JEE CHEMISTRY .
BY AMIT JHAKAL
GASEOUS STATE
1. Gases ::
1.1 Properties of Gases
(a) They do not have definite shape and volume.
(b) They can occupy whole space open to then.
(c) Gases have unlimited dispensability and high
compressibility.
(d) They have very low densities because of
negligible intermolecular forces.
(e) Gases exerts pressure on the wall of the
container with perfectly elastic collisions.
(f) They diffuse rapidly through each other to form
homogeneous mixture against the electric,
magnetic and gravitational field.
1.2 Parameters of Gases
The characteristics of gases are described in terms of
four measurable parameters and it is also called as
measurable properties of gases which are
(a) Mass
(b) Volume
(c) Pressure and
(d) Temperature
(a) Mass (m)- The mass of a gas is denoted by 'm'
which is related to the no of moles 'n'.
Therefore,
m(mass in grams)
n (no of moles) =
M(Molar mass)
so, m = n × M
(b) Volume V(i) Gases occupy whole space available to then.
The volume occupy by a gas is simply the
volume of container in which it is filled.
(ii) The volume of a gas is denoted by 'V' and it is
measured in units of litre or cubic metre (m3) or
cm3 or dm3.
(iii) 1 litre = 1 dm3 = 1000 cm3 = 1000 ml
(c) Pressure (P)(i) It is force acting per unit area. A confined gas
exerts uniform pressure on the walls of its
container in all the direction.
(ii) It is denoted by 'P' and specified in pascal (P a).
(iii) Other units of pressure are atm, cm hg, mmHg,
N/m2, bar, torr.
(iv) 1 atm = 76 cm Hg = 760 mm
Hg = 1.013 × 105 N/m2 = 1.013 × 105
Pa = 1.013 bar = 760 torr.
F(Force)
(v) P (Pressure) =
A(Area)
=
Mass  Acceleration
Area
2nd Floor Rajcastle, Above Dominos Pizza , Ellorapark
(vi) Pressure exerted by a gas is due to kinetic
energy (K.E. = ½ mv2) of the gases molecules.
(vii) K.E. of the gas molecules increases, as the
temperature is increased so, pressure of a gas is
directly proportional to temperature. P  T
(d) Temperature (T) (i) The temperature of a gas is denoted by 'T' and it
is measured in the unit of kelvin (K).
(ii) Other units of temperature are , ºC, ºF, ºR.
(iii) K = ºC + 273.15
xC
( yf – 32)
(iv)
=
5
9
2. Gas Laws ::
The certain laws which relate the four parameters
are called gas laws.
2.1 Boyle's Law
(a) It states that at constant temperature, the volume
of a given mass of a gas is inversely proportional
to the pressure.
(b) Mathematically 1
P
(at constant temperature)
V
or PV = K or P1V1 = P2V2
(c) Graphical representations
1
(i) P vs V
(ii) P vs
V
(iii) log P vs, log V
(iv) PV vs P
log P
P/T
log V
T
2.2 Charle's law
(a) This law states that at constant pressure, the
volume of a given mass of a gas is directly
proportional to its absolute temperature.
(Absolute temperature = ºC + 273.15)
GASEOUS STATE
1
AMIT JHAKAL .
JEE
(b) Mathematically - V  T
V
T
V
(at constant pressure)
= volume of gas
= Absolute temperature
= KT or
(iv)
V
= K
T
Hence, if the volume of a gas of mass is V1 at
temperature T1 changes to V2 at T2, pressure
remaining constant,
V
V
then 1 = 2 = constant
T1
T2
or log V – log T = constant
(c) For each degree change of temperature the
volume of sample of a gas changes by the
1
fraction of
its volume at 0 ºC
273.15
 273.15  t 
Vt = V 0 

 273.15 
This equation is called Charles-gay-lussac
equation.
where
Vt = volume of gas at temperature t ºC
V0 = volume of gas at 0 ºC temperature
t = temperature in ºC.
so
(d) Graphical representations -

(i) V vs T V
(ii) V vs
V
vs T
T
2.3 Gay-Lussac's Law or Amonton's Law
(a) It states that at constant volume, the pressure of
a given mass of a gas is directly proportional to
its absolute temperature.
(b) Mathematically - P  T (at constant volume)
where P = pressure of gas
T = Absolute temperature
P = KT
P
or
=K
T
Hence , if the pressure of a gas is P1 at
temperature T1 changes to P2 at T2, volume
remaining constant.
P1
P
then
= 2 = constant
T1
T2
log P – log T = constant.
t 

(c) Pt = Po 1 

 273.15 
where Pt = Pressure of gas at t ºC
Po = Pressure of gas at 0 ºC
t = Temperature in ºC.
(d) Graphical representation 1
1
(i) P vs T (ii) P vs
or T vs
T
P
1
1
or T vs
T
V
P
P
V
T
(iii) log P vs log T
1/T
1/T
P
(iv)
Vs T
T

log P
(iii) log V vs log T, log V
P/T
T
log T
76
log T —
GASEOUS STA
JEE CHEMISTRY .
Examples
based on
Ex.1
Sol.
Ex.2
Gas Laws
A gas occupies a volume of 2.4 litres at a
pressure of 740 mm of mercury. Keeping the
temperature constant, calculate its volume at
standard pressure (A) 2.4 litres
(B) 2.34 litres
(C) 2.5 litres
(D) None of these
(Ans. B)
Initial volume (V1) = 2.4L, Initial pressure (P1)
= 740 mm. Final volume (V2) = ?
Final pressure (P2) = 760 mm.
740 2.4
P1V1 = P2V2 ;
 V2 =
760
= 2.34 litres.
BY AMIT JHAKAL
2.4 Ideal gas equation
(a) It correlate all the four parameters of a gas.
(b) It is the combination of Boyle's and charle's law.
(c) PV = nRT
m
RT
M
The equation is called as ideal gas equation.
Where
n = no of moles of the gas
m = mass of the gas
M = Mol. wt. of the gas
R = Molar gas constant.
(d) For 1 mole of the gas n = 1
PV = RT
PV =
So
Where P1, V1, T1 are the initial pressure, volume
and temperature and P2, V2 T2 are final.
The above equation is called as ideal gas
equation.
(e) The unit of R is the unit of work or energy per
degree per mole as -
A gas occupies a volume of 580 ml at 17 ºC. It
is heated to 100 ºC at constant pressure.
Calculate the volume of the gas (A) 746 ml.
(B) 760 ml.
(C) 773 ml.
(D) 780 ml.
(Ans.A)
R=
Sol.
Initial volume (V1) = 580 ml,
T1 = 17 + 273 = 290 K
Final Volume (V2) = ?
T2 = 100 + 273 = 373 K
Ex.3
Sol.
V2 = T 2 ×
R=
(f)
(i)
(ii)
(iii)
(iv)
(v)
(vi)
V1
580
= 373 ×
= 746 ml.
T1
290
A gas occupies 3 litres at 32 ºC and one
atmospheric pressure. What volume will it
occupy if the temperature is changed to 18 ºC,
the pressure remaining constant (A) 2.91 litres
(B) 2.86 litres
(C) 2.30 litres
(D) None of these
(Ans. B)
According to Charle's law,
Ex.4
or V2 =
3 291
= 2.86 litres.
305
2nd Floor Rajcastle, Above Dominos Pizza , Ellorapark
Sol.
work (energy)
mole temperature
Numerical values of R in different units R = 0.0821 litre atm. deg.–1 mole–1
R = 62.4 litres mm. deg.–1 mole–1
R = 8.314 × 107 ergs deg.–1 mole–1
R = 82.05 C.C.atm. deg.–1 mole–
R = 2 cals. deg.–1 mole–
R = 8.314 J K–1 mole–1
Examples
based on
V1
V
VT
= 2 or V2 = 1 1
T1
T2
T2
V2
3L
=
(273  32)K
(273  18)K
PV
Pressure  volume
=
nT
mole  Temperature
force
 volume
force length
Area
R=
=
mole  Temperature mole temperature
V1
V
= 2 ;
T1
T2

P1V1
PV
PV
PV
= 2 2 = R or = 1 1 = 2 2
T1
T2
T1
T2
The ideal gas equation
How many litres would 5 moles of H2 occupy at
25 ºC and 2 atm pressure (A) 61.20 litres
(B) 61.09 litres
(C) 30.50 litres
(D) 30.60 litres
(Ans.B)
PV = nRT
(5)(0.082)(298)
2
= 61. 09 litres.
=
GASEOUS STATE
77
AMIT JHAKAL .
JEE
Ex.5
Sol.
What is the pressure of a mixture of 1 g of
hydrogen and 1.4 g of nitrogen stored in a 5 litre
vessel at 127 ºC (A) 5.50 atm.
(B) 3.61 atm.
(C) 4.40 atm.
(D) 4.50 atm.
(Ans.B)
1
moles of H2 =
= 0.5
2
1.4
moles of N2 =
= 0.05
28
 Total number of moles of gas (n) = 0.55
Using PV = nRT
P
nRT
=
V
0.55 0.0821 400
5
= 3.61 atm.
=
partialpressure of gas
Totalpressure of themixtureof gas
(b) % of a gas in a mixture
=
partialpressure of gas
× 100
Totalpressure
(c) Pressure of dry gas which is collected over the
water isPTotal = Pmoist air = Pdry gas + Pwater vapour
(Note : Pwater
vapour
Dalton's law of Partial Pressure ::
(a) According to this law, when two or more than
two chemically inert gases are kept in a closed
container, the total pressure exerted by the
gaseous mixture is equal to the sum of the
partial pressures of individual gases - i.e.P = P1 + P2 + P3 + ..... + Pn
(b) Let n1 & n2 be the no. of moles of two inert
gases A and B which is filled in a container of
volume 'V' at temperature T. So the total
pressure of container 'P' may be calculated as PV = (n1 + n2) RT ...... (i)
Partial pressure of individual gas calculates at PAV
= n1 RT .... (ii)
PBV
= n2 RT .... (iii)
so Pdry gas = PTotal – PWater vapour
(Note: Aqueous tension is directly proportional to
absolute temperature)
Examples
based on
Ex.6
Sol.
Dalton's law of partial pressure
0.333 grams of alcohol displaced in a Victor
Meyer apparatus 171 c.c. of air measured over
water at 15 ºC. The barometric pressure was
773 torr. Calculate the molecular weight of
alcohol-(Aqueous tension at 15 ºC = 13 torr.)
(A) 33.34 g/ mol.
(B) 28.80 g/ mol.
(C) 46.0 g/ mol
(D) 13.0 g/mol.
(Ans. C)
Pdry gas = 773 – 13 = 760 torr
=
760
= 1 atm
760
 PV = nRT
On the addition of eq. (ii) & (iii) we get (PA + PB) V = (n1 + n2) RT .... (iv)
171
0.333
=
× 0.0821 × 288
M.wt .
1000
M = 46 g. per mol.
1×
On the comparison of eq. (i) & (iv)
P = PA + PB
Dividing by equation (ii) by (i), we get
PA
n1
=
= xA
P
n1  n 2
Ex.7
Atmospheric air contains 20% O2 and 80% N2
by volume and exerts a pressure of 760 mm.
Calculate the partial pressure of each gas (A) 152 mm, 608 mm
(B) 608 mm, 152 mm
(C) 760 mm both
(D) None of these
(Ans. A)
Sol. Partial Pressure
= Mole fraction × Total pressure
= Vol. fraction × Total pressure
 PO 2 = 0.2 × 760 = 152 mm
P A = xA × P
where xA = mole fraction of 'A'
Similarly dividing (iii) by (i), we get
PB = xB × P
so
Partial pressure of a component
= mole fraction × total pressure

