GASEOUS STATE
IIT-JEE Syllabus
1. Gaseous, liquid and solid states
2. Absolute scale of temperature
3. Ideal gas equation
4. Kinetic theory of gases
5. Average
6. r.m.s. and most probable velocities and their relation with temperature
7.
Law of partial pressures.
8. Ideal gas equation
9. Deviation from ideality (Compression factor z)
10. vander waal's equation
11. diffusion of gases.
Total No. of questions in Gaseous State are :
Solved examples…....…………………………..…33
Exercise # 1 …….……………………………….…40
Exercise # 2 …….……………………………….…38
Exercise # 3 …….……………………………….…45
Exercise # 4 ……………………………………..…21
Exercise # 5 ……………………………………..…29
Total No. of questions………………..206
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*** Students are advised to solve the questions of exercises in the same sequence or as
directed by the faculty members.
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Index : Preparing your own list of Important/Difficult Questions
Instruction to fill
(A) Write down the Question Number you are unable to solve in column A below, by Pen.
(B) After discussing the Questions written in column A with faculties, strike off them in the
manner so that you can see at the time of Revision also, to solve these questions again.
(C) Write down the Question Number you feel are important or good in the column B.
EXERCISE
NO.
COLUMN :A
COLUMN :B
Questions I am unable
to solve in first attempt
Good/Important questions
1
2
3
4
5
Advantages
1. It is advised to the students that they should prepare a question bank for the revision as it is
very difficult to solve all the questions at the time of revision.
2. Using above index you can prepare and maintain the questions for your revision.
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GASEOUS STATE
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KEY CONCEPTS
1. Indroduction
Matter as we know broadly exists in three states.
There are always two opposite tendencies between particles of matter which determine the state of matter
•
Inter molecular attractive forces.
•
The molecular motion/random motion.
Properties
Solid State
• Large
• Almost zero
• Fixed volume
• Definite
• Less
• High
Attractive force
Thermal motion
Volume
Shape
Compressibility
Density
Matter
Liquid state
• Smaller
• Greater
• Fixed volume
• Not definite
• Inter mediate
• Inter mediate
Gaseous state
• Almost zero
• Random motion
• Varies with container
• Not definite
• High
• Less
2. Units
Volume
Volume of the gas is the
volume of the container
S.I.
unit cm3
C.G.S. unit  cm3
Pressure
Pressure = N/m2 = Pa  S.I. unit
C.G.S. unit = dyne-cm2
convert 1N/m2 into dyne/cm2
1  = 10–3 m3
1m 2 10 4 cm 2
1 N/m2 = 10 dyne/cm2
1 atm = 1.013 × 105 N/m2
1N
1 = 103 cm9
1dm3 = 1
 = 10–3 cm3

10 5 dyne
1ml = 10–3  = 1 cm3
Temperature
Kelvin scale  Boling point = 373 K
ice point = 273 K
Fahrehneit scale  B.P. = 212º F
ice point = 32ºF
K  273
F  32
C0
=

100  0 373  273 212  32
R  R ( 0)
=
R (100)  R (0)
where R = Temp. on unknown scale
Atmospheric pressure :
The pressure exerted by atmosphere on earth's surface at sea level is called 1 atm.
1 atm = 1.013 bar 1 atm = 1.013 × 105 N/m2 = 1.013 bar = 760 torr
3. Instruments Pressure Calculations
Vacuum
(A) Barometer : P =
Ahdg
A
P
where d = density of fluid
h = vertical height
g = acceleration due to gravity
X
h
Y
atm
(B) Manometer : Pgas = Patm + hdg
gas
Patm
Pgas
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Real gases
Correction for
intermolecular attraction
(+)ve
•Z>1
• Repulsive forces
• Difficult to compress
• Difficult to Liquefy
Correction for
molecular volume
Standard
temperature
T = 0ºC
Standard
pressure
P = 1 atm
Under STP
conditions
Standard molar
volume = 22.4 L
Gas constant, R
used for
Dalton's law
Ptotal = P1 + P2 + P3 +…
P1 = x1 P
x1 = mole fraction
P = total pressure
Using
behaviour explained by
KE = 3/2 PV
= 3/2 RT
Kinetic energy, Ex
Root mean square
Effusion
Average speed
Most probable speed
Velocity, u
Molecular speed
Diffusion
applied to
gives rise to
Temp., T
Mass, M
relates
molecular
Kinetic molecular
theory
Temperature, T
in kelvins, K
R = 8.314 J/Kmol 25/3
= 1.987 cal/mol 2
=.0821 Latm/molK 1/2
containing
Charles law
V/T = k
Molar amount, n
in moles, mol
Ideal gas law
PV = nRT
combined
in the
Boyle's law
PV = k
Including
Gas laws
related by
Volume, V
in liters, L
Condition defined by
Gases
Pure gases Gas mixtures
Avogadro's law
V/n = k
Pressure, P in
1. Pascals, Pa
2. mm Hg (torr)
3. Atmospheres, atm
which gives
(+)ve
•Z<1
• Repulsive forces
• Difficult to compress
• Difficult to Liquefy
• Zero volume
• Zero attractive
• PV = nRT
•Z=1
• non-zero volume
• some intermolecular force
Ideal gases
4. Table
graphs are Isochor
5. Some laws
I.
Boyle's law :
V
nR
tan  
V

TK
1
V
(T, constant ; n constant) P1V1 = P2V2
P
graphs are Isotherms
T
log RT/P
PV
nRT
V
nR/V
45º
log P
P
P/T
log T
IV. Avogadro's law :
P
V  n (T, P constant; n constant)
log P
45º
V1 V2

n1 n 2
log nRT P
 [tan = nRT]
1/V
log V
V
RT
tan  
P

n
II. Charle's law :
V  T (P, constant ; n constant)
V1 V2

T1 T2
V/n
RT/P
n
45º
log V
log nR/P
graphs are Isobars
log T
Combined gas law :
V
tan  

nR
P
V/T
nR/P
T
TK
P1V1 P2 V2

T1
T2
Equation of state :
PV = nRT
d = density of gas
log V
 PV =
45º
w
RT ; R = Universal gas constant
M
= 0.0821 atm litres/Kelvin/mol
log nR/P
PM = dRT = 8314 J/K = 2 cal/K/mol
log T
Dalton's law of particle pressure :
Ptotal = PA + PB + ……….. ;
III. Gay Lussac's law :
P  T (V, constant ; n constant)
Pwet gas = Pdry gas + PH 2O vapuor i.e. aq. tension
P1 P2

T1 T2
PA , PB are partial pressures ;
PA = mole fractionA × Total pressure
and % of gas in mixture
=
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Partial pressure
 100
Total pressure
GASEOUS STATE
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Amagat's law :
Solving using Graham's law, x =
The total volume of a mixture of gases is equal
to the sum of the partial volumes of the
constituent gases, at same temperature and
pressure.
(B) Payload / lifting power [based on Buoyancy]
L.P. of balloon = V(d – d1) g – Mg
Graham's law of Diffusion or Effusion :
r 
1
or r 
d
1
M
r
V = Volume of balloon
P
M
d = density of outside gas
d = density of gas in the balloon
[For gases effusing at different pressures]
(r is rate of diffusion of any gas.)
M = Mass of balloon
(C) Analysis of a reaction involving gaseous
r1
d2
M2
volume / time
M2
;



r2
d1
M1
volume / time
M1
A(g) + B(g)  C(g)
 What happens to pressure as reaction
proceeds (in a closed container)
(d is density at same temperature)
r


moles diffused
time taken
(D) Vapour density and degree of dissociation
distance travelled in a narrow tube
time taken
=
Kinetic theory of gases :
Barometric pressure distributor in a gas [To
calculate pressure at various height in a gas]
P2 Mg

[H 2  H1 ]
P1
RT
Dd
( n  1)d
8. Kinetic theory of gases
Pr essure drop I
Pr essure drop II
6. Barometric pressure distributor in a gas
ln
P2 = P1
Mg

[ H 2  H1 ]
e RT
PV =
1
1
m N u2 = M u2 (for 1 mole)
3
3
Types of velocities :
u2 =
u12  u 22  ....  u 2N
N
u = root mean square speed.
M = molecular wt.
3RT
3PV
3P


M
M
d
7. Isotropic Separation Factor
(A) Isotropic Separation factor :
f
n11 / n12
n1 / n 2
Average speed =
f
n1, n2 and n11 , n12 are the concentration of two
isotopes before and after processing.
Theoretical separation factor f ' =
M2
M1
If required enrichment of species (1) is attained
after 'x' times, then ;
(f ' ) x 
n11 / n12
n1 / n 2
2 log f
M 
log 2 
 M1 
u12  u 22  ....  u 2N
=
N
most probable speed =
8RT
M
2RT
M
most probable : average : r.m.s.
1 : 1.13 : 1.22  Re lationship between three

8
 2.
: 3
types of speeds


urms > uav > ump
f
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Average kinetic energy of a single molecule
=
greater the value of 'a' more easily the gas is
liquefiable ;
3 R
3
. .T  kT
2 N
2
greater the value of 'b' greater the molecular size.
k = Boltzman constant = 1.3806 × 10–16 erg deg–1
Total kinetic energy for one mole of a gas =
kinetic energy of n moles of a gas = n ×
Real Gases : Deviation from ideal behavior
The curve for the real gas a tendency to coincide
with that of an ideal gas at low pressures when the
volume is large. At higher pressure however
deviations are observed.
3
RT .
2
3
RT
2
Ideal
H2
P
Maxwell distribution laws :
 M 
dNu = 4N

 2RT 
 M 
= 4N

 2RT 
3/ 2
[exp(– Mu2 / 2RT)]u2du
3/ 2
T2 > T1
1 dN
N du
[exp(– Mu2 / 2kT)]u2du
T1
V
Compressibility factor :
z=
T2
PV volume observed

nRT
volume ideal
Z
u1 u2
t = 0ºC
H2
1.0
u
N2
CH4
Ideal gas
CO2
Collision frequency and Mean free path :
0
d  d 2  .....  d n
Mean free path  = 1
n
=
Average velocity / RMS velocity
kT

collision number or frequency
2 2 P
2  2 uN * [collisions made by one molecule
Z11 =
1
 2 uN *2
2
THE REAL PATH
1.0
0
T1 > T2 > T3 > T4
T3
T2
T1
ideal gas
200 400 600
p/101.325 kPa
Boyle Temperature :
Vander Walls equation of state :
a 2

 P  2 .n  (V – nb) = n RT
V


T4
Z
k = Boltzman constant ;  = collision diameter.
Z1 =
100 200 300
p/101.325 kPa
TB =
a
bR
Inversion Temperature :
a, b are Vander Waals constants ; different for each gas
unit of a  atm L2 mol–1 ;
Ti =
2a
bR
S.I. unit  Pa m6 mol–2
unit of b  L mol–1 ;
S.I. unit  m3 mol–1
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Interpretation of deviation from vander waals equation :
(i) At low pressure z =
PV
a
1
RT
VRT
(ii) At high pressure z =
PV
Pb
1
RT
RT
Vibrational Motion :
E vib 
(per degree of freedom)
10. Some other equation of state
PV
a
(iii) At extremely low pressure z =
 1; Pb 
RT
V
Dieterici equation :
Rena/VRT × (V – nb) = nRT
9. Heat capacities
Berthelot equation :
CP = Molar heat capacity at constant pressure

n 2a 
P 
 (V – nb) = nRT

TV 2 

CV = Molar heat capacity at constant volume
 CP – CV = R

1
1
KT  KT  KT
2
2
CP
  ,  = 1.66 (monoatomic) ; 1.4 (diatomic)
CV
(a and b are Berthlot's constant different from
vander waal's constant)
T4 > T3 > Tc > T2 > T1
Molar specific heat :
p d
= specific heat × molecular mass
CP– CV =
Cp
Cv
R CP
;
 1.66 for monoatomic
J CV
 1.4 for diatomic
Degree of freedom :
Three for monoatomic gas ;
c
For a molecule having N atoms, total are 3N
When pressure is increases at constant temp volume
of gas decreases
Translational :
D
for all types [at all temp.] [Each contributing
1/2 KT]
P P3
T4
TC
P2
T3
[Each contributing 1/2 KT]
3 for Non-linear
P1
C
Vibration :
3N  5 for linear

Each contributing KT
3N  6 for Non  linear 
Law of Equipartition of energy :
Translational motion :
E trans 
3
KT
2
Rotational motion :
E rot  KT (linear)
E rot 
X
11. Critical constant of a gas
Five for diatomic gas.
Rotational :
2 for linear
T4
T3
Tc
T2
b a T1
Y
B
(T3 > T2)
Isotherm
T2 > T1
A (T1)
VC V m
AB  gases
BC  vapour + liquid
CD  liquid
Critical point :
At this point all the physical properties of
liquids phase will be equal to physical
properties in vapour such as
density of liquid = density of vapour
3
KT ( Non  linear)
2
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TC or critical temp :
Temperature above which a gas can not be
liquidied
PC or critical pressure :
minimum pressure which must be applied at
critical temp to convert the gas into liquid.
VC or critical volume :
Volume occupied by one mole of gas at TC and PC
Critical constant using vander wall equations :

a 
 P  2  (Vm – b) = RT

Vm 

(PVm2
 a ) (Vm – b) =
RT Vm2
Vm3  aVm  PbVm2  ab  RTVm2  0
But at critical point all three roots of the equation
should be equal, hence equation should be :
Vm3  3Vm2 VC  3Vm VC2  VC3  0
…(2)
comparing with equation (1)
b
RTC
 3VC
PC
…(i)
a

