Gaseous State

Chapter 2 : Gaseous States

2.1 Ideal Gas 04

2.2 Diffusion 09

2.3 Daltons Law 12

2.4 Eudiometry 13

2.5 Kinetic Theory of Gases 14

2.6 Maxwell’s Law of Distribution of Molecular Speeds 15

2.7 Molecular Speeds 16

2.8 Real Gases 18

2.9 Critical State 22

Solved Example 28

Exercises 33

Previous Years’ IITJEE Questions 38

Answer Key 39

2Gaseous States

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Chemist

Robert Boyle

Interesting FactAlthough Boyle’s chief scientif ic interest waschemistry, his first published scientific work, NewExperiments Physico-Mechanicall, Touching theSpring of the Air and Its Effects (1660), concernedthe physical nature of air, as displayed in a brilliantseries of experiments in which he used an air pumpto create a vacuum. The second edition of this work,published in 1662, delineated the quantitativerelationship that Boyle derived from experimentalvalues, later known as “Boyle’s law”: that the volumeof a gas varies inversely with pressure.

“Education is

what remains

after one has

forgotten

everything he

learned in

school.”Albert Einstein

The Age of Plastics

In 1907 the Belgian chemist Leo Baekelandreacted together the organic molecules Phenoland formaldehyde to produce a hard, mouldablematerial that his company named Bakelite. Thiswas the first plastic that retained its shape afterbeing heated, and it heralded a new age of usefulsynthetic substance.

GASEOUS STATES IITJEE COURSE

CHEMISTRY| 4 IIT STUDY CIRCLE

Section - 2.1 Ideal Gas

2.1.1 Boyles Law

For a fixed amount of an ideal gas kept at a fixed temperature pressure (P) and volume (V) are inversly proportional to each other.

1P

V∝ or 2

1P k

V=

or 2PV k (constant)= 1 1 2 2 3 3P V P V P V ....= = =

Alte

zza

mdr

cri (

polli

ci)

Volume (Pollici Cubi)

0 10 20 30 40 50

20

0

40

60

80

100

120

Diagramma p-V dei dati originali di Robert Boyle

Modello pV=K Dati di Boyle

0 12 24 36 48 60 72 84 96 108 120

Pressure

0.00

1.00

2.00

3.00

4.00

5.00

6.00

7.00

8.00

9.00

10.00

Pressure vs. 1/Volume

1/Volum

e . (× 100)

PV

P (cm)

1

100 200 300 400 500 600 700

Illustration 1:

Calculate the minimum pressure required to compress 700 mL gas at 1.75 bar to 200 mL. Assume that temperatureremains constant.

Solution :

According to Boyle’s law 1 1 2 2P V P V= (n, T constant)

Given that 1P 1.75= bar 1V 700= mL

2P ?= 2V 200= mL

21.75 700 P 200× = ×

2P 6.125= bar

IITJEE COURSE GASEOUS STATES

IIT STUDY CIRCLE CHEMISTRY| 5

2.1.2 Charle’s Law

At constant pressure, the volume of a given mass of an ideal gas is directly proportional to the absolute scale of temperature.

V T∝ or 1V k T= where k1 is a constant it can also be written as

1

Vk ( constant)

T= or

31 2

1 2 3

VV V...

T T T= = =

0.0 20.0 40.0 60.0 80.0 100.0

Temperature/degrees C

Vo

lum

e/l

iters

22.0

24.0

26.0

28.0

30.0

Vol

ume

Temperature (°C) 0 –273

P1 < P2 < P3

P1 P2

P3

Illustration 2:

The volume of a gas is 0.30 L at 25°C. Calculate the final volume at a temperature of 315°C.

Solution :

According to Charles’ law1 2

1 2

V V

T T= (n, T constant)

Given that, 1V 0.30L= 1T 298K=

2V ?= 2T 588=

22

V0.30V 0.59 L

298 588= ⇒ =

2.1.3 Gay- Lussac’s Law

The pressure of a gas of fixed mass and fixed volume is directly proportional to the absolute temperature of the gas

P T∝ or 3P k T= or 3

Pk

T= (constnat)

31 2

1 2 3

PP P....

T T T= = =

V1

V2

V3

V4

V1 < V2 < V3 < V4

Temperature (K)

Pres

sure

(ba

r)

Pressure vs temperature (K) graph

(Isochores) of a gas

0 100 200 300 400

Gay-Lussac’s law is actually the law of combining volumes.

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Illustration 3:

A gas kept in closed vessel is heated by 1°C During the process, pressure of gas incrreases by 0.4%. Assuming the gasto be ideal, calculate its final temperature.

Solution :

Using the law P = T law 1 2

1 2

P P

T T=

Let the initial pressure is P. then P1 = P, T

1 = T

2

0.4P P P,

100= +

2T T 1= +

P 100.4P

T 100(T 1)=

+ ⇒ T = 250 K

Final temperature = T + 1 = 251 K.

2.1.4 Avogadro’s Law

Equal volumes of ideal or perfect gases, at same temperature and pressure contain same number of molecules

V n∝ 4

Vk

n= (constant)

Illustration 4:

An acetylene tank provides 9340 L C2H

2 for producing oxyacetylene flame, at 0°C and 1 atm pressure then oxygen

cylinder attached provides 7.0×10–3 L O2 at 0°C and 1 atms pressure. Calculate the number of tanks required for complete

combustion of acetylene provided by the tank.

Solution :

2 2C H burns as

2 2 2 2 22vol 5vol

2C H 5O 4CO 2H O+ → + (at same P & T)

Thus O2 required of 4

2 2 2

59340 LC H 9340 2.34 10 L O

2= × = × and

So number of tanks 4

3

2.34 103.34

7.0 10

×= =

× tank ≈ 4 tank.

2.1.5 Ideal Gas Equation

In a gas, the molecules have enough kinetic energy so that the effect of intermolecular forces is small (or zero for an ideal gas),and the typical distance between neighbouring molecules is much greater than the molecular size. A gas has no definite shapeor volume, but occupies the entire container in which it is confined. A liquid may be converted to a gas by heating at constantpressure to the boiling point, or else by reducing the pressure at constant temperature.

It is also called perfect gas equation or equation of state. It is given by

PV nRT= where P = pressure of the gas

1 atm = 1.013 bar = 1.013 × 105 Pa = 76 cm of Hg = 760 mm of Hg = 760 torr



1 bar = 105 Pa = 100 kPa; 1 Pa = 1 Nt/m2; 1 mm of Hg = 1 torr

V = Volume occupied by the gas (volume of the container)

1 L = 1000 cc = 1000 mL = 10–3m3 = 1 dm3 1 mL = 1 cc

n = Number of moles of the gas n0

g given mass in gramsM molar mass in grams

= =

T = Temperature in Absolute scale (t°c + 273.15) = T K

R = Universal gas constant, whose value is same for all the gases

R = 0.0821 L atm/ mole K = 0.0831 L bar / mole K = 8.314 J/mole-K.

= 8.314 × 107 erg/mole K = 1.89 Cal/mole K.

It is equivalent to Boltzmann Constant (KB)

BA

RK

N= = 231.38 10 J / K−× ⇒

1 1

23 1

8.314J mole KK

6.022 10 mole

− −

−=

×

Combining all the law’s

We get ideal gas equationPV

nT= constant = R i.e., PV nRT=

• According to IUPAC STP (standard conditions of temperature and pressure) is at 0°C and 100 kPa (1 bar)

• Molar volume of an ideal gas at 0°C and 1 bar (100 kPa ) is 22.7 L/mole and at 0°C and 1 atm it is 22.4 L/mole

2.1.6 Various Forms of Ideal Gas Equations

Ideal gas equation can be written in many forms what ever is the form it carries the same meaning.

(i) PV nRT= N → no. of gas molecules

(ii) BPV NK T= KB → Boltzmann constant

(iii) 0PM RT= ρ 0M → Molar mass of the gas

(iv) 0PVM gRT= g → mass of the gas smaple.

ρ → density of the gas sample

Illustration 5:

A 2.0 mL bubble of gas is released at the bottom of a lake where the pressure is 6.5 atm. and temperature is 10°C. Whatis the volume of gas when it reaches to the surface where pressure is 0.95 atm and the temperature is 24°C.

Solution :

By using 1 1 2 2

1 2

P V P V

T T=

Given that,P1 = 6.5 atm, 2P 0.95 atm=

1T 10 C= ° = 283 K, 2T 24 C 297K° =

1V 2.0mL= 2V ?=

6.5 2.0 0.95 V

283 297

× ×= ⇒ V 14.36= mL

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Illustration 6:

Calculate the weight of CH4 in a 9 dm3 cylinder at 16 bar and 27°C.

Solution :

Given : 3P 16bar; V 9dm ; T 300K= = =

4

3 1 1CHM M 16; R 0.083bar dm K mol− −= =

∵

wPV RT

M

=

∴ 16 9 (w /16) 0.083 300× = × × ⇒ w 92.5= g.

Illustration 7:

At 0°C, the density of a gaseous oxide at 2 bar is same as that of nitrogen at 5 bar. What is the molecular mass of theoxide?

Solution :

M

RTρ = ; ∴

oxideoxide

2 M5 28M 70g / mol

RT RT

××= ⇒ =

Molecular mass of gaseous oxide.