78
is called aqueous tension)
3.2 Limitations of Dalton's law of partial pressure
(a) It is applicable only inert gases like N2 and O2,
N2 and Cl2 etc.
(b) It is not applicable for chemically reactive gases
like H2 and Cl2, CO and Cl2 etc.
=
3.
3.1 Applications of Dalton's Law of Partial pressure
(a) mole fraction of a gas in a mixture of gas
GASEOUS STA
PN 2 = 0.8 × 760 = 608 mm
JEE CHEMISTRY .
Ex.8
Sol.
Equal weights of ethane and hydrogen are
mixed in an empty container at 25ºC. Determine
the fraction of the total pressure exerted by
hydrogen (A) 16/30
(B) 15/16
(C) 30/16
(D) 16/15
(Ans. B)
Let the wt. of ethane and H2 = w gm
 No. of moles of C2H6 =
Ex.9
Sol.
BY AMIT JHAKAL
5.
w
30
No. of moles of H2 =
w
2
Total no. of moles
=
w
w
+
2
30
Mole fraction of H2
=
w
30
15
×
=
2
16w
16
=
16w
30
Graham's Law of Diffusion or Effusion ::
Diffusion: It is the ability of a gas to mix
spontaneously and to form a homogenous mixture is
known as diffusion.
Effusion: It is a process in which a gas is allowed
to escape under pressure through a fine orifice from
closed container.
LAW
(a) This law was proposed by Thomas Graham.
(b) According to this law, at constant temperature
and pressure, the rate of diffusion or effusion of
a gas is inversely proportional to the square root
of its density. Thus and directly proportional to
its pressure.
Rate of diffusion (r) 
d
p
r1
= 1
p2
r2
2.8 g of N2, 2.8 g CO, 4.4 g CO2 are found to
exert a pressure of 700 Torr. Find partial
pressure of N2 gas in the mixture (A) 280.8 Torr
(B) 233.3 Torr
(C) 300 Torr
(D) None of these
(Ans. B)
pN2 = its mole-fraction × Ptotal
r1
=
r2
2nd Floor Rajcastle, Above Dominos Pizza , Ellorapark
d2
d1
where r1 & r2 are rates of diffusion of two gases
and d1 & d2 are densities.
 2 × vapour density = Molecular mass

r1
=
r2
M2
M1
where M1 & M2 are the molecular masses of
two gases.
=
4.
d2
d1
It pressure is constant
2.8 / 28
=
× 700
2 .8 2 .8 4 .4


28 44 44
0.1
 700
0.3
= 233.3 Torr.
Avagadro's law ::
(a) According to this law under the same condition
of temperature and pressure, equal volumes of
all gas contains equal no. of molecules.
V  n (At constant temperature & pressure)
Where V = volume
n = no of molecules
(b) Molar Volume or gram molecular volume - 22.4
litres or 22400 ml of every gas at NTP is the
volume occupied by its one gram mole and it is
called molar volume or gram molecular volume.
(c) The mole Concept - Mole is defined as the total
amount of substance that contains as many basic
units as there are atoms in 12 g of the isotopes
of carbon -12. Thus a mole will be defined as
the Avogadro no of particles which is equal to
6.023 × 1023.
(d) Loschmidt number - It the no of molecules
present in the volume of a gas at S.T.P. Its value
is 2.617 × 1019 per c.c.
P
(c) Rate of diffusion =
Volume of gas diffused
Time takenfor diffusion
5.1 Applications of graham's law of diffusion
(a) To the Detection of marsh gas in mines.
(b) Separation of isotopes
(c) Determination of density and molecular mass of
gases.
Examples
based on
Ex.10
Sol.
Graham's Law of diffusion
Calculate the relative rates of diffusion of
235
UF6 and 238UF6 in the gaseous state (At. mass
F = 19).
(A) 1.0043 : 1.0000 (B) 1.0000 : 1.0043
(C) 1.349 : 1.352
(D) 1.352 : 1.349
(Ans. A)
Mol. mass 235UF6 = 235 + 6 × 19 = 349
Mol. mass 238UF6 = 238 + 6 × 19 = 352
GASEOUS STATE
79
AMIT JHAKAL .
JEE
r1
=
r2
M2
=
M1
7.
352
= 1.0043
349
r1 : r2 : : 1.0043 : 1.0000
Ex.11
The densities of CH4 and O2 are in the ratio
1 : 2. Calculate the ratio of rates of diffusion of
oxygen and methane.
(A) 1.414 : 1
(B) 1: 1.414
(C) 1.614 : 1
(D) 1.614: 1
(Ans.B)
rO 2
Sol.
rCH 4
=
 d CH 4

 dO
2


 =


 8 
  =
 16 
1
 
2
= 1 : 1.414
Ex.12
Calculate the molecular weight of a gas which
diffuses through a porous plug at 1/6th of the
speed of hydrogen under same conditions.
(A) 36
(B) 76
(C) 72
(D) 63
(Ans.C)
 M gas  Rate of diffusion of H 2

=
 M H  Rate of diffusion of gas
2 

Sol.
capacity per mole at constant volume and at
constant pressure.
 M gas   1 


 2  =  1 / 6 


= 2 × 36 = 72
=
or Mgas
Ex.13
Sol.
(e) Ratio for Molar Heats (i) For monoatomic gases
The vapour density of gas A is thrice that of the
gas B. If the molecular weight of B is M, then
calculate the molecular weight of A.
(1) M
(2) 3 M
(3) M/3
(4) None of these
(Ans. B)
3
VD A
MA
MA
=
 =
, So Mol. wt of A
1
VD B
MB
MB
(MA) = 3 M.
6.
Explanation Of Deviation ::
Kinetic theory of gases do not hold good at all
condition mostly these two assumptions
(a) The force of attraction between gaseous
molecules are negligible
(b) The volume occupied by the gaseous molecules
is negligible compared to total volume of gas at
high pressure. Both the assumptions do not hold
good hence deviating from ideal gas.
80
SPECIFIC HEAT OF GASES ::
(a) Specific Heat - It is the amount of heat required
to raise the temperature of one gm of the
substance through one degree centigrade. It is
usually expressed in "Calories".
(b) Calorie - A calorie is defined as the amount of
heat required to raise the temperature of one gm
of water through 1ºC (more accurately from
15.5 ºC to 16.5 ºC.).
The heat so supplied is used up in increasing the
internal energy of the molecule, i.e. in raising
the temperature or in raising the average K.E. of
the molecules.
(c) Specific Heat of a gas at constant volume(Cv) The amount of heat required to raise the
temperature of one gm of the gas by 1 ºC when
the volume is kept constant and pressure is
allowed to increase.
(d) Specific Heat of a gas at constant Pressure (Cp) The amount of heat required to raise the
temperature of one gm of the gas by 1 ºC when
the pressure is kept constant and volume is
allowed to increase.
CV = cV × M and CP = cP × M
Where CV and CP are molar specific heat or heat
GASEOUS STA
=
CP
=g
CV
5.0
= 1.66
3.0
 Molar heat capacity at constant volume (CV)
= K.E. =
3RT
2
at 1 ºC  K.E = 3/2 R
K.E. =
3
× 2 = 3.
2
Therefore  CV = 3 calories.
& CP = Increase in K.E. + work done
3
5
=
R + R = R (Since R = 2)
2
2
CP =
5
× 2 = 5 calories.
2
JEE CHEMISTRY .
(ii) For diatomic gases
CP
5 x
=
CV 3  x
In case of diatomic gases x in many
cases is 2.
CP
7
=
= 1.40
CV
5
C
53
(iii) For triatomic gases P =
CV
33
BY AMIT JHAKAL
.9.
BEHAVIOUR ::
(i)
DEVIATION FROM IDEAL GAS BEHAVIOUR ::
(i)
The gas show maximum deviation from ideal gas
(ii) The volume of a molecule is not negligible in
comparision of total volume of gas.
V1  0
10. VANDER WAAL 'S GAS EQUATION ::
2

 P  an
2

V

behaviour at low temperature and high pressure.
(ii) Compressibirity factors (Z) : The ratio of molar
observed volume to the ideal gas volume is called as
compressibility factor.
Z=
The intermolecular force of attraction between
molecules is not negligible.
Fatt  0
8
=
= 1.33
6
8.
REASON OF DEVIATION FROM IDEAL GAS
At T, P, it act as ideal gas equation.
10.1 Vander waal gas constant (a) :
Where a is a vander waal gas constant which
indicate the inter molecular force of attraction.
P ( Vm ) obs
(Vm ) obs
=
RT
Vi
Pressure =
Case IIf Z = 1
an 2
V2
unit of a = atm lit2/mol2
P ( Vm ) obs
=1
RT
P1 = P +
P(Vm)obs = RT
an 2
V2
Pi > PR
Then, the gas show the ideal gas behaviour
Case IIIf Z > 1

( V – nb)  nRT


a,Fatt, liquification.
10.2 Vander waal gas constant (b)
P ( Vm ) obs
>1
RT
Volume = nb
P(Vm)obs > RT
Unit of b =
The gas show the positive deviation generally at
b = correction volume
high pressure from ideal gas behaviour and the gas
b = vibratory volume
will be less compressible than ideal gas.
b = co-volume
Case IIIIf Z < 1
b = excluded volume
b = incompressible volume
P ( Vm ) obs
<1
RT
b = 4Va × NA
Va = 4/3 r3 = volume of 1 molecule which is rest.
P(Vm)obs < RT
r = radius of 1 molecule
The gas show the negative deviation generally at
Vi > V R
low pressure from ideal gas behaviour and the gas
will be highly compressible than ideal gas
lit.
mol.
If b, effective size of molecule

Incompressible volume 
compressible volume 
2nd Floor Rajcastle, Above Dominos Pizza , Ellorapark
GASEOUS STATE
81
AMIT JHAKAL .
JEE
Special note :
"a" and "b" depend on the nature of gas but not
depend on the temperature and pressure.
Examples based on Vander Waals equation :
Ex. 1
Vander Waal's equation of state is obeyed by
real gases. For n moles of a real gas, the
expression will be –
 P na  V 
(1)   2 
 = RT
 n V  n – b 
a 