 3VC2 ....(ii )
PC

VC  3b
ab
 VC3 ....(iii ) 

PC
PC 
a
a
a
substituting PC =

3(3b) 2 27 b 2
3VC2
and TC 
8a
27 Rb
RT  a
ab

Vm3  Vm2  b 
0
  Vm 
P
P
P


Cubic can hence there will be three roots of
equation at any temperature of pressure.
At critical point all three roots will coincide and
will give singles dx = VC at critical point Vander
Waal equation will be

RTC  a
ab
 
Vm3  Vm2  b 
Vm 
 0 ….(1)
P
P
P
C 
C
C

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SOLVED EXAMPLES
Ex.1
Which of the following curves does not
represent Boyle's law P
(A)
(B)
log P
log V
V
32 gm of oxygen and 3 gm of hydrogen are
mixed and kept in a vessel of 760 mm pressure
and 0º C. The total volume occupied by the
mixture will be nearly (A) 22.4 lit
(B) 33.6 lit
(C) 56 lit
(D) 44.8 lit
Sol.(C) 32 gm O2 = 1 mole O2
3 gm hydrogen =
P
(D) P
(C)
1/V
K
V
Curve A represent rectangular hyperbola which
satisfied the relation PV = K
 PV = K
or log P + log V = log K
or log P = log K – log V
log P Vs log V data represent a straight line
with negative slope which is same as (B)
PV = K
P vs
P=
3
= 1.5 mole H2
2
Hence total moles of gas present = 2.5
volume of total 2.5 moles of gas mix at
STP = 2.5 × 22.4 = 56 lit
V
Sol.(D) As per Boyle's law
PV = K (at constant temp, for a definite mass of gas)
P =
Ex.3
Ex.4
A closed vessel contains equal number of
nitrogen and oxygen molecules at pressure of P
mm. If nitrogen is removed from the system,
then the pressure will be (A) P
(B) 2P
(C) P/2
(D) P2
Sol.(C) Equal no. of molecules, that means equal no. of
moles of gas present.
 n O 2 = n N 2 = x say
P1V = ( n O 2 + n N 2 ) RT = 2x RT
K
Cƒ y = mx
V
P2V = n O 2 RT = x RT

1
data represent a straight line as
V
P1
=2
P2
or
P1 = 2P2
when nitrogen is removed, final pressure
will be P/2
passing through the origin which is same as (C)
P vs V curve is rectangular hyperbola which is
not a straight line as represented in curve (D)
Ex.5
Ex.2
If the pressure of a given mass of gas is
reduced to half and temperature is doubled
simultaneously, the volume will be (A) Same as before
(B) twice as before
(C) Four time as before
(D) One fourth as before
Sol.(C) As per equation of state
P1V1
PV
P
= 2 2 P2 = 1
T1
T2
2
or V2 =
Since the atomic weights of C, N and O are 12,
14 and 16 respectively, among the following
pair, the pair that will diffuse at the same rate is
(A) carbon dioxide and nitrous oxide
(B) carbon dioxide and nitrogen peroxide
(C) carbon dioxide and carbon monoxide
(D) nitrous oxide and nitrogen peroxide
Sol.(A) Mol. wt. of CO = 28 Mol wt of CO2 = 44
T2 = 2T1
T2 P1
2T
P
×
× V1 = 1 × 1 ×V1
P1
T1 P2
T1
2
Mol. wt. of N2O = 28 + 16 = 44
Mol. wt. of NO2 = 14 + 32 = 46
CO2 and N2O having same mol. wt.
therefore, rate of diffusion for both the gases
are same.
= 4V1
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Ex.6
2 gms of hydrogen diffuses from a container in
10 minutes. How many gms of oxygen would
diffuse through the same time under similar
conditions ?
(A) 0.5 gm (B) 4 gm (C) 6 gm (D) 8 gm
Sol.(D) rH 2 =

or

rH 2
rO 2
rO 2
rH 2
VH 2
=
VO 2
rO 2
t
VH 2
t
 n O 2 
 m O 2 =
1
n O2
=
VO 2
Ex.9
1
16

Now for helium atom,
1

4
K.E. =
1
× 32 = 8 gm
4
C1 =
KE =
C2
C1
=
3 C1 =
molecules
Ex.10 The ratio of average molecular kinetic energy
of UF6 to that of H2, both at 300 K is (A) 1 : 1
Sol.(A) K.E. for UF6 =
K.E. for H2 =
(KE)2 =
3
× R × 400
2
3
× R × 800
2
(KE) 2

=2
(KE)1
or
1
3RT
3
× M UF6 ×
= RT
2
2
M UF6
1
3RT
3
× M H2 ×
=
RT
2
2
M H2
K.E. of UF6 to that H2 is 1 : 1
3 × 104 cm/sec
3
RT
2
 (K.E.)1 =
(B) 7 : 2 (C) 176 : 1 (D) 2 : 7
3
The temperature of a sample of gas is raised
from 127 ºC to 527 ºC. The average kinetic
energy of the gas (A) Does not changes
(B) Is doubled
(C) Is halved
(D) Cannot be calculated
Sol.(B) K.E. =
2
1
1
3RT
3
M H2 C =
× M H2 ×
= RT
2
2
H2
2
K.E. of He atom is same as it is for H2
3R  27
M
or C 2 =
Ex.8

2
1
1
3RT
3
M He C =
M ×
= RT
2
2 He
M He
2
Again for H2 molecules
The rms speed of a gas molecules at
temperature 27 K and pressure 1.5 bar is 1 × 104
cm/sec. If both temperature and pressure are
raised three time, the rms speed of the gas will
be (A) 9 × 104 cm/sec
(B) 3 × 104 cm/sec
(C) × 104 cm/sec
(D) 1 × 104 cm/sec
Sol.(C) Do remember rms speed does not depend upon
the pressure.
3R  3  27
M
2
1
MC
2
Sol.(B) K.E. =
Ex.7
C2 =
A helium atom is two times heavier than a
hydrogen molecule at 298 K, the average
kinetic energy of helium is (A) Two times that of hydrogen molecules
(B) Same as that of hydrogen molecules
(C) Four time that of hydrogen molecules
(D) Half that of hydrogen molecules
Ex.11 A mono atomic gas, diatomic gas and triatomic
gas are mixed, taking one mole of each
for the mixture is (A) 1.40
(B) 1.428 (C) 1.67
Sol.(B)
Gas
CP in cals/mole
CP
CV
(D) 1.33
CV in cals/mole
monoatomic
5
3
diatomic
7
5
triatomic
8
6
when we are mixing one mole of each gas,
then total CP = 5 + 7 + 8 = 20 cals/3moles
(KE)2 = 2 (KE)1
CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000
then total CV = 3 + 5 + 6 = 14 cals/3moles

CP
20
=
= 1.428
CV
14
GASEOUS STATE
16
Ex.12 According to kinetic theory of gases, for a
diatomic molecule (A) The pressure exerted by the gas is
proportional to the mean velocity of the
molecule.
(B) The pressure exerted by the gas is
proportional to the root mean square
velocity of the molecule.
(C) The root mean square velocity of the
molecule is inversely proportional to the
temperature
(D) The mean translational kinetic energy of
the molecule is proportional to the absolute
temperature
2
1
Sol.(D) P =  C 
3

(C) 25 seconds : CO
Sol.(B) rH 2 =


2
p  C and P  C avg.
P  C and C =

Ex.15 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 ideal condition is (A) 10 seconds : He (B) 20 seconds : O2
C
3RT
M
T & C 

1
T
Mean Translational energy =
3
KT
2
 Mean translational energy  T
Ex.14 At low pressure, the vander waals equation is
written as 

a 
RTV 
(B) 1 –


a 
RTV 
(D) 1 
(A) 1 –
(C) 1 

Sol.(A)  P 





RTV 
a 
RTV 
a 
a 
 V = RT
V2 
a
or PV +
= RT
V
or Z = 1–
rX =
x t
× =
5 x
or t =
x
t
Mx

M H2
Mx
× 5 or t =
M H2
Mx
×5
2
For He Mx = 4
t=5 2
For CO Mx = 28
For O2 Mx = 32
t = 5 14
t = 20
For CO2 Mx = 44
t = 5 22
Ex.16 The valves X and Y are opened simultaneously.
The white fumes of NH4Cl will first form at :
(A) A
(B) B
(C) C
(D) A, B and C simultaneously
Ex.13 The values of vander waals constant 'a' for the
gases O2, N2, NH3 and CH4 are 1.36, 1.39, 4.17
and 2.253 lit2 atm mol–2 respectively. The gas
which can most easily be liquefied is (A) O2
(B) N2
(C) NH3 (D) CH4
Sol.(C) More the 'a' value of the gas, more will be the
inter molecular attraction between the gas
molecules, therefore, easier will be the
liquefaction.
x
5
(D) 55 seconds : CO2
HCl
NH3
X B A
CY
Sol.(C) Since rate of diffusion of NH3 is more than HCl
due to its comparatively lower mol. wt,
therefore, fumes of NH4Cl will be formed at C.
Ex.17 To which of the following gaseous mixtures
Dalton's law not applicable.
(A) Ne + He + SO2
(B) NH3 + HCl + HBr
(C) NO + O2 + CO2
(D) N2 + H2 + O2
Sol.(B) and (C)
Dalton's law is applicable for non reacting
gases only.
PV
a
or
+
=1
RT
RTV
a
RTV
CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000
 NH3 + HCl + HBr are reacting gas mixture
and since, NO + O2 + CO2 are reacting gas
mixtures, therefore Dalton's law is not
applicable for those gas mixtures.
GASEOUS STATE
17
Ex.18 Given reaction
Sol.(A), (C)
C (S) + H2O(g)  CO(g) + H2(g)
Calculate the volume of gaseous mixture
obtained at STP from 48 gm. of carbon and
excess H2O (A) 179.2 lit
(C) 44.8 lit
n N2 =
wt. of N 2
0.76
=
= 0.027
molecular wt of N 2
28
n O2 =
0.24
= 0.0075
32
 Total moles = n N 2 + n O 2 = 0.0345
(B) 89.6 lit
(D) 22.4 lit
Now Total Pressure P =
48
Sol.(A) 48 gm. C =
= 4 moles of C. From 4 moles
12
=
of C, 4 moles of CO (g) and 4 moles of H2 (g)
is obtained.
nRT
V
0.0345  0.0821 293
1
= 0.83 atm.
Partial pressure of
 total moles of gas produced = 8
volume of product gas mixture at
STP = 8 × 22.4 = 179.2 lit.
N2 =
=
Ex.19 Urea, CO(NH2)2 is made by the reaction of
Mole of N 2
× Total pressure
Total mole
0.027
× 0.83 = 0.65 atm.
0.0345
carbon dioxide and ammonia :
CO2(g)+2NH3(g) CO(NH2)2 (s) + H2O (g)
To produce 2.5 kg urea at 200 atm and
450 ºC, the volume of (A) CO2 required is 12.4 L
(B) CO2 required is 24.8 L
(C) NH3 required is 12.4 L
(D) NH3 required is 24.8 L
Sol.(A), (D)
From ideal gas equation PV = nRT
VNH 3 =
Ex.21 Carbon monoxide gas is purchased in a 425 mL
bottle at a pressure of 5 atm. and 23ºC. Choose
the correct alternatives (A) Mass of the gas purchased is 1.45 g.
(B) Mass of the gas purchased is 2.45 g.
(C) Density of the gas in the bottle is 5.76 g/L.
(D) Density of the gas in the bottle is 2.88 g/L.
Sol. (B), (C)
PV = nRT 

2.5  103  2
723
× .0821×
200
60
wt =
2.5 10 3
723
×.0821×
60
200
= 12.4 L
Acc. to Avogadro's principle V  n
 24.8 L of NH3
wt
× 0.0821 × 296.
28
5  425  28
= 2.45 g and density of
1000  0.0821 296
gas in the bottle is 5.76 g/L.
= 24.8 L
VCO 2 =
 5 × 425 / 1000 =
Ex.22 The density of an ideal gas A is 1.43 gm/lit at
STP. Determine the density of A at 17º C and
700 torr (mm)
Sol.
PV = nRT
or
Ex.20 A 1.0 g sample of air consists of approximately
0.76 g of nitrogen and 0.24 g of oxygen. This
sample occupies a 1.0 L vessel at 20º C. Then
(A) the partial pressure of N2 is 0.65 atm
(B) the partial pressure of O2 is 0.36 atm
(C) the total pressure is 0.83 atm
(D) the total pressure is 1.05 atm.
CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000

W PM
=
V RT
or
P
2
T
= 2 × 1
1
T2
P1
or 2 =
W
RT
M
PM
=
RT
or PV =
or 2 =
P2 T1
×