Illustration 8:

When 4 g of gaseous substance X is introduced into an initially evacuated flask kept at 25°C, the pressure is found tobe 1 atm. The flask is evacuatd and 6 g of y is introduced. The pressure is found to be 0.5 atm at 25°C. Calculate the ratio

of x

y

M(M-

Mmolecular weight).

Solution :

wPV RT

M= ⇒

yx x

y x y

MP w

P M w= × (∵ V, R and T are constant)

y

x

M1 4

0.5 M 6= × ⇒ x

y

M 4 0.5 1

M 6 3

×= =

2.1.7 Applications of Ideal Gas Equations

One important point about pressure is when ever two region’s are connected physically the pressure in both the regions will besame.



This principle can be applied to an open vessel also i.e., Pinside = Pexternal = Constant usually external (atmospheric pressure)pressure will be considered constant.

Similarly volume of the gas (container) is also constant

∴ Applying ideal gas equation to the gas inside this open vessel

PV nRT=

PVnT

R= = constant

∴ nT constant=

i.e., on heating an open vessel. T raises but (nT) should remain constant, so ‘n’ decreases i.e., some amount (moles) ofgas will escape out.

Similar effect will be observed in the case of colling.

Illustration 9:

Two flasks of equal volume connected by a narrow tube (of negligible volume) are at 27°C and contain 0.70 mole of H2

at 0.5 atm pressure. One of the flask is then immersed into a bath kept at 127°C, while the other remains at 27°C. Calculatethe final pressure and the number of moles of H

2 in each flask.

Solution :

n1 = 0.3; n

2 = 0.4

= 0.4

0.40.35

× atm = 0.56 atm

Section - 2.2 Diffusion

Diffusion is migration of matter down a concentration gradient.

Consider a vessel divided into two halves with the help of a partion. Let us enclose a gas (A) in chamber 1 and chamber 2 isempty initially. So there is a concentration difference in both the chambers.

Now as soon as the partition is removed the gas spreads uniformly throught the vessel.till the concentration becomes ‘C’ everywhere.

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This movement of gas (matter) from a region of high concentration toregion of low concentration till the concentration becomes uniform (equal)every where is called as diffusion.

It is a natural process in the case of gases i.e., it occurs on its own.

If a gas is enclosed in a container with a small pin hole. Then it escapesout due to the process of diffusion. So, when diffusion is occuring throughsmall hole we will call it as effusion.

Rate at which gas is effusing out = Rate of collision of the gas with thewall of the container.

2.2.1 Graham’s law of diffusion

This rate of effusion was summarized by Graham’s law of effusion, which state that “rate of effusion is inversly proportional tothe square root of the molar mass”. Rate of effusion is also applicable to rate of diffusion.

When a gas at a pressure p and temperature T is separated from a vaccume by a small hole, the rate of escape of its moleculesis equal to the rate at which they strike the area (A

0) of the hole.

Rate of effusion = 0 A

0

pA N

2 M RTπwhen p, T and A

0 are constants.

Rate of effusion (r) = 0

constant

Mi.e.,

0

1r

M∝ This is Graham’s law of effusion

The rate of effusion of a gas can be measure in various ways.

moles effused outr =

time taken

∆=

n

t

pressure drop

time taken

∆=

p

t

volume change

time taken

∆=

v

t

distance travelled by the gas

time taken= =

l

t

Comparision of rate of effusion of two gases 1 2

2 1

=r M

r M(at consat T and p)

1 1 2

2 2 1

=r p M

r p M (at constant T only)

where 1p and 2p are partial pressures of the gases 1 and 2 respectively..

2.2.2 Importance of Graham’s Law

It forms the basis for separating the isotopes of same elements.

It provides a simple method for determining the densities and molecular weights of unknown gases by comparing their rates ofdiffusion or effusion with those of known gases. It is also useful in separating gases having different densities.

Illustration 10:

The pressure in a vessel that contained pure oxygen dropped from 2000 torr to 1500 torr in 55 minute as the oxygenleaked through a small hole into a vacuum. When the same vessel is filled with another gas, The pressure dropped from2000 torr to 1500 torr in 85 minute. What is the molecular mass of gas.



Solution :

2

2

Og

O (g)

Mr

r M= ⇒ 2

2

Og

g O (g)

tLoss in P 32

t Loss in P M× =

(g)

500 55 32

85 500 M× = ⇒ (g)M 76.43= g/mol.

Illustration 11:

A vessel contains 4 : 1 molar mixture of He and CH4 at a pressure of 20 bar. Due to a hole in the vessel, the gas mixture leaks

out. Find out the composition of the gas that leaks out initially.

Solution :

Molar ratio of He and CH4

= 4 : 1

Ratio of partial pressure of the gas in the vessel = 4 : 1

= 16 : 4

We know that

Rate of effusion of gas 1

M∝ and also Rate of effusion of gas∝ P

Thus4

4 4

CHHe He

CH CH He

Mr P

r P M= ×

16 16.

4 4=

328 :1

4= =

The composition of He and CH4 effusing out initially will be in the ratio of 8 : 1.

Illustration 12:

A tube with a porous wall allows 0.53 litre of N2 to escape per minute from a pressure of 1 atm to an evacuated chamber.

What will be the amount escaping under the same conditions for He, CCl4 vapour and UF

6?

(He = 4, N = 14, C = 12, Cl = 35.5, F = 19, U = 328)

Solution :

Using equation 9 for He,

2

2

NHe He

N He

MV V 28; 7

V M 0.53 4= = =

HeV 0.53 7 1.40= × = litre per minute

Similarly, for CCl4 vapour and UF

6,

4 2

4

2 4

CCl NCCl

N CCl

V M 28.V 0.227

V M 152= = = litre per minute

6 2

6

2 6

UF NUF

N UF

V M 28; V 0.149

V M 352= = = litre per minute

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Section - 2.3 Daltons Law

Daltons law states that the total pressure exerted by a gaseous mixture is equal to the sum of the partial pressures of eachindividual component in a gas mixture.

Mathematically

n

total i 1 2 3 ni 1

P p p p p .....p=

= = + + +∑

where 1 2 3 np , p , p ....p represent the partial pressure of each component

Where partial pressure is defined as the pressure that each gas would exert if it occupied the same container alone at the sametemperature partial pressure of a gas (p

i) can be obtained interms of mole fraction of the gas as

i i totalp X P=

i.e, partial pressure of a gas is mole fraction of that component multiplied by total pressure.

Illustration 13:

In a gaseous mixture 1.0 g H2 and 5.0 g He are mixed to a volume 5.0 L at 20°C. Calculate

(a) total pressure of the mixture (b) partial pressure of H2 and He.

Solution :

Total moles of gases 2H He(n) n n= +

total

1.0 5.0n 1.75

2 4= + = mole

(a) Using ideal gas equation PV nRT= . Given that, V 5.0L,= R 0.0821L= atm K–1 mol–1

totalT 293 K, n 1.75 mole= =

total

1.75 0.0821 293P

5.0

× ×= = 8.4 atm

(b) Now,

∴Hn 2

P ' P x PH total H total n2 2 total= × = ×

0.58.4 2.4

1.75= × = atm and

He

1.25P ' 8.4 6.0

1.75= × = atm



In-Chapter Exercise - 1

1. The ratio of speeds of diffusion of two gases A and B is 1 : 4. If the ratio of their mass present in the mixture is 2 : 3, thenwhich of the following is the ratio of their mole-fractions?

(a) 24 : 1 (b) 1 : 24 (c) 32 : 1 (d) 3 : 17

2. The vapour density of gas is 11.2. The volume occupied by 11.2 g of this gas at S.T.P. is :

(a) 22.4 L (b) 11.2 L (c) 1 L (d) 2.24 L

3. The density of the mixture of nitrogen and oxygen is 1.15 g/L at 750 mm of Hg at 27°C. Calculate the percentagecomposition of these gases in the mixture. Assume the gases behave ideally.

4. A cooking gas cylinder can withstand a pressure of 15 atm. At 27°C the pressure of the gs in cylinder is 12 atm. Theminimum temperature above which it will burst out is ___________________ °C.

Section - 2.4 Eudiometry

It is based on Gay Lussac’s law of combining Volumes. This method is used to analyze the gaseous mixture of hydrocarbons andto determine their molecular formulae. In this process the combustible gases, like hydrocarbon , are exploded in a tube with theexcess of O

2 so that carbon and hydrogen in the gas are converted to CO

2 (g) and H

2O (g) respectively. After cooling and

contraction ,the volume of contents of the tube are measured (this does not include H2O as it condense). At this stage ,the

content include CO2 (g) ,unused O

2 (if any left) and N

2 (if any in the air).

Now NaOH is used to separate out CO2 by the following reaction. It also absorbs Cl

2 .

2NaOH + CO2 → Na

2CO

3 +H

2O

As a result a further contraction in volume takes place. After this ,the unused O2 is left which generally absorbed by the

pyrogallol solution .In general after cooling ,the contraction in volume is given as :

V∆ = VVR– V

P (V

= Volume of reactant, V

P= Volume of products after cooling) various reagents used for absorption of gases-

Water Vapour - conc. H2SO

4, fused CaCl

2

O3 - turpentile oil

O2

- alkaline pyrogallol

NO - FeSO4 soln.