(2)  P  2  (V – b) = nRT
V 

na 

(3)  P  2  (nV – b) = nRT
V 


n 2a 
(4)  P  2  (V – nb) = nRT
V 

Sol. (4) Vander Waal's equation is
2

 P  an

V2

Ex.2

 (V – nb) = nRT


Vander Waal's constant 'a' and 'b' are related
with……respectively –
(1) Attractive force and bond energy of
molecules
(2) Volume and repulsive force of molecules
(3) Shape and repulsive force of molecules
(4) Attractive force and volume of the molecules
Sol. (4) ' a' is related to attractive forces and 'b' to the
volume of the molecules
82
GASEOUS STA
JEE CHEMISTRY .
BY AMIT JHAKAL
SOLVED EXAMPLES
Ex.1
A gas occupies 300 ml at 27ºC and 730 mm pressure. What would be its volume at STP(A) 162.2 ml
(B) 262.2 ml
(C) 362.2 ml
(D) 462.2 ml
Sol. (B)
Given at T1 = 300 K, T2 = 273 K (STP)
 300 
 730 
V1 = 300 ml = 
 litre, P1 = 
 atm. P2 = 1 atm.,
 1000 
 760 
PV
PV
 = 1 1= 2 2 ,
T1
T2
V2 = ?
1 V2
730 300
=
7601000 300
273
 V2 = 0.2622 litre = 262.2 ml.
Ex.2
A truck carrying oxygen cylinders is filled with oxygen at -23ºC and at a pressure of 3 atmosphere in Srinagar,
Kashmir. Determine the internal pressure when the truck drives through Madras. Tamilnadu. Where the
temperature is 30ºC –
(A) 2.64 atm.
(B) 1.64 atm.
(C) 1 atm.
(D) 3.64 atm.
Sol. (D)
P1 = 3 atm., P2 = ?
T1 = – 23 + 273 = 250 K
T2 = 273 + 30 = 303 K.
P1
P
= 2 ,
T1 T2
Ex.3
P
3
= 2
250 303
3 303
P2 =
= 3.64 atm.
250
The density of a gas at -23ºC and 780 torr is 1.40 gram per litre. Which one of the following gases is it (A) CO2
(B) SO2
(C) Cl2
(D) N2
Sol. (D)
PV = nRT
780
× 1 = n × .082 × (273 - 23)
760
780
n=
760 0.082 250
n = 0.0501 moles.
 .0501 moles of the gas weight = 1.40 gm.
1.40
 1 mole of gas weight =
= 28 gram.
.0501
So gas is N2.
Ex. 4
Calculate the weight of CH4 in a 9 litre cylinder at 16 atm and 27ºC (R = 0.08 lit. atm/K) (A) 96 gm
(B) 86 gm
(C) 80 gm
(D) 90 gm
Sol. (A)
Given P = 16 atm, V = 9 litre.
T = 300 K, m CH 4 = 16, R = 0.08 litre atm/k.
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AMIT JHAKAL .
JEE CHEMISTRY
Ex. 5
PV = w/m × R × T
w
16 × 9 =
× 0.08 × 300
16
w = 96 gm.
What is the density of sulphur dioxide (SO2) at STP (A) 2.86 gm/lit.
(B) 1.76 gm/lit
(C) 1.86 gm/lit
(D) None of these.
Sol. (A)
The gram molecular weight of SO2 = 64 gm/mole.
Since 1 mole of SO2 occupies a volume of 22.4 litres at S.T.P.
Density of SO2 =
Ex.6
64
= 2.86 gm/lit.
22.4
5gm of XeF4 gas was introduced into a vessel of 6 litre capacity at 80ºC. What is the pressure of the gas in
atmosphere (A) 0.21 atm
(B) 0.31 atm
(C) 0.11 atm
(D) 0.41 atm.
Sol. (C)
Ex.7
Given
V
=
6
litre,
T
=
353
K,
W = 5gm. m = 207.3
W
PV =
×R×T
m
5
P×6=
× 0.082 × (273 + 80)
207.3
5  .082 353
P =
= .11 atm.
6  207.3
Calculate the temperature at which 28 gm N2 occupies a volume of 10 litre at 2.46 atm(A) 300 K
(B) 320 K
(C) 340 K
(D) 280 K
R=
0.082,
Sol. (A)
Given w N 2 = 28gm , P = 2.46 atm, V = 10 litre. m N 2 = 28, 


w
 PV =   RT
m
 28 
2.46 × 10 =   × 0.0821 x T
 28 
T = 300 K.
Ex.8
A mixture of gases at 760 mm pressure contains 65% nitrogen, 15% oxygen and 20% Carbondioxide by volume.
What is the partial pressure of each in mm (A) 494, 114, 252
(B) 494, 224, 152
(C) 494, 114, 152
(D) None of these.
Sol. (C)
P'N2
P'O2
P'CO2
Ex.9
2
65
= 494 mm
100
15
= 760 ×
= 114 mm
100
20
= 760 ×
= 152 mm.
100
= 760 ×
0.45 gm of a gas 1 of molecular weight 60 and 0.22 gm of a gas 2 of molecular weight 44 exert a total pressure of
75cm of mercury. Calculate the partial pressure of the gas 2 (A) 30 cm of Hg
(B) 20 cm of Hg
(C) 10 cm of Hg
(D) 40 cm of Hg.
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JEE CHEMISTRY .
BY AMIT JHAKAL
Sol. (A)
w1
0.45
=
m1
60
= 0.0075
0.22
w
No. of moles of gas 2 = n2 = 2 =
44
m2
= 0.0050
Total no. of moles = n1 + n2
= 0.0075 + 0.0050 = 0.0125
0.0050
P2 partial pressure of gas 2 =
× 75
0.0125
= 30 cm of Hg.
No. of moles of gas 1 = n1 =
Ex. 10 The total pressure of a sample of methane collected over water is 735 torr at 29ºC. The aqueous tension at 29ºC is
30 torr. What is the pressure exerted by dry methane (A) 605 torr
(B) 205 torr
(C) 405 torr
(D) 705 torr
Sol. (D)
Ptotal = Pdry methane + Pwater
735 = Pdry methane + 30
 Pdry methane = 735 - 30 = 705 torr.
Ex.11
The odour from a gas A takes six seconds to reach a wall from a given point. If the molecular weight of gas A is
46 grams per mole and the molecular weight of gas B is 64 grams per mole. How long will it take for the odour
from gas B to reach the same wall from the same point. Approximately (A) 6 Sec
(B) 7 Sec
(C) 8 Sec
(D) 9 Sec
Sol. (B)
rA
=
rB
MB
MA
rA
64
=
= 1.18
46
rB
Time taken for the odour of B to reach the wall = 1.18 x 6 = 7.08 sec 7 sec.
Ex.12
1 litre of oxygen effuses through a small hole in 60 min. and a litre of helium at the same temperature and pressure
effuses through the same hole in 21.2 min. What is the atomic weight of Helium (A) 2.99
(B) 3.99
(C) 2.08
(D) 1.99
Sol. (B)
rO 2
rHe
=
1000/ 60
21.2
=
=
1000/ 21.2
60
Squaring both of sides =
M He
=
M O2
( 21.2) 2
(60)
2
=
M He
32
M He
32
(21.2)  32
2
MHe =
= 3.99
(60) 2
Since Helium is monoatomic so
Atomic weight = Molecular weight = 3.99
Ex.13
What is the temperature at which oxygen molecules have the same r.m.s. velocity as the hydrogen molecules at
27ºC -
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AMIT JHAKAL .
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(A) 3527ºC
(B) 4227ºC
(C) 4527ºC
(D) 4000ºC
Sol. (C)
r.m.s. velocity C =
3RT
M
3RT
32
For oxygen CO2 =
For C H 2 at 27ºC =
3  R  300
2
When Co2 = CH2
3RT
=
32
3  R  300
2
3RT 3R  300
=
2
32
T=
300 32
= 4800 K.
2
Then in ºC
Ex.14
T = 4800 - 273 = 4527ºC
Calculate the total kinetic energy in joules, of the molecules in 8 gm of methane at 27ºC(A) 1770.5 Joule
(B) 1870.5 joule
(C) 1970.5 joule
(D) 1670.5 joule
Sol. (B)
3
RT = 3741 Joule.
2
Total energy of 8 gm of methane
1
=
mole of methane
2
3741
=
= 1870.5 Joule.
2
EK (For 1 mole) =
Ex.15
Calculate the root mean square velocity of SO2 at S.T.P. (A) 3.26 × 104 cm/sec
(B) 1.26 × 102 cm/sec
(C) 1.26 × 104 cm/sec
(D) 3.26 × 102 cm/sec
Sol. (A)
 vrms of SO2 =
=
Ex.16
3RT
M
3  8.314  107  273
= 3.26 x 104 cm/sec.
64
Calculate the number of atoms in 1 g of helium (A) 1.506 × 1023 atoms
(B) 1.605 × 1032 atoms
(C) 1.056 × 1025 atoms
(D) None of these
Sol. (A)
Atomic mass of He = 4
4
2nd Floor Rajcastle, Above Domino’s Pizza, Ellorapark
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4
JEE CHEMISTRY .
BY AMIT JHAKAL
4 g of He contain = 6.023 x 1023 atoms
 1 g of He contains =
6.0231023
= 1.506 x 1023 atoms.
4
Ex.17
7.00 g. of a gas occupies a volume of 4.1 litres at 300 K and 1 atmosphere pressure. Calculate the molecular mass
of the gas (A) 40 g mol–1
(B) 42 g mol–1
–1
(C) 48 g mol
(D) 45 g mol–1
Sol. (B)
n=
PV = nRT  n = PV/RT
(1atm)(4.1L)
–1
–1
(0.082 L atm K mol )(300K )
now n =
=
1
mol
6
Mass of gas
1
=
Mol. mass of gas 6
7
Mol. mass of gas
Thus, molecular mass of gas = 7 × 6
= 42 g mol–1
=
Ex.18
Calculate density of ammonia at 30ºC and 5 atm. pressure (A) 3.03 g/litres.
(B) 3.82 g/litres
(C) 3.42 g/litres.
(D) 4.42 g/litres.
Sol. (C)
PV = nRT, or PV
or P = d ×
RT
M
d = MP/RT ; d =
Ex.19
Sol.
m
m RT
= RT or P =

M
V M
17  5
= 3.42 g/litres.
0.082 303
5g of ethane are confined in a bulb of 1 lit. capacity. The bulb is so weak that it will burst it the pressure exceeds
10 atm. At what temperature will be the pressure of gas reach the bursting value.
wt = 5g, V = 1 lit., P = 10 atm., Mw= 30
PV = nRT
wt
or PV =
RT
Mw
5
× 0.0821 × T
30
T = 730.81 K = 457.81 °C
10 × 1 =
Ex.20
Sol.
A gas occupies 600 ml at 27°C and 730 mm pressure. What would be its volume at STP.
At STP
V2 = ?,
T2 = 0°C = 273 K,
P2 = 1 atm.
600
V1 =
lit = 0.6 lit
1000
T1 = 27 + 273 = 300 K
730
P1 =
atm.
760
P2 V2
PV
= 1 1
T2
T1
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AMIT JHAKAL .
JEE CHEMISTRY
Ex.21
Sol.
1 V2
730
0 .6
=
×
273
760
300
V2 = 0.5244 lit.
3.7 g of a gas at 25°C occupy the solid volume as 0.184 g H 2 at 17°C same pressure. What is molecular weight of
gas.
wt1 = 3.7g , T1 = 25 + 273 = 298 K, M w1 = ?
For H2 gas wt2 = 0.184 g,
M w2 = 2
T2 = 17 + 273 = 290 K,
wt 1
wt 2
T1 =
× T2
Mw2
M w1
Ex.22
Sol.
Ex.23
Sol.
or
0.184
3.7
× 298 =
× 290
2
M w1
or
M w1 = 41.326
Calculate the density of CO2 at 100°C and 800 mm Hg pressure.
800
 44
PM w
d=
= 760
RT
0.0821 373
or d = 1.5124 gm/lit.
An open vessel at 27°C is heated until 3/5th of the air in it has been expelled. Assuming that the volume of the
vessel remains constant, find out.
(i) The temperature at which vessel was heated
(ii) The air escape out if vessel is heated to 900 K.
(iii) The temperature at which half of the air excape out.
(i) n1T1 = n2T2
n1 = n
or n × 300 = 2/5 n T2
n2 = n – 3/5n = 2/5n
or T2 = 750 K
T1 = 300 K = 27 + 273
(ii) n1T1 = n2T2
n × 300 = n2 × 900
n2 = n/3
T1 = 300 K = 27 + 273
T2 = 900 K
n1 = n
Number of moles escaped out
= n1 – n2 = n – n/3 = 2n/3 mol
(iii) n1T1 = n2T2
T1 = 300 K = 27 + 273
n
n × 300 =
T2
n1 = n
2
T2 = 600 K
n1 – n2 = n/2
or n2 = n – n/2 = n/2 mol
Ex. 24 An automobile tyre is inflated to less pressure in
summer than in winter. Why ?
Sol.
To decrease the pressure inside the tyre because
in summer the pressure of gas inside increase
due to increase temperature.
Ex.25
Sol.
What is an Isotherm ?
The graph plotted between P and V for a gas at
constant temperature is known as isotherm.
Ex.26
4 litre C2H4(g) burns in oxygen at 27°C and 1
atm to produce CO2(g) and H2O (g).
6
2nd Floor Rajcastle, Above Domino’s Pizza, Ellorapark
GASEOUS STATE
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JEE CHEMISTRY .
BY AMIT JHAKAL
Calculate the volume of CO2 formed at
(a) 127°C and 1 atm, (b) 27°C and 2 atm.
C2H4(g) + 2O2(g)  2CO2(g) + 2H2O(g)
Under same conditions of P and T, volume of gases react in their mole ratio and produce in the same molar ratio.
Thus, at 27°C and 1 atm.
1 volume or 1 mole of C2H4 gives
= 2 volume CO2
 4 volume of C2H5 gives = 2 × 4 volume
CO2 = 8 litre CO2
(a) Now, at 127°C and 1 atm :
1 8 1 V2
P1V1 P2V2
or
=

