CG – 1
GASEOUS STATE
C1
The gaseous state of the matters is characterised by their following properties :
(1)
They have neither fixed volume not fixed shape. i.e., their volume and shape depends upon the
size and shape of the container.
(2)
They can expand indefinitely and uniformly to the shape available to them.
(3)
They may be compressed by the application of external pressure.
(4)
They can diffuse and mix with each other to form mixture of any composition
(5)
At high pressure and low temperature, they can be converted into their liquid state
(6)
They can exert pressure on the surrounding.
(7)
The properties of gases can be fully described in terms of pressure, temperature, volume and
amount of the gas.
All of these properties of the gaseous state are due to their very weak intermolecular force (nearly
negligible).
C2A GAS LAWS
The Laws which co-relates the variable while defining the behaviour of the gases are known as gas laws.
They are :
BOYLE’S LAW :
At constant temperature, the volume of definite mass of a gas is inversely proportional to its pressure.
Mathematically, it is expressed as
V  1/P or PV = constant
Graphical Representation of Boyle’s Law :
Isotherms : They refers the graphical variation of volume with pressure, when the temperature remains
constant.
CHARLE’S LAW :
At constant pressure for a fixed amount of a gas, the volume of a given gas is directly proportional to the
absolute temperature
V  T......
(temperature is in Kelvin),
V1 V2

T1 T2
Graphical Representation of Charle’s Law :
GAY LUSSAC’S LAW :
An expression similar to Charles law exists between pressure and temperature of the gas at a fixed volume.
This law can be expressed as
PT
Einstein Classes,
or
P/T = constant
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CG – 2
AVOGADRO’S HYPOTHESIS :
According to this equal volume of all the gases contain equal number of molecules, under similar conditions of temperature and pressure. It implies that
or
VN
(at constant temperature and pressure)
Vn
(where N is number of molecules and n is number of moles of
gas)
Dalton’s Law of Partial Pressure :
The total pressure exerted by a mixutre of non-reacting gases in a definite volume (closed container) is
always equal to the sum of the their individual pressures which each gas would exert if it occupies the same
space (volume) at a constant temperature.
If pA, pB, pC are individual pressures (partial pressures) of the non-reacting gases in the gaseous mixture and
PT is the pressure exerted by the mixture, then
PT = pA + pB + pC + .....
Also
pA = PTXA......
(XA is the mole fraction of A and PT is the total pressure of
the mixture)
Practice Problems :
1.
A flask contains 12 g of a gas of relative molecular mass 120 at a pressure of 100 atm was evacuated
by means of a pump until the pressure was 0.01 atm. Which of the following is the best estimate of
the number of molecules left in the flask (N0 = 6 × 1023 mol–1) :
(a)
2.
3.
(a)
remain unchanges
(c)
increase to four-fold
(c)
6 × 1017
(d)
6 × 1013
(b)
be doubled
(d)
be reduced to 1/4 th
Under constant pressure, a certain gas at 0 C was cooled until its volume was reduced to one-half.
The temperature at this stage is
0K
(b)
136.5 K
(c)
136.50C
(d)
–2730C
A gas of volume 100 ml is kept in a vessel at pressure 104 Pa maintained at temperature 240C. If now
the pressure is increased to 105 Pa, keeping the temperature constant, then the volume of the gas
becomes
10 ml
(b)
100 ml
(c)
1 ml
(d)
1000 ml
A vessel of 120 mL capacity contains a certain amount of gas at 350C and 1.2 bar pressure. The gas
is transferred to another vessel of volume 180mL at 350C. The pressure would be
(a)
6.
6 × 1018
0
(a)
5.
(b)
If the absolute temperature of a gas is doubled and the pressure is reduced to one-half, the volume of
the gas will
(a)
4.
6 × 1019
0.3 bar
(b)
0.8 bar
(c)
1.8 bar
(d)
1.3 bar
0
N2 + 3H2  2NH3. 1 mol N2 and 4 mol H2 are taken in 15 L flask at 27 C. After complete conversion
of N2 into NH3, 5 L of H2O is added. Pressure set up in the flask is :
(a)
3  0.0821 300
atm
15
(b)
2  0.0821 300
atm
10
(c)
1 0.0821 300
atm
15
(d)
1 0.0821 300
atm
10
[Answers : (1) b (2) c (3) b (4) a (5) b (6) d]
C2B
IDEAL GAS EQUATION :
The combination of Boyle’s law and Charless law leads to the expression
PV = nRT.
where n is the amount of gas and R is universal gas constant whose value in different units are given below.
Einstein Classes,
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CG – 3
–1
–1
–1
–1
–1
R = 0.0821 L atm K mol or 8.314 J K mol or 2 cal mol K
–1
Also, ideal gas equation can be expressed in terms of density (d) and vapour density (D) as follows :
PM = dRT, 2PD = dRT
Practice Problems :
1.
A gas in an open container is heated from 270C to 1270C. The fraction of the original amount of gas
escaped from the container will be :
(a)
2.
(b)
1/2
(c)
1/4
(d)
1/8
(c)
2730C, 1 atm
(d)
2730C,2 atm
Then density of neon will be highest at
(a)
3.
3/4
S.T.P.
00C, 2 atm
(b)
At 00C; the density of a gaseous oxide at 2 bar is same as that of nitrogen at 5 bar. The molecular
mass of the oxide is
(a)
40
(b)
50
(c)
60
(d)
70
[Answers : (1) c (2) b (3) d]
C3A DIFFUSION AND EFFUSION :
The diffusion is defined “as a process of intermixing of two or more gases, irrespective of their density
without the help of any external agency.”
The diffusion of gases takes place due to rapid movement of gaseous molecules and due to the presence of
large intermolecular space available to them.
If a gas is taken in a pot and is allowed to leak through a small hole is diffuses into the atmosphere and the
process is known as Effusion.
Hence, effusion of a gas refers the passage of gas through a tiny opening into a large space.
C3B
GRAHAM’S LAW OF DIFFUSION AND EFFUSION :
The rate of diffusion of a gas is inversely proportional to the square root of its density or molar mass.
Mathematically, it is expressed as
r
1