CO - ammonical cuprous chloride

CO2,SO

2- alkali solution

Cl2

- water/ alkali soln.

For the standard hydrocarbon combustion the following formula can help

Cx H

y +

x y y

4 4

+ +

CO

2= xCO

2 +

y

2

H2O

Illustration 14:

What volume of oxygen will be required for the complete combustion of 18.2 litres of propane at NTP?

Solution :

3 8 2 2 2C H 5O 3CO 4H O+ → +

At NTP we can assume V nα

Hence to combuse 18.2 litre propane we need = 18.2 × 5 = 91 L

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Illustration 15:

20 mL of CO was mixed with 50 mL of oxygen and the mixture was exploded. On cooling, the resulting mixture was shakenwith KOH. Find the volume of the gas that is left.

Solution :

2 2

1CO O CO

2+ →

At const. temperature and pressure, V nα ; 1 ml reacts with 1

2 ml of O

2

∴ 20 ml reacts with 10 ml O2; KOH absorbs 20 ml formed.

Hence 40 ml O2 left after the reactions

Section - 2.5 Kinetic Theory of Gases

Kinetic theory (or kinetic-molecular theory of gases) is the theory that all matter is made up of a large number of small particles(atoms or molecules), all of which are in constant, random motion.

This theory is the proof of ideal gas equation. It is based on certain postulates (assumptions)

(i) The gas consists of very small particles, all with non-zero mass(ii) The number of molecules are large.(iii) These molecules are in constant, random motion. The rapidly moving particles constantly collide with each other and

with the walls of the container.(iv) The collisions of gas particles with the walls of the container are perfectly elastic.(v) Except during collisions the interactions among molecules are negligible.(vi) The total volume of individual gas molecules added up is negligible compared to the volume of the container.(vii) Molecules are perfectly spherical in shape, and elastic in nature(viii) Average kinetic energy of the gas particles depends only on the temperature of the system

K.E T∝

(ix) The time during collision of molecule with the wall is negligible as compared to the time between successive collisions.

2.5.1 Kinetic Gas Equation

Based on the above assumustions it derives an equation called as kinetic gas equation

i.e.,21

PV mNC3

=

where P and V are the pressure at which the gas is stored and, the volume of the container in which gas is storedrespectively.

m → mass of one gas particle

N → no. of gas particles

C → R.M.S. velocity of gas particles.

Ideal gas equation → BPV NK T=

Kinetic gas equation → 21

PV mNC3

=



Comparing both of them we get 2B

1NK T mNC

3= ; 2

B3K T mC=

2B

3 1K T mC

2 2= = average translational kinetic energy of a molecule.

∴ average translational kinetic energy of a molecule (particle) = B

3K T

2

average translational kinetic energy of N molecules (particles) = B

3NK T

2

average translational kinetic energy of 1 mole is = A B

3 3N K T RT

2 2=

average translational kinetic energy of n moles is = 3

nRT2

.

Illustration 16:

Calculate the pressure exerted by 1023 gas molecules, each of mass 10–22 g in a container of volume one litre. The rmsspeed is 105 cm sec–1.

Solution :

According to kinetic gas equation 2rms

1PV mnu

3= . Given than 23 22 3 3n 10 , m 10 g, V 1 L 10 cm−= = = =

5 1rmsu 10 cm sec−=

∴3 22 23 5 21

P 10 10 10 (10 )3

−× = × × ∴ 7P 3.3 10= × Dyne cm–2.

Section - 2.6 Maxwell’s Law of Distribution of Molecular Speeds

Maxwell and boltzmann theoretically predicted the shape of distribution curves of molecular speeds in a given sample of gas ata given temperature. They assumed that molecular collisions are completely random and involve all possible values of molecularspeeds as a result of exchange of kinetic energy in between molecules. Note that it is only the kinetic energy of a gas (notindividual kinetic energy of molecule) that remains constant with temperature. It is thus evident that even a single moleculecannot have a constant speed for any duration of time because of continuous interchange of momentum between the moleculesdue to frequent collisions. Infact the speed of molecules changes within a span of less than 10–9 seconds and therefore it isdifficult to know the exact speed of molecules. However, Maxwell predicted probability distribution curves by assuming that fora given mass of a gas at a constant temperature, there must be a certain fraction of the total number of molecules, that have aparticular speed. On the basis of law of probability (Maxwell proposed an equation for the distribution of molecular speeds) as:

Distribution curves for molecular speeds

Speed umpuavurms

Frac

tion

of M

olec

ule

T1 < T2

T2

T1

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The curve at any temperature is parabolic near the origin,since the factore u2 is dominate in this region ,the exponential functionbeing approximately equal to unity.At high value of u ,however ,the exponential factore dominates the behaviour of the function,causing it to decrease rapidely in value.

As per Maxwell’s speed distribution :

23 bc 2CN 4 Na e c c−∆ = π ∆ ...(i)

where CN∆ = No. of molecules having speeds between c and c + c∆

B B

m ma ; b

2 k T 2k T= =

π

m = Mass 1 molecule of the gas; kB = Boltzmann constnat

23R1.38 10 J / K

N°

≡ = ×

N = Total number of gas molecules.

The fraction of molecules having either very low speeds or very high speeds are small in number.The majority of molecules havespeeds which cluster around most probable speeds in the middle of the range of u.Since the total number of molecules is thesame at both temperature both the curve will have the same area.Since increase of temperature increases the kinetic energy ofmolecules ,it follows that the fraction of molecules having lower speed range decreases whereas the fraction of moleculeshaving higher speed range increases on increasing the temperature of the gas.

The speed distribution also depends on the mass of the molecules .At the same temperature a heavy gas has a narrower

distribution of speeds than a light gas.In general the disribution depends upon the value of M

T.Thus the distribution will be the

same for a gas of molar mass 2M at temperature 2T since the ratio remains the same.for example, the distribution of O2 molecules

at temperature T will be the same as those of SO2 molecules at temperature 2T.

The sharpness of maxima in curves decreases with increase in temperature which reveals that number of molecules having their

speeds in the vicinties of mpu increase.

Also mp

2RTu

M

=

Section - 2.7 Molecular Speeds

Three speeds terms were obtained by Maxwell’s distribution curves, which are :

(a) Average speed av(u ) (b) Most probable speeds mp(u ) (c) Root mean square speed rms(u )

2.7.1 Average Speed (uav

)

It is simply arithmetic mean of the various speeds of the molecules. Let n molecules are present in a gas sample and each one is

moving with speed 1 2 3 nu , u , u .... u , then 1 2 3 nav

u u u ...uu

n

+ + +=

From kinetic equation, av

8RTu

M

= π =

88=

πρBK TP

m

M = molar mass of the gas, m = mass of one gas molecule



2.7.2 Most probable speed (ump

)

It is the speed attained by majority of molecules at that temperature.

mp

2RTu

M

=

22= =

ρB

K TP

m

2.7.3 Root Mean square speed (urms

)

It is defined as the square root of the mean of squares of speeds of molecules at a given temperature. Let the individual speed

of molecules be 1 2 3u , u , u ,...un .

then2 2 21 2 3

rms

u u u ...u

n

+ + +=

Also it has been derived from kinetic equation, that

Brms

3K T3RT 3Pu

M m

= = = ρ

2.7.4 Relation between uav

, urms

and ump

mp av rms

2RT 8RT 3RTu : u : u : : : :

M M M π

:: 2 : (8 / ) : 3π

::1:1.128 :1.224

Illustration 17:

Calculate urms

, uav

, ump

of

(a) O2 at STP. (b) Ethane at 27°C and 720 mm of Hg pressure.

Solution :

(a) At STP T 273K=

∴ 2

7

rsm O

3RT 9 8.314 10 273(u )

M 32

× × × = =

= 4 14.61 10 cmsec−×

and 4 4 1mpu 4.61 10 0.816 3.76 10 cm sec−= × × = ×

4 4 1avu 4.61 10 0.9213 4.25 10 cm sec−= × × = ×

(b) 2 6

7

rms C H

3RT 3 8.314 10 300(u )

M 30

× × × = =

(∵ M = 30, T = 27 + 273 = 300 K)

4 1rmsu 4.99 10 cm sec−= ×

4 1MP

4 1av

u 4.07 10 cm sec

u 4.60 10 cmsec

−

−

= ×

= ×

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Section - 2.8 Real Gases

2.8.1 Deviations from Ideal Gas Behaviour

A gas which obeys the general gas equation (P V = n R T) at all temperatures and pressures is called an ideal gas. However, noneof the actually known gases called real gases obey this equation over the entire range of temperature and pressure. These gasesshow ideal behaviour at low pressures and high temperatures. As the pressure becomes high and temperature becomes low,more and more deviations are observed from general gas equation and other gas laws.

Amagat , Andrews and others made extensive study of the behaviour of various gases and concluded that real gases do notobey the general gas equation under all conditions of temperature and pressure.

A convenient way to study deviations of real gases from ideal behaviour is to plot a graph between(PV/nRT) and P. The quantity (PV / nRT) is called compressibility factor and is denoted by Z . For ideal gases , Z = 1 under allconditions.Compressibility factor is the ratio of actual molar volume of a gas to molar volume of it, if it were an ideal gas at that temperatureand pressure.

m

m

V , realZ=

V , ideal

From the figure we find that the curves obtained on plotting Z against pressure for different gases have the followingcharacteristics:

All curves approach the ideal value as the pressure approaches zero. Thus at low pressures, all gases behave as ideal gases.