P1 T2 1
700 273
×
×1.43 = 1.2399 gm / lit
760 290
 The density of the gas A at 17ºC and
700 torr = 1.2399 gm/lit
GASEOUS STATE
18
= 107 × d
Ex.23 A balloon filled with helium rises to a certain
height at which it gets fully inflated to a volume
of 1 × 105 litre. If at this altitude temperature and
pressure is 268K and 2 × 10–3 atm respectively,
what weight of helium is required to fully
inflate the balloon ?
Sol.
V = 1 × 105 lit
T = 268 K
W
PV =
RT
M
or W =
d = 1.018 gm/lit.
This is the density of air at height h
 Using barometric distribution formula
=
P = 2 × 10–3 atm
–3
2 10  10  4
800
=
0.082  268
0.082  268
values in the above expression, h = 1.396 km.
Ex.25 At what temperature would N2 molecules have
 to fully inflate the balloon 36.403 gm He is
required.
the same average speed as He atoms at 300 K ?
Sol.
Ex.24 A balloon having a capacity 104 m3 is filled
with helium at 20º C and 1 atm pressure. If the
balloon is loaded with 80% of the load that it
can lift at ground level, at what height will
balloon come to rest ? Assume that volume of
the balloon is constant, the atmosphere is
isothermal, 20º C, the mol. wt. of air is 28.8
and the ground level pressure is 1 atm. The
mass of the empty balloon is 1.3 × 106 gm.
Sol.
Given : volume of balloon V = 104 m3= 107 lit.
Temperature T = 293 K
pressure P = 1 atm
from pay load calculation
Wt. of gas + empty balloon + pay load
= Wt. of displaced air
= volume of displaced air × density
110 7
× 4 + 1.3 × 106 + pay load
0.0821 293
= 107 ×
1 28.8
= 1.1972
0.0821 293
Substituting the values of d, d0, M, g, R, T
5
= 36.403 gm.
– Mgh
RT
where d0 =
PVM
W=
RT
or
d
= e
d0
1 28.8
0.0821 293
Cavg for N2 =
8RT
M N 2
Cavg for He =
8R  300
M He
8RT
=
M N 2
8R  300
M He

or
8R  300
8RT
=
M N 2
M He
or T = 300 ×
M N2
M He
= 300 ×
28
= 2100
4
at 2100 K temperature would N2 molecules
have the same average speed as He atoms at
300 K
Ex.26 A gas of molecular weight 40 has a specific
heat 0.075 cal/gm/deg at what is the CV value
for it and what is the atomicity of the gas.
Sol.
Molar specific heat CV = mol wt × CV
CV = 40 × 0.075 = 3 cals/mol
 pay load = 9.024 ×
gm
The balloon is loaded with 80 % of lifting
capacity
106
Applied load = 0.8 × 9.024 × 106
= 7.2194 × 106 gm
The balloon will come to rest when
Wt. of balloon + Applied load = Vol. of
displaced air × density
It is known to us,
C P – CV = R
or CP = R + CV = 2 + 3 = 5 cals/mol
 =
CP
5
= = 1.66
CV
3
 = 1.66, therefore gas is monoatomic.
1.6628 × 106 + 1.3 × 106 + 7.2194 × 106
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GASEOUS STATE
19
Ex.27 At 627 ºC and 1 atm, SO3 partially dissociates
into SO2 and O2. One litre of the equilibrium
mixture weighs 0.94 under the above
conditions. Calculate the partial pressures of
the constituent gases in the mixture.
Sol.
T = 900
P = 1 atm
V = 1 lit.
n =
Sol.
or Mexperimental =
and  = 0.656
1
0
a–x
x
a +
0 At initial point
x
2
x
= total no. of moles = 0.01355 … (1)
2
(a – x) 80 + x × 64 +
apparent mol. wt. of the gas mixture is 55.56
Ex.29 A mixture of H2 and O2 in 2 : 1 volume ratio is
allowed to diffuse through a porous diaphragm.
Calculate the composition of gases coming out
initially.
Sol.
 n SO 3 = 0.1175 – 0.0036 = 0.00815 moles
n SO 2 = 0.0036 moles
n O 2 = 0.0018 moles
PO 2
PH 2
PO 2
PH 2
PO 2
=
rH
2
 2 =
1
rO 2
32
2
4 2 8
×
=
=
1
1
1
2
L
HCl
at 1.2
atm
Sol.
Ex.28 The degree of dissociation of N2O4 according
to the equation N2O4
2NO2 at 70 ºC and
atmospheric pressure is 65.6 %. Calculate the
apparent molecular weight of N2O4 under the
above conditions.
CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000

=
NH3
at 1 atm
200–x
X
0.0036
× 1 = 0.2659 atm
0.01355
0.0018
=
× 1 = 0.13269 atm
0.01355
M H2
×
x
0.00815
=
× 1 = 0.6015 atm
0.01355
PSO 2 =
M O2
Ex.30 A straight glass tube has two inlets X and Y at
the two ends. The length of the tube is 200 cm.
HCl gas at 1.2 atm is sent through inlet X and
NH3 gas at 1 atm through intel y is allowed to
enter the tube at the same time. White fumes
first appears at point L inside the tube. Find the
distance L from X.
n = 0.01355 moles
 PSO 3
=
The composition of the gases coming out
initially is VH 2 : VO 2 = 8 : 1
Putting the value of a in equation (1) we get,
or x = 2 × 0.0018 = 0.0036
rO 2

0.94
a=
… (2)
80
x
0.94
=0.01355 –
= 0.1355–0.01175 = 0.0018
2
80
rH 2
volume ratio = moles ratio
x
× 32 = 0.94
2
or 80 a – 80 x + 64 x + 16 x = 0.94
or 80a = 0.94 or
92
92
=
= 55.56
1  0.656 1656
Mexperimental =
Let initially a moles of SO3 was taken
SO3
M normal
1  (n – 1)
Have, n = 2, Mnormal = 28 + 64 = 92
PV
1 1
1
=
=
= 0.01355
RT
0.082  900
73.3
1
SO2 + O2
2
M normal
= 1 + (n – 1) 
M exp erimental
x
=
200 – x
Y
200 cm
M NH 3
M HCl
×
1.2
1
17
× 1.2 = 0.818953
36.5
 x = 163.7905 – 0.818953 x
or x = 90.047 cm.
white fumes first appears at 90.047 cm from
the end X.
GASEOUS STATE
20
Ex.31 By how many times the absolute temperature of
a gas when Urms of a gas in a container of fixed
volume is increased from 5 × 104 cm/sec to
10 × 104 cm/sec.
Sol.
Urms at temp T1 =
3RT1
M
Urms at temp T2 =
3RT2
M

or
U rms at temp. T2
=
U rms at temp. T1
10  10
5  10
4
4
=
Ex.33 Find the temperature at which 3 moles of SO2
will occupy a volume of 10 litres at a pressure
of 15 atms.
a = 6.71 atm lit2 mol–2
b = 0.0564 lit mol–1
Sol.
P = 15 atm V = 10 lit
n=3

T2
T1
T2
or
T1
T2
4
=
1
T1
temperature is to be increased 4 times
Ex.32 The critical temperature and pressure of CO2
gas are 304.2 K and 72.9 atm respectively.
what is the radius of CO2 molecule assuming it
to behave as vander waals gas ?
Sol.
TC = 304.2 K PC = 72.9 atm
TC =
8a
27Rb
PC =
  P 


or 15 

n 2 a 
(V –nb) = nRT
V 2 
9  6.71 
 (10 – 3×0.0564) =3×0.082× T
100 
or 15.6039 × 9.8308 = 3 × 0.082 × T
or T = 623. 5724 K
or T = 350.5724 ºC
The given gas at the given condition occupy
volume of 10 litres at 350.5724 ºC
a
27b 2
8a
TC
8a
27 b 2
8b

= 27Rb =
×
=
a
PC
R
27Rb
a
27b 2
or b =
RTC
1 0.082  304.2
= ×
= 0.04277 lit
8PC
8
72.9
b = 4 NA ×
4 × NA ×
or r3 =
4 3
r = 42.77 cm3
3
4 3
r = 42.77
3
3  42.77  10 –23
16  6.023  3.14
or r3 = 0.424 × 10–23 = 4.24 × 10–24
or r3 = (4.24)1/3 × 10–8 cm = 1.62 × 10–8 cm
radius of CO2 molecule = 1.62 Å
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GASEOUS STATE
21
EXERCISE # 1
Questions
based on
Q.1
Q.2
Boyle's law
As per Boyle’s law which of the following
is/are kept constant ?
(A) Pressure
(B) Mass
(C) Temperature
(D) Mass and Temperature both
(C) log P
log V
(D) log P
log V

Fig II
entrapped by a column of Hg of length 8 cm. In
figure-II length of same air column at the same
temperature is 2. The
Questions
based on
(A) 1 +
2
× cos 
19
(B) 1 +
(C) 1 +
2
× sin 
21
(D)
Fig I
(C)
1000
ml
3
As per Charles law which of the following
is/are correct (A) Pressure remain definite
(B) Mass remain definite
(C) volume is proportional to the absolute
temperature
(D) All of the above are correct
Q.7
Which of the following graph is/are correct as
per Charles law ?
Fig II
15000
(B)
ml
17
(D)
21
19
Q.6
In the above figures a doll is entrapped within a
piston and cylinder containing gas. Initial and
final conditions are shown Figure-I and
Figure-II respectively. The volume of doll is (A) 1000 ml
2
× sin 
19
Charles law
P1>P2
P1
P2

P= 7atm
V= 1000ml
P= 10atm
V= 800ml
1
is 2
(1 atm = 76 cm of Hg)
(A) V
Q.4
8cm
In figure-I an air column of length 1, is
log V
At the definite temperature the volume of a
definite mass of gas is 10 L at 5 atm pressure,
at the same temperature if the pressure of the
gas in decreased to 1 atm, the volume of same
gas become (A) 50 L
(B) 2 L
(C) 5 L
(D) 0.5 L
2
8cm
(B) log P
log V
1
Fig I
Which of the following represent log P vs log
V variation as per Boyle’s law ?
(A) log P
Q.3
Q.5
1000
ml
15
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T
P2>P1
P1
P2

(B) V
T

(C) log V
log T
(D) Both (B) and (C) are correct
GASEOUS STATE
22
Q.8
When an open container of volume V is heated
from normal temperature T1 K to T2 K, the
volume of expelled air at temperature T1 K is
V. Therefore value V/V is (A) 1 –
T2
T1
(B) 1 –
(C) 1 +
T1
T2
(D)
Q.13
d
P
For the different ideal gases   versus P
variations at definite temperature is 
(A) d
P
T1
T2
T2
–1
T1
M1
M2
M3 M3 > M2 > M1
P

Questions
based on
Q.9
Q.10
Q.11
Ideal gas equation
P
For a definite amount of gas pressure and
volume are increased to triple of the initial
amount. Therefore (A) Temperature increased to nine time of its
initial value
(B) Temperature increased to thrice of its
initial value
(C) Temperature remain unaltered
(D) Temperature reduced to thrice of its initial
value
Which of the following is/are incorrect
regarding the universal gas constant (R) ?
(A) R is independent on pressure
(B) R is independent on temperature
(C) R is independent on volume of gas
(D) R is dependent on nature of gas
To determine the value of R, which of the
PV value is considered to be equal for every
gas at 273 K ?
(A) lim (PVm )
(B) lim (PVm )
(C) lim (PVm )
(D) lim (PVm )
P1atm
P 
Q.12
(B) d
(C) 20.5 L
M3
M2 M3 > M2 > M1
M1
P

(D) d
P
P
Questions
based on
Partial pressures
Q.14
Dalton's law of partial pressures are applicable
to (A) Non reacting gases
(B) Ideal gases
(C) Temperature of the component gases in the
mixture remain same
(D) All of the above
Q.15
1000 ml of a gas A at 600 torr and 500 ml of
gas B at 800 torr are placed in a 2L flask. The
final pressure will be (A) 2000 torr
(B) 1000 torr
(C) 500 torr
(D) 1400 torr
Q.16
Which of the following is/are true regarding
vapoure pressure ?
(A) Vapour pressure is surface property of the
solvent
(B) Vapour pressure is independent on
temperature
(C) The saturation vapour pressure is
corresponding to the liquid vapour
equilibrium
(D) Both (A) and (C) are correct
V0
(D)
M1 > M2 > M3
M1 M2
P