or 
1
M ..... (P & T constant)
Comparison of the rate of diffusion of two gases when the pressure is not constant, then
r2 P1

r1 P2
M1
......(T constant) M1 and M2 are molecular mass of two gases.
M2
(Rate of diffusion of gas)
rd =
V( volume of gas diffused)
t(time of diffusion)
rd 
A(area of cross  section of tube)  d
.....
t(time of diffusion)
(d is the distance
travelled by the gas till
the point of diffusion)
Practice Problems :
1.
2.
A bottle of dry ammonia and a bottle of dry hydrogen chloride connected through a long tube are
opened sumultaneously at both ends, the white ammonium chloride ring first formed will be
(a)
at the centre of the tube
(b)
near the hydrogen chloride bottle
(c)
near the ammonia bottle
(d)
throughout the length of the tube
Equal volumes of two gases A and B diffuse through a porous pot in 20 and 10 seconds respectively.
If the molar mass of A be 80, the molar mass of B is
(a)
20
Einstein Classes,
(b)
30
(c)
40
(d)
50
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CG – 4
3.
240 ml of a hydrocarbon diffuses through a porous membrane in 20 min while 120 ml of SO2 under
identical conditions diffuses in 20 min. The molecular weight of hydrocarbon is
(a)
4.
6.
(b)
46
(c)
30
(d)
16
20 dm of SO2 diffuse through a porous partition in 60s. The volume of O2 will diffuse under similar
conditions in 30s is
(a)
5.
44
3
14.14
(b)
25
(c)
0.707
(d)
0.25
X ml of H2 gas effuses through a hole in a container in 5 seconds. The time taken for the effusion of
the same volume of the gas specified below under identical condition is
(a)
10 seconds : He
(b)
20 seconds : O2
(c)
25 seconds : CO
(d)
55 seconds : CO2
One mole of nitrogen gas at 0.8 atm takes 38s to diffuse through a pinhole, whereas one mole of an
unknown compound of xenon with fluorine at 1.6 atm takes 57 s to diffuse through the same hole.
The molecular formula of the compound is [At. Mass Xe = 131.30, F = 19.0]
(a)
XeF4
(b)
XeF2
(c)
XeF6
(d)
none
[Answers : (1) b (2) a (3) d (4) a (5) b (6) c]
C4A KINETIC THEORY OF GASES :
Following are the different assumptions of kinetic theory which true only for the ideal gases :
C4B
(i)
The volume occupied by a molecule is negligible in comparison to the total volume.
(ii)
The molecules are in state of rapid, straight-line motion, colliding with each other. The collision
between them is perfectly elastic.
(iii)
The absolute temperature of gas is a measure of the average K.E. of all the molecules present in
it.
KINETIC GAS EQUATION :
PV 
1
mnC 2 ......
3
[P  Pressure of gas
V  Volume of gas
m  mass of a molecule
C  root mean square velocity]
For one mole of a gas PV = RT and n = NA (Avogadro’s Number)
PV 
Kinetic energy per mole =
1
1
mN A C 2 , RT  MC 2
3
3
3
3
RT , Average kinetic energy = kT
2
2
Practice Problems :
1.
At what temperature will hydrogen molecules have the same KE as nitrogen molecules at 280 K :
(a)
280 K
(b)
40 K
(c)
400 K
(d)
50 K
[Answers : (1) a]
C4C DIFFERENT TYPES OF VELOCITIES OF GASEOUS MOLECULES :
ROOT MEAN SQUARE SPEED :
u rms
 u 2  u 22  ....  u 2N
 1