At moderate pressures, Z < 1 i.e., there is negative deviation. This means that the attractive forces are dominating repulsiveforces and the gas is more compressible than expected from ideal behaviour.

At high pressures , Z > 1 , i.e., there is a positive deviation. It means that the repulsive forces and dominating and the gas is lesscompressible than expected from ideal behaviour.

For H2 and He, Z is always greater than one. This means that these gases are less compressible than expected from ideal

behaviour at all pressures because there gases have ineligible charge separation at moderate pressure.

2.8.2 Causes of Deviation from Ideal Behaviour

The causes of deviations from ideal behaviour may be attributed to the two faulty assumptions of kinetic theory of gases. Theassumptions are :

• The volume occupied by the gas molecules is negligibly small as compared to the total volume occupied by the gas.

• The forces of attraction between gas molecules are negligible.

The kinetic theory assumes that the volume occupied by molecules of a gas is negligible as compared to the total volume of thegas. This assumption is quite valid high temperature and low pressure. For example, it has been calculated that the volume of themolecule is about 0.1% of the total volume of the gas under ordinary conditions of temperature and pressure. However as thetemperature is considerably decreased or pressure is considerably increased the total volume of the gas decreases appreciablywhereas the volume occupied by molecules remain the same because the molecules are incompressible. Therefore the fractionof the volume of the molecules cannot be neglected under these conditions. This proves the invalidity of this postulate at highpressure and low temperature.



According to kinetic theory, the forces of attraction between the gas molecules are negligible. This assumption is quite valid atlow pressure and high temperature because , the molecules lie apart from one another. On the other hand at high pressure andlow temperature, the volumes of the gas is quite small. Consequently, the distance between the molecules decreases andattractive forces, though very small, exist between them. The direct proof of the presence of intermolecular forces is that thetemperature of the gas falls when compressed gas is allowed to expand under adiabatic conditions. This fall in temperature is dueto the fact that some work has to be done to overcome the attractive forces between molecules during expansion. Thus, thispostulate is also invalid at low temperature and high pressure.

2.8.3 Van der Waal’s Equation

To account for the behaviour of real gases it is necessary to apply suitable corrections to the ideal gas equation, P V = n R T, soas to make it applicable to real gases. The ideal gas equation was modified by van der Waal who incorporated the idea of finitemolecular volume and intermolecular forces. The modified equation of the state which explains the deviations of real gases fromideality is known as Van der Waal’s Equation of state. He applied the following corrections to the ideal gas equation.

Volume correction

The volume occupied by the molecules cannot be neglected in comparison to total volume. This means that the moleculesare not free to move in the whole volume V, but the free volume is less than the observed volume (volume of thecontainer). In other words, the ideal volume of the gas is less than the observed volume. It was therefore suggested byvan der Waal that a suitable correction term may be subtracted from the observed volume. The correction term for onemole of gas is ‘b’ where ‘b’ is constant depending upon the nature of the gas and independent on temerature and isknown as ‘ excluded volume’ or co-volume . The excluded volume is the volume within which the molecules cannot move.The co-volume has found to be four times the actual volume of the molecules. Thus, if V be the other volume occupied by1 mole of the gas, then the volume available to the molecules for the movement i.e. the corrected volume is :

Vcorrected

= ( V – b ) for one mole.

Vcorrected

= ( V – n b ) for n moles of the gas

Magnitude of ‘b’

The repulsive interaction between molecules are taken into account by supposing that they cause the molecules tobehave as small but impenetrable spheres. When repulsive forces are dominate the closest distance of two hard sphere

molecules of radius r , and volume Vmolecule

= 34

πr3

,is 2r so the volume excluded is 34π(2r)

3 or 8V

molecule.The excluded

volume per molecule is half of this volume ,or 4Vmolecule

.b = 4 VN

A= Excluded Volume of one mole of molecules

Excluded volume is equal to four times the actual volume possessed by one mole of molecules.

Illustration 18:

Calculate molecular diameter of He if b (Van der Waals’ constant for volumecorrection) for He is 24 mL mol–1

Solution :

b 4= × Volume occupied by the molecules in one mole of gas 3424 4 N r

3

= × × π

∴8

23

3 24r 1.355 10 cm

2216 6.023 10

7

−

× = = ×

× × ×

∴ Diameter 82 r 2 1.355 10 cm 2.71Å−= × = × × = .

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Pressure Correction

Some attractive forces do exist between the molecules of real gases . Consider a molecule of a gas in the middle of thecontainer. The molecule being attracted uniformly by neighbouring molecules. The attractive forces neutralise oneanother and there is no resultant pull on the molecule. On the other hand , when a molecule approaches the wall of thecontainer, it experiences an inward pull as a result of attractive forces exerted by the neighbouring molecules inside thevessel. Thus, the molecule strikes the wall with lesser force than it would have done if there were no attractive forces.Therefore, the observed pressure is less than the ideal pressure. Consequently, some correction term should be added tothe observed pressure in order to calculate ideal pressure.

Hence, corrected pressure, P corrected

is given by : Pcorrected

= P + p where P is the observed pressure and p is the correctionterm.

Inward attactive force on the gas molecules near the wall is prportional to the concentration and density, hence

i.e.,2

2

np α

Vtherefore p = a n2 / V2 where a is a constant.

Substituting the values of the corrected pressure and volume in an ideal gas equation, P V = n R T we get,

∴ P corrected = 2

2

+

n aP

v

⇒2

2

n aP (V nb) nRT

V

+ − =

Ven der waal’s equation

The constants ‘a’ and ‘b’ are also called van der Waal’s constants for pressure correction and volume correction,respectively.

Significance and units of ‘a’ and ‘b’

The constants ‘a’ and ‘b’ are called Van der Waal’s constants and their values depend upon the nature of the gas. Theconstant ‘a’ measures the forces of attraction between the molecules of a gas. Greater the value of ‘a’ greater is thestrength of van der Waal’s interactions. The value of ‘a’ for the gas with lower molecular mass such as H

2 and He is quite

low whereas the values of ‘a’ for gases with high molecular mass is relatively high. It may be noted that the greater thevalue of ‘a’ of a gas also reflects its greater tendency to be liquefied. Therefore, the gases having higher values of ‘a’ willbe easily liquefied. For example most of the aromatic compounds like benzene, phenols etc. exist is liquid state at normalconditions of T & P because of their higher molar mass.

The constant ‘b’ relates to incompressible volume of molecules and measures the effective size of gas molecules. It maybe noted that ‘b’ is not equal to actual volume of the molecule but it is four times the actual volume of molecules.

Units of ‘a’ and ‘b’

We know : P = a n2 / V2

ora = P V2 / n2 = 2

2

(pressure ) (volume )

( moles)

Therefore the units of ‘a’ are atm L2 moL–2

Similarly, V = n b or b = (V / n)

Therefore , the units of ‘b’ are L mol–1

2.8.4 Application of Van der Waal’s Equation

Exceptional Behaviour of H2 and He

The behaviour of H2 and He is exceptional because the compressibility factor always increases with increase in pressure.

This is due to the fact that ‘a’ for hydrogen and helium are very small indicating that forces of attraction in these gasesare very weak. Therefore, (an2 / V2) is negligible at all pressures so that Z is always greater than one.



In their case we can take ‘a’ ≈ o

∴ Van der Waal’s equation for them becomes

mP(V b) RT− = or mPV PbZ 1

RT RT= = +

Illustration 19:

Calculate the temperature of gas if it obeys van der Waals’ equation from the following data :

A flask of 25 L contains 10 mole of a gas under 50 atm. Given a = 5.46 atm L2 mol–2 and b = 0.031 L mol–1.

Solution :

According to van der Waals’ equation for n moles of gas

2

2

n aP [V nb] nRT

V

+ − =

P 50 atm= , V 25 L, n 10= =

2 2 1a 5.46atm L mol , b 0.031 L mol− −= =

100 5.4650 [25 10 0.031] 10 0.0821 T

625

× + − × = × ×

∴ T 1529.93K 1256.93 C= = °

Van der Waal’s equation in low or moderate pressure

At moderate pressure volume is very high only attractive forces are effective because it is long ranged forces. Hencevolume correction can be neglected because it is effective only at high pressure.

∴ ( )m mV b V− ≈ ∴ ( )m2m

aP V RT

V

+ ≈

mm

aPV RT

V+ ≈ ⇒ m

m

aPV RT

V≈ −

m

m

PV a1

RT V RT≈ − i.e.

m

aZ 1

V RT≈ −

Van der Waal’s equation in high pressure

At high pressure molecules are close enough hence repulsive forces dominates the attractive forces.Hence a can be

neglected.