T1
T2
300
400
 V2 = 10.67 litre
(b) Similarly at 27°C and 2 atm
V = 16 litre
Sol.
Ex. 27 How many gram of CaCO3 be decomposed to
produce 20 litre of CO2 at 750 torr and 27°C.
Sol.
CaCO3  CaO + CO2
It is thus evident that 1 mole of CO2 is obtained by decomposition of 1 mole of CaCO3
PV
750 20
Also moles of CO2 =
=
RT
760 0.0821 300
Thus, mole of CaCO3 required = 0.8
Also amount of CaCO3 required = 0.8 × 100
= 80 g
Ex. 28 The size of the weather balloon becomes large
and larger as it ascends up into higher altitude.
Assign reason.
Sol.
The volume of the gas is inversely proportional to the pressure at a given temperature according to Boyle's law. As
the weather balloon ascends, the pressure tends to decrease. As a result, the volume of the gas inside the balloon
or the size of the balloon is likely to increase.
Ex.29
Sol.
A human adult breathes in approximately 0.50 mL of air at 1.00 atm with each breath. If an air tank holds 100 L of
air at 200 atm, how many breaths the tank will supply ? Assume that the temperature is 37°C
From the given data :
Volume of air in the tank (V1) = 100 L
Pressure of air in the tank (P1) = 200 atm.
Pressure of air in each breath (P2) = 1 atm
Volume of air at 1 atm (V2)
=?
Since temperature is constant, Boyle's Law is applicable.
P1V1 = P2V2
or
V2 =
P1V1
P2
V2 =
(200atm)  (100L)
= 20,000
(1atm)
Volume of air breathed in each breath = 0.50 L
Total number of breaths the tank will supply
=
Totalvolume of air
Volume of air in each breath
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AMIT JHAKAL .
JEE CHEMISTRY
=
Ex.30
Sol.
20000
= 40,000 breaths.
0.50
A gas cylinder containing cooking gas can withstand a pressure of 14.9 atmospheres. The pressure gauge of the
cylinder indicated 12 atmospheres at 27°C. Due to sudden fire in the building, the temperature starts rising. At
what temperature will the cylinder explode ?
From the available data :
P1 = 14.9 atmospheres ; T1 = ?
P2 = 12.0 atmospheres ; T2 = 27 + 273 K
Since the gas is kept in the cylinder, its volume remains constant. According to Gay – Lussac's law.
P1 T1
P
or T1 = 1 × T2

P2 T2
P2
(14.9 atm)  (300K)
= 372.5 K
(12 atm)

Cylinder will explode at a temperature
= 372.5 – 273 = 99.5°C
20 dm3 of SO2 diffuses through a porous partition in 120 sec. What volume be O 2 will diffuse similar conditions in
60 sec.
For SO2 gas
For O2 gas
M w1 = 64
M w 2 = 32
T1 =
Ex.31
Sol.
V1 = 20 dm3 = 20 lit.
t1 = 120 sec.
r2
V /t
= 2 2 =
r1
V1 / t1
or
t2 = 60 sec.
M w1
Mw2
64
V2 / 60
=
32
20 / 120
V2 = 14.14 dm3 = 14.14 lit.
Ex.32
Sol.
Gases change to liquid state at low temperature and high pressure.
On decreasing the temperature, the kinetic energy of molecules decreases. Also, on increasing the pressure, the
forces of attractions increases. This ultimately brings in liquefaction.
Ex.33
Sol.
H2 and He show exceptional behaviour in deviation from ideal gas nature. Explain.
H2 and He are the lightest gases having almost negligible attractive forces among their molecules and thus shows
exceptional behaviour.
Ex.34
Sol.
Ex.35
Tea of coffee is sipped from a saucer when it is quite hot. Why ?
in saucer, tea or coffee possesses larger surface area for evaportaion and thus loses its temperature.
The volumes of ozone and chlorine diffusing during the same time are 35 ml and 29 ml respectively. If the
molecular weight of chlorine is 71. Calculate molecular mass of ozone.
r (O 3 )
M (Cl 2 )
=
M (O 3 )
r (Cl 2 )
Sol.
Rate of diffusion in same time is proportional to volume diffused so
V(O3 )
M (Cl 2 )
=
M (O 3 )
V(Cl 2 )
35
=
29
8
71
M (O 3 )
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GASEOUS STATE
8
JEE CHEMISTRY .
(35) 2
( 29)
Ex.36
2
=
71
: M(O3) = 48.74
M(O3 )
A fluoride of phosphorous in gaseous state was found to diffuse 2.12 times more slowly than nitrogen under
similar conditions. Calculate the molecular mass and molecular formula of this fluoride which contains one atom
of phosphorous per molecule.
r (fluoride)
=
r (nitrogen)
Sol.
BY AMIT JHAKAL
M(N 2 )
M ( fluoride)
28
M ( fluoride)
1
=
2.12
2
Ex.37
Sol.
Ex.38
28
 1 

 =
M ( fluoride)
 2.12 
M(fluoride) = 126
Further as the molecule contains one phosphorous atom 31 + 19 x = 126
x=5
So the formula of the compound must be PF5.
Which gas will effuse faster, NH3 or CO ? What are their relative rates of effusion ?
Ammonia effuses faster because of its lower molecular mass.
r ( NH 3 )
28
=
= 1.28
17
r (CO )
i.e. ammonia effuses 1.28 times faster than that of CO.
If 1.40 L of an unknown gas requires 57 second to diffuse and if the same volume of N 2 gas takes 84 second to
diffuse at the same temperature and pressure. what is the molecular mass of the unknown gas ?
r1
=
r2
Sol.
M2
V /t
: 1 1 =
M1
V2 / t 2
1.4
84
×
=
57
1.4
28

19
M2
M1
28
M1
28
Squaring both sides
M1
28
28 28
1919
=
M1 =
= 12.9
M1
19 19
28
Molecular mass of unknown gas = 12.9
Ex.39
Sol.
A beaker contains N2 and a gas 'X' at 300 K the gaseous mixture is diffused into empty container of 5.0 L volume.
After a certain time pressure of container becomes 5.5 atm. If 0.9 moles of N 2 are diffuses then calculate
molecular mass of 'X'
From Dalton's law
RT
P = (nA+ nB)
V
0.082 300
5.5 = (0.9 + nx)
5.0
nx = 0.2 moles
From Graham's Law
M2
r1
n
= 1 =
M1
r2
n2
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AMIT JHAKAL .
JEE CHEMISTRY
Mx
0.9
=
28
0.2
0.9  0.9  28
Mx =
= 56.7
0.2  0.2
Ex. 40 Gas 'A' has
1
rate of diffusion than gas 'B'. If molecular mass of gas A is 16 than calculate molecular mass of gas
2
'B'
Sol.
rA
=
rB
MB
MA
MB
1

2 1
16
1 16
MB =
=4
4
nd
10 2 Floor Rajcastle, Above Domino’s Pizza, Ellorapark
GASEOUS STATE
10
JEE CHEMISTRY .
BY AMIT JHAKAL
EXERCISE # 1
Q.1
Q.2
At constant temperature, in a given mass of an
ideal gas (1) The ratio of pressure and volume always
remains constant
(2) Volume always remains constant
(3) Pressure always remains constant
(4) The product of pressure and volume always
remains constant
Q.7
In the gas equation PV = nRT, the value of
universal gas constant would depend only on (1) The nature of the gas
(2) The pressure of the gas
(3) The temperature of the gas
(4) The units of measurement
Q.8
8.2 L of an ideal gas weight 9.0 gm at 300 K
and 1 atm pressure. The molecular mass of gas
is(1) 9
(2) 27
(3) 54
(4) 81
Q.9
Energy in an ideal gas is (1) Completely kinetic
(2) Completely potential
(3) KE + PE
(4) All the above.
Q.10
A 0.5 dm3 flask contains gas 'A' and 1 dm3 flask
contains gas 'B' at the same temperature. If
density of A = 3.0 gm dm–3 and that of B = 1.5
gm dm–3 and the molar mass of A = 1/2 of B,
then the ratio of pressure exerted by gases is(1) PA/PB = 2
(2) PA/PB = 1
Three flasks of equal volumes contain CH4,
CO2 and Cl2 gases respectively. They will
contain equal number of molecules if (1) the mass of all the gases is same
(2) the moles of all the gas is same but
temperature is different
(3) temperature and pressure of all the flasks
are same
(4) temperature, pressure and masses same in
the flasks
Q.3
A certain mass of a gas occupies a volume of 2
litres at STP. Keeping the pressure constant at
what temperature would the gas occupy a
volume of 4 litres (1) 546ºC
(2) 273ºC
(3) 100ºC
(4) 50ºC
Q.4
At 100 ºC a gas has 1 atm. pressure and 10 L
volume. Its volume at NTP would be (1) 10 litres
(2) Less than 10 litres
(3) More than 10 litres
(4) None
Q.5
If 500 ml of a gas 'A' at 1000 torr and 1000 ml
of gas B at 800 torr are placed in a 2L container,
the final pressure will be(1) 100 torr
(2) 650 torr
(3) 1800 torr
(4) 2400 torr
Q.6
Two flasks A and B of 500 ml each are
respectively filled with O2 and SO2 at 300 K
and 1 atm. pressure. The flasks will contain(1) The same number of atoms
(2) The same number of molecules
(3) More number of moles in flask A as
compared to flask B
(4) The same amount of gases
2nd Floor Rajcastle, Above Domino’s Pizza ellorapark
(3) PA/PB = 4
(4) PA/PB = 3.
Q.11
The density of a gas is equal to ?
(P = pressure ; V = volume ; T = temperature,
R = gas constant, n = number of moles and
M = molecular wt) (1) nP
(2) PM/RT
(3) P/RT
(4) M/V
Q.12
One litre of an unknown gas weighs 1.25 gm at
N.T.P. which of the following gas pertains to
the above data (1) CO2
(2) NO2
(3) N2
Q.13
(4) O2
If the density of a gas A is 1.5 times that of B
then the molecular mass of A is M. The
molecular mass of B will be(1) 1.5 M
(2) M/1.5
(3) 3M
(4) M/3
GASEOUS STATE
11
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AMIT JHAKAL .
JEE CHEMISTRY
Q.14
When the pressure of 5L of N2 is doubled and
Q.22
its temperature is raised from 300K to 600K, the
final volume of the gas would be(1) 10 L
(2) 5 L
(3) 15 L
(4) 20 L
Q.15
The value of gas constant R is 8.314 X. Here X
is represent (1) Litre atm K–1 mol–1
(2) Cal mol–1 K–1
(3) J K–1 mol–1
(4) None of the above.
Q.16
The value of gas constant per mole is
approximately(1) 1 cal
(2) 2 cal
(3) 3 cal
(4) 4 cal
Q.17
A gas is found to have a formula [CO]x. If its
the pressure is 750 mm of Hg, the partial
pressure of O2 is (1) 157.5 mm of Hg
(2) 175.5 mm of Hg
(3) 315.0 mm of Hg
(4) 257.5 mm of Hg
Q.23
Which is not correct in terms of kinetic theory
of gases(1) Gases are made up of small particles called
molecules
(2) The molecules are in random motion
(3) When molecules collide, they lose energy
(4) When the gas is heated , the molecules
moves faster
Q.24
The kinetic energy of 1 mole of gas is equal to 3
RT
2
RT
(3)
2
(1)
vapour density is 70 the value of x is(1) 2.5
(2) 3.0
(3) 5.0
(4) 6.0
Q.18
A cylinder is filled with a gaseous mixture
containing equal masses of CO and N2. The
Q.19
(4) PCO =
A gas 'A' having molecular weight 4 diffuses
thrice as fast as the gas B. The molecular weight
of gas B is(1) 36
(2) 12
(3) 18
(4) 24
Q.26
The increasing order of effusion among the
gases, H2, O2, NH3 and CO2 is-
1
P
2 N2
At room temperature Dalton's law of partial
pressure is not applicable to (1) H2 and N2 mixture
(1) H2, CO2, NH3, O2
(2) H2, NH3, O2, CO2
(2) H2 and Cl2 mixture
(3) H2, O2, NH3, CO2
(3) H2 and CO2 mixture
(4) CO2, O2, NH3, H2
(4) None
Q.20
The total pressure of a mixture of two gases is (1) The sum of partial pressures of each gas
(2) The difference in partial pressures
(3) The product of partial pressures
(4) The ratio of partial pressures.
Q.21
Equal masses of SO2, CH4 and O2 are mixed in
empty container at 298 K, when total pressure is
2.1 atm. The partial pressures of CH4 in the
mixture is (1) 0.5 atm
(3) 1.2 atm
(2) 0.75 atm
(4) 0.6 atm
nd
12 2 Floor Rajcastle, Above Domino’s Pizza, Ellorapark
3
KT
2
2R
(4)
3
(2)
Q.25
ratio of their partial pressure is(1) PN2 = PCO
(2) PCO = 0.875 PN2
(3) PCO = 2 PN2
Air contains 79% N2 and 21% O2 by volume. If
Q.27
The rate of diffusion of methane at a given
temperature is twice that of a gas X. The
molecular weight of X is (1) 64
(2) 32
(3) 4
(4) 8
Q.28
A gas X diffuses three times faster than another
gas Y the ratio of their densities i.e., Dx : Dy is(1)
(3)
1
3
1
6
(2)
1
9
(4)
1
12
GASEOUS STATE
12
JEE CHEMISTRY .
Q.29
BY AMIT JHAKAL
In which of the following pairs the gaseous
species diffuse through a porous plug with the
same rate of diffusion -
Q.36
Most probable speed, average speed and RMS
speed are related as (1) 1 : 1.128 : 1.224
(1) NO, CO
(2) NO, CO2
(2) 1 : 1.128 : 1.424
(3) NH3, PH3
(4) NO, C2H6
(3) 1 : 2.128 : 1.224
(4) 1 : 1.428 : 1.442
Q.30
Which of the following expression does not
give root mean square velocity1
1
 3P  2
(2) 

