sol. The reaction that takes place in ozonizer as 3O2

6$6 2O3.

If Vml of O2out of 1000 ml is ozonized then vol of O

3obtained =

3

2V

) Final vol of ozonized oxygen

= (1000 V +3

2V) mL = 820 mL

. 1000 3

1V = 820

. 3V

= 180 ) V = 540

) O2remaining = (1000 540) mL (" O3is absorbed in oil of turpentine)= 460 mL Ans.

Example. A gaseous mixture containing CO, methane CH4 & N

2gas has total volume of 40 ml. This mixture

is exploded with excess of oxygen on cooling this mixture a contraction of 30 ml is observed & when

this mixture is exposed to aqueous KOH a further contraction of 30 ml is observed .find the composition

of the mixture.

Sol. Let vol of CO be x mL

vol of CH4be y mL

vol of N2be z mL

On explosion with excess of oxygen the following reactions takes place


33/34

Page # 33

CO(g) +2

1O

2(g) $ CO

2(g) (By Gaylussac`s law of combing volume)

x mL x mL

CH4(g) + O

2(g) 6$6 CO

2(g) + 2H

2O(g)

y mL y mL 2 ymL

N2remains unreacted

On cooling H2O (g) liquifies hence volume reduction of 30 mL is observed) 2y = 30 ) y = 15

But, vol of CO2obtained = (x + y) mL

This is absorbed in KOH & vol reduction of 30 mL is observed.

) x + y = 30 .x = 30 y = (30 15) = 15and, x + y + z = 40 . z = 40 x y = 40 15 15 = 10

) Composition of mixture isvol of CO = 15 mL Ans.

vol of CH4= 15 mL Ans.

vol of N2= 10 mL Ans.


34/34

CHEMISTRY LECTURE NOTES

COURSE - VIKAAS (A)

(LECTURE No. 1 TO 10)

TOPIC : GASEOUS STATE

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Gaseous State A

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