2m

aP P

v

+ ≈

⇒ ( )mP V b RT− ≈

⇒ mPV RT Pb≈ + ⇒mPV Pb

1RT RT

≈ +

⇒Pb

Z » 1+RT

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Section - 2.9 Critical State

In chemistry, a critical point, also called a critical state, specifies the conditions (temperature, pressure and sometimes composition)at which a phase boundary ceases to exist. As the critical temperature is approached, the properties of the gas and liquid phasesapproach one another, resulting in only one phase at the critical point: a homogeneous supercritical fluid. The heat of vaporizationis zero at and beyond this critical point, so there is no distinction between the two phases. Above the critical temperature a liquidcannot be formed by an increase in pressure.At and above critical temperature the sample has a single phase that occupies theentire volume of the container such a phase is by definition a gas. The single phase that fills the entire volume when T is greaterthan T

C may be much denser than we normally consider typical of gases, and the name supercitical fluid is preferred. At critical

point transition from liquide to gaseous state (or vice versa) takes place and thus it is not possible to state whether the substanceis in gaseous form or in the liqude form.In fact both the states becomes indistinguishable at the critical point.The surface ofseparation disappeares.At this point the various physical properties such as density ,refractive index ,etc have identical valuesfor both the states.

Some important data about the critical point can be understood from the P-V diagram of real gases

The P-V diagram for a real gas looks as follows

Family of P-v isotherms for a pure component

V

E

F

G

P

Bubble Point

Dew Point

In this case temperature is being held constant and gas is undergoing an isothermal compression process. Starting at E (all-vapor condition), an increase in pressure will result in a significant reduction in volume since the gas phase is compressible. Ifwe keep compressing isothermally, we will end up at point F, where the gas will be saturated and the first droplet of liquid willappear. We have come to the two-phase condition, where liquid (L) and vapor (V) co-exist in equilibrium, for the first time,

While we keep on compressing by decreasing the volume (path F-G) the pressure of the system remains constant; this conditioncontinues until all the vapor has become liquid. Point G represents the last condition of liquid and vapor coexistence. Once wehave only liquid, if we keep on compressing we will observe a rapid increase in pressure, as indicated by the steep slope in theP-V diagram. This is because liquid is virtually incompressible.

Family of P-v isotherms for a pure component

V

P1

P2

P3

P4

P5

PC

P

Critical Point

Critical Isotherm (T = TC)

T5

T4

T3

T2

T2



Point F, where only a tiny quantity of liquid exists in an otherwise completely gaseous system is the dew point of the system atthe given temperature. Similarly, Point G is the bubble point; where only an infinitesimally small bubble of vapor exists in anotherwise liquid system.

Now, if we want to generate all the possible points that make up the vapor pressure curve We would end up with a family ofisotherms (each similar to the one presented in the above figure. This is represented in the following Figure.

P-v Diagram And Phase Envelope Of A Pure Substance

V

P

Bubble Point Line

Liquid Region

Critical Point

Vapor Region

Dew Point Line

Two-Phase Region

The zone where the isotherms become flat delineates the two-phase region.

If we now draw a line through all the bubble points in the above Figure, and then draw a line connecting all the dew points, wewill end up with the bubble point curve and the dew point curve, respectively. It is clear that the two curves meet at the criticalpoint (P

c, T

c).

If we carefully follow the trend of the critical isotherm we will realize that it has a point of inflexion (change of curvature) at thecritical point. Furthermore, the critical point also represents the maximum point (apex) of the P-V envelope. Mathematically, thisinformation is conveyed by the expressions:

2

2, ,

0 ∂ ∂

= = ∂ � �

C C C CBP T P T

P P

V dV

which are usually known as the criticality conditions. These conditions are always satisfied at the critical point.

Van der Waals equation in cubic form:

For one mol, ( )m2m

aP V b RT

V

+ − =

, opening the brackets m 2

m m

a abPV Pb RT

V V− + − =

Multiplying through out by 2mV

3 2 2m m m mPV PbV aV ab RTV 0− + − − =

Dividing through out by P

3 2 2mm m m

aV ab RTV bV V 0

P P P− + − − =

3 2 mm m

aVRT abV b V 0

P P P

− + + − =

At critical point the equation becomes a perfect cube.

( )3

m cV -V =0 3 2 2 3m m c m c cV –3V V +3V W –V =0

comparing coefficients of similar terms in both the equations coefficient of 3mV 1=1→

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of 2 3→ = +m c

RTV V b

P

coefficient of 23→ =m c

aV V

P

constant term 3→ =c

abV

P

Solving for Pc, V

c and T

c we get

3=c

V b

⇒ 227=

c

aP

b⇒

827

=c

aT

Rb

which is the cribical point at critical point the compressibility factor (Z) for any real gas will be 3/8 i.e., less than 1.

Illustration 20:

The critcal constant for water are 374°C, 218 atm and 0.0566 L mol–1. Calculate a, b and R.

Solution :

Given that, cT 374 C 647K= ° = , 1c cP 218 atm, V 0.0566 L mol−= =

∵1CV 0.0566

b 0.0189 L mol3 3

−= = =

2 2C Ca 3P V 3 218 (0.0566)= = × × = 2.095 L2 atm mol–1

1 1C C

C

P V8 8 218 0.0566R 0.0508L atm K mol

3 T 3 647− −× ×

= = =×

Van der Waal’s equation in virial form

Any equation which can be written inm

2 3m m m

PV B C DA ......

RT V V V= + + + + ∞

orPVm = RT(1 + Bp + Cp2 +........) are called as its virial form

where 1st virial coefficient = A

2nd virial coefficient = B

3rd virial coefficent = C and so on. These coefficients are temperature dependent.The first virial coefficient is 1 ,the thirdcoefficient C is less important than the second coefficient B. We can use the virial equation to demonstrate the importantpoint that although the equation of state of a real gas may concide the perfect gas law as p approaches zero,not all theproperties necessarily coincide with those of a perfect gas in that limit.Consider fo example the value of dZ/dp. The slopeof the graph of compression factore against pressure.for a perfect gas dZ/dp = 0 (becaose Z =1 at all pressure), but for areal gas from above equation we get

dZ/dp = B + apC +....... approaches B as we decreases pressure.

however B is not necessarily approached with the perfect gas values at low pressure.

Because the virial coefficients are temperature dependent ,there may be a temperature at which Z approaches 1with zeroslope at low pressure or high molar volume.At this temperature ,which is called the Boyl temprature T

B ,the properties of

the real gas do coincide with of a perfect gas as pressure approaches zero over a more extended range of pressure thanat other temperature because the first term after 1 in the virial equation is zero and remaining terms are negligible. Forhelium gas T

B is 22.64 K, for air it is 346.8 K. At Boyle temperature B = 0.



Van der Waal’s equation is

( )m2m

aP V b RT

V

+ − =

⇒

2mm

a RTP

V bV

+ =

−

2m m

RT aP

V b V= −

−⇒ m

mm m

RTV aPV

V b V= −

−

m m

m m

PV V a

RT V b V= −

−⇒ m m

m m

PV V a

RT V b V RT= −

−

m

m

m

PV 1 abRT V RT1

V

= − −

⇒ 2 3 411 x x x x ...

1 x

= + + + + + ∞ −

Using the above mathematical result we can write

2 3 4

2 3 4m m m m

m

1 b b b b1 ......

b V V V V1V

= + + + + + ∞

−

Substituing the above equation in m

m

m

PV 1 abRT V RT1

V

= −

−

we can write 2 3

m2 3

m m m

PV b b b1 .....

RT V V V= + + + + ∞

m

a

V RT−

Collecting both the m

1

V terms we can write

2 3m

2 3m m m

PV a 1 b b1 b ....

RT RT V V V

= + − + + + ∞

Comparing it with the actual virial equation. We can see that the second virial coefficient is a

bRT

−

, and this becomes

zero at boyles temperature

∴B

ab 0

RT− = or B

aT

Rb=

As we have discussed that at low pressure dZ/dp = B = a

bRT

−

which is the initial slope of the plot of Z versus p.

Now if b is greater than a/RT, the initial slope is positive and the size effect (b-factore) will dominate the behavior of thegas.However if b is less than a/RT ,the initial slope is negative and the effect of attractive forces will dominate.Thus theVan der Waals equations ,which includes both the effect of size and of intermolecular forces, can interpret both thepositive and negative slopes of the Z versus p plots.If the temperature is low enough, the term a/RT will be larger thanb and so the initial slope of Z versus p is negative. As the temperature rises, a/RT becomes smaller and at a suffcient hightemperature it becomes less than b, and the initial slpoe of Z versus p turns positive.

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Compressibility Factor of Air

Temperature range 250 – 1000°C

0

50

100

150

200

250

300

350

400

450

500

Pressure, Bars

0.90

0.95

1.00

1.05

1.10

1.15

1.20

1.25

1.30

Z

250 °K 300 °K 350 °K 400 °K 450 °K 500 °K 600 °K 800 °K 1000 °K

N2

CH4

He

All at 300 K

Pressure, bar

200K 300K 600K

300 0

0.0

0.5

1.0

1.5

2.0

Z

CH4 at 3 temperatures

600

600 300 0

0.0

0.5

1.0

1.5

2.0

Z

Pressure, bar

Van der Waal’s equation in reduced form

An important general technique in science for comparing the properties of objects is to choose a related fundamentalproperty of the same kind and to set up a relative scale on that basis.We have seen that the critical constants arecharectristic properties of gases, so it may be that a scale can be set up by using them as yardsticks.We thereforeintroduce the dimensionless reduced variables of a gas by dividing the actual variable by the corresponding criticalconstant.