 DM 
 3RT  2
(1) 

 M 
1
1
 3P  2
(3)  
D
Q.31
Q.32
Q.33
(2) Carbon dioxide
(3) Sulphur dioxide
(4) Carbon monoxide.
(2) 172.6 K
(3) 596 K
(4) 1192 K
(2)
Q.39
The correct expression for the vander waal's
equation of states is-
 3RT 


 M 
(4) All are correct.
The term that accounts for intermolecular force
in vander Waal's equation for non ideal gas is (1) RT
(2) V - b
(3) (P + a / V2)
(4) [RT]-1
The critical temperature of a substance is (1)The temperature above which the substance
undergoes decomposition
(2)The temperature above which a substance
can exist only as a gas
 3PV 


 M 
(3) Boiling point of the substance
(4) All are wrong
Q.41
the
gas
the
(4) At which one mole of it occupies volume of
22.4 L
Q.42
Molecular attraction and size of the molecules
in a gas are negligible at (1) Critical point
(2) HBr > O2 > N2 > H2
(2) High pressure
(3) HBr > H2 > O2 > N2
(3) High temperature and low pressure
(4) N2 > O2 > H2 > HBr
(4) Low temperature and high pressure
2nd Floor Rajcastle, Above Domino’s Pizza ellorapark
is
(3) At which it occupies 22.4 L of volume
At S.T.P. the order of mean square velocity of
(1) H2 > N2 > O2 > HBr
of
(2) Above which it cannot be liquified
(4) Cl2
molecules H2, N2, O2 and HBr is -
Critical temperature
temperature-
(1) Below which it cannot be liquified
Among the following gases which one has the
lowest root mean square velocity at 25ºC(3) O2
(4) Non-real gases.
(4) (p + an2/V2) (V - nb) = nRT
Q.40
 3P 
 
 d 
(2) N2
(2) Real gases
(3) Vapours
(3) (p + an2/V2) (V - b) = nRT
The RMS velocity at NTP of the species can be
calculated from the expression -
(1) SO2
(1) Ideal gases
(2) (p + an2/V2) (V - nb) = nRT
If the r.m.s. velocity of nitrogen molecules is
5.15 ms–1 at 298K, then a velocity of 10.30ms–1
will be possessed at a temp(1) 149 K
The Vander Waals' equation explains the
behaviour of -
(1) (p + a/n2 V2) (V - nb) = nRT
(1) Oxygen
(3)
Q.35
Q.38
 3PV  2
(4) 

 M 
Which one of the following gases would have
the highest R.M.S. velocity at 25ºC -
(1)
Q.34
Q.37
GASEOUS STATE
13
13
AMIT JHAKAL .
JEE CHEMISTRY
Q.43
The units of the Van der Waal’s constant ‘a’ are (1) atm L2 mol–2
(3) atm L
Q.44
Q.45
mol–1
Q.47
(2) atm L–2 mol–2
(4) atm
is -
mol L–2
The units of the van der Waal’s constant ‘b’ are (1) atmosphere
(2) joules
(3) L mol–1
(4) mol L–1
For H2 gas, the compressibility factor,
The correct relationship between T C, TB and Ti
Q.48
(1) TC < TB < Ti
(2) TC > TB > Ti
(3) TC < Ti < TB
(4) Ti < TB < TC
If the Vander Waal’s constants of gas A are
given as a (atm L2 mol2) = 6.5
Z = PV/n RT is -
b (L mol–1) 0.056
(1) equal to 1
than ciritical pressure of A is
(2) equal to 0
(1) 56.24 atm
(2) 76.77 atm
(3) always greater than 1
(3) 42.44 atm
(4) 36.42 atm
(4) initially less than 1 and then becomes greater
than 1 at high pressures
Q.46
The Van der Waal’s parameters for gases W, X,
Y and Z are -
Gas
W
X
Y
Z
a (atm L2 mol2 ) b(Lmol1 )
4.0
0.027
8.0
0.030
6.0
0.032
12.0
0.027
Which one of these gases has the highest critical
temperature ?
(1) W
(2) X
(3) Y
(4) Z
nd
14 2 Floor Rajcastle, Above Domino’s Pizza, Ellorapark
GASEOUS STATE
14
JEE CHEMISTRY .
BY AMIT JHAKAL
EXERCISE # 2
Q.1
V versus T curves at constant pressure P1 and
Q.6
A flask of methane (CH4) was weighed.
Methane was then pushed out and the flask
P2 for an ideal gas are shown in fig. Which is
again weighed when filled with oxygen at the
correct -
P1
V
same temperature and pressure. The mass of
P2
oxygen would be (1) The same as the methane
(2) Half of the methane
Q.2
T
(3) Double of that of methane
(1) P1 > P2
(2) P1 < P2
(4) Negligible in comparison to that of
(3) P1 = P2
(4) All
If pressure of a gas contained in a closed vessel
methane
Q.7
A balloon filled with methane (CH4) is pricked
is increased by 0.4% when heated by 1ºC its
with a sharp point and quickly plunged into a
initial temperature must be -
tank of hydrogen at the same pressure. After
(1) 250K
(2) 250ºC
sometime, the balloon will have -
(3) 2500K
(4) 25ºC
(1) Enlarged
(2) Shrinked
Q.3
At a constant pressure, what should be the
(3) Remain unchanged in size
percentage increase in the temperature in kelvin
(4) Ethylene (C2H4) inside it
for a 10% increase in volume (1) 10%
(2) 20%
(3) 5%
(4) 50%
Q.8
Contaniers X, Y and Z of equal volume contain
oxygen, neon and methane respectively at the
same temperature and pressure. The correct
Q.4
There is 10 litre of a gas at STP. Which of the
incereasing order of their masses is -
following changes keeps the volume constant -
(1) X < Y < Z
(2) Y < Z < X
(1) 273 K and 2 atm
(3) Z < X < Y
(4) Z < Y < X
(2) 273ºC and 2 atm
Q.9
(3) 546ºC and 0.5 atm
Two flasks X and Y have capacity 1L and 2L
respectively and each of them contains 1 mole
(4) 0ºC and 0 atm
of a gas. The temperature of the flask are so
Q.5
The density of oxygen gas at 25ºC is 1.458
adjusted that average speed of molecules in X is
mg/litre at one atmosphere. At what pressure
twice as those in Y. The pressure in flask X
will oxygen have the density twice the value-
would be -
(1) 0.5 atm/25ºC
(2) 2 atm/25ºC
(1) Same as that in Y
(3) 4 atm/25ºC
(4) None
(2) Half of that in Y
(3) Twice of that in Y
(4) 8 times of that in Y
2nd Floor Rajcastle, Above Domino’s Pizza ellorapark
GASEOUS STATE
15
15
AMIT JHAKAL .
JEE CHEMISTRY
Q.10
The partial pressure of hydrogen in a flask
Q.16
does not obey Dalton's law of partial pressure-
containing 2gm of H2 & 32gm of SO2 is -
Q.11
Q.12
(1)
1
of total pressure
16
(2)
1
of total pressure
2
(3)
2
of total pressure
3
(4)
1
of total pressure.
8
A gas can be liquefied by (1) Cooling
(2) Compressing
(3) Both
(4) None of these.
Q.17
Q.18
(4) SO2 and SO3
The rate of diffusion of hydrogen is about (1) One half that of He
The velocity possessed by most of the gaseous
(2) Most probable velocity
(3) R.M.S. velocity
(4) None of these.
Q.19
A 2.24L cyclinder of oxygen at N.T.P. is found
to develop a leakage. When the leakage was
(4) Intermolecular forces
plugged the pressure dropped to 570 mm of Hg.
The number of moles of gas that escaped will be -
A box of 1L capacity is divided into two equal
compartments by a thin partition which are
(1) 0.025
(2) 0.050
(3) 0.075
(4) 0.09
The pressure in each compartment is recorded
Q.20
(1) P
(3) P/2
At constant temperature 200 cm3 of N2 at 720
mm and 400 cm3 of O2 at 750 mm pressure are
as P atm. The total pressure when partition is
removed will be -
put together in a one litre flask. The final
pressure of mixture is -
(2) 2P
(4) P/4
(1) 111 mm
(3) 333 mm
(2) 222 mm
(4) 444 mm
Average K.E. of CO2 at 27ºC is E. The average
kinetic energy of N2 at the same temperature
Q.21
will be(1) E
(3) E/22
Q.15
(3) CO and CO2
molecules is (1) Average velocity
filled with 2g H2 and 16gm CH4 respectively.
Q.14
(2) NH3 and HCl
(4) Four times that of He
(3) Increase in KE of molecules
Q.13
(1) NO2 and O2
(2) 1.4 times that of He
(3) Twice that of He
The P of real gases is less than the P of an ideal
gas because of (1) Increase in number of collisions
(2) Finite size of molecule
Which mixture of gases at room temperature
the same mass of gas at the same pressure is (1) 273 K
(2) -273ºC
(2) 22E
(4) E/
Helium atom is twice times heavier than a
hydrogen molecule. At 25ºC the average K.E. of
helium atom is -
(3) 273ºC
Q.22
(4) Half that of hydrogen
(4) 546ºC
26 c.c. of CO2 are passed over red hot coke. The
volume of CO evolved is (1) 15 c.c
(3) 32 c.c.
(1) Twice that of hydrogen
(2) Same as that of hydrogen
(3) Four times that of hydrogen
10 gm of a gas at NTP occupies 5 litres. The
temp. at which the volume becomes double for
Q.23
(2) 10 c.c.
(4) 52 c.c.
If temperature and volume are same, the
pressure of a gas obeying Vander Waals
equation is -
nd
16 2 Floor Rajcastle, Above Domino’s Pizza, Ellorapark
GASEOUS STATE
16
JEE CHEMISTRY .
BY AMIT JHAKAL
(1) Smaller than that of an ideal gas
(2) Larger than that of an ideal gas
(3) Same as that of an ideal gas
(4) None of these
Q.24
(3) 533K
Q.30
Q.25
(2) Very strong
(3) Weak
(4) Very weak
Q.31
(4) CO2
(3) 2.4 m sec–1
(4) 3.0 m sec–1
Average velocity is equal to -
(3) 0.9602 RMS velocity
(4) 0.9813 RMS velocity
CO2 = 3.59 , SO2 = 6.71 , Cl2 = 6.49)
(3) SO2
(2) 0.9 m sec–1
(2) 0.9 RMS velocity
liquefied ? (Given value of 'a' for NH3 = 4.17,
(2) Cl2
(1) 0.6 m sec–1
(1) 0.9213 RMS velocity
Which of the following can be most readily
(1) NH3
The RMS velocity of an ideal gas at 27 ºC is 0.3
m sec–1. Its RMS velocity at 927 ºC is -
In case of hydrogen and helium the Vander
Waals forces are (1) Strong
(4) 300 K
Q.32
Which of the following expression gives the
variation of density of ideal gas with changes in
temperature ?
Q.26
Which of the following is true -
Q.27
(1) urms >  > .
(2) urms <  < .
(3) urms >  < .
(4) urms <  > .
Hydrogen and Argon are kept in two separate
Q.33
(1)
d 2 P2 T1