N

Einstein Classes,
½

 and is given by the expression u
rms 


3RT
M
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CG – 5
AVERAGE SPEED :
u av 
u 1  u 2  ....  u N
, u av 
N
8RT
M
MOST PROBABLE SPEED :
u mp 
2RT
, Comparing the three speeds listed above, we find that u rms : u av : u mp :: 3 : 8 /  : 2
M
urms > uav > ump
Practice Problems :
1.
The average velocity of gas molecules is 400 m/sec. The rms velocity at the same temperature is
(a)
2.
(b)
300 m/sec
(c)
4.34 m/sec
(d)
3.00 m/sec
0
The average velocity of an ideal gas molecule at 27 C is 0.3 m/sec. The average velocity at 9270C will
be
(a)
3.
434 m/sec
0.6 m/sec
(b)
0.3 m/sec
(c)
0.9 m/sec
(d)
3.9 m/sec
The average speed at T1K and the most probable speed at T2K of CO2 gas is 9.0 × 104 cm/s. The
values of T1 and T2 are respectively
(a)
T1 = 1000 K, T2 = 2000 K
(b)
T1 = 1682 K, T2 = 2143.4 K
(c)
T1 = 1082 K, T2 = 2140.4 K
(d)
T1 = 3282 K, T2 = 4140.4 K
[Answers : (1) a (2) a (3) b]
C5A VAN DER WAAL’S EQUATION OF STATE :
Van der Waal pointed out that the deviations shown by real gases are due to the following two facts.
1.
The volume occupied by the molecules is not negligible in comparison to the total volume of the gas.
2.
There exist forces of attraction between the molecules.
Van der Waal systematically corrected the ideal gas equation in the light of the above two facts. The corrected equation is


 P  a ( Vm  b)  RT
2 

vm


Units of a and b :
Unit of a : We know that
P
an 2
V
(P = pressure connection), a 
2
PV 2
n2
so, that unit of a is atmL2mol–2.
C5B
Unit of b : It is the volume correction per mole of gas, so its unit is litres mol–1.
Significance of a and b :
(i)
The value of a is a measure of the intermolecular forces of attraction. An easily liquefiable has
has greater intermolecular foces of attraction. So, greater is the value of a for a gas, greater will
be the ease of its liquification. The easily liquefiable gases (SO2 > NH3 > H2S > CO2) have
higher values of a than the permanent gases like N2, H2, O2 and He. The constant a is dependent
on temperature.
(ii)
The value of b has constant value over a wide range of temperature and pressure, this indicates
that the gas molecules are incompressible.
VANDER WAAL’S EQUATION AT LOW PRESSURE :
Van der Waals equation accounts for the behaviour of real gases. At low pressures, the gas equation can be
written as


 P  a  Vm  RT
2 

vm


Einstein Classes,
or
Z
pVm
a
 1
RT
Vm RT
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CG – 6
where Z is known as compressibility factor.
C5C VANDER WAAL’S EQUATION AT HIGH PRESSURE :
At high pressure the real gas equation can be written as :
P(Vm – b) = RT, Z 
*
PVm
Pb
 1
RT
RT
At very low pressure and at very high temperature real gases behave ideally i.e., Z = 1.
Practice Problems :
1.
2.
3.
The compressibility factor of gas is less than unity at STP, therefore :
(a)
Vm (molar volume) > 22.4 L
(b)
Vm < 22.4 L
(c)
Vm = 22.4 L
(d)
Vm = 44.8 L
For the non-zero value of force of attraction between gas molecules, gas equation will be
(a)
n 2a
PV  nRT 
V
(b)
PV = nRT + nbP
(c)
PV = nRT
(d)
P
nRT
Vb
The compression factor (compressibility factor) for one mole of a van der Waals gas at 00C and 100
atmosphere pressure is found to be 0.5. Assuming that the volume of a gas molecular is negligible,
the van der Waals constant(a) is
(a)
30.55
(b)
1.253
(c)
25.55
(d)
none
[Answers : (1) b (2) a (3) d]
C6
*
Important Points for Gases (Real Gases)
All gases do not follow the ideal gas behaviour (PV = nRT) over a range of pressure and
temperature. They show the ideal behaviour at high temperature and very low pressure. But at high
pressure and low temperature they are not obeying the behaviour of ideal gases.
All gases condense at high pressure and sufficiently low temperature.
The deviation of real gas from ideal gas is measured by compressibility factor (Z)
Z
1.
2.
PVm
PV
V
or
(Vm = molar volume =
)
nRT
RT
n
Graphs are at 273 K
Conclusions from Graph :
Z = 1 for ideal gas (No force of attraction or repulsion)
All gases have 1 at very low pressure.
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CG – 7
3.
4.
5.
1.
Z > 1 means repulsive forces dominant and the gases are more difficult to compress. Z > 1 is at high
pressure.
At intermediate P, some of the gases have Z < 1 indicating that attractive forces are dominant and
favour compression.
For H2, Z > 1 at all ‘P’ at 273 K. Although below 273K, for H2 Z is less than 1
Liquification of gases and Critical Point :
Critical Temperature : The maximum temperature at which a gas can be liquified i.e. the
temperature above which a liquid cannot exist.
2.
Critical Pressure and Critical Volume : The minimum pressure necessary to liquify and gas at its
critical temperature is called critical pressure (PC) and the corresponding volume by one mole of gas
is called critical volume (VC)
3.
4.
5.
At critical point densities of a substance in liquid and gaseous state is same.
Generally gases below their TC are called vapours.
Boyle’s Temperature : Temperature at which real gases obeys the gas laws over a wide range of
pressure is called Boyle’s temperature.
Gases which are easily liquified have a high Tb whereas which are difficult to liquify have a low Tb.
Practice Problems :
1.
Under critical states of a gas for one mol of a gas, compressibility factor is :
(a)
3
8
(b)
8
3
(c)
1
(d)
1
4
[Answers : (1) a]
Einstein Classes,
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
CG – 8
SINGLE CORRECT CHOICE TYPE
1.
2.
3.
4.
5.
6.
7.
The compressibility factor of gas is less than unity
at STP, therefore :
(a)
Vm (molar volume) > 22.4 L
(b)
Vm < 22.4 L
(c)
Vm = 22.4 L
(d)
Vm = 44.8 L
A quantity of gas is collected in a graduated tube
over the mercury. The volume of the gas at 200C is
50.0 mL and the level of the mercury in the tube is
100 mm above the outside mercury level. The
barometer reads 750 mm. Volume at STP is
(a)
39.8 mL
(b)
40 mL
(c)
42 mL
(d)
60 mL
When CO2 under high pressure is released from a
fire extinguisher, particles of solid CO2 are formed,
despite the low sublimation temperature (–770C) of
CO2 at 1.0 atm. It is due to :
(a)
the gas does work pushing back the
atmosphere using KE of molecules and
thus lowering the temperature
(b)
volume of the gas is decreased rapidly
hence temperature is lowered
(c)
both correct
(d)
none of correct
At what temperature will hydrogen molecules have
the same KE as nitrogen molecules at 280 K :
(a)
280 K
(b)
40 K
(c)
400 K
(d)
50 K
For the non-zero value of force of attraction
between gas molecules, gas equation will be
PV  nRT 
(b)
(c)
PV = nRT + nbP
PV = nRT
(d)
P
10.
11.
12.
13.
nRT
Vb
A 0.20 mol sample of a hydrocarbon CxHy after
complete combustion with excess O2 gas yields,
0.80 mol of CO 2 and 1.0 mol of H 2 O. Hence
hydrocarbon is :
(a)
C4H10
(b)
C4H8
(c)
C4H5
(d)
C8H16
At low pressure, the vander Waals equation is
as
9.
n 2a
V
(a)
written
8.
a 