If the reduced pressure of a gas is given , we can easily calculate its actual pressure by using p = pr × p

c and like wise

for the volume and temperature. Van der Waals, who first tried this procedure ,hoped that gases confined to the samereduced volume at the same reduced temperature would exert the same reduced pressure. The hope was largelyfulfilled. The observation that real gases at the same reduced volume and reduced temperature exert the same reducedpressure is called principle of corresponding states.The principle is only an approximation .It works best for gasescomposed of spherical molecules; it fails, sometimes badly, when the molecules are non - spherical or polar.

i.e., Reduced variable Actual variable

Critical constant=

mr r r

C C

VP TP V T

P V T= = = or r c m r c r CP P P V V V T T T= = =

Putting the three values of P, V and T in Van der Waals equation for 1 mol of gas

( )m2m

aP V b RT

V

+ − =

⇒ ( )r C r C r C2 2r C

aP P V V b RT T

V V+ − =

By putting the values of C C2

aP V 3b

27b= = and C

8aT

27Rb= in the above equation

( )r r r2 2r

a a aP 3V b 1 8T

27Rb27b V 9b

+ − = ×

⇒ rv r2

r

8T3 1P V

3 3V

+ − =

This is the reduced form of gas equation one notable feature of this equation is, it is independence of ‘a’ and ‘b’ unlikeVan der Waal’s equation.i.e., if isotherm’s are plotted in terms of the reduced variables same curves are obtained, whatever may be the gas. Thisis precisely the content of the principle of corresponding states ,so the van der waals equationis compatible with it.



The importance of the principle is then not so much its theoretical interpretation but the way that it enables theproperties of a range of gasesto be coordinated on to a single diagram, for example graph

2.9.1 Effect of temperature Ven der Wals Gas Equation

1= −m

aZ

V RTor 1= +

PbZ

RT

as T → ∞ (high temperature) as → ∞T (high temperature)

→ ∞m

a

V RT0→

b

RT

∴ 1=Z ∴ 1=Zi.e., real gas apporaches ideal gas from both sides similarly we can check for low pressure condition also.

1= −m

aZ

V RT1= +

PbZ

RT

as 0→P as 0→P

→ ∞mV 1→Z

0→m

a

V RT 1=Z

In-Chapter Exercise - 2

*1. NH3 gas is liquefied more easily than N

2, thereby

(a) Van der Waal’s constant a and b of NH3 is higher than that of N

2

(b) Van der Waal’s constant a and b of NH3 is less than that of N

2

(c) ‘a’ of NH3 > ‘a’ of N

2, but ‘b’ of NH

3 < ‘b’ of N

2(d) ‘a’ of NH

3 < ‘a’ of N

2, but ‘b’ of NH

3 > ‘b’ of N

2

*2. A mixture of O2 and He kept in a container at 27°C. Which of the following statements is/are correct?

(a) O2 molecule will hit the wall of the container with smaller average speed as compared to He

(b) O2 molecules will hit the wall of the container with greater average speed as compared to He

(c) O2 molecule will hit the wall of the container with the greater kinetic energy as compared to He

(d) O2 molecule will hit the wall of the container with equal kinetic energy as compared to He

3. The average speed of an ideal gas molecule at 27°C is 0.3 ms–1. Calculate average speed at 927°C?

(a) 0.6 ms–1 (b) 6 ms–1 (c) 60 ms–1 (d) 8 ms–1

4. The temperature of an ideal gas is increased from 140 K to 560 K. If at 140 K the root mean square velocity of the gasmolecules is V, at 560 K it becomes

(a) 5 V (b) 2 V (c) V/2 (d) V/4

5. The density of oxygen at S.T.P. is 0.00143 g/mL. Find the rms velocity, average velocity and most probable velocity.

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Solved Examples

ciseExample 1:

1 litre flask contains Nitrogen, water vapour and a small amount of liquid water at a pressure of 200 mm Hg. If this isconnected to another 1 litre evacuated flask, what will be the final pressure of gas mixture at equilibrium? Assume thetemperature to be 323 K. Aqueous tension of water at 323K is 93 mm Hg.

Solution :

Initial pressure due to water vapour in the tube = 93 mm

Initial pressure due to nitrogen in the tube = 200 – 93 mm = 107 mm

Final pressure of the gas is

1 1 2 2(Initial) (Final)

P V P V=

1 12

2

P V 107 1P 53.5m m

V 2

×= = =

Final pressure of water vapour = 93 mm (Since liquid water will vapourise and vapour pressure of water will remainconstant, so long temperature is constant). Thus total pressure of the mixture = 53.5 + 93 = 146.5 mm

Example 2:

A bulb of unknown volume V contains an ideal gas at a pressure of 1 atm. This bulb was connected to anotherevacuated bulb of 0.5 litre volume through a stop cock. When the stop cock is opened the pressure at each bulbbecame 530 mm of mercury. Find the value of V. Assume that the temperature remains constant.

Solution :

Initial volume of the gas = V litre

Final volume of the gas = V + 0.5 litres

Initial pressure of the gas = 1 atm = 760 mm

Final pressure of the gas = 530 mm

Since temperature is constant 1 1 2 2(Initial) (Final)

P V P V=

Thus 760 × V = 530 × (V + 0.5)

760V 0.5 V

530+ = ×

V + 0.5 = 1.434V

0.434 V = 0.5

V = 1.152 litres

Example 3:

When 4 g of gaseous substance X is introduced into an initially evacuated flask kept at 25°C, the pressure is found tobe 1 atm. The flask is evacuated and 6 g of y is introduced. The pressure is found to be 0.5 atm at 25°C. Calculate the ratio

of x

y

M

M (M - molecular weight).



Solution :

mPV RT

M=

yx x

y x y

MP m

P M m= × (∵V, R and T constant)

y

x

M1 4

0.5 M 6= ×

x

y

M 4 0.5 1

M 6 3

×= =

Example 4:

A mixture of gases consists of 20% N, 30% O2and 50% He (all V/V) at 760 mm of Hg. What is the partial pressure of each

gas?

Solution :

Say total volume is 100 litres

Total pressure is 760 mm

Since the gases are kept at same pressure and same temperature

Volume percent = mole percent (Avogadro’s Law)

Further Partial pressure of a gas = Total numberof moles

No. of moles of the gas × Total pressure

Thus2N

20P 760 152 mm of Hg

100= × =

2O

30P 760 228mm of Hg

100= × =

He

50P 760 388mm of Hg

100= × =

Example 5:

The rate of effusion of two gases a and b under similar conditions of pressure and temperature are found to be in theratio of 2 : 1. Calculate the ratio of r. m.s. velocity of their molecules if Ta and Tb are in the ratio of 2 : 1.

Solution :

We know according to Graham’s Law of diffusion

a b

b a

r M 2

r M 1= = Further, rms

TU

Mα

Thusrms(a ) b a

rms(b) a b

U M T.U M T

= 2 2.1 1

= Thus Urms (a)

: Urms (b) = 2 2 :1

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Example 6:

0.15 amperes current is passed for 150 minutes in an electrolyte. The mass of metal deposited was 0.783 g. If atomic massof the metal is 112, calculate equivalent mass and valency of metal in the salt.

Solution :

Since, Q = I × t = 0.15 × 150 × 60 C = 1350 coulombs. Now 1350 coulombs of electricity deposit = 0.783 g metal

Example 7:

When a certain quantity of oxygen was ozonised in a suitable apparatus the volume decreased by 4 ml on addition ofturpentine the volume further decreased by 8 ml. All volume were measured at the same temperature and pressure. Fromthese data find out the formula of ozone.

Solution :

Let the formula of ozone is On. The initial volume of oxygen is ‘a’ ml.

2 nnO 2O

Initial volume a ml 0

Volume after the reaction (a nx) ml 2x ml−

��⇀↽��

Total volume of (ozone + oxygen) = (a – nx + 2x) ml

After the reaction according to condition

a – (a – nx + 2x) = 4 ...(i)

nx – 2x = 4

Since the turpentine oil absorbs the ozone gas, on addition of turpentine oil volume decrease by 8 ml it means volume ofozone in ozonised oxygen is 8 ml.

therefore, 2x = 8

x = 4 ml

on solving we get n = 3

∴ formula of ozone is O3

Example 8:

10 ml of a mixture of CH4, C

2H

4 and CO

2 was exploded with excess of air. After explosion there was a contraction of

17 ml and after treatment with KOH there was a further reduction of 14 ml. What was the composition of the mixture?(Volume measurement refer to the same P-T conditions)

Solution :

In the explosion the reactant CO2 does not change while CH

4 and C

2H

4 changes to CO

2 and H

2O.

Let the volume of CH4 = x ml. Volume of C

2H

4 = y ml

∴ volume of CO2 = (10 – x – y) ml

4 2 2 2

2 4 2 2 2

CH (g) + 2O (g) CO (g) + 2H O(g)

1ml 2ml 1ml 2ml

x ml 2x ml x ml 2x ml

C H (g) 3O (g) 2CO (g) 2H O(g)

1ml 3ml 2ml 2ml

y ml 3y ml 2y ml 2y ml

→

+ → +



Since after the explosion there was contraction 17 ml. It means when the explosion takes place at that time water is ingaseous form after the explosion water vapours converted into liquid form and the volume of liquid water assumed tobe zero.