d1 P1T2
(3)
d 2 T2

d1 T1
(2) d 2 
(4) d 2 
d1T1
T2
d 2 P2 T2
P1T1
In the below experiment, the value of P is –
1 mol gas A 2 mol gas B
Pressure
Partition


removedat
250 mm
250 mm
P
but identical vessels at constant temperature and
pressure -
sametem
perature(T )
(1) Both contain same number of atoms.
(2) The number of atoms of argon is half that of
(1) 250 mm
(2) 500 mm
(3) 300 mm
(4) 400 mm
hydrogen.
(3) The number of atoms of argon is double that
Q.34
of hydrogen
(1) CO
(4) None of these
Q.28
Q.35
(4) C2H6
The density of a gas at 27°C and 1 atm pressure
.
ratio 1 : 2 . The ratio of rates of diffusions of O2
is
and CH4 at same P and T is -
temperature at which its density is 0.5  is-
(1) 1 : 2
(1) 200 K
(2) 400 K
(3) 600 K
(4) 800 K
(3) 1 : 1.41
(2) 2 : 1
(4) 1 : 4.14
Q.36
Pressure
remaining
constant,
the
At a given temperature (X) = 2(Y) and
At what temperature will be total kinetic energy
(KE) of 0.30 mole of He be the same as the total
M(Y) = 3 M(X), where  and M stand
KE of 0.40 mole of Ar at 400K-
gases X and Y, then the ratio of their pressures
(1) 400K
will be -
(2) 373 K
2 Floor Rajcastle, Above Domino’s Pizza ellorapark
nd
(2) CH4
(3) CO2
The vapour densities of CH4 and O2 are in the
Q.29
Which of the following molecule has the lowest
average speed at 273 K ?
respectively for density and molar mass of the
GASEOUS STATE
17
17
AMIT JHAKAL .
JEE CHEMISTRY
Q.37
Q.38
(1) p(X) / p(Y) = 1/4 (2) p(X) / p(Y) = 4
(2) Z = 1 + a / Vm RT
(3) p(X) / p(Y) = 6
(3) Z = 1 + pb / RT
(4) p(X) / p(Y) = 1/6
The total kinetic energy of 0.6 mol of an ideal
gas at 27° C is (1) 1122 J
(2) 1681 J
(3) 2245 J
(4) 2806 J
(4) Z = 1 – pb / RT
Q.43
The most favourable conditions to liquefy a gas
are (1) High T and Low P
For a fixed mass of a gas at constant pressure,
which of the following is correct -
(2) High T and High P
(3) Low T and Low P
(1) Plot of volume versus Celsius temperature is
(4) Low T and High P
linear with intercept zero
(2) Plot of volume versus kelvin temperature is
linear with a nonzero intercept
(3) Plot of V/T versus T is linear with a positive
slope
(4) Plot of V/T versus T is linear with a zero
slope
Q.39
The density of carbon monoxide at STP is (1) 0.625 g L–1
(2) 1.25 g L–1
(3) 2.5 g L–1 (4) 1.875 g L–1
Q.40
A real gas is expected to exhibit maximum
deviations from ideal gas laws at (1) Low T and High P
(2) Low T and Low P
(3) High T and High P
(4) High T and Low P
Q.41
Which
of
the
following
expressions
of
compression factor Z (= pVm / RT) of a real gas
is applicable at high pressure (1) Z = 1 – a / Vm RT
(2) Z = 1 + a / Vm RT
(3) Z = 1 + pb / RT
(4) Z = 1 – pb / RT
Q.42
Which
of
the
following
expressions
of
compression factor Z (= pVm / RT) of a real gas
is applicable at low pressure (1) Z = 1 – a / Vm RT
EXERCISE # 3(A)
nd
18 2 Floor Rajcastle, Above Domino’s Pizza, Ellorapark
GASEOUS STATE
18
JEE CHEMISTRY .
Q.1
BY AMIT JHAKAL
The rate of diffusion of hydrogen is about –
(2) The number of molecules in one gram of gas
[NCERT-74]
(1)
(3) The number of molecules contained in 12
grams of 12C isotop
1
that of Helium
2
(4) The number of molecules in 22.4 litres of a
gas at S.T.P.
(2) 1.4 times that of Helium
(3) twice that of H
Q.8
(4) Four times that of Helium
The V.D. of a gas is 11.2. The volume occupied
by 11.2 g of this gas at N.T.P is -
Q.2
The kinetic theory of gases predicts that total
(1) 22.4 litres
(2) 11.2 litres
kinetic energy of a gaseous assembly depends on –
(3) 1 litres
(4) 2.24 litres
[NCERT-74]
(1) Pressure of the gas
Q.9
(3) Volume of the gas
is known as -
(4) Pressure, temperature and volume of the gas
The vapour density of a gas is 35.5. The volume
occupied by 3.55 g of the gas at N.T.P. is –
[NCERT-74]
Q.4
(1) 1.12 litres
(2) 11.2 litres
(3) 22.4 litres
(4) 44.8 litres
Q.10
(2) 1/5
(3) 1/4
(4) 1
Q.11
[NCERT-75]
(3) 1/8
(4) 1/2
Q.12
Q.6
(3) Avogadro's law
(4) Dalton's law
One litre of a gas weighs 2g at 300 K and 1 atm
pressure. If the pressure is made 0.75 atm at
(1) 450 K
(2) 600 K
(3) 800 K
(4) 900 K
The density of a gas at 27°C and 1 atm is d.
Pressure remaining constant, at which of the
following temperatures will its density become
0.75 d ?
[CBSE-92]
weight = 128) as compared to oxygen is –
(2) 1/4
(2) Charle's law
[CBSE-86]
The relative rate of diffusion of a gas (molecular
(1) 2 times
(1) Boyle's law
litre of the same gas weigh one gram ?
The relative rate of diffusion of a gas (Mol wt. =
98) as compared to hydrogen will be –
(1) 1/7
[CPMT-86]
which of the following temperatures will one
[NCERT-72]
Q.5
The total pressure exerted by a number
nonreacting gases is equal to the sum of partial
pressure of the gases under the same conditions
(2) Temperature of the gas
Q.3
[CPMT-74]
Since the atomic weights of carbon nitrogen and
oxygen are 12, 14 and 16 respectively, among
(1) 20°C
(2) 30°C
(3) 400 K
(4) 300 K
Select one correct statement. In the gas equation :
PV = nRT
[CBSE-92]
(1) n is the number of molecules of a gas
the following pairs of gases, the pair that will
(2) n moles of the gas have a volume V
diffuse at the same rate is -
(3) V denotes volume of one mole of the gas
[NCERT-81]
(1) Carbon dioxide and nitrous oxide
(4) p is the pressure of the gas when only one
mole of gas is present
(2) Carbon dioxide and nitrogen peroxide
(3) Carbon dioxide and carbon monoxide
(4) Carbon dioxide and nitric oxide
Q.13
3.2 g of oxygen (At wt. = 16) and 0.2 g of
hydrogen (At wt. = 1) are placed in a 1.12 L
flask at 0°C. The total pressure of the gas
Q.7
One mole of a gas refers to -
[NCERT-80]
(1) The number of molecules in one litre of gas
2nd Floor Rajcastle, Above Domino’s Pizza ellorapark
mixture will be GASEOUS STATE
[CBSE-92]
19
19
AMIT JHAKAL .
JEE CHEMISTRY
Q.14
(1) 1 atm
(2) 4 atm
(3) 3 atm
(4) 2 atm
Q.19
The average kinetic energy of an ideal gas per
molecule in SI units at 25°C will be - [CBSE-96]
50 ml of a gas A diffuse through a membrane in
(1) 6.17 × 10–21 KJ
(2) 6.17 × 10–21 J
(3) 6.17 × 10–20 J
(4) 7.16 × 10–20J
the same time as for the diffusion of 40 ml of a
gas B under identical pressure temperature
Q.20
The compressibility factor of an ideal gas is –
conditions. If the molecular weight of A = 64,
that of B would be -
[AIIMS-97]
[CBSE-92]
(1) 100
(2) 250
(3) 200
(4) 80
Q.21
(1) 0
(2) 1
(3) 2
(4) 4
Which of the following expressions correctly
represents the relationship between the average
Q.15
The temerpature of the gas is raised from 27°C
molar kinetic energy of CO and N2 molecules at
to 927°, the root mean square velocity is -
the same temperature ?
[CBSE-94]
(1)
[CPMT-2000]
(1) KE of CO = KE of N2
(2) KE of CO > KE of N2
927
times the earlier value
27
(3) KE of CO < KE of N2
(4) Cannot predict
(2) same as before
(3) halved
Q.22
(4) doubled
An ideal gas will have maximum density when [CPMT-2000]
Q.16
(1) P = 0.5 atm T = 600 K
50 ml of hydrogen diffuses through a small hole
(2) P = 2 atm T = 150 K
from vessel in 20 minutes time. Time taken for
(3) P = 1 atm T = 300 K
40 ml of oxygen to diffuse out under similar
conditions will be –
(4) P = 1 atm T = 500 K
[CBSE-94]
(1) 12 min.
(2) 64 min
(3) 8 min.
(4) 32 min.
Q.23
The rate of diffusion of a gas having molecular
weight just double of nitrogen gas is 56 ml per
Q.17
sec the ratio of diffusion of nitrogen gas will be -
The densities of two gases are in the ratio of 1 :
[CPMT-2000]
16. The ratio of their rates of diffusion is –
[CBSE-95]
(1) 16 : 1
(2) 4 : 1
(3) 1 : 4
(4) 1 : 16
Q.24
Q.18
The pressure of 2 moles of ideal gas at 546 K
having volume 44.8 L is (1) 2 atm
(2) 3 atm
(3) 4 atm
(4) 1 atm
[CBSE-95]
nd
20 2 Floor Rajcastle, Above Domino’s Pizza, Ellorapark
(1) 79.19 ml/sec
(2) 112 ml/sec
(3) 56 ml/sec
(4) 90 ml/sec
Density ratio of O2 and H2 is 16 : 1. The ratio of
their rms velocities will be (1) 4 : 1
(2) 1 : 16
(3) 1 : 4
(4) 16 : 1
[AIMS-2000]
GASEOUS STATE
20
JEE CHEMISTRY .
Q.25
BY AMIT JHAKAL
Vander waal's real gas, act as a ideal gas at
of the mixture was found 1 atmosphere, the
which conditions -
partial pressure of the nitrogen (N2) in the
[CPMT-2000]
(1) High temperature , Low pressure
mixture is :
(2) Low temperature, High pressure
(1) 1 atm
(2) 0.5 atm
(3) High temperature , high pressure
(3) 0.8 atm
(4) 0.9 atm
Q.29
(4) Low temperature , low pressure
Q.26
By what factor does the average velocity of a
gaseous
molecule
increase
when
the
temperature (in Kelvin) is doubled ?
Q.27
(2) 2.0
(3) 2.8
A bubble of air is underwater at temperature
15°C and the pressure 1.5 bar. If the bubble
rises to the surface where the temperature is
25°C and the pressure is 1.0 bar, what will
happen to the volume of the bubble?
[AIPMT-2011]
(1) 1.4
[AIPMT-2011]
(4) 4.0
Two gases A and B having the same volume
diffuse through a porous partition in 20 and 10
seconds respectively. The molecular mass of A
[AIPMT MAINS-2011]
(1) Volume will become greater by a factor of 2.5
(2) Volume will become greater by a factor of 1.6
(3) Volume will become greater by a factor of 1.1
(4) Volume will become smaller by a factor of 0.70
is 49 u. Molecular mass of B will be :
[AIPMT-2011]
Q.28
(1) 25.00 u
(2) 50.00 u
(3) 12.25 u
(4) 6.50 u
A gaseous mixture was prepared by taking
equal mole of CO and N2. If the total pressure
EXERCISE # 3(B)
Q.1
The vapour density of pure ozone would be [Delhi PMT-82]
2nd Floor Rajcastle, Above Domino’s Pizza ellorapark
(1) 48
(3) 24
(2) 32
(4) 16
GASEOUS STATE
21
21
AMIT JHAKAL .
JEE CHEMISTRY
Q.2
According to kinetic theory of gases there are [AP-80]
(1) intermolecular attractions
(2) Molecules which have considerable volume
(3) no intermolecular forces of attraction
(4) The velocity of molecules decreases for each
collision
Q.3
At the same temperature and pressure which of
the following gas will have highest K.E. per
mole [MLNR-91]
(1) Hydrogen
(2) Oxygen
(3) Methane
(4) All the above
Q.4
Equal masses of methane and oxygen are mixed
in an empty container at 25°C. The fraction of
the total pressure exerted oxygen is - [IIT-81]
(1) 1/3
(2) 1/2
1 273
(3) 2/3
(4) ×
3 298
Q.5
Q.6
Q.8
The density of Neon will be highest at –[IIT-90]
(1) STP
(2) 0°C, 2 atm
(3) 273°C, 1 atm
(4) 273°C, 2 atm
Q.9
According to kinetic theory of gases – [IIT-91]
(1) the pressure exerted by a gas is, proportional
to mean square velocity of the molecules
(2) the pressure exerted by the gas is proportional
to the root mean square velocity of the
molecules
(3) the root mean square velocity is inversely
proportional to the temperature
(4) the mean translational K.E. of the molecule
is directly proportional to the absolute
temperature
Q.10
A real gas most closely approaches the behaviour
of an ideal gas at [CEE Karnataka-92]
(1) 15 atm and 200 K (2) 1 atm and 273 K
(3) 0.5 atm and 500 K (4) 15 atm and 500 K
Q.11
If rate of diffusion of A is 5 times that of B,
what will be the density ratio of A and B [AFMC-94]
(1) 1/25
(2) 1/5
(3) 25
(4) 5
Q.12
Which of the following statement is false [BHU-94]
(1) The product of pressure volume of fixed
amount of a gas is independent of
temperature
(2) Molecules of different gases have the same
K.E. at a given temperature
(3) The gas equation is not-valid at high
Equal masses of methane and hydrogen are
mixed in an empty container at 25°C. The
fraction of the total pressure exerted by
hydrogen is [IIT-84]
(1)
1
2
(2)
8
9
(3)
1
9
(4)
16
17
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-88]
(1) at the centre of the tube
(2) near the hydrogen chlrodie bottle
(3) near the ammonia bottle