(C) d
P
P0
The volume 10 mole of an ideal gas at 10 atm
and 500 K is (A) 82 L
(B) 41 L
M3
82
L
3
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GASEOUS STATE
23
Questions
based on
Distribution of molecules with velocity is
represented by the curve :
Velocity corresponding to point A is -
(A)
Q.18
Questions
based on
Q.19
Q.20
3RT
M
(B)
Q.21
8RT
(D)
M
the
intermolecular
(D) All of the above are correct
Q.22
Regarding compressibility factor, Z which of
the following is/are correct ?
(A) For most of the real gases Z decreases with
P at the lower pressure
(B) For most of the real gases Z increases with
P at the higher pressure
Diffusion of gas
Which of the following differentiate between
diffusion and effusion ?
(A) Diffusion is the intermixing of the gas
molecules at any direction effusion is the
reverse of diffusion
(B) Diffusion is the property of the gas
molecules effusion is the property of the
gas container only
(C) Diffusion occur at any direction whereas
effusion occur in under the potential
difference
(D) Diffusion is the intermixing of the gas
molecules whereas effusion is the passage
of gas molecules through the pores in one
direction
(C) For H2(g) and He(g) Z increases with P at
all the pressure & at room temperature
(D) All of the above are correct
Q.23
Which of the following satisfies the greater
compressibility of real gas ?
(A) Z < 1
(B) At the higher pressure
(C) Above the Boyle’s temperature
(D) Lesser the value of ‘‘a’’ but higher value of ‘‘b’’
Q.24
To a given container having a pore of definite
size, gas A (mol. wt = 81) is filled till the final
pressure become 10 atm. It was seen in 50
minutes 10 gm of A was effused out. Now the
container was completely evacuated and filled
with gas B (mol. wt = 100) till the final
pressure become 20 atm. In 75 minutes how
many gm of B will be effused out ?
100
gm
6
200
(C)
gm
3
on
(C) ‘‘a’’ and ‘‘b’’ are the characteristic
constants not the universal gas constant
RT
M
Temperature at which most probable speed of
O2 becomes equal to root mean square speed of
N2 is - [Given : N2 at 427ºC]
(A) 732 K
(B) 1200 K
(C) 927 K
(D) 800 K
(A)
Regarding the vanderwaals constant which of
the following is/are correct ?
(B) ‘‘b’’ depends on the size of the gas
molecules
u
2RT
(C)
M
Real gas
(A) ‘‘a’’ depends
interactions
A
molecules
Q.17
Questions
based on
Kinetic theory of gas
100
gm
3
250
(D)
gm
3
(B)
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Below the critical temperature which of the
following represent vanderwaals gas ?
(A)
(B)

P

P
Vm 

Vm 
(D) 
(C) P
P
Vm 
GASEOUS STATE
V m
24
Q.25
Q.26
Q.27
Q.28
Q.29
Which of the following characterises the
critical point ?
(A) At the critical point both liquid and solid
phase coexist
(B) At the critical point solid, liquid and gas
phase coexist
(C) At the critical point liquid and gas phase
coexist together
(D) At the critical point liquid and gas phase
have unequal density
Which of the following is true at the critical
point ?
(A) At the critical point three roots of
vanderwaals equation are equal
(B) Below the critical point two roots of the
vanderwaals equation are equal and
imaginary but one root is real
(C) Above the critical point density of gas is
greater than density of liquid
(D) Above the critical point three roots of
vanderwaals equation are real but unequal
Which of the following is most suitable for
liquefication ?
(A) T > TC & P > PC (B) T < TC & P < PC
(C) T < TC & P > PC (D) T < TC & P = 0
Critical temperature and critical pressure values
of four gases are given Gas
C.Temp (K) C. Pressure (atm.)
P
5.1
2.2
Q
33
13
R
126
34
S
135
40
Which of the gas/gases cannot be liquefied at a
temperature 100 K and pressure 50 atmospheres ?
(A) S only
(B) P only
(C) R and S
(D) P and Q
Q.30
For which of the following gas/gases.
close to 0.22 ?
(A) Cl2
(C) C2H4
PC VC
RTC
(B) CH3OH
(D) CH4
 True or false type questions
Q.31
Effusion is a one directional process
Q.32
For ideal gas, density 
Q.33
For ideal gas mean free path 
Q.34
For ideal gas, collision frequency Z11   
Q.35
For He, Cp,m/Cv,m = 1.66
PM
RT
P
T
P
T
2
 Fill in the blanks type questions
Q.36
Vanderwaals reduced equation of state is


   3  (3 – 1) = 
2 

 

Q.37
At the higher pressure compressibility factor Z
increases with pressure because of ……….. of
gas molecules…….
Q.38
‘‘b’’ = …………….
Q.39
Above…………….compressibility
increases with pressure.
Q.40
factor
Z
TB
 ………………..
TC
Under critical states of a gas for one mole of a
gas, compressibility factor is (A)
3
8
(B)
8
3
(C) 1
(D)
1
4
EXERCISE # 2
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GASEOUS STATE
25
Q.6
Part-A
Q.1
Only single correct answer type
questions
A gas at a pressure of 5.0 atm is heated from 0º
to 546 ºC and simultaneously compressed to
one-third of its original volume. Hence final
pressure is (A) 10.0 atm
(B) 30.0 atm
(C) 45.0 atm
(D) 5.0 atm
Q.2
An open vessel containing air is heated from
300 K to 400 K. The fraction of air originally
present which goes out of it is at 400 K (A) 3/4
(B) 1/3
(C) 2/3
(D) 1/8
Q.3
What weight of hydrogen at STP could be
contained in a vessel that holds 4.8 g oxygen at
STP ?
(A) 4.8 g
(B) 3.0 g
(C) 0.6 g
(D) 0.3 g
Q.4
N2 + 3H2  2NH3. 1 mol N2 and 4 mol H2
are taken in 15 L flask at 27 ºC. After complete
conversion of N2 into NH3, 5 L of H2O is
added. Pressure set up in the flask is -
Q.5
(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
A sample of air contains only N2, O2 and H2O.
It is saturated with water vapours and total
pressure is 640 torr. The vapour pressure of
water is 40 torr and the molar ratio of
N2 : O2 is 3 : 1. The partial pressure of N2 in
the sample is (A) 540 torr
(B) 900 torr
(C) 1080 torr
(D) 450 torr
A vessel is filled with mixture of oxygen and
nitrogen. At what ratio of partial pressure will
the mass of gases be identical (A) p(O2) = 0.785 p(N2) (B) p(O2) = 8.75 p(N2)
(C) p(O2) = 11.4 p(N2) (D) p(O2) = 0.875 p(N2)
Q.7
A cylinder of compressed gas bears no label is
supposed to contain either ethylene (C2H4) or
propylene (C3H6). Combustion of the gaseous
sample shows that 12 mL of gas required 54
mL of oxygen for complete combustion. This
indicates that the gas is (A) only ethylene
(B) only propylene
(C) 1 : 1 mixture of two gases
(D) some unknown mixture of two gases
Q.8
In an effusion experiment, it required 40 s for a
certain number of moles of a gas of unknown
molar mass to pass through a small orifice into
a vacuum. Under the same conditions, 16 s
were required for the same number of moles of
O2 to effuse. What is the molar mass of the
unknown gas ?
(A) 5.1 g/mol
(B) 12.8 g/mol
(C) 80 g/mol
(D) 200 g/mol
Q.9
Two glass bulb A and B are connected by a
very small tube (of negligible volume) having
stop cock. Bulb A has a volume of 100 cm3 and
contains certain gas while bulb B is empty. On
opening the stop cock, the pressure in ‘A’ fell
down by 60%. The volume of bulb B must be (A) 200 mL
(B) 150 mL
(C) 250 mL
(D) 100 mL
Q.10
A certain gas effuses out of two different
vessels A and B. A has a circular orifice while
B has a square orifice of length equal to the
radius of the orifice of vessel A. The ratio of
rate of diffusion of the gas from vessel A to
that from vessel B is -
Q.11
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(A)  : 1
(B) 1 :  (C) 1 : 1 (D) 3 : 2
A balloon with volume 4200 m3 is filled with
helium gas at 27ºC, 1 bar pressure and is found
to weigh 700 kg. if density of air is 1.2 kg m–3,
the pay load of ballon iS GASEOUS STATE
26
(A) 5040 kg
(C) 3500 Kg
Q.12
one. 0.8gm of argon gas had to be removed to
maintain original pressure. The temperature T is–
(A) 510 K
(B) 200 K
(C) 100 K
(D) 73 K
(B) 4340 kg
(D) 5740 kg
Two glass bulbs A (of 100 mL capacity), and B
(of 150 ml capacity) containing same gas are
connected by a small tube of negligible
volume. At particular temperature the pressure
before opening the valve
Q.17
2 gm of a gas ‘A’ in a closed vessel at room
temperature showed a pressure of 1 atm. When
3 gm of another gas ‘B’ was mixed with gas
‘A’ in the vessel. The pressure rose to 1.5 atm
at the same temperature. The ratio of the molar
masses of the gases ‘A’ and ‘B’ is (A) 3 : 1
(B) 1 : 3
(C) 4 : 3
(D) 2 : 1
Q.18
Specific heats of certain elementry gas at
constant volume is 315 J kg–1 K–1 and that at
constant pressure is 441 J Kg–1 K–1.
7.0 g of the gas is found to occupy a volume of
4.1 L at 27ºC and 1 atm. pressure. What is the
atomic mass of the gas ?
(A) 21 U
(B) 42 U
(C) 63 U
(D) 10.5 U
Q.19
Van der Waal’s constant ‘b’ and the
corresponding values of critical temperature for
three gases P, Q, R are given below Gas
Critical
Van der Waal’s
temperature (Tc) constant (b)
PA
20
=
. The
PB
1
stopcock is opened without changing the
temperature. The pressure in A will (A) drop by 75%
(B) drop by 57%
(C) drop by 25%
(D) will remain same
Q.13
A gas with formula CnH2n+2 diffuses through
the porous plug at a rate one sixth of the rate of
diffusion of hydrogen gas under similar
condition. The formula of gas is (A) C2H6
(B) C10H22
(C) C5H12
(D) C9H14
Q.14
A gaseous mixture of three gases A, B and C
has a pressure of 10 atm. The total number of
moles of all the gases is 10. If the partial
pressure of A and B are 3.0 and 1.0 atm
respectively and if C has mol. wt. of 2.0, what
is the weight of C in g present in the mixture ?
(A) 6
(B) 8
(C) 12
(D) 3
Q.15
Q.16
A 1 : 1 mixture (by weight) of hydrogen and
helium is enclosed in a one litre flask at
temperature 0º C. Assuming ideal behaviour,
the partial pressure of helium is found to be
0.42 atm then concentration of hydrogen would
be (A) 0.0375
(B) 0.028
(C) 0.0562
(D) 0.0187
At a temperature T K, the pressure of 4.0 gm
argon in a bulb is P. The bulb is put in a bath
having temperature higher by 50K then the first
P
–200 ºC
0.03 litre mole–1
Q
–100 ºC
0.02 litre mole–1
R
+ 50 ºC
0.01 litre mole–1
Which of the gas/gases is/are liquefiable at a
temperature – 110 ºC by application of increasing
pressure ?
(A) P, Q, R
(B) P, Q
(C) Q, R
(D) None of P, Q, R
Q.20
If molecules of the gas are spherical of radius
1 Å, the volume occupied by the molecules in
1 mol of a gas is (A) 22400 mL
(B) 22.4 L
(C) 2.514 mL
(D) 4.22 mL
Q.21
NH3 gas is liquefied more easily than N2.
Hence (A) Van der Waals constants a and b of
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GASEOUS STATE
27
NH3 > that of N2
(B) Van der Waals constants a and b of
NH3 < that of N2
(C) a(NH3) > a (N2) but b (NH3) < b (N2)
(D) a (NH3) < a (N2) but b (NH3) > b (N2)
Part-B
Q.22
Q.23
Q.24
Q.25
(C) The value of r.m.s increases with rise in
temperature.
(D) Area under the curve gives the total
number of molecules.
Q.26
Which of the following mixtures of gases at
room temperature follow Dalton’s law of
partial pressure ?
(A) NH3, HCl
(B) H2, O2
(C) NO, O2
(D) SO2, O2
Q.27
Which sample of gas given below contains
Avogadro number of atom at S.T.P. ?
(A) 1 mol of Helium
(B) 11.2 L of carbon monoxide
(C) 11.2 L of sulphur dioxide
(D) 1 mol of phosphine
Q.28
The temperature of ideal gas can be increased
by (A) decreasing the volume and pressure but
keeping the amount constant
(B) increasing the pressure but keeping the
volume and amount constant
(C) decreasing the amount but keeping the
volume and pressure constant
(D) increasing the amount but keeping the
volume and pressure constant
Q.29
Which of the following matchings
incorrect?
(A) C6H6 (g)
....... a = 0.217
(B) C6H5CH3 (g) ....... a = 18
(C) Ne
...... a = 5.464
(D) H2O(g)
....... a = 24.06
Q.30
The critical temperature and pressure for NO
are 177 K and 6.485 MPa respectively, and for
CCl4 these are 550 K and 4.56 KPa,
respectively. which gas.
(i) has the smaller value for vander waals
constant b ?
(ii) has smaller value of constant a ?
(iii) is most nearly ideal in behaviour at 300 K
and 1.013 MPa ?
One or more than one correct
answer type questions
Which of the statements are false ?
(A) Gases and liquids have viscosity as
common property
(B) Gases and liquids have pressure as
common property
(C) Gases cannot be directly condensed into
solids without passing through liquid state
(D) Particles in all the three states have random
translational motion.
Which of the following do not pertain to the
postulates of kinetic theory of gases ?
(A) The gas molecules are perfectly elastic
(B) Gas molecules move at random with ever
changing speeds.
(C) Molecular collisions against the wall are
responsible of gas pressure.
(D) KE of a gas is given by the sum of 273 and
temperature in celsius scale.
Four gas balloons A, B, C, D of equal volumes
containing, H2, N2O, CO, CO2 respectively were
picked with needle and immersed in a tank
containing CO2. Which of them will shrink
after some time ?
(A) A
(B) B
(C) C
(D) both A and D
According to Maxwell-Boltzman distribution
of speeds among gas molecules, what is false?
(A) The maxima in plot of
N
s speeds
N
pertains to average speed
(B) The plot of
N
s speeds is straight line
N
with slope > 0
are
Part-C Assertion-Reason type questions
Choose any one of the following four responses.
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GASEOUS STATE
28
(A) If both Assertion and Reason are true
and the Reason is correct explanation of
the Assertion.
(B) If both Assertion and Reason are true
but the Reason is not correct
explanation of the Assertion.
(C) If Assertion is true but the Reason is
false.
(D) If Assertion is false & Reason is true.
Part-D Column Matching type questions
Q.36
Column-I
(A)
(B)
Q.31
Q.32
Q.33
Q.34
Q.35
Assertion : The mixture of ideal gases A and B
on their liquefaction gives ideal solution of A
and B.
Reason : The ideal gas can not be liquefied.
Assertion : At a particular temperature, the
value of mean free path increases with decrease
in pressure.
Reason : All the gas molecules at a particular
temperature possess same speed.
Assertion : Dry air is heavier than moist air.
Reason : The vapour density of moist air lies
between 9 and 14.4.
Assertion : Heat capacity of a diatomic gas is
higher than that of monoatomic gas.
Reason : Monoatomic gases are non-polar in
nature.
Assertion : Noble gases can be liquified.
Reason : Attractive forces can exist between
nonpolar molecules.
Match the entries in column-I with entries in
Column-II and then give the correct answer.
(C)
(D)
Q.37
(A)
Column-II
1
vs P for ideal gas at
V2
constant T and n
(P)
1
for ideal gas at
T
constant P and n
(Q)
log P vs log V for ideal
gas at constant T and n
(R)
1
for ideal gas at
P2
constant T and n
(S)
V vs
V vs
Match the column :
Column-I