∴ 2x + 2y = 17 ...(i)

Since KOH absorbs CO2 gas when the gaseous mixture is treated with KOH further reduction in volume is 14 ml,

therefore volume of CO2 in gaseous mixture is 14 ml.

10 – x – y + x + 2y = 14 ...(ii)

10 + y = 14

y = 4 ml

on putting y = 4 in equation (i) we get,

x = 4.5

Volume of CH4 = 4.5 ml

Volume of C2H

4 = 4.0 ml

Volume of CO2 = (10 – 4 – 4.5) = 1.5 ml

Example 9:

10 mL of a gaseous hydrocarbon was burnt completely in 80 mL of O2 at NTP. The remaining gas occupied 70 mL at NTP.

This volume became 50 mL on treatment with KOH solution.. What is the formula of the hydrocarbon?

Solution :

Let the hydrocarbon Be Cx H

y.

We know that the hydrocarbon on combustion gives CO2 and H

2O. As said in the beginning, since the volume occupied

by water is neglected, we have,

Volume of CO2 produced + unreacted O

2 = 70 mL and

Volume of unreacted O2 (CO

2 absorbed by KOH) = 50 mL.

∴ Volume of O2 reacted with 10 mL of hydrocarbon = (80 – 50) mL = 30 mL

and volume of CO2 produced by 10 Ml of hydrocarbon and 20 mL of O

2 = (70 – 50) mL = 20 mL.

Now the reaction is

Cx H

y+ O

2 → CO2

+ H2O

10 mL 30 mL 20 mL

or 10 moles 30 moles 20 moles

Applying POAC for O atoms, we get,

2 × moles of O2 = 2 × moles of CO

2 + 1 × moles of H

2O

2 × 30 = 2 × 20 + moles of H2O

∴ moles of H2O = 20

Applying POAC for C atoms, we get

x × moles of CxH

y = 1 × moles of CO

2

x × 10 = 1 × 20

∴ x = 2

Applying POAC for H atoms,

y × moles of CxH

y = 2 × moles of H

2O

y × 10 = 2 × 20

∴ y = 4

The formula of the hydrocarbon is C2H

4.

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Example 10:

16 mL of a hydrocarbon gas was exploded with excess of oxygen. On cooling, the volume of the resulting gaseous mixturewas reduced by 48 mL. When KOH was added, there was a further decrease of 48 mL in the volume. Find the molecularformula of the compound.

Solution :

Since water vapour condenses to practically zero volume of water, the decrease in volume on cooling is the volume ofwater vapour. CO

2 is absorbed by KOH and so volume of CO

2 is equal to 48 mL.

Thus,

CxH

y (Say) + O

2 → CO2 + H

2O

16 mL 48 mL 48 mL

or 16 moles 48 moles 48 moles

Since C and H are completely converted to CO2 and H

2O respectively, applying POAC for C and H atoms we get

respectively,

x × moles of CxH

y = 1 × moles of CO

2

16 x = 48. ∴ x = 3

y × moles of CxH

y = 2 × moles of H

2O

16 y = 2 × 48 ∴ y = 6

Hence, the formula of the hydrocarbon is C3H

6.



Exercises

Level - 1

Single Choice Questions

1. A gas is found to have a formula (CO)x. If its vapour denisty of 70, the value of x is

(a) 2 (b) 3 (c) 5 (d) 6

2. 4.4 g of CO2 contains how many litre of CO

2 at S.T.P.?

(a) 2.4 L (b) 2.24 L (c) 44 L (d) 22.4 L

3. The temperature of 20 litre of Nitrogen was increased from 100 K to 300 K at a constant pressure. The change in volumewill be

(a) 80 litre (b) 60 litre (c) 40 litre (d) 20 litre

4. At a constant pressure, what should be the percentage increase in the temperature in kelvin for a 10% decrease involume?

(a) 10% (b) 20% (c) 5% (d) 50%

5. Equal mass of methane and hydrogen are mixed in an empty container at 250C°. The fraction of total pressure exerted byhydrogen is

(a) 1

2 (b)

8

9(c)

1

9(d)

16

17

6. 32 g of oxygen and 3 g of hydrogen are mixed and kept in a vessel of 760 mm pressure and 0°C. The total volume occupiedby the mixture will be nearly.

(a) 22.4 litre (b) 11.2 litre (c) 44.8 litre (d) 56 litre7. A gas is found to have a density of 1.80 g/litre at 1 atm pressure and 27°C. The gas will be

(a) N2

(b) CO (c) CO2

(d) SO0

8. The ratio of the root mean square velocity of H2 at 50 K and that of O

2 at 800 K is

(a) 4 (b) 2 (c) 1 (d) 1/4

9. The average velocity of gas molecule is 400 m/s. The rms velocity at the same temperature will be

(a) 550 m/s (b) 434 m/s (c) 750 m/s (d) 350 m/s

10. The translational kinetic energy of 80 g methane at 127°C will be (R = 2 Cal K–1 mol–1)

(a) 6 kcal (b) 96 kcal (c) 1.905 kcal (d) 3.048 kcal

11. The ratio (a/b) (a and b being the van der Waal’s constants of real gases) has the dimensions of

(a) atm mol–1 (b) L mol–1 (c) atm L mol–1 (d) atm L mol–2

12. Which is not true in case of an ideal gas?

(a) It can be converted into a liquid at high pressure (b) There is no interaction between the molecules

(c) All molecules of the gas move with same speed

(d) At a given temperature PV is proportional to the amount of the gas.

13. Air contains 79% N2 and 21% O

2 by volume. If the barometric pressure is 750mm Hg the partial pressure of oxygen is

(a) 157.7mmofHg (b) 175.5mmofHg (3)315.0mmofHg (d) None

14. At what temperature will be total kinetic energy (KE) of 0.30 mole of He be the same as the total KE of 0.40 mole of Ar at400K

(a) 400K (b) 373 K (c) 533K (d) 300 K

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Multiple Choice Questions

15. The correct relation is

(a) C

8aT

27Rb= (b) C 2

aP

27b= (c) CV 3b= (d)

C C

C

P V 3RT 8

=

16. The correct statement(s) regarding compressibility factor (Z) is/are

(a) For ideal gas, Z = 1 (b) If Z > 1 then gases are more compressible

(c) If Z < 1 then gases are less compressible (d) For real gas, Z ≠ 1

17. Which is/are correct for molecular speed of gases?

(a) rms

2PVu

M= (b) rms avu u> (c) u u

mp av> (d) mp

8RTu

M=

π

18. Regarding H2 gas, the correct statement(s) is/are

(a) For H2 gas, Z > 1 at 273 K

(b) When H2 gas expands at above inversion temperature, then it shows heating effect

(c) The critical temperature of H2 gas is very high

(d) The value of vander Waal’s constant ‘a’ is very low for H2

Subjective Type Questions

19. 7.5 mL of a gaseous hydrocarbon was exploded with 36 mL of oxygen. The volume of gases on cooling was found to be 28.5mL, 15 mL of which was absorbed by KOH and the rest was absorbed in a solution of alkaline pyrogallol. If all volumes aremeasured under the same conditions, deduce the formula of the hydrocarbon.

20. 1 litre of a mixture of CO and CO2 is taken. This mixture is passed through a tube containing red hot charcoal. The volume

now becomes 1.6 litres. The volumes are measured under the same condition. Find the composition of the mixture byvolume.

Assertion and Reason Type

This section contains 2 questions. Each question contains Statement-1 (Assertion) and Statement-2 (Reason). Eachquestion has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct.

21. Satateemnt - 1 : A bottle of dry ammonia and a bottle of dry hydrogen chloride connected through a long tube are openedsimultaneously at both ends, the white ammonium chloride ring first formed will be near the hydrogen chloride bottle.

because

Satateemnt - 2 : Rate of diffusion is inversely proportional to molecular mass.

(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

(c) Statement-1 is True, Statement-2 is False

(d) Statement-1 is False, Statement-2 is True

Comprehension Type

The Van der Waal’s equation of state for 1 mole real gas is

2

aP (V b) RT

V

+ − =

The virial equation for 1 mole real gas is as follows:

2 3

x y zPV RT 1 ... to higher power on n

V V V

= + + +

where x, y and z are constants which are known as second, third and fourth virial coefficients respectively.

The temperature at which real gas obeys ideal gas equation i.e. (PV = nRT) is known as Boyle’s temperature.



Choose the correct answer :

22. The third virial coefficient of a He gas is 4 × 10–2 (litre/mole)2 then what will be volume of 2 mole He gas at NTP?

(a) 44.4 l (b) 44.6 l (c) 44.8 l (d) 45.2 l

23. If the critical temperature of the gas be C

8aT

27Rb= and BT is the Boyle’s temperature, then which of the following is the

correct relation between TC and T

B?