(4) throughout the length of the tube
pressure and low temperature
(4) The gas constant per molecule is known as
Boltzmann constant
Q.13
Two spearate bulbs contain ideal gases A and B.
The density of A is twice as that of gas B. The
Q.7
The values of vander waals' constant 'a' for the
gases O2, N2, NH3 and CH4 are 1.360, 1.390,
4.170 and 2.253 L atm mol–2 respectively. The
gas which can most easily be liquefied is –
[IIT-89]
(1) O2
(2) N2
(3) NH3
(4) CH4
nd
22 2 Floor Rajcastle, Above Domino’s Pizza, Ellorapark
molecular mass of gas A is half as that of B. If
two gases are at same temperature, the ratio of
the pressure of A to that of B is(1) 2
(2) 1/2
(3) 4
(4) 1/4
[BHU-94]
GASEOUS STATE
22
JEE CHEMISTRY .
Q.14
BY AMIT JHAKAL
(3) Boyles Law
The molecular velocities of two gases at same
(4) Gay Lussac Law
temperature are u1 and u2, their masses are m1
and m2 respectively, which of the following
expression is correct ?
(1)
m1
u12
=
m2
u 22
m
m
(3) 1 = 2
u1
u2
Q.20
(in Kelvin) is x cm/sec. At what temperature
[BHU-94]
(in kelvin) the rms velocity of nitrous oxide
would be 4x cm/sec.
[EAMECT-2000]
(2) n1u1 = n2u2
(4) m1 u 12 = m2 u 22
Q.21
Q.15
The density of methane at 2.0 atmosphere
pressure at 27°C is -
Q.16
(2) 0.26 gL–1
(3) 1.30 gL–1
(4) 26.0 gL–1
If the volume of 2 moles of an ideal gas at 540
Q.22
[AFMC-98]
(2) 2 atmosphere
(3) 3 atmosphere
(4) 4 atmosphere
Q.23
Q.17
(2) 2 T
(3) 4 T
(4) 32 T
If the four tubes of a car are filled to the same
pressure with N2, O2, H2 and we separately then
(1) N2
(2) O2
(3) H2
(4) He
If the average velocity of N2 molecules is 0.3
m./sec. at 27°C, then the velocity of 0.6 m/sec
will take place at [Manipal-2001]
K is 44.8 litre then its pressure will be(1) 1 atmosphere
(1) 16 T
which one will be filled first - [Manipal-2001]
[BHU-94]
(1) 0.13 gL–1
The rms velocity of CO2 at a temperature T
Densities of two gases are in the ratio 1 : 2 and
(1) 273 K
(2) 927 K
(3) 1000 K
(4) 1200 K
Which of the following has maximum root
mean square velocity at the same temperature ?
their temperatures are in the ratio 2 : 1 then the
[CPMT-2002]
ratio of their respective pressure is [BHU-2000]
Q.18
(1) 1 : 1
(2) 1 : 2
(3) 2 : 1
(4) 4 : 1
Q.24
Gas equation PV = nRT is object by -
(1) SO2
(2) CO2
(3) O2
(4) H2
For an ideal gas number of moles per lit in
terms of its pressure P, gas constant R and
temperature T is -
[BHU-2000]
(1) Only isothermal process
(2) Only adiabatic process
(3) Both
[AIEEE-2002]
(1)
PT
R
(2) PRT
(3)
P
RT
(4)
RT
P
(4) None
Q.19
The following graph illustrates [JIPMER-2000]
Under identical conditions of temperature and
pressure the ratio of the rates of effision of O2
and CO2 gases is given by - [Kerla CEE-2002]
V
Temp(°C)
(1) Dalton's Law
Q.25
(1)
ratio of effusion of oxygen
= 0.87
ratio of effusion of CO 2
(2)
ratio of effusion of oxygen
= 1.17
ratio of effusion of CO 2
(2) Charles Law
2nd Floor Rajcastle, Above Domino’s Pizza ellorapark
GASEOUS STATE
23
23
AMIT JHAKAL .
JEE CHEMISTRY
(3)
ratio of effusion of oxygen
= 8.7
ratio of effusion of CO 2
(1) (PA/PB) (MB/MA)1/2
(4)
ratio of effusion of oxygen
= 0.117
ratio of effusion of CO 2
(3) (PA/PB) (MA/MB)1/2
(5)
ratio of effusion of oxygen
= 11.7
ratio of effusion of CO 2
(2) (PA/PB)1/2 (MB/MA)
(4) (PA/PB)1/2 (MA/MB)
Q.30
The molecular velocity of any gas is –
[AIEEE-2011]
Q.26
If the volume of 2 moles of an ideal gas at 540
K is 44.8 litre then its pressure will be –
[AFMC-98]
(1) 1 atmosphere
(2) 2 atmosphere
(3) 3 atmosphere
(4) 4 atmosphere
(1)
inversely
proportional
to
absolute
temperature
(2)
directly
proportional
to
square
of
temperature
(3) directly proportional to square root of
Q.27
If 10 4 dm 3 of water is introduced into a 1.0
dm
3
flask at 300 K, how many moles of water
are in in the vapour phase when equilibrium is
temperature
(4) inversely proportional to the square root of
temperature
established ? (Given : Vapour pressure of
H 2 O at 300 is 3170 pa; R = 8.314 J K 1 mol 1 )
[AIEEE-2010]
(1) 1.27 x 10 3 mol (2) 5.56 x 10 3 mol
(3) 1.53 x10 2 mol (4) 4346 x 10 2 mol
Q.28
'a' and 'b' are van der Waals' constants for gases.
Chlorine is more easily liquefied than ethane
because:
[AIEEE-2011]
(1) a and b for Cl2 > a and b for C2H6
(2) and b for Cl2 < a and b for C2H6
(3) a for Cl2 < a for C2H6 but b for Cl2 > b for
C2H6
(4) a for Cl2 > a for C2H6 but b for Cl2 < b for
C2H6
Q.29
When r, P and M represent rate of diffusion,
pressure and molecular mass, respectively, then
the ratio of the rates of diffusion (r A/rB) of two
gases A and B, is given as -
[AIEEE-2011]
EXERCISE # 4
(ASSERTION AND REASON TYPE)
nd
24 2 Floor Rajcastle, Above Domino’s Pizza, Ellorapark
GASEOUS STATE
24
JEE CHEMISTRY .
BY AMIT JHAKAL
These questions of two statements each,
printed as Assertion and Reason. While
answering these Questions you are required
to choose any one of the following four
responses.
(A) If both Assertion & Reason are true &
the Reason is a correct explanation of the
Assertion.
(B) If both Assertion and Reason are true
but Reason is not a correct explanation
of the Assertion.
(C) If Assertion is true but the Reason is
false.
(D) If Assertion & Reason both are false.
Q.5
Assertion : At very high pressures, compressibility
factor is greater than 1.
Reason : At very high pressure, 'b' can be
neglected in Van der Waal's gas equation.
(1) A
Q.6
(2) B
(3) C
(D) 4
Assertion : Greater is the critical temperature,
more difficult is to liquify the gas.
Reason : Stronger are the intermolecular forces
lower would be the critical temperature of that
gas.
Q.1
Q.2
Assertion : The equation PV = nRT does not
apply to real gases at all conditions of
temperature and pressure.
Reason : For real gases the attractive forces
between the molecules cannot be neglected.
(1) A
(2) B
(3) C
(D) 4
Assertion : The value of Vander Waal constant
a is higher for NH3 than for N2.
Reason : Hydrogen bonding is present in
ammonia.
(1) A
(2) B
(3) C
(D) 4
(1) A
Q.7
factor (Z) is less than unity.
Reason : Z = PV/RT, if PV < RT then Z < 1.
(1) A
Assertion : K.E. of all the gases approach zero
as their temperature approach zero kelvin.
Reason : Molecular motion ceases at
absolute zero.
(1) A
(2) B
(3) C
(D) 4
Assertion : Helium shows only positive
deviation from ideal behaviour.
Reason : Helium is chemically inert noble gas.
(1) A
(2) B
(3) C
(D) 4
(D) 4
Assertion : For a real gas if molar volume is
(2) B
(3) C
(D) 4
Assertion : Vapour pressure of the liquid
increases with increase in temperature.
Reason : At elevated temperature kinetic
energy of the liquid molecules increases.
(1) A
Q.4
(3) C
less than 22.4 litres, at S.T.P., compressibility
Q.8
Q.3
(2) B
Q.9
(2) B
(3) C
(D) 4
Assertion : At constant temperature, the gas
density is directly proportional to pressure.
Reason : More is the pressure on the gas, the
denser it becomes.
(1) A
2nd Floor Rajcastle, Above Domino’s Pizza ellorapark
(2) B
(3) C
GASEOUS STATE
(D) 4
25
25
AMIT JHAKAL .
JEE CHEMISTRY
EXERCISE # 5
Q.4
Q.1
Sol.
What will be the minimum pressure required to
Pressure of 1 g of an ideal gas A at 27°C is
found to be 2 bar when 2 g of another ideal gas
compress 500 dm3 of air at 1 bar to 200 dm3 at
30°C ?
B is introduced in the same flask at same
temperature the pressure becomes 3 bar. Find a
V1 = 500 dm3 = 500 lit.
relationship between their molecular masses.
P1 = 1 Bar
Sol.
For gas A
V2 = 200 dm3 = 200 lit.
wt1 = 1 g
P2V2 = P1V1 (at constant T)
T1 = 27 + 273 = 300 K
or P2 =
P1V1
1 500
=
= 2.5 Bar.
V2
200
P1 = 2 Bar V1 = V
For gas B
wt2 = 2g
Q.2
A vessel of 120 mL capacity contains a certain
amount of gas at 35°C and 1.2 bar pressure. The
T1 = 27 + 273 = 300 K
gas is transferred to another vessel of volume
180 mL at 35°C. What is the molecular mass of
P2 = 3 – P1 = 3 – 2 = 1 Bar, V2 = V
P1 + P2 = 3 Bar
the oxide ?
Sol.
PV = nRT =
V1 = 120 mL
Mw
P2
wt 2
2
=
×
M w1
P1
wt 1
P1 = 1.2 Bar
V2 = 180 mL
P2V2 = P1V1
(at constant T)
1
1 M w1
=
2
2 Mw2
PV
1.2 120
or P2 = 1 1 =
= 0.8 Bar.
V2
180
Q.3
M w 2 = 4 M w1
MwB = 4 MwA
At 0°C the density of a gaseous oxide at 2 bar
is same as that of nitrogen at 5 bar. What is the
molecular mass of the oxide ?
Sol.
Q.5
T = constant,
The drain cleaner, Drainex contains small bits
of aluminum which react with caustic soda to
P1 = 2 Bar
produce hydrogen. What volume of hydrogen at
d 1 = d 2,
20°C and one bar will be released when 0.15 g
of aluminum reacts ?
P2 = 5 Bar,
M w 2 = 28
d=
PM w
RT
M w2
d2
P
= 2 ×
M w1
d1
P1
or 1 =
or
wt
RT
Mw
5
28
×
2 M w1
M w1
5  28
=
= 70 g/mol.
2
nd
26 2 Floor Rajcastle, Above Domino’s Pizza, Ellorapark
Sol.
 Na2AlO2
Al + 2NaOH 
moles n =
0.15
27
0.15
mol
27
PV = nRT
V=
nRT
0.15 1.0821 293
=
×
P
1
27
V = 0.13364 lit.
= 133.64 mL
GASEOUS STATE
26
JEE CHEMISTRY .
Q.6
Sol.
BY AMIT JHAKAL
What will be the pressure exerted by a mixture
of 3.2 g of methane and 4.4 g of carbon dioxide
contained in a 9 dm3 flask at 27°C ?
V = 9 dm3 = 9 lit.
T = 27 + 273 = 300 K,
or PV =
M w wt ×
Mw =
nt = n CH 4 + n CO 2
4 .4
3.2
+
= 0.3 mol.
44
16
PV = nRT
P × 9 = 0.3 × 0.0821 × 300
P = 0.821 atm.
Q.10
Q.7
Q.8
Sol.
Density of a gas is found to be 5.46 g/dm3 at
27°C at 2 bar pressure. What will be its density
of STP ?
T1 = 27 + 273 = 300 K
P1 = 2 Bar
d1 = 5.46 g/dm3
At STP
T2 = 0°C = 273 K, P2 = 1 atm = 1 Bar
d=
PM w
RT
1 300
d2
P
T
d
= 2 × 1  2 = ×
2 273
d1
P1
T2
5.46
d2 = 3 g/dm3
Q.9
Sol.
34.05 mL of phosphorus vapour weighs 0.0625
g at 546°C and 0.1 bar pressure. What is the
molar mass of phosphorus ?
V = 34.05 mL = 34.05 × 10–3 lit.
wt = 0.0625 g, T = 546 + 273 = 819 K
P = 0.1 Bar = 0.1 atm
PV = nRT
2nd Floor Rajcastle, Above Domino’s Pizza ellorapark
RT
V
0.0625 0.0821 819
34.05 10– 3
= 123.42 g/mol.
nt =
What will be the pressure of the gas misture
when 0.5 L of H2 at 0.8 bar and 2.0 L of oxygen
at 0.7 bar are introduced in a 1 L vessel at 27°C ?
Sol.
For H2 gas
for O2 gas
V1 = 0.5 lit
V2 = 2 lit
P1 = 0.8 Bar
P2 = 0.7 Bar
Vmix = 1 lit
T = 27 + 273 = 300 K
Pmix Vmix = P1V1 + P2V2
Pmix × 1 = 0.8 × 0.5 + 0.7 × 0.2
or Pmix = 1.8 Bar.
wt
RT
Mw
Sol.
A student forgot to add the reaction mixture to
the round bottomed flask at 27°C but put it on
the flame. After a lapse of time, he realized his
mistake, using a pyrometer he found the
temperature of the flask was 477°C. What
fraction of air would have been expelled out ?
n = 4 mol, V = 5 dm3 = 5 lit.
P = 3.32 Bar = 3.32 atm.
R = 0.083 atm lit K–1 mol–1
PV = nRT
T=
Q.11
Sol.
3.32  5
= 50 K
4  0.083
Calculate the total pressure in a mixture of 8 g
of oxygen and 4g of hydrogen confined in a
vessel of 1 dm3 at 27°C. (R = 0.083 bar dm3 K–1
mol–1).
nt = n O 2 + n H 2
4
8
+
= 2.25 mol.
2
32
V = 1 dm3 = 1 lit.
T = 27 + 273 = 300 K
PV = nRT
PtV = ntRT
Pt × 1 = 2.25 × 0.083 × 300
Pt = 56.025
nt =
Q.12
Calculate the volume occupied by 8.8 g of CO2
at 31.1°C and 1 bar pressure.
Sol.
wt = 808 g,
mw = 44
T = 31.1 + 273 = 304.1 K
P = 1 Bar = 1 atm
R = 0.083 lit atm mol–1 K–1
PV = nRT
wt
or PV =
RT
Mw
8 .8
× 0.083 × 304.1
44
or V = 5.05 lit.
or 1 × V =
GASEOUS STATE
27
27
AMIT JHAKAL .
JEE CHEMISTRY
Q.13
Sol.
2.9 g of a gas at 95°C occupied the same
Q.15
Equal volumes of two gases A and B diffuse
volume as 0.184 g of hydrogen at 17°C, at the
through a porous pot in 20 and 10 seconds
same pressure. What is the molar mass of the
respectively. If the molar mass of A be 80, find
gas ?
the molar mass of B.
wt1 = 2.9 g,
rA
=
rB
Sol.
T1 = 95 + 273 = 368 K
M w2 = 2
wt2 = 0.184 g
or PV =
wt
RT
Mw
Q.16
V 10
=