 P  2  (V) = RT, hence
V 

compressibility factor is :
Einstein Classes,
14.
(a)
1
a
RTV
(b)
1
RTV
a
(c)
1
a
RTV
(d)
1
RTV
a
The molecular weight of O2 and SO2 are 32 and 64
respectively. If one litre of O2 at 150C and 750 mm
presure contains N molecules, the number of
molecules, the number of molecules in two litres of
SO2 under the same conditions of temperature and
pressure will be
(a)
N/2
(b)
N
(c)
2N
(d)
4N
When an ideal gas undergoes unrestrained
expansion, no cooling occurs because the molecules
(a)
are above the inversion temperature
(b)
exert no attractive force on each other
(c)
do work equal to loss in kinetic energy
(d)
collide without loss of energy
The average velocity of an ideal gas molecule at 270C
is 0.3 m/sec. The average velocity at 9270C will be
(a)
0.6 m/sec
(b)
0.3 m/sec
(c)
0.9 m/sec
(d)
3.9 m/sec
In Van der Wall’s equation of state for a non-ideal
gas, the term that accounts for intermolecular force
is
(a)
(V – b)
(b)
RT
(c)
[p + (a/V2)]
(d)
(RT)–1
The value of Van der Waal’s constant ‘a’ for the
gases O2, N2, NH3 and CH4 are 1.360, 1.390, 4.170
and 2.253 litre2atm.mole–2. The gas which can most
easily be liquefied is
(a)
O2
(b)
N2
(c)
NH3
(d)
CH4
16 g of oxygen and 3 g of hydrogen are mixed and
kept in 760 mm pressure at 00C. The total volume
occupied by the mixture will be nearly ?
(a)
22.4 l
(b)
33.6 l
(c)
448 litres
(d)
44800 ml
The three states of matter are solid, liquid and gas.
Which of the following statements are correct about
them
(a)
Gases and liquids have viscosity as a
common property
(b)
The molecules in all the three states
possess random translational motion
(c)
Gases cannot be converted into solids
without passing through the liquid phase
(d)
Solids and liquids have vapour pressure
as a common property
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CG – 9
15.
16.
17.
How much time would it take to distribute one
Avogadro number of wheat grains, if 1010 grains
are distributed each second ?
(a)
6.02 × 1023
(b)
6.02 × 1013
3
(c)
6.02 × 10
(d)
none
The total kinetic energy of the molecules in Joules
in 8.0 g of methane at 270C is
(a)
1222.22 J
(b)
1870.56 J
(c)
2321.44 J
(d)
4425.33 J
The curve shows Maxwell’s distribution of
velocities with increase of temperature
(a)
most probable velocity increases
(b)
Fraction of molecules possessing most
probable velocity decreases
(c)
A true, B wrong
(d)
A and B are true.
22.
23.
24.
18.
19.
20.
21.
An iron cylinder contains helium at a pressure of
250 kPa at 300 K. The cylinder can withstand a
pressure of 1 × 106 Pa. The room in which cylinder
is placed catches fire. Predict whether the cylinder
will blow up before it melts or not.
(M.P. of the cylinder = 1800 K).
(a)
cylinder will blow up
(b)
it will not effect
(c)
cylinder will not blow
(d)
cannot be predicted
Through the two ends of a glass tube of length
200 cm hydrogen chloride gas and ammonia are
allowed to enter. The distance at which ammonium
chloride will first appear is
(a)
80.8 cm
(b)
50 cm
(c)
95 cm
(d)
119 cm
An evacuated glass vessel weighs 50.0 g when empty,
148.0 g filled with a liquid of density 0.98 g ml–1
and 50.5 when filled with an ideal gas at
760 mm Hg at 300 K. The molecular weight of the
gas is
(a)
200.12 g mol–1 (b)
123.15 g mol–1
–1
(c)
184.22 g mol
(d)
333.21 g mol–1
The vapour pressure of water at 200C is 17.5 torr.
The mol of water present in one litre of air at 200C
and 45% relative humidity is