a 
 P 

4V 2 

(P)
Column-II
Real gas
(2V–b = RT)
(B)
(C)
3P / d
Cp
Cv
= 1.66
(Q)
(R)
van der waal’s eqn for
0.5 mole gas
Root mean square
velocity
(D)
2a
Rb
(S)
Monatomic gas
(E)
Z > or < 1
(T)
Inversion temperature
Q.38
Match the column :
(A)
Column-I
Liquefaction of gas
Column-II
(P) T < TC
(B)
(C)
(D)
Impossible liquefaction of gas
Condensed vapour
Super critical fluid
(Q)
(R)
(S)
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GASEOUS STATE
T > TC
P > PC
P < PC
29
EXERCISE # 3
Part-A Subjective Type Questions
Q.5
Find the number of diffusion steps required to
separate the isotopic mixture initially
containing some amount of H2 gas and 1 mol of
D2 gas in a container of 3 lit capacity
maintained at 24.6 atm and 27ºC to the final
 wD 
1
2 
mass ratio 
equal to .
 wH 
4
2 

Q.6
Calculate the pressure of a barometer on an
aeroplane which is at an altitude of 10 Km.
Assume the pressure to be 101.325 Kpa at sea
level and the mean temperature 243 K. Use the
average molar mass of air (80% N2, 20%O2)
Q.7
The ratio of velocities of diffusion of gases A
and B is 1 : 4. If the ratio of their masses
present in the mixture is 2 : 3, calculate the
ratio of their mole fractions.
Q.8
A gaseous mixture of He and O2 is found to
have a density of 0.518 gL–1 at 25º C and 720
torr. what is the percent by mass of helium in
this mixture ?
Q.9
What is the ratio of the number of molecules
having speeds in the range of 2ump and 2ump + du
to the number of molecules having speeds in
the range of ump and ump + du ?
Q.10
Two gases A and B having molecular weights
60 and 45 respectively are enclosed in a vessel.
The wt. of A is 0.50 g and that of B is 0.2 g.
The total pressure of the mixture is 750 mm.
Calculate partial pressure of the two gases.
Q.11
A jar contains a gas and a few drops of water at
T K. The pressure in the jar is 830 mm of Hg.
The temperature of the jar is reduced by 1%.
The vapour pressure of water at two
temperatures are 30 and 25 mm of Hg.
Calculate the new pressure in the jar.
Into a gas bulb of 2.83 litres, are introduced
0.174 g of H2 and 1.365 g of N2 which can be
assumed to behave ideally. The temperature is
0ºC. What are the partial pressure of H2 and N2
and what is the total gas pressure ? What are
the mole-fractions of each gas ? What are
pressure fractions ?
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GASEOUS STATE
Q.1
Calculate the number of moles of gas present in
the container of volume 10 lit at 300 K. If the
manometer containing glycerin shows 5m
difference in level as shown in diagram. Given:
dglycerin = 2.72 gm/ml, dmercury = 13.6 gm/ml.
Patm = 1atm
5m
Q.2
A manometer attached to a flask contains NH3
gas have no difference in mercury level initially
as shown in diagram. After the sparking into
the flask, it have difference of 19 cm in
mercury level in two columns. Calculate %
dissociation of ammonia.
Patm = 760 mm
NH3 gas
Q.3
Q.4
A large irregularly shaped closed tank is first
evacuated and then connected to a 50 litre
cylinder containing compressed nitrogen gas.
The gas pressure in the cylinder, originally at
21.5 atm, falls to 1.55 atm after it is connected
to the evacuated tank. Calculate the volume of
the tank.
30
Q.12
Show that at low densities, the vander waals
equation
Q.19
partition in 60 S, what volume of O2 will
diffuse under similar conditions in 30 S.

a 
 p  2 (Vm  b)  RT and the Dieterici's eqn

Vm 

p(Vm – b) = RT exp (–a/RTVm) give essentially
the same value of p.
Q.13
Q.14
A certain volume of H2 effuses from an
apparatus in one minute. The same volume of
ozonized oxygen (O3 + O2 mix) took 246 s to
effuse from the apparatus under identical
conditions. Find the % composition of the
ozonized oxygen.
One mole of an ideal gas is subjected to a
1
process in which P =
V where P is in atm
8.21
and V in litre. If the process is operating from 1
atm to finally 10 atm (no higher pressure
chieved during the process) then what would be
the maximum temperature obtained and at what
instant will it occur in the process.
Q.16
Reduced temperature for benzene is 0.7277 and
its reduced volume is 0.40. Calculate the
reduced pressure of benzene.
Q.17
Q.18
Q.20
Calculate the total pressure in a 10 litres
cylinder which contains 0.4 g of helium, 1.6 g
of oxygen and 1.4 g of nitrogen at 27 ºC. Also
calculate the partial pressure of helium gas in
the cylinder. Assume ideal behaviour of gases.
(R = 0.082 l- atm K–1 mol–1)
Q.21
A compound exists in the gaseous state both as
monomer (A) and dimer (A2). The molecular
weight of the monomer is 48. In an experiment,
96 g of the compound was confined in vessel of
volume 33.6 L and heated to 273 ºC. Calculate
the pressure developed, if the compound exists
as a dimer to the extent of 50% by weight
under these conditions.
Q.22
A long rectangular box is filled with chlorine
(at. wt. : 35.45) which is known to contain only
35
Cl and 37Cl. If the box could be divided by a
partition and the two types of chlorine
molecules put in the two compartments
respectively, calculate where should the
partition be made if the pressure on both sides
are to be equal. Is this pressure the same as the
original pressure ?
Q.23
At 627º C and 1 atm pressure, SO3 undergoes
partial dissociation into SO2 and O2
SO3
SO2 + 1/2O2
A mixture consisting of 80 mole per cent
hydrogen and 20 mole per cent deuterium at
25º C and a total pressure of 1 atm is permitted
to effuse through a small orifice of area
0.20 mm2. Calculate composition of the initial
gas that pass through.
Q.15
A gas has a density of 1.2504 g / l at 0ºC and a
pressure of 1 atm, Calculate the rms, average
and the most probable speeds of its molecules
at 0º C.
If the observed density of the equilibrium
mixture is 0.925 g / l, calculate degree of
dissociation of SO3.
Q.24
The kinetic molecular theory attributes an average
kinetic energy of
20 dm3 of SO2 diffuse through a porous
3
KT to each particle. What
2
rms speed would a mist particle of mass 10–12 g
have at room temperature (27ºC) according to the
kinetic molecular theory ?
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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 the number of mol of O2 gas
in each flask.
GASEOUS STATE
31
Q.25
One of the best rocket fuels is dimethyl
hydrazine (DNH) when mixed with N2O4 (l), it
reacts according to the equation
Q.30
(CH3)2 N2H2 () + 2N2O4 () 
 
3N2 (g) + 4H2O(g) + 2CO2 (g)
If 2.5 mol of DMH react completely with N2O4
and products formed are collected at 27 ºC in
250 L tank, calculate total pressure formed in
the tank. [H2O (g) is liquefied at 27 ºC]
Q.26
Q.27
Q.28
Q.29
A mixture containing 1.12 litre of H2 and 1.12
litre of D2 at NTP is taken inside a bulb
connected to another bulb a stop cock with a
small opening. The second bulb is fully
evacuated, the stop cock opened for certain
time and then closed. The first bulb is found to
contain 0.05 g of H2. Determine the percentage
composition by volume of the gases in the
second bulb.
Part-B Passage based objective questions
Passage-1 (Question 31 to 35)
Consider the diagram given below about
Maxwell’s distribution of speeds at two
temperatures T1 and T2 and answer the
following question :
A cylinder containing 5.0 litre O2 at 25 ºC was
leaking. When the leakage was detected and
stopped there was a change in the pressure of
the gas from 3.0 atm to 2.235 atm. How much
oxygen in g has been leaked during this period
? Also report the volume leaked if collected at
1 atm and 25 ºC.
A mixture of H2Ov, CO2 and N2 was trapped in
a glass apparatus with a volume of 0.731 mL.
The pressure of total mixture was 1.74 mm of
Hg at 23ºC. The sample was transferred to a
bulb in contact with dry ice (–75ºC) so that
H2Ov are forzen out. When the sample returned
to normal value of temperature, pressure was
1.32 mm of Hg. The sample as then transferred
to a bulb in contact with liquid N2 (–95ºC) to
freeze out CO2. In the measured volume,
pressure was 0.53 mm of Hg at orginal
temperature. How many moles of each
constituent are in mixture ?
A gas present in a container connected to
frictionless, weightless piston operating always
at one atmosphere pressure such that it permits
flow of gas outside (with no adding of gas).
The graph of n vs T (Kelvin) was plotted and
was found to be a straight line with coordinates of extreme points as (300, 2) and
(200, 3). Calculate
(i) relationship between n and T
(ii) relationship between V and T
(iii) Maxima or minima value of 'V'
y
x
c
d
T1
N
N
a b z
T2
speed
Q.31
Area marked as a b d c represents (A) Number of molecules having speeds
between a and b
(B) Number of molecules having speed less
than z
(C) Number of molecules having r.m.s. at T1
(D) Number of molecules having speed
between c and d
Q.32
Total area under the curve T1 is (A) equal to that under curve T2
(B) less than that under curve T2
(C) greater than that under curve T2
(D) can be greater or less than that under
curve T2
Radius of a spherical molecule of a gas is
2 × 10–8 cm. Calculate (a) Co–volume per molecule
(b) Co–volume per mole
(c) Cirtical volume
CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000
GASEOUS STATE
32
Q.33
Q.34
Q.35
The relation ship between temperatures T1 and
T2 is (A) T1 = T2
(B) T1 > T2
(C) T1 < T2
(D) unpredictable from the diagram
In a certain sample of gas at 25ºC, the number
of molecules having speeds between 4 km sec–1
and 4.1 km sec–1 is N . If the total number of
gas molecules at the same temperature are
doubled what will happen ?
(A) Value of most probable velocity will
change
(B) Area under the Maxwell’s curve for
distribution of speeds will increase by four
times
(C) No. of molecules between 4 km sec–1 and
4.1 km sec–1 will become 2 N
(D) No. of molecules between 4 km sec–1 and
4.1 km sec–1 will remain same.
According to Maxwell-Boltzman distribution
of speeds among gas molecules, what is false?
(A) The maxima in plot of N/N s speeds
pertains to average speed.
(B) The plot of N/N s speeds is straight line
with slop > 0
(C) The value of r.m.s. increases with rise in
temperature
(D) Area under the curve gives the total
number of molecules
Passage-2 (Question 36 to 38)
On the recently discovered 10th planet it has
been found that the gases follow the
relationship PeV/2 = nCT where C is constant
other notation are as usual (V in lit., P in atm
and T in Kelvin). A curve is plotted between
P and V at 500 K and 2 moles of gas as shown
in figure.
Q.36
The value of constant C is (A) 0.01
(B) 0.001
(C) 0.005
(D) 0.002
Q.37
Find the slope of the curve plotted between
P vs T for closed container of volume 2 lit.
having same moles of gas (A)
e
2000
(C) 500 e
Q.38
(B) 2000 e
(D)
2
1000e
If a closed container of volume 200 lit. of O2
gas (ideal gas) at 1 atm and 200 K is taken to
planet. Find the pressure of oxygen gas at the
planet at 821 K in same container 10
e100
(C) 1 atm
(A)
20
e50
(D) 2 atm
(B)
Passage-3 (Question 39 to 43)
Any gas can be liquified by decreasing it’s
temperature and by increasing it’s pressure.
Decreasing of the temperature decreases the
average kinetic energy of molecules &
increasing of the pressure decreases the average
distance between molecules. When the
molecules are close together, their kinetic
energy is lowered. They do not posses enough
energy to overcome the force of attraction
between molecules & liquid is formed. For
each substance, however there exist a
temperature, above which the substance can not
be liquified, no matter how great the applied
pressure.
Q.39
1 atm
Temperature above which gas can not be
liquified is called (A) Critical temperature
(B) Boyle’s temperature
(C) Inversion temperature
(D) triple point
P
V (lit)
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GASEOUS STATE
33
Q.40
An ideal gas obeying kinetic gas equation can
be liquified if (A) it’s temperature is more than critical
temperature
(B) It’s pressure is more than critical
temperature
(C) it’s pressure is more than critical pressure
but it’s temperature is less than critical
temperature
(D) it can not be liquefied at any value of P&T
Passage-4 (Question 44-45)
Under a given condition, it is found that two
separate gases effuse out of two separate
container in such a way that they follows the
dN
dN
equation
  K1N and
  K 2 N , K1 =
dt
dt
6.93 × 10–3 sec–1 , K2 = 6.93 × 10–5 sec–1, where
N is no. of molecules remaining in the
container.
Q.44
Q.41
Q.42
Q.43
Critical temperature of CO2 is 31ºC. CO2 is (A) a gas at 35ºC & vapour at 25ºC
(B) a gas at 35ºC as well as at 25ºC
(C) Vapour at 35ºC as well as at 25ºC
(D) a gas at 35ºC & liquid at 25ºC
To liquefy a gaseous substance whose critical
temperature is below room temperature,
requires (A) high pressure & lowering of temperature
(below Tc)
(B) low pressure & raising of temperature
(above Tc)
(C) high pressure & raising of temperature
(above Tc)
(D) low pressure & lowering of temperature
(below Tc)
1
(B) t = 0
1
(C) t = 0
1
(D) t = 0
1
Q.45
Which of the following statements on critical
constants of gases are correct T 
A. Larger the value of  c  of a gas, larger
 Pc 
would be excluded volume
B. Critical temperature (Tc) of a gas is greater
than Boyle temperature (Tb)
C. At critical point in the van der waals gas
 P 
isotherm 
 =0
 V  Tc
(A) A and B
(C) B and C
Which of the following may represent fraction
of no. molecules present after the given interval
for gas-I ?
t = 100sec
t = 200sec
(A) t = 0
(B) A and C
(D) A,B and C
CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000
1
2
1
8
t = 100sec
t = 200sec
1
8
1
16
t = 100sec
t = 200sec
1
2
1
4
t = 100sec
t = 200sec
1
4
1
16
Identify the correct option regarding sequence
of (True) and (False) statements
(i) The time required for moles of gas I to get
reduced to half of original and that of gas II
to reduced to half of original is
independent of initial moles of gas I and
gas II.
(ii) The rate at which initially molecules will
come out in gas I as compared to gas II
will be greater in gas II if initial no. of
molecules are same
(iii) The time required for moles to get reduced
from 1 to 0.8 in gas I and 2 to 1.6 in gas II
will be same
(iv) For the two gases, moles remaining on the
container after same interval should be in
Geometrical Progression.
(A) TFFT
(B) TFTT
(C) FTFT
(D) TTFF
GASEOUS STATE
34
EXERCISE # 4
 Old IIT-JEE Objective type questions
Q.1
Q.2
At 100 ºC and 1 atm, if the density of liquid
water is 1.0 g cm–3 and that of water vapour is
0.0006 g cm–3, then the volume occupied by
water molecules in 1 litre of steam at that
temperature is [IIT-2000]
3
3
(A) 6 cm
(B) 60 cm
3
(C) 0.6 cm
(D) 0.06 cm3
The rms velocity of hydrogen is 7 times the
rms velocity of nitrogen. If T is the temperature
of the gas, then [IIT-2000]
(A) T( H 2 ) = T( N 2 )
(B) T( H 2 ) > T( N 2 )
(C) T( H 2 )
Q.3
Q.4
<
T( N 2 )
(D) T( H 2 ) 
(D)
Q.6
Q.7
• (38.8L,373 K)
(14.2L,373 K)
•
In the case of positive deviation from ideal gas,
(A) Interactions in molecules,
PV
>1
nRT
(B) Interactions in molecules,
PV
<1
nRT
(C) Finite size of molecules,
PV
>1
nRT
(D) Finite size of molecules,
PV
<1
nRT
For 1 mole of gas the average kinetic energy is
given as E. The Urms of gas is [IIT-2004]
1
 2E  2
(C) 