(a) C B

8T T

27= (b) C B

27T T

8= (c) C B

4T T

27= (d) C B

27T T

4=

24. Which of the following is the correct statement about the Boyle’s temperature (TB)?

(a) Temperature at which second virial coefficient becomes zero(b) Temperature at which first virial coefficient becomes zero

(c) The value of TB is equal to

aRb

(d) Both (1) & (3)

Matrix Match

25. Column - I Column - II

(A) Force of attraction is dominating (p) Z < 1

(B) Force of repulsion is dominating (q) Z > 1

(C) Volume of gas molecules is negligible (r) PV = RT + Pb

(D) Pressure of CH4 gas is low (s) PV = RT

a

V−


(A) Molar volume of gas (p) Temperature dependent

(B) Translational K.E. of gas molecules (q) Temperature independent

(C) Vapour density of gas (r) Pressure dependent(D) Density of a gas (s) Pressure independent

Level - 2

Single Choice Questions

1. The internal pressure of one mole of a Van der Waal gas is equal to

(a) zero (b) b2 (c) 2

a

v(d)

ab

RT−

2. Solid CO2 is called ‘dry ice’ because

(a) at 25°C and 1 atm, only solid and vapour phases of CO2 are in equailibrium

(b) the critical temperature of CO2 is above 25°C

(c) the boiling point of liquid CO2 is above 100°C

(d) the melting point of solid CO2 is 0°C

3. The triple point of water is at

(a) 273.16 K (b) 273.16 K and 760 torr (c) 273.16 K and 4.58 torr (d) 760 torr

4. The ratio of the temperature at which the average velocity of ethane (T1) is equal to the root mean square velocity of

benzene (T2) is

(a) 0.151 (b) 0.48 (c) 1.0209 (d) 0.389

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5. Two separate bulls contain ideal gases A and B respectively. The density of gas A is twice that of gas B and molecularweight of gas A is half of that of gas B. The ratio of pressure of gas A to that of gas B is

(a) 3 (b) 6 (c) 4 (d) 1

6. Compared to C2H

6, the value of van der Waal’s constants “a” and “b” for He will be

(a) both will be smaller (b) “a” will be larger but “b” will be smaller

(c) “b” will be larger but “a” will be smaller (d) both will be larger

7. The rmsv of a gas at 300 K is 30 R1/2. The molar mass of the gas,in kg mol–1, is

(a) 1.0 (b) 1.0 × 10–1 (c) 1.0 × 10–2 (d) 1.0 × 10–3

8. The triple point for water is

(a) unique (b) depends on p but is independent ot T

(c) depends on T but is independent of P (d) depends on both p and T.

9. As per the kinetic theory of ideal gases, which of the following statements is NOT correct?

(a) gas particles have mass but no volume (b) particles are in a Brownian motion between collisions

(c) during the collision, the system does not lose energy

(d) particles exert same force per unit area on all sides of the container

Multiple Choice Questions

10. Which of the following are the characteristics of a real gas?

(a) The molecules attract each other

(b) It obeys the ideal gas law at low temperature and high pressure

(c) The mass of molecule is negligible (d) It shows deviation from the ideal gas law

11. Which statement is correct regarding van der Waal’s constant ‘a’ and ‘b’?

(a) ‘a’ is the measure of force of attraction in between the particles

(b) ‘b’ is the excluded or co-volume of the gas

(c) Higher is the value of ‘a’, easier is the liquifaction of the gas

(d) Lower is the value of ‘b’, easier is the liquifaction of the gas

12. Which of the following graphs represents Boyle’s law correctly? (n1 and n

2 = number of moles)

(a) (b)

(c) (d)

13. Which of the following statement is incorrect ?

(a) At 273°C and 1 atm pressure the volume of a given mass of gas will be twice the volume at 0°C and 1 atm pressure

(b) At –136.5°C and 1 atm. pressure, the volume of a given mass of gas will be half of its volume at 0°C and 1 atm

(c) The mass ratio of equal volumes of NH3 and H

2S under similar conditions of temperature and pressure is 1 : 2

(d) The molar ratio of equal masses of CH4 and SO

2 is 1 : 4



Subjective Type Questions

14. The molar volume of helium at 10.1325 MPa and 273 K is 0.011 075 of its molar volume at 101.325 kPa at 273 K. Calculatethe radius of helium atom. The value of a may be neglected.

15. Caculate the pressure exerted by 2 mol of CO2(g) confined to 855 cm3 volume at 300 K using (a) ideal gas equation and (b)

van der Waals equation. Given: 3 12b(CO ) 42.8cm mol−= and 2

2a(CO ) 3.61L= atm mol–2.

16. The total pressure of a mixture of H2 and O

2 is 1.00 bar. The mixture is allowed to react to form water, which is completely

removed to leave only pure H2 at a pressure of 0.35 bar. Assuming ideal gas behaviour and that all pressure measurements

were made under the same temperature and volume conditions, calculate the composition of the original mixture.

17. A mixture of ethane (C2H

6) and ethene (C

2H

4) occupied 35.5 L at 1.0 bar and 405 K. The mixture reacted completely with

110.3 g of O2 to produce CO

2 and H

2O. What was the composition of the orgiginal mixture? Assume ideal gas behaviour.

Matrix Match

18. Column II gives the values of critical temperatures of a few gases which are shown in Column I. Identify the gas with thecorresponding T

C value

Column - I Column - II

(A) H2

(p) 33.3 K

(B) He (q) 304 K

(C) O2

(r) 5.3 K

(D) CO2

(s) 154 K

(t) 647 K


(A) Rate of diffusion (p) Directly proportional to pressure of gas

(B) Partial pressure of gas in closed vessel (q) Directly proportional to mole fraction

(C) Kinetic energy of gas (r) Inversely proportional to square root of molecular mass

(D) Average velocity (s) Increases with temperature

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Previous Years’ IITJEE Questions

1. If a gas is allowed to expand at constant temperature then [I.I.T.1986]

(a) Number, of molecules of the gas decreases (b) Tbe kinetic energy of gas molecules remains the same

(c) The,kinetic energy of gas molecules increases (d) The kinetic energy of gas molecules decreases

2. 3.2g of oxygen (At. wt. = 16) and 0.2 gm of hydrogen (At.wt = 1) are placed in a 1.12 litre flask at 0ºC. The total pressureof the gas mixture will be : [IIT 1993]

(a) 1 atm (b) 4 atm (c) 3 at m (d) 2 atm

3. Equal weights of ethane and hydrogen are mixed in an empty container at 25ºC. The fraction of the total pressure exertedby hydrogen is : [IIT 1993]

(a) 1 : 2 (b) 1 : 1 (c) 1 : 16 (d) 15 : 16

4. One mole of N2O

4(g) at 300 K is kept in a closed container under one atmospheric pressure. It is heated to 600 K when 20%

by mass of N2O

4(g) decomposes to NO

2(g). The resultant pressure is [IIT 1996]

(a) 1.2 atm (b) 2.4 atm (c) 2.0 atm (d) 1.0 atm

5. The ratio between the root mean square speed of H2 at 50 K and that of O

2 at 800 K is [IIT 1996]

(a) 4 (b) 2 (c) 1 (d) 1/4

6. The compressiblility of a gas is less than unity at STP therefore [IIT 2000]

(a) Vm > 22.4 litres (b) V

m < 22.4 litres (c) V

m = 22.4 litres (d) V

m = 44.8 litres

7. The root mean square velocity of an ideal gas at constant pressure varies with density (d) as [IIT 2001]

(a) d2 (b) d (c) d (d)1

d

8. Which of the following volume (V) - temperature (T) plots represents the behaviour of one mole of an ideal gas at oneatmospheric pressure? [IIT 2004]

(a) (b)

(c) (d)

9. X mL of H2 gas effuses through a hole in a container in 5 secons. The time taken for the effusion of the same volume of

the gas specified below under identical conditions is [IIT 1996]

(a) 10 second : He (b) 20 seconds : O2

(c) 25 seconds : CO (d) 55 seconds : CO2



Answer Key

In-Chapter Exercises - 1

1. (b) 2. (b) 3. % N2 = 82.5, % O

2 = 17.5 4. 102°C

In-Chapter Exercises - 2

1. (c) 2. (a, d) 3. (a) 4. (b)

5. Crms

= 460.13 m/s, Cavg

= 429.9 m/s, CMP

= 375.9 m/s

Level - 1

1. (c) 2. (b) 3. (c) 4. (a) 5. (b) 6. (d)

7. (c) 8. (c) 9. (b) 10. (a) 11. (c) 12. (c)

13. (a) 14. (c) 15. (a, b, c, d) 16. (a, d) 17. (a, b) 18. (a, b, d)

19. [Ans. C2H

4] 20. [Ans. 0.4 litre] 21. (c) 22. (d) 23. (a) 24. (d)

25. (A) → (p, s), (B) → (q, r), (C) → (p, s), (D) → (p, s)

26. (A) → (p, r), (B) → (p), (C) → (q, s), (D) → (p, r)

Level - 2

1. (c) 2. (a) 3. (c) 4. (b) 5. (c) 6. (c)

7. (a) 8. (d) 9. (b) 10. (a, d) 11. (a, b, c, d) 12. (c)

13. (a, b, d) 14. [Ans. 81.34 10 cm 134 pm−× =

15. [Ans. (a) 57.61 atm, (b) 44.27 atm] 16. [Ans. 0.22]

17. [ Ans. 1n 0.484mol= and 2n 0.57 mol= ]

18. (A) → (p); (B) → (r); (C) → (s); (D) → (q)

19. (A) → (p, q, r); (B) → (q, s); (C) → (s); (D) → (r, s), s)

Answer Key: Previous Years’ IITJEE Questions

1. (b) 2. (b) 3. (d) 4. (b) 5. (c) 6. (b)

7. (d) 8. (c) 9. (b)

Gaseous State

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