Partial pressure of H 2 O in air 
Re lative Humidity 

Vapour pressure of H 2 O 

(a)
4.3 × 10–4 mol (b)
3.3 × 10–4 mol
–4
(c)
2.1 × 10 mol (d)
5.2 × 10–4 mol
Einstein Classes,
25.
If the van der Waals constant a = 3.592 dm6 atm
mol–2. The pressure exerted by one mole of CO2 gas
at 273 K is
(a)
P = 0.9935 atm (b)
P = 2.99 atm
(c)
P = 4.56 atm
(d)
P = 5.333 atm
A mixture of ethane (C 2H 6) and ethene (C 2H 4)
occupies 40 L at 1.00 atm and at 400 K. The
mixture reacts completely with 130 g of O 2 to
produce CO 2 and H 2 O. Assuming ideal gas
behaviour, the mol fractions of C2H6 and C2H4 in
the mixture is
(a)
X C 2 H 6  0.67, X C 2 H 4  .33
(b)
X C 2 H 6  0.33, X C 2 H 4  .67
(c)
X C 2 H 6  0 .2 , X C 2 H 4  . 8
(d)
none
The radius of an Xe atom is 1.3 × 10–8 cm. A 100
cm3 container is filled with Xe at a pressure of
1 atm and a temperature of 273 K. The fraction of
the volume that is occupied by Xe atoms is
(a)
2.47%
(b)
0.247 %
(c)
2.47 × 10–2 %
(d)
none
When 2 g of a gaseous substance A is introduced
into an initially evacuated flask at 25 0C the
pressure is found to be 1 atm. 3 g of another
gaseous substance B is then added to it at the same
temperature . The final pressure is found to be
1.5 atm. Assuming ideal gas behaviour, the ratio of
the molecular weights of A to B is
(a)
1:1
(b)
1:2
(c)
1:3
(d)
1:4
ANSWERS (SINGLE CORRECT
CHOICE TYPE)
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
b
a
a
a
a
a
c
c
b
a
c
c
d
a
15.
16.
17.
18.
19.
20.
b
b
d
a
a
b
21.
22.
23.
24.
25.
a
a
a
c
c
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CG – 10
EXCERCISE BASED ON NEW PATTERN
1.
2.
3.
COMPREHENSION TYPE
MATRIX-MATCH TYPE
Comprehension-1
Matching-1
60 ml of a mixture of equal volumes of chlorine and
an oxide of chlorine heated and then cooled back
to the original temperature. The resulting mixture
was found to have a volume of 75 ml. On treatment
with caustic soda solution the volume contracted
to 15 ml. Assuming that all the measurements were
made at the same temperature and pressure.
The oxide of chlorine on heating decomposes
quantitatively to oxygen and chlorine.
Column - A
Column - B
The compressibility
(P)
The volume of oxygen is
(a)
60 ml
(b)
50 ml
(c)
15 ml
(d)
25 ml
The simple formula of oxide is
(a)
Cl2O7
(b)
Cl2O
(c)
Cl2O3
(d)
Cl2O5
(a)
absorbs Cl2
(b)
absorbs O2
(c)
absorbs oxide of chlorine
(d)
absorbs all of the above
5.
6.
7.
(B)
At low pressure
(Q)
Vm(molar
volume at STP)
< 22.4 L
(C)
For the non-zero value
of force of attraction
between gas molecules
(R)
4 v/N0
(D)
v is the volume of
1 molecule then b is
(S)
4 v × N0
(T)
P
RT
( V  b)
Matching-2
Column-A
The density of the vapour of a substance at 1 atm
pressure and 500 K is 0.36 kgm–3. The vapour fuses
through a small hole at a rate of 1.33 times faster
then oxygen under the same condition.
a 