 3M 
Q.8
1
 3E  2
(B)  
M
1
 3E  2
(D) 

 2M 
Rate of diffusion of helium with respect to
methane at same temperature and pressure [IIT-2005]
V(L)
(20.4L,•
273 K)
(22.4L,•
273 K)
1
T(K)
(B)
V(L)
 2E  2
(A)  
M
Which
of
the
following
volume
(V) – temperature (T) plots represent the
behaviour of one mole of an ideal gas at one
atmospheric pressure [IIT-2002]
V(L)
(A) (22.4L,•
273 K)
•(30.6L,373 K)
[IIT-2003]
(D) 1/ d
d
(22.4L,•
273 K)
T(K)
7 T( N 2 )
The root mean square velocity of an ideal gas at
constant pressure varies with density (D) as[IIT-2001]
2
(A) d
(B) d
V(L)
T(K)
The compressibility of a gas is less than unity
at S.T.P. Therefore,
[IIT-2000]
(A) Vm > 22.4 litres
(B) Vm < 22.4 litres
(C) Vm = 22.4 litres
(D) Vm = 44.8 litres
(C)
Q.5
(C)
•(28.6L,373 K)
(A) 1/2
(B) 2
(C) 4
(D) 1
T(K)
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GASEOUS STATE
35
Q.9
Which of the following is the only incorrect
statement ?
[IIT-2006]
Q.12
C
A
Ideal gas
Z
B
P
 Old IIT-JEE Subjective type questions
(A) For the gas (A) ; a = 0 and it dependence
linearly with P at all pressure value
Q.13
Calculate the pressure exerted by one mole of
CO2 gas kept in one litre vessel at 273 K if the
vander Waal's constant a = 3.592 dm6 atm mol–2.
Assume that the volume occupied by CO2
molecules is negligible.
[IIT-2000]
Q.14
The compression factor (compressibility factor)
for one mole of a vander Waal's gas at 0ºC and
100 atm pressure is found to be 0.5. Assuming
that the volume of a gas molecule is negligible,
calculate the vander Waal's constant (a)
[IIT-2001]
Q.15
The density of the vapour of a substance at
1 atm pressure and 500 K is 0.36 kg m–3. The
vapour effuses through a small hole at a rate of
1.33 times faster than oxygen under the same
condition.
(a) Determine (i) Molecular weight of
substance (ii) Molar volume of vapour (iii)
Compression factor (Z) of vapour (iv)
Type of force dominent in gas molecules,
attraction or repulsion.
(b) If vapour shows ideal behaviour at 1000 K,
then calculate average translational kinetic
energy.
[IIT-2002]
Q.16
The average velocity of gas molecules is
400 m/sec. Calculate its rms velocity at the
same temperature.
[IIT-2003]
Q.17
For a real gas which obeys van der waal’s
equation a graph is plotted between PV m
(y-axis) and P(x - axis), where Vm is molar
volume. Find the y - intercept of the graph.
[IIT-2004]
(B) For the gas (B), b = 0 & its dependence
linearly with pressure at all pressure value
(C) C represent a typical real gas for which
neither a nor b is equal to 0. If minima &
the point of intersection is known with
Z = 1, a & b can be calculated.
(D) At high pressure, the slope is positive for
all real gases
Q.10
A gas described by van der Waals equation[IIT-2008]
(A) behaves similar to an ideal gas in the limit
of large molar volumes
(B) behaves similar to an ideal gas in the limit
of large pressures
(C) is characterised by van der Waals
coefficients that are dependent on the
identity of the gas but are independent of
the temperature
(D) has the pressure that is lower than the
pressure exerted by the same gas behaving
ideally
Q.11
The term that correct for the attractive forces
present in a real gas in the van der Waals
equation is [IIT-2009]
(A) nb
(C) –
an 2
V2
According to kinetic theory of gases (A) collisions are always elastic [IIT-2011]
(B) heavier
molecules
transfer
more
momentum to the wall of the container
(C) only a small number of molecules have
very high velocity
(D) between collisions, the molecules move in
straight lines with constant velocities
an 2
(B) 2
V
(D) – nb
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GASEOUS STATE
36
N2 is adsorbed in 20% of the surface sites. N2
gas evolved on heating was collected at 0.001
atm and 298 K in a container of volume
2.46 cm3. Find out the no. of surface sites
occupied per molecule of N2. If the density of
surface sites is 6.023× 1014/cm2 and surface
area is 1000 cm2.
[IIT-2005]
Q.21
To an evacuated vessel with movable piston
under external pressure of 1 atm., 0.1 mol of
He and 1.0 mol of an unknown compound
(vapour pressure 0.68 atm, at 0ºC) are
introduced. Considering the ideal gas
behaviour, the total volume (in litre) of the
gases at 0ºC is close to.
[IIT-2011]
Q.19
Match gases under specified conditions listed
in Column-I with their properties / laws in
Column-II.
[IIT-2007]
Column-I
Column-II
(A) hydrogen gas
(P) compressibility
Q.22
For one mole of a van der Waals gas when b =
0 and T = 300 K, the PV vs. 1/V plot is shown
below. The value of the van der Waals constant
a (atm. liter2 mol–2)
[IIT-2012]
(P = 200 atm,
factor 1
T = 273 K)
(B) hydrogen gas
(Q) attractive forces
(P ~ 0, T = 273 K)
are dominant
(C) CO2 (P = 1 atm, (R) PV = nRT
Q.20
T = 273 K)
(D) real gas with very (S) P (V - nb) = nRT
large molar volume
At 400 K, the root mean square (rms) speed of
a gas X (molecular weight = 40) is equal to the
most probable speed of gas Y at 60 K. The
molecular weight of the gas Y is - [IIT-2009]
CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000
PV (liter-atm mol–1)
Q.18
24.6
23.1
21.6
20.1
0
(A) 1.0
(C) 1.5
[Graph not to scale ]
2.0 3.0
1
(mol liter–1)
V
(B) 4.5
(D) 3.0
GASEOUS STATE
37
EXERCISE # 5
 Old IIT-JEE Objective type questions
Q.1
When an ideal gas undergoes unrestrained
expansion, no cooling occurs because the
molecules.
[IIT-1984]
(A) are above the inversion temperature
(B) exert no attractive forces on each other
(C) do work equal to loss in kinetic energy
(D) collide without loss of energy
Q.2
Equal weights of methane and hydrogen are
mixed in an empty container at 25ºC. The
fraction of the total pressure exerted by
hydrogen is [IIT-1984]
(A)
Q.3
Q.4
Q.5
1
2
(B)
8
9
(C)
1
9
(D)
A bottle of dry ammonia and a bottle of dry
hydrogen chloride connected through a long
tube are opened simultaneously at both ends the
white ammonium chloride ring first formed
will be [IIT-1988]
(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
Q.7
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 L2 atm mol–2 respectively.
The gas which can most easily be liquified is [IIT-1989]
(A) O2
(B) N2
(C) NH3
(D) CH4
Q.8
The density of neon will be highest at [IIT-1990]
(A) STP
(B) 0ºC, 2 atm
(C) 273ºC, 1 atm
(D) 273ºC, 2 atm
Q.9
The rate of diffusion of methane at a given
temperature is twice that of a gas X. The
molecular weight of X is [IIT-1990]
(A) 64.0
(B) 32.0
(C) 4.0
(D) 8.0
Q.10
According to kinetic theory of gases, for a
diatomic molecule [IIT-1991]
(A) the pressure exerted by the gas is
proportional to mean velocity of the
molecule
(B) the pressure exerted by the gas is
proportional to the root mean velocity of
the molecule
(C) the root mean square velocity of the
molecule is inversely proportional to the
temperature
(D) the mean translational kinetic energy of the
molecule is proportional to the absolute
temperature
16
17
A liquid is in equilibrium with its vapour at it’s
boiling point. On the average, the molecules in
the two phases have equal [IIT-1984]
(A) inter-molecular forces
(B) potential energy
(C) kinetic energy
(D) total energy
Rate of diffusion of a gas is [IIT-1985]
(A) directly proportional to its density
(B) directly proportional to its molecular
weight
(C) directly proportional to the square root of
its molecular weight
(D) inversely proportional to the square root of
its molecular weight
The average velocity
at 27ºC is 0.3 m/s.
927ºC will be (A) 0.6 m/s
(C) 0.9 m/s
Q.6
of an ideal gas molecule
The average velocity at
[IIT-1986]
(B) 0.3 m/s
(D) 3.0 m/s
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GASEOUS STATE
38
Q.11
Q.12
Q.13
At constant volume, for a fixed number of
moles of a gas the pressure of the gas increases
with rise of temperature due to - [IIT-1992]
(A) increase in average molecular speed
(B) increase rate of collisions amongst
molecules
(C) increase in molecular attraction
(D) decrease in mean free path
Equal weights of ethane and hydrogen are
mixed in an empty container at 25ºC. The
fraction of the total pressure exerted by
hydrogen is [IIT-1993]
(A) 1 : 2
(B) 1 : 1
(C) 1 : 16
(D) 15 : 16
Q.17
The temperature of an ideal gas is increased
from 120 K to 480 K. If at 120 K the rootmean-square velocity of the gas molecules is v,
at 480 K it becomes [IIT-1996]
(A) 4v
(B) 2v
(C) v/2
(D) v/4
Q.18
The compressibility factor for an ideal gas is [IIT-1997]
(A) 1.5
(B) 1.0
Q.19
A vessel contains 1 mole of O2 gas (molar mass
32) at a temperature T. The pressure of the gas
is P. An identical vessel containing one mole
of He gas (molar mass 4) at a temperature 2T
has a pressure of [IIT-1997]
(A) P/8
(B) P
(C) 2P
(D) 8P
Q.20
According to Graham’s law, at a given
temperature the ratio of the rates of diffusion
rA/rB of gases A and B is given by- [IIT-1998]
A gas is heated from 0ºC to 100ºC at 1.0 atm
pressure. If the initial volume of the gas is
10.0 , its final volume would be - [REE-1995]
Q.14
(A) 7.32 
(B) 10.00 
(C) 13.66 
(D) 20.00 
The ratio between the root square velocity of
H2 at 50 K and that of O2 at 800 K, is -
(A) (PA/PB) (MA/MB)1/2
(B) (MA/MB) (PA/PB)1/2
[IIT-1996]
(A) 4
(C) 1
Q.15
(C) 25 seconds : CO
Q.16
(C) (PA/PB) (MB/MA)1/2
(B) 2
(D) 1/4
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 conditions is [IIT-1996]
(A) 10 seconds : He (B) 20 seconds : O2
(D) (MA/MB) (PB/PA)1/2
(where P and M are pressures and molecular
weights of gases A and B respectively)
Q.21
A gas will approach ideal behaviour at [IIT-1999]
(A) low temperature and low pressure
(B) low temperature and high pressure
(C) high temperature and low pressure
(D) high temperature and high pressure
Q.22
At a temperature T K, the pressure of 4.0 g
argon in a bulb is p. The bulb is put in a bath
having temperature higher by 50 K than the
first one. 0.8 g of argon gas and to be removed
to maintain original pressure. The temperature
T is equal to [REE-1999]
(A) 510 K
(B) 200 K
(C) 100 K
(D) 73 K
(D) 55seconds:CO2
One mole of N2O4(g) at 300 K is kept in a
closed container under one atmosphere. It is
heated to 600 K when 20% by mass of N2O4
(g) decompress to NO2 (g). The resultant
pressure is [IIT-1996]
(A) 1.2 atm
(B) 2.4 atm
(C) 2.0 atm
(D) 1.0 atm
(D) 
(C) 2.0
CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000
GASEOUS STATE
39
Q.23
Q.24
Q.25
A mixture of ethane (C2H6) and ethene (C2H4)
occupies 40 litres at 1.00 atm and at 400 K.
The mixture reacts completely with 130g of O 2
to produce CO2 and H2O. Assuming ideal gas
behaviour, calculate the mole fractions of C2H4
and C2H6 in the mixture.
[IIT-1995]
The composition of equilibrium mixture
(Cl2
2Cl) which is attained at 1200ºC, is
determined by measuring the rate of effusion
through a pinhole. It is observed that at 1.80
mm Hg pressure, the mixture effuses 1.16 times
as fast as krypton effuses under the same
condition. Calculate the fraction of chlorine
molecules dissociated into atoms
(At.wt. of Kr. = 84)
[IIT-1995]
Q.27
Calculate the mol fraction of N2O5(g)
decomposed at a constant volume and
temperature, if the initial pressure is 600 mm
Hg and the pressure at any time is 960 mm Hg.
Assume ideal gas behaviour.
[IIT-1998]
Q.28
One mole of nitrogen gas at 0.8 atm takes 38s
to diffuse through a pin hole, whereas one mole
of an unknown compound of xenon with
fluorine at 1.6 atm takes 57s to diffuse through
the same hole. Calculate the molecular formula
of the compound.
[IIT-1999]
Q.29
The pressure exerted by 12 g of an ideal gas at
temperature tºC in a vessel of volume V litre is
one atm. When the temperature is increased by
10 degrees at the same volume, the pressure
increases by 10%. Calculate the temperature t
and volume V. (Molecular wt. of gas is 120)
[IIT-1999]
One way of writing the equation of state for
–