20 V

10 10  80
= 20 g mol–1
20  20
So
or
or
mol–1).
T
wt
=
× 2
Mw
T1
M w1
Calculate the total average kinetic energy of 32
g methane molecules at 27°C. (R = 8.314 J K–1
Mw  wt.T
Mw2
MB 
volume diffused 
 r 

time
80 


T2 = 17 + 273 = 290 K
PV = nRT
MB
MA
Sol.
2
0.184
290
=
×
M w1
2.9
368
n=
wt
32
=
= 2 mol
M2
16
T = 27 + 273 =200
For total KE
M w1 = 40 g/mol.
KET = 3/2 nRT or KET = 3/2 × 2 × 8.314 × 300
Q.14
Through the two ends of a glass tube of length
200 cm hydrogen chloride gas and ammonia are
allowed to enter. At what distance ammonium
chloride will first appear ?
KET = 7482.6 J mol–1
For average KE
Method (I) :
Ring of NH4Cl
Sol.
KEav =
HCl
NH3
(200 –x)cm
x cm
or KEav =
200 cm
r
1
M w1
Mw2
=
2 / t2 
 t2 = t1
 1 / t1 
M w1

2
=
or HCl =
Mw2
 NH3
1
x
=
200 – x
3
8.314
×
× 300
2
6.0231023
KEav = 6.21 × 10–21 J molecule–1
Mw
r2
=
r1
3
3 R
KT =
T
2
2 NA
M w NH
Method (II) :
KEav =
3
RT
2
or
KEav =
3
× 8.314 × 300
2
3
M w HCl
KEav = 3741.3 J mol–1
17
36.5
x = 81.12 cm.
length from HCl end (x) = 81.12 cm.
nd
28 2 Floor Rajcastle, Above Domino’s Pizza, Ellorapark
GASEOUS STATE
28
JEE CHEMISTRY .
Q.17
Sol.
A mixture of hydrogen and oxygen at one bar
Critical temperature for carbon dioxide and
methane are 31.1°C and –81.9°C respectively.
Calculate the partial pressure of hydrogen.
Which of these has stronger intermolecular
WtH2 = 20 gm, n H 2 =
20
= 10 mol
2
wt O 2 = 80 gm, n O 2 =
80
mol
32
PH 2 =
Q.18
Q.19
pressure contains 20% by weight of hydrogen.
Pt = 1 Bar = 1 atm.
or
BY AMIT JHAKAL
PH 2 =
n H2
n H2  n O2
forces and why ?
Sol.
Higher critical temperature, more easily the gas
can be liquified i.e. greater are the inter
molecular forces attraction. Hence CO2 has
stronger intermolecular forces than CH4.
Pt
10
× 1 = 0.80 atm.
80
10 
32
In terms of Charles's law explain why –273°C is
the lowest possible temperature.
Sol.
At –273°C, volume of the gas becomes equal to
zero
2nd Floor Rajcastle, Above Domino’s Pizza ellorapark
GASEOUS STATE
29
29
AMIT JHAKAL .
JEE CHEMISTRY
ANSWER KEY
EXERCISE # 1
Q.No.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Ans.
4
3
2
2
2
2
4
2
1
3
2
3
2
2
3
2
3
1
2
1
Q.No.
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
Ans.
3
1
3
1
1
4
1
2
4
2
4
4
4
4
1
1
2
4
3
2
Q.No.
41
42
43
44
45
46
47
48
Ans.
2
3
1
3
3
4
1
2
EXERCISE # 2
Q.No.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Ans.
2
1
1
2
2
3
1
4
4
3
3
4
1
1
2
2
2
2
1
4
Q.No.
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
Ans.
3
4
1
4
3
1
2
3
3
1
1
1
1
3
3
3
3
4
2
1
Q.No.
41
42
43
Ans.
3
1
4
EXERCISE # 3(A)
Q.No.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Ans.
2
2
1
1
4
1
4
2
4
1
3
2
2
1
4
2
2
1
2
2
Q.No.
21
22
23
24
25
26
27
28
29
Ans.
1
2
1
3
1
1
3
2
2
EXERCISE # 3(B)
Q.No.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Ans.
3
3
4
1
2
2
3
2
4
3
1
1
3
4
3
2
1
3
2
1
Q.No.
21
22
23
24
25
26
27
28
29
30
Ans.
3
4
4
3
2
2
1
4
1
3
EXERCISE # 4
Q.No.
1
2
3
4
5
6
7
8
9
Ans.
1
1
1
2
3
4
1
1
2
nd
30 2 Floor Rajcastle, Above Domino’s Pizza, Ellorapark
GASEOUS STATE
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