 P  2  V  RT
V 

of gas is less than
unity
On passing the mixture through the KOH solution,
the volume contraction is because of
Comprehension-2
4.
(A)
Column-B
0
(A)
Average velocity at 27
is 0.3 m/s, then at 9270
it is
(P)
3/8
(B)
Under critical state of
1 mol of a gas Z is
(Q)
1.83 × 103
(C)
Root mean square
velocity of H2 molecule
having density 0.09 g/L
at 1 atm is
(R)
2.463
(D)
(S)
0.6
The molar mass of the unknown gas is
(a)
50.25 × 10–3L
1 mol of N2 and 4 mol
of H2 taken in a 15 L
flask after complete
conversion of N2 to NH3
pressure is
(b)
50.25 × 10–3cm3
MULTIPLE CORRECT CHOICE TYPE
(c)
50.25 × 10–3m3
(d)
All the above
(a)
18.09
(b)
30.00
(c)
42.00
(d)
90.00
The molar volume of the gas is
1.
The compression factor (Z) of the vapour is
(a)
1
(b)
1.224
(c)
0.5
(d)
none
If the vapour behaves ideally at 1000 K, the
average translation kinetic energy a molecule is
(a)
0.07 × 10–19 J
(b)
7 × 10–20 J
(c)
17 × 10–20 J
(d)
2.07 × 10–20 J
Einstein Classes,
2.
At constant volume and temperature condition, the
rates of diffusion rA and rB of gases A and B having
densities A and B are related by the expression
(a)
rA = [rB.A/B]½
(b)
rA = [rB(B/A)½]
(c)
rA = [rB.B/A]½
(d)
rA = [rB(A/B)½]
A gas will approach ideal behaviour at
(a)
low temperature, low pressure
(b)
low temperature, high pressure
(c)
high temperature, low pressure
(d)
high temperature, high pressure
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CG – 11
3.
4.
For one mole of a gas at moderate pressure Vander
Waal’s equation is
(a)
a 

 P  2  v  RT
v 

(b)
a 

 P  2 ( v  b )  RT
V 

(c)
A and B are true
(d)
A and B are false
The root mean square velocity of an ideal gas at
constant pressure varies
(a)
d2
(b)
d
1
(c)
5.
6.
7.
8.
Assertion-Reason Type
Each question contains STATEMENT-1 (Assertion)
and STATEMENT-2 (Reason). Each question has
4 choices (A), (B), (C) and (D) out of which ONLY
ONE is correct.
M
1.
1
(d)
d
Which of the following pair of gases will have same
rate of diffusion under similar conditions
(a)
H2 and He
(b)
CO2 and N2O
(c)
CO and C2H4
(d)
NO and CO
One mole of following (s) will have 22.4L at NTP
(a)
He
(b)
H2 O
(c)
CCl4
(d)
SO2
The numerical value of Gas constant R are
(a)
0.0821 L atm K–1M–1
(b)
8.31 JK–1M–1
(c)
0.821 L atm K–1M–1
(d)
1.99 Cal K–1M–1
Which of the following pair of gases will have same
rate of differsion under similar conditions
(a)
Ne and He
(b)
CO2 and N2O
(c)
CO and C2H4 (d)
CO2 and CO
(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
STATEMENT-1 : For one mole of a gas at
moderate pressure Vander Waal’s equation is
a 