real gases is P V  RT 1 

B

 ....... where B is
V

a constant. Derive an approximate expression for
B in terms of Vander Waal's constants a and b.
[IIT-1997]
Q.26
For the reaction
N2O5(g)
2NO2(g) + 0.5 O2(g).
Using van der waal’s equation, calculate the
constant 'a' when two moles of a gas confined
in a four litre flask exert a pressure of 11.0
atmosphere at a temperature of 300 K. The
value of ‘b’ is 0.05 lit mol–1
[IIT-1998]
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GASEOUS STATE
40
ANSWER KEY
EXERCISE # 1
Qus.
Ans
Qus.
Ans
1
D
16
C
30.
True
31.
35.
8
36.
40.
Constant.
2
D
17
B
3
A
18
B
True
4
C
19
B
5
B
20
D
32.
Size
6
D
21
D
7
B
22
A
8
B
23
C
9
A
24
C
10
D
25
A
False
11
B
26
C
33.
37. 4 × NA ×
4 3
r
3
38.
12
B
27
D
13
C
28
A
14
D
29
B
15
C
False
34. True
Boyle's temperature
39.
27
8
EXERCISE # 2
(Part – A)
Qus. 1
Ans C
Qus. 16
Ans B
2
B
17
B
3
D
18
A
4
D
19
C
5
D
20
C
6
D
21
C
7
B
8
D
9
B
10
A
11
B
27
A,B
28
B,C
29
A,C
12
B
13
C
14
C
15
A
(Part – B)
Qus. 22
Ans C,D
30. (i) b NO  b CCl 4
23
D
24 25 26
A,C A,B B,D
(ii) a NO  a CCl 4
(iii) NO
(Part – C)
Qus. 31
Ans D
36. A  R ; B  S ; C  P ; D  Q
32
C
33
A
34
B
35
A
(Part – D)
37. A  Q ; B  R ; C  S ; D  T ; E  P
38. A  P,R ; B  Q ; C  P,S ; D  Q,R
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GASEOUS STATE
41
EXERCISE # 3
1.
2.
3.
0.94 mol
1/4
In cylinder, V1 = 50 L,
V2 = ?,
From P 

P1 = 21.5 atm
P2 = 1.55 atm

8.
X
nB
24
n
1
1
=
 A =
 A =
nA
nB
1
24
XB
24
Let the wt. of the mixture be 100 g & the wt. of
He be x g
Then the number of moles He in the mixture
= 100 – x g
1
V
and the no. of moles of He in the mixture =
P1
V
= 2
P2
V1
And the no. of moles of O2 in the mixture
PV
21.5  50
 V2 = 1 1 =
P2
1.55
=
V2 = 693.548 L
Gas expelled into the tank

Now from PV = nRT =
 V2 = 693.548 – 50
= 643.548 L
It is the vol. of tank.
4.


Here P1(gas) = 830 mm of Hg – 30 mm of Hg

(gas) = 800 mm of Hg
Then from P  T (at const. V.)


dRT
M
dRT
0.518  0.0821 298
 M=
=
P
(720 / 760)
 M = 13.34 (Average molecular wt.)
800
T
=
P2
0.99T
Net new pressure = 792 + 25
 V=
wRT
PM

 V=
100  0.0821 298
(720 / 760)  13.34

 V = 193.56 L
Now again from PV = nRT


720
 x 100 – x 
 193.56 =  
 × 0.0821 × 298
32 
760
4


7.496 =
= 817 mm of Hg
5. 4
6. 25.027 KPa
7.


MB
MB
1

=
MA
MA
16
Given
wB
3
=
wA
2
Dividing (ii) by (i), we get
wB
M
3
16
24
× A =
×
=
MB
wA
2
1
1
x
100 – x
+
4
32



MB
MA
1
=
4
wRT
M

 P2 = 792 mm of Hg (of gas)
r
From A =
rB
w
RT
M
Now again from PV = nRT =
T1 = T K, T2 = 0.99 T K
100 – x
4
 PV =
P2 = ?
P1
T
= 1
P2
T2
x
4
……(i)
.….(ii)
 239.88 = 8x + 100 – x
 139.88 = 7x
 19.9 = x
Hence the mass % of He in the mixture = 19.9%
9.
0.199
10.
Given WA = 0.50 g, MA = 60
WB = 0.20 g, MB = 45, P = 750 mm Hg

 nA = 0.50/60 = 0.0083
nB = 0.20/45 = 0.0044

 nA + nB = 0.0083 + 0.0044
= 0.0127
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GASEOUS STATE
42
Partial pressures :
PA =
13.
Let the molecular wt. of the mixture (O3 + O2)
be M2
then from
nA
0.0083
×P=
× 750 mm
nA  nB
0.0127
V / t1
=
V / t2
= 490.16 mm of Hg
& PB =
11.
0.0044
× 750
0.0127
= 259.84 mm Hg
Vol. of gas bulb V = 2.83 L
 nH2 =
wt. of H2 = 0.174 g
0.174
2


t2
=
t1


246
=
60

 M2 = 33.62 (average mol. Wt. of O3 + O2)
Let the mol % of O3 be a then the mol % of
O2 = 100 – a
= 0.087
 n H 2 =
Wt. of N2 = 1.365 g
1.365
28
M2
M1
 Total no. of moles
n = n H 2 + n N 2 = 0.087 + 0.04875

 n = 0.13575
Temperature T = 0º C = 273 K
Then pressure inside the bulb
P=
 33.62 =


 a = 10.125
 b = 89.875
14.
 
=
0.087
× 1.075
0.13575
= 0.689 atm
0.04875
× 1.075
0.13575
Px ( H 2 ) =
Px ( N 2 ) =
P
PN 2
P
0.689
=
= 0.64
1.075
=
0.386
= 0.36
1.075
PD 2
4
4
×
1
2
10000 K
16.
Given Tr = 0.7277
Vr = 0.40
Pr = ?
From van der waals equation for reduced state
[Pr + 3/Vr2] [3Vr – 1] = 8Tr


 [3 × 0.4 – 1] = 8 × 0.7277
(0.40) 
3

 Pr 

 Pr +


 Pr + 18.75 = 29.108
 Pr = 10.358 atm
Pressure fractions
PH 2
PH 2
15.
0.087
= 0.64
0.013575
X N 2 = 1 – X H 2 = 1–0.64 = 0.36
M H2
×
= 5.656 : 1
= 0.386 atm.
Mole fractions X H 2 =
M D2
=
nD2
Partial pressures
PN 2 =
Molar ratio of H2 & D2 = 80 : 20
Partial pressure ratio of H2 & D2 = 4 : 1
(total pressure = 1 atm)
nH2
= 1.075 atm
a  48  (100 – a )32
100
a  48  (100 – a )32
100

nRT
0.13575  0.0821 273
=
V
2.83
PH 2 =
M2
2
then average mole wt. =
= 0.04875

M2
M1
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
2
3
= 29.108
0.16
GASEOUS STATE
43
Given d = 1.2504 g/L = 1.2504 × 10–3 g/cm3
P = 1 atm = 1.01325 dyne cm–2
Then
17.
Vrms =
3p
=
d
21.
From given,
m.wt. of A = 48
then m.wt of A2 = 96
Given wt. of compound in vessel = 96 g
Given wt. of A2 in the compound = 50% of 96
= 48 g
3  1.01325  10 6
1.2504  10 – 3
= 4.93 × 104 cm/sec
Vrms =
8P
=
d
Then the no. of A moles of A2 =
8 1.01325  106
22
1.2504 10 – 3
7
Given wt. of A in the compound = 50% of 96
= 48 g
Then the no of moles of A =
8
= 20.63  10
= 4.54×108 cm/sec
Vrms =
2p
=
d
2  1.01325  10




the vessel n = 1 +
6
3 × 1.38 × 10–23 × 300 = 10–15 × C2
1242 × 10–8 = C2
12.42×10–6 = C2
3.52 × 10–3 m/sec = C
0.352 cm/sec = C
1
3
=
2
2
Given V = 33.6 L, T = 273º C = 546 K
1.2504  10 – 3
3
1
KT =
mc2
2
2





48
=1
48
Then the total no. of moles of the compound in
Then from
= 16.207 108
= 4.02×104 cm/sec
18.
48
1
=
2
96
nRT
V
3 / 2  0.0821 546
=
33.6
P=
= 2.0 atm
.
Let the no of moles of Cl35 be n1 & that of Cl37
be n2
22.
Then Average mol. Wt. =
35  n1  37n 2
n1  n 2
= 35.45
r
V /t
From 1 = 1 1 =
r2
V2 / t 2
19.
M2
M1


20 dm 3 / 60 s
10

=
V2 / 30 s
V2


200 = V2 = 14.14 dm
20.

Now. From V  moles (At const. P1T)
1
2
V1
n
3.44
= 1 =
V2
n2
1
3
i.e., partition should be in ratio
3.44
pressure
1
in this condition is same as the original
pressure (  V, T & n are const.)
Given V = 10 L
Total no. of moles n =
0 .4
1.6
1.4
+
+
4
32
28
T = 27ºC
= 0.1 + 0.05 + 0.05
= 0.2
 300 K
Then from
P=
 P =

n1
3.44
=
n2
1
= 0.4926 atm
Partial pressure of He = XHe . P
Given, observed density of SO3, d = 0.925 g/L
d=
PM
1 80
 d =
× 1.08 g/L
RT
0.0821 900
Now, degree of dissociation
nRT
V
0.2  0.0821 300
10
=
23.
=

0.1
× 0.4926
0 .2
(D – d)
(n – 1)d
1.08 – 0.925
= 0.335
3 
 – 10.925
2 
= 33.5 %
 =
= 0.246 atm.
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GASEOUS STATE
44
24.
Constant


I
II
V
V
P1 = 0.5 atm
n1 = 0.30
T1 = 300 K
T2 = 300 K
From P =
Let moles diffused from II to I be n on heating
II at 600 K
Then n2 in flask I = 0.30 + n
& n2 in flask II = 0.30 – n
Now for flask I,
P2 × V = n2 RT2
 P2 × V = (0.30 + n)× R × 300
…..(i)
& for flask II
P2 × V = n2RT2
 P2 × V = (0.30 – n) × R × 600
…..(ii)
P = 1.2375 atm.
26
At Initial,
In Bulb I :- Vol. of H2 = 1.12 L
In Bulb II :- Vol. of P2 = 1.12 L
After diffusion
In Bulb II :- wt. of H2 = 0.05 g
=




0.30 + n = 0.60 – 2n
3n = 0.30
n = 0.10
In flask I, n2 = 0.30 + n
= 0.30 + 0.10
= 0.40
& In falsk II, n2 = 0.30 – n
= 0.30 – 0.10
= 0.20
At initial condition
From PV = nRT
 0.5 × 2V = 0.60 × 0.0821 × 300
 V = 14.778 L
 Vol. of each flask = 14.778 L
Now from (ii)
P2 × 14.778 = 0.20 × 0.0821 × 600
P2 = 0.667 atm
25.
From the reaction given, it is clear that. 1 mol
of DMH(l) gives 3 mol of N2(g) & 2 mol of
CO2(g).
i.e., effused vol. of H2 = 0.56 L
Now from
VH 2

VD 2


VD 2 = 0.396 L
0.56
=
=
M H2
M D2
2
0.56
 VD 2 =
4
1.414
Now total Vol. of gas in bulb II
= 0.56 + 0.396
= 0.956 L

 Vol % of H2 =
0.56
× 100 = 58.57 %
0.956
& Vol. % of D2 =
27.
0.396
× 100 = 41.43 %
0.956
At P1 = 3.0 atm
n1 =
=
P1V
RT
3V
RT
At P2 = 2.235 atm
 Total moles of gaseous products
n2 =
n = 7.5 + 5
= 12.5
VD 2

So, 2.5 mol of DMH (g) will give 7.5 mol of N 2
(g) & 5 mol of CO2 (g)

0.05
× 22.4 L
2
= 0.56 L
0.30  n
1
(i) ÷ (ii)
×
=1
2
0.30 – n




12.5  0.0821 300
250
P=
P1 = 0.5 atm
n1 = 0.30
T1 = 300 K
T2 = 600 K
nRT
V

P2 V
2.235 V
=
RT
RT
 No. of moles leaked n = n1 – n2
T = 27º C = 300 K, V = 250 L
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GASEOUS STATE
45
n=
= 0.765
=
= 2.1 × 10–8
2.235 V
3V
–
RT
RT
For CO2 : n =
V
RT
= 3.1 × 10–8
0.765  5
0.0821  298
For H2O : n =

 wt. of O2 leaked = 0.156 × 32 –~ 5 g

 Volume of O2 leaked = 0.156 × 22.4
29.
= 3.49 L
Volume of glass apparatus = 0.731 mL
=4×
Temperature = 23 + 273 = 296 K
Initial pressure = P 'H 2O + P 'CO 2 + P' N 2 ]
=1.74 mm
At – 75º C H2O is frozen out & the mixture on
retaining 23ºC has pressure
……….(ii)
At – 95º C, CO2 is also frozen out and the
mixture on retaining 23ºC has pressure
P' N 2 = 0.53 mm
……….(iii)
By (ii) and (iii) P 'CO 2 = 1.32 – 0.53 = 0.79 mm
n
30.
By (i) and (ii) P 'H 2O = 1.74 – 1.32 = 0.42 mm
Now using PV = nRT for each gas separately
For N2 = n =
Qus. 31
Ans A
4
22
×
× (2 × 10–8 )
3
7
= 1.34 × 10–22 cm3
(b) Co-Volume per mole
b = 4 × NA × v
= 1.34 × 10–22 × 6.023 × 1023
= 80.71 cm3
(c) Critical volume
Vc = 3b
Vc = 3 × 80.71 cm3
= 242.13 cm3
………(i)
P = P 'CO 2 + P' N 2 = 1.32 min
PV
0.42  0.731
=
RT
760  0.0821 296
= 1.7 × 10–8
Radius of a spherical molecule
r = 2 × 10–8 cm
(a) Co-volume per molecule
= 4 × Vol. of one molecule
=4×v
= 0.156
28.
PV
0.79  0.731
=
RT
760  0.084  296
T
 RT 2
 5, V 
 5RT, 51.3125 l
100
100
PV
0.53  0.731
=
RT
760  0.0821 296
32
A
33
C
34
C
35
A,B
36
B
37
C
38
C
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39
A
40
D
41
A
42
A
43
B
44
C
45
A
GASEOUS STATE
46
EXERCISE # 4
Qus.
Ans.
13.
1
C
2
C
3
B
4
D
5
C
6
C
7
A
Given :
A = 3.592 dm3 atm mol–2
T = 273 K, V = 1 L
n=1
P=?
from van der waal's equation if b = 0 (given)

12  3.592 
 P 
1 = 1 × 0.0821 × 273
2

 P = 18.82 atm.


Z=
=
14.

=


16.
a 
 V = RT
V2 

 100 


 0.112 = 0.0821 × 273
(0.112) 2 
a

a
 11.2 + 0.112 = 22.4133
 a = 1.256 atm L2 mol–2
15.
Given:- P = 1 atm
d = 0.36 kg /m3 = 0.36 g/L
T = 500 K
(a)


rs
rO 2
M O2

Ms
 MS = 18
(b) Molar volume (V/n) =
=
32
= 1.33
Ms
……(i)
mol. wt.
Density
18
0.36
= 50 L
1
× 50
0.0821 500
3
KT
2
3
× 1.38 × 10–23 J/K molecule × 1000
2

Vrms =
3RT
M
Vrms =
8RT
M
Vrms
=
Vav
3
= 1.1786 = 1.0854
8
Vrms
= 1.0854
400
Vrms = 434.17 m/s.
17.
= 1.33
= 1.33 
12
A,D
= 2.07 × 10–20 J
100  V
= 0.5  V = 0.112 L
1 0.0821 273
  P 
11
B
PV
P
V
=
×
nRT
RT
n
(b) K.E. =
PV
= 0.5
nRT
Z=
10
A,C,D
= 1.218 T.E/molecule
(iv)  Z > 1
 Dominant force = Repulsion


1
9
B
(iii) compression factor (z)
2 

 P  n a  V = nRT

V 2 


8
B


For n = 1


 P  a  (Vm – b) = RT
2 

Vm 

a
ab
 PVm – Pb +
– 2 = RT
Vm
Vm
 PVm = RT + Pb –
a
ab
+ 2
Vm
Vm
At P = 0, Vm 


 PVm = RT + Pb

intercept = RT
…….(ii)
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GASEOUS STATE
47
18.
No. of moles of N2 occupying the 20%
19.
PV
Surface area = n =
RT
=
0.001 (2.46  10 –3 )
0.0821 298
20.
A  (P, S)
B  (R)
C (P, Q)
D  (R)
Crmsx = Cmy
3R  400
=
40
= 1.0055 × 10–7 mole
= 6.056 × 1016 molecules
2R  60
2  60
or 30 =
or M = 4
M
M
21.[7]
Total surface area = 1000 cm2
Pext = 1 atm
14
2
Density of surface sites = 6.023 × 10 /cm

 No. of sites in total surface area
He +
compounds
14
= 6.023 × 10 × 1000

= 6.023 × 1017
 No. of sites occupied by 6.056 × 1016 N2
Molecules = 6.023 × 1017 ×

Vapour pressure of compound = 0.68
 PHe = 1 – 0.68 = 0.32
 By PV = nRT, for He
n RT 0.1 0.0821 273
V = He

PHe
0.32
20
100
= 12.046 × 1017
 No. of sites occupied by 1 N2 molecule
=
V~7L
12.046  1017
22. (C)
6.056  1016
= 1.9  2
EXERCISE # 5
Qus.
Ans.
Qus.
Ans.
23.
1
B
18
B
2
B
19
C
3
C
20
C
4
D
21
C
5
A
22
B
6
B
7
C
8
B
9
A
Let, the volume of C2H6 = x L so, the volume
of C2H4 = (40 – x) L
Combustions of ethane and ethene are
represented in the form of following thermo
chemical equations,
2C2H6 (g) + 7O2 (g)  4CO2 (g) + 6H2O(l)
C2H4 (g) + 3O2 (g)  2CO2 (g) + 2H2O(l)
For 2 L of C2H6 required oxygen = 7 L
so, for x L of C2H6 required oxygen =
7x
L
2
Similarly, total required oxygen for combustion of
C 2H 4 =
(40 – x )  3
L
1
 7x
Total volume of oxygen =  
 2
(40 – x )  3 

1

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10
D
11
A
12
D
13
C
14
C
15
B
16
B
17
B
x  240
 7 x  (80 – 2x )  3 
=
L

2
2


= 
For oxygen,
Given that, PV = nRT
P = 1 atm, T = 400 K, R = 0.0821 L atm/K/mol

1×
( x  240)
= n × 0.0821 × 400
2

n=
( x  240)
2  0.0821 400
now, n =

mass of oxygen
130
=
molecular mass of oxygen
32
130
x  240
=
32
2  0.0821 400
GASEOUS STATE
48
or
x=
130
× 2 × 0.0821 × 400 – 240
32
= 26.5
moles of C2H6 = 26.5 and moles of C2H4 = 13.5
Total moles = 26.5 + 13.5 = 40
 mole fraction of C2H6 =
24.

25.

13.5
×100 = 33.75
40
84
1.16 =
M mix
4

a
4
 11.0   (3.9) = 49.26
rmix
=
rKr

 a = 1.63 × 4 = 6.52 atm L2mol–2
M Kr
M mix
i

27.
N2O5(g)
2NO2 (g) + 0.5 O2(g).
Initial pressure 600
0
0
Final pressure 600 – P
2P
0.5 P
Thus 600 – P + 2P + 0.5 P = 960
 P = 240 mm of Hg
  P  n (at constant V & T)
M normal
M exp .
71
62.425

 1 +  = 1.137   = 0.137 = 13.7 %

 P=
RT
a
– 2
( V – b)
V


–1
2

b
a 
 ....... –
V
VRT 
 PV = RT 1 

 

a  1
 PV = RT 1   b –
.  .......
RT  V
 


 B=b–

M
 M = 252
28


t = – 173ºC = 100 K
Now from,
PV = nRT


1×V=
b
b
=1+
+   …….
V
V

M2
M1
Given; T1 = tº C = t + 273
T2 = (t + 10)ºC = t + 283
P1 = 1 atm, P2 = 1 + 10% of 1 = 1.1 atm
[w = 12 g, M = 120]
W, & V are constant.
So, from
P1
T
1
t  273
= 1 
=
P2
T2
1 .1
t  283
a 
 V
 PV = RT 
–

V
–
b
VRT


b

1 – 
 V
1 / 38
0 .8
=
1 / 57
1 .6
240
= 0.4
600
Thus the required compound is XeF6 Because it
can't have 2 Xe atoms
 2 × At. Wt. of Xe > 252
29.
–1

b
a 
 PV = RT 1 –  –

VRT 
 V 
From binomial theorem
X N 2O5 decomposed =
n1 / t1
P
= 1 ×
P2
n2 / t2
28.
According to vander waals equation
VRT
a
PV =
–
V–b
V





On multiplying by V,

2
= 2 × 0.0821 × 300

a 
 P 
 (V – b) = RT
V2 





2 a
 11.0  2  (4 – 2 × 0.05)
and mole fraction of C2H4 =
1 +  (n – 1) =

2 

 P  n a  (V – nb) = nRT

V 2 

26.5
× 100 = 66.25
40
Mmix = 62.425
For Cl2
2Cl
1
0
1 – 

26. From van der waal's equation
12
× 0.0821 (– 173 + 273)
120
 V = 0.821 L
a
RT
CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000
GASEOUS STATE
49