 P  2  v  RT .
v 

STATEMENT-2 : The root mean square velocity
of an ideal gas at constant pressure varies
inversely with square root of temperature.
2.
STATEMENT-1 : The equation PV = nRT does
not apply to real gases.
STATEMENT-2 : ‘a’ and ‘b’, the vander waal’s
constants should be zero for ideal gas.
3.
STATEMENT-1 : The pressure of the fixed
amount of an ideal gas is proportional to its
temperature.
STATEMENT-2 : Frequency of collisions and
their impacts, both increase in proportion to the
square root of temperature.
4.
STATEMENT-1 : A vessel contains a mixture of
two gases X and Y having the molecules in the
ratio of 2 : 3. The pressure of the mixture is 5 atm.
If the gas X is completely removed from the
mixture, then the pressure of the gas left in the
vessel at the same temperature will be 3 atm
STATEMENT-2 : The ratio of velocities of
diffusion of two gases A and B is 1 : 4. If the ratio
of their masses in the mixture is 2 : 3, the ratio of
their mole fraction is 1/2.
(Answers) EXCERCISE BASED ON NEW PATTERN
COMPREHENSION TYPE
1.
c
2.
b
3.
a
7.
d
MATRIX-MATCH TYPE
1.
A-Q, P; B-Q, P; C-Q, P; D-S
2.
MULTIPLE CORRECT CHOICE TYPE
1.
d
2.
c
3.
a
7.
a, b, d
8.
b, c
ASSERTION REASON TYPE
1.
C
2.
B
3.
A
Einstein Classes,
4.
a
5.
c
6.
b
b, c
6.
a, d
A-S; B-P; C-Q; D-R
4.
c, d
4.
B
5.
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CG – 12
INITIAL STEP EXERCISE
(SUBJECTIVE)
1.
2.
3.
4.
5.
6.
7.
8.
9.
What will be the volume of gas at 800mm pressure
if it occupies 40 litre at 700 mm pressure at
constant temperature 300 K.
Calculate the temperature of 2 moles of a gas
occupying a volume of 5 litres at 2.46 atm
Oxygen is present in a one litre flask at a pressure
of 7.6 × 10–10 mm Hg. Calculate the number of
oxygen molecules in the flask at 00C.
A 20 lt flask contains 4.0 gm of O2 & 0.6 gm of H2 at
373 K. If the contents are allowed to react to form
water vapours at 373 K, find the contents of flask
and their partial pressures.
Two flasks of equal volume have been joined by a
narrow tube of negligible volume. Initially both
flasks are at 300 K containing 0.60 mol of O2 gas at
0.5 atm pressure. One of the flasks is then placed in
a thermostat at 600 K. Calculate final pressure and
and the number of mol of O2 gas in each flask.
A certain dry gas at 270C and 760 mm pressure has
density 28 g/L. What will be density at 70C and 740
mm pressure.
One litre of a gaseous mixture is effused in 5 minute
11 second, while 1 litre of oxygen takes 10 minutes.
The gaseous mixture contains methane and
hydrogen. Calculate the percentage composition of
the mixture.
An open vessel at 270C is heated until three-fifth of
the air in it has been expelled. Assuming that the
volume of the vessel remains constant, find the
temperature to which the vessel has been heated ?
The pressure exerted by 12 g of an ideal gas at
temperature t0C in a vessel of volume V litre is one
atm. When the temperature is increased by 100 at
the same volume, the pressure increases by 10%.
Calculate the temperature t and volume V
(Molecular weight of the gas = 120).
10.
11.
12.
13.
14.
15.
16.
A compound exists in the gaseous state both as a
monomer (A) and dimer (A2). The molecular weight
of the monomer is 48. In an experiment, 96 g of the
compound was confined in a vessel of volume 33.6
litres and heated to 2730C. Calculate the pressure
developed, if the compound exists as a dimer to the
extent of 50 per cent by weight, under these
conditions.
An inflated balloon has a volume of 0.55 litre at sea
level (1.0 atm) and is allowed to rise to a height of
0.5 km, where the pressure is about 0.40 atm.
Assuming that the temperature remains constant,
what is the final volume of the balloon ?
A meteorological balloon had a radius of 1.0 m when
released at sea level at 200C and expanded to a
radius of 3.0 m at an altitude when the
temperature was –20 0C. What is the pressure
inside the balloon at that temperature ?
The drail cleaner, Drainex contains small bits of
aluminium which react with caustic soda to
produce hydrogen. What volume of hydrogen at
200C and one bar will be released when 0.15g of
aluminium reacts ?
What will be the pressure of gas mixture when 0.5L
of H2 at 0.8 bar and 2.0L of oxygen at 0.7 bar are
introduced in a 1L vessel at 270C ?
2.9 g of a gas at 950C occupied the same volume as
0.184 g of hydrogen at 170C, at the same pressure.
What is the molar mass of the gas ?
A mixture of hydrogen and oxygen at one bar
pressure contains 20% by weight of hydrogen.
Calculate the partial pressure of hydrogen.
FINAL STEP EXERCISE
(SUBJECTIVE)
1.
A mixture of NH3(g) and N2H4(g) is placed in a
sealed container at 300 K. The total pressure is
0.5atm. The container is heated to 1200 K at which
time both substances decompose completely according to the equation
2NH3(g)  N2(g) + 3H2(g), N2H4(g)  N2(g) + 2H2(g)
After decomposition is complete the total pressure
at 1200 K is found to be 4.5 atm. Find the percent
of N2H4(g) in the original mixture.
Einstein Classes,
2.
A mixture containing 1.12 litres of H2 and 1.12 litres
of D2 (deuterium) at N.T.P. is taken inside a bulb
connected to another bulb by a stop-cock with a
small opening. The second bulb is fully evacuated,
the stop-cock is opened for a certain time and then
closed. The first bulb is now found to contain 0.05
g of H2. Determine the percentage composition by
weight of the gases in the second bulb.
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CG – 13
3.
Oxygen gas generated in the decomposition of
potassium chlorate is collected over water. The
volume of the gas collected at 240C and atmospheric
pressure of 762 mm Hg is 128 mL. Calculate the
mass (in grams) of oxygen gas obtained. The
pressure of the water vapor at 240C is 22.4 mm Hg.
4.
A spherical balloon of 21 cm diameter is to be filled
with hydrogen at N.T.P. from a cylinder containing
the gas at 20 atmosphere at 270C. If the cylinder
can hold 2.82 litres of water, calculate the number
of balloons that can be filled up.
5.
A given volume of ozonised oxygen containing 20%
by volume of ozone required 175 sec to diffuse while
an equal volume of O2 took 167 sec. under the same
condition. Find out the vapour density of ozone.
6.
What is the difference in the density of dry air at
1 atm and 250C and moist air with 50% relative
humidity under the same condition ? The vapour
pressure of water at 250C is 23.7 torr and air has
75.5 % N2 and 24.5 % O2.
Re lative Humidity 
7.
Partial pr. of H 2 O in air
vapour pressure of H 2 O
At 12000C, mixture of Cl2 and Cl atoms (both in
gaseous state) effuses 1.16 times as fast as Krypton
effuses under identical conditions. Calculate the
fraction of chlorine molecules dissociated into
atoms. M (Kr) = 83.8 g mol–1.
11.
Eudiometry
12.
10 ml of a mixture of methane, ethylene and
carbondioxide was exploded with excess of air.
After explosion there was contraction of 16 ml and
after treating with KOH there was a further
reduction of 14.5 ml. What was the composition of
the mixture ?
13.
5 ml of a gaseous hydrocarbon was exposed to 30
ml of O2. The resultant gas, on cooling is found to
measure 25 ml of which 10 ml are absorbed by
NaOH and the remainder by pyrogallol. Determine
molecular formula of hydrocarbon. All
measurements are made at constant pressure and
temperature.
14.
The weight of one litre sample of ozonised oxygen
at NTP was found to be 1.5 g. When 100 ml of this
mixture at NTP were treated with terpentine oil,
the volume was reduced to 90 ml. Hence calculate
the molecular weight of ozone.
15.
A mixture containing 20 mL of CO, CH4 and N2
was burnt in excess of O2 resulting in the reduction
of 13mL of volume. The residual gas was then
treated with KOH solution which led to the
contraction of 14 mL in volume. Assume all
measurements to be made at constant temperature
and pressure. Find the volume of CO, CH4 and N2
in the mixture.
16.
A mixture of CH 4(g) and C 2 H 6 (g) has a total
pressure of 0.53 atm. Just enough O 2 (g) is
introduced in the mixture to bring its complete
combustion to CO 2 (g) and H 2O(g). The total
pressure of these two gasses is found to be 2.2 atm.
Assuming constant volume and temperature,
determine the fraction of CH4 in the mixture.
17.
A student forgot to add the reaction mixture to the
round bottomed flask at 270C but put it on the
flame. After a lapse of time, he realised his mistake,
using a pyrometer he found the temperature of the
flask was 4770C. What fraction of air would have
been expelled out ?
18.
Pay load is defined as the difference between the
mas of displaced air and the mass of the balloon.
Calculate the pay load when a balloon of radius
10 m, mass 100 kg is filled with helium at 1.66 bar
at 270C. (Density of air = 1.2 kg m–3 and R = 0.083
bar dm3 K–1 mol–1).
3
8.
A 100 dm flask contains 10 mol each of N2 and H2
at 777 K. After equilibrium was reached, partial
pressure of H2 was 1 atm. At this point 5 L of H2O(l)
was injected and gas mixture was cooled to 298 K.
Find out the gas pressure. [NH3(g) is absorbed by
H2O(l)]
9.
Colourless NO comes in contact of oxygen of air to
produce brown NO2. In an experiment NO and O2
gases were introduced at the two ends of a one metre
long tube simultaneously (the glass tube has
uniform cross section). At what distance from NO
brown fumes of NO2 will be seen first inside the
tube ?
10.
120 mL of NH3 at 250C and 750 Torr was mixed
with 165 mL of O 2 at 50 0C and 635 Torr and
transferred to a 300 mL reaction vessel where they
are allowed to react according to the equation :
4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g).
What will be that the total pressure in the reaction
vessel at 1500C after the reaction is over, assuming
the reaction goes to completion ?
Einstein Classes,
The total pressure of a mixture of H2 and O2 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 measurement were
made under the same temperature and volume
conditions, calculate the composition of the
original mixture.
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CG – 14
19.
10 ml of a gaseous organic compound containing
C, H and O only was mixed with 100 ml of oxygen
and exploded under conditions which allowed the
water formed to condense. The volume of the gases
after explosion was 90 ml. On treatment with
potash solution, a further contraction of 20 ml in
volume was observed. Given that the vapour
density of the compound is 23, deduce the
molecular formula. All volume measurements were
carried out under the same conditions.
20.
At room temperature following reaction goes to
completion
2NO + O2  2NO2  N2O4
Dimer N2O4 at 262 K is solid. A 250 ml flask and a
100 ml flask are separated by a stop cock. At 300
K, the nitric oxide in the larger flask exerts a
pressure of 1.053 atm and the smaller one contains
O2 at 0.789 atm. The gases are mixed by opening
the stop cock and after the end of the reaction the
flask are cooled to 220 K. Neglecting the vapour
pressure of dimer find out the pressure and
composition of gas remaining at 220 K.
ANSWERS SUBJECTIVE (INITIAL STEP EXERCISE)
1.
35 litres
2.
75 K
3.
2.69 × 1010 molecules
4.
P( H 2 )  .076 atm , P( H 2O )  0.383 atm
5.
P = 0.66 atm
6.
29.21 g/lt
7.
mole % CH4 = 47.14, mol% H2 = 52.86
8.
4770C
9.
0.82 L, –1730C
10.
2 atm
11.
1.4 L
12.
0.032 atm
13.
0.2 L
14.
1.8 bar
15.
40 g/mol
16.
0.8 bar
ANSWERS SUBJECTIVE (FINAL STEP EXERCISE)
1.
25%
2.
% H2 = 41.67, % D2 = 58.33
3.
0.164 gm
4.
10
5.
23.85
6.
0.007 kg m–3
7.
14 %
8.
2.26 atm
9.
50.8 cm
10.
974.7 Torr
11.
X H 2  0.78, X O 2  0.22
12.
CH4 = 3.5mL, C2H4 = 4.5 mL, CO2 = 2mL
14.
48.2
15.
CO = 10 mL, CH4 = 4 mL, N2 = 6 mL
Einstein Classes,
13.
C2H4
16.
0.4245
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111