Chapter 5 States of Matter the Gaseous State by TEACHING CARE Online coaching and tuition classes
Chapter 5 States of Matter the Gaseous State by TEACHING CARE Online coaching and tuition classes
The state of matter in which the molecular forces of attraction between the particles of matter are minimum, is known as gaseous state. It is the simplest state and shows great uniformity in behaviour.
 Gases or their mixtures are homogeneous in
 Gases have very low density due to negligible intermolecular
 Gases have infinite expansibility and high compressibility.
 Gases exert
 Gases possess high
 Gases do not have definite shape and volume like
 Gaseous molecules move very rapidly in all directions in a random manner e., gases have highest kinetic energy.
 Gaseous molecules are loosely packed having large empty spaces between
 Gaseous molecules collide with one another and also with the walls of container with perfectly elastic collisions.
 Gases can be liquified, if subjected to low temperatures (below critical) or high
 Thermal energy of gases >> molecular
 Gases undergo similar change with the change of temperature and In other words, gases obey certain laws known as gas laws.
 The characteristics of gases are described fully in terms of four parameters or measurable properties :
 The volume, V, of the
 Its pressure, P
 Its temperature, T
 The amount of the gas (e., mass or number of moles).
 Volume : (i) Since gases occupy the entire space available to them, the measurement of volume of a gas only requires a measurement of the container confining the
(ii) Volume is expressed in litres (L), millilitres (mL) or cubic centimetres (cm^{3} ) or cubic metres(m^{3} ) .
(iii) 1L = 1000 mL ; 1 mL = 10 ^{3} L
1 L = 1dm^{3} = 10^{3} cm^{3}
1m^{3} = 10^{3} dm^{3} = 10^{6} cm^{3} = 10^{6} mL = 10^{3} L
 Mass : (i) The mass of a gas can be determined by weighing the container in which the gas is enclosed and again weighing the container after removing the gas. The difference between the two weights gives the mass of the
 The mass of the gas is related to the number of moles of the gas e.
moles of gas (n) = Mass in grams = m
Molar mass M
 Mass is expressed in grams or kilograms, 1 Kg = 10^{3} g
 Temperature : (i) Gases expand on increasing the If temperature is increased twice, the square of the velocity (v ^{2} ) also increases two times.
 Temperature is measured in centigrade degree ( ^{o} C)
Temperature is also measured in Fahrenheit (F^{o}).
or celsius degree with the help of thermometers.
 I. unit of temperature is kelvin (K) or absolute degree.
K = ^{o} C + 273
 Relation between F and
^{o} is
o C = F o – 32
C 5 9
 Pressure : (i) Pressure of the gas is the force exerted by the gas per unit area of the walls of the container
in all directions. Thus, Pressure (P) = Force(F) = Mass(m) ´ Acceleration(a)
Area(A) Area(a)
 Pressure exerted by a gas is due to kinetic energy
(KE = 1 mv^{2} )
2
of the molecules. Kinetic energy of
the gas molecules increases, as the temperature is increased. Thus, Pressure of a gas µ Temperature (T).
 Pressure of a pure gas is measured by manometer while that of a mixture of gases by
 Commonly two types of manometers are used,
 Open end manometer; (b) Closed end manometer
 The I. unit of pressure, the pascal (Pa), is defined as 1 newton per metre square. It is very small unit. 1Pa = 1Nm2 = 1kg m1 s 2
 G.S. unit of pressure is dynes cm^{2} .
 K.S. unit of pressure is absolute).
kgf / m2 . The unit
kgf / cm2
sometime called ata (atmosphere technical
 Higher unit of pressure is bar, KPa or MPa.
1bar = 10^{5} Pa = 10^{5} Nm^{2} = 100KNm^{2} = 100KPa
 Several other units used for pressure are,
Name  Symbol  Value 
bar  bar  1bar = 10^{5} Pa 
atmosphere  atm  1atm = 1.01325 ´ 10^{5} Pa 
Torr  Torr  1Torr = 101325 Pa = 133.322 Pa 
760  
millimetre of mercury  mm Hg  1mm Hg = 133.322 Pa 
 The pressure relative to the atmosphere is called gauge pressure. The pressure relative to the perfect vacuum is called absolute pressure.
Absolute pressure = Gauge pressure + Atmosphere pressure.
 When the pressure in a system is less than atmospheric pressure, the gauge pressure becomes negative, but is frequently designated and called For example, 16 cm vacuum will be
76 – 16 ´ 1.013 = 0.80 bar .
76
 If ‘h’ is the height of the fluid in a column or the difference in the heights of the fluid columns in the two limbs of the manometer d the density of the fluid (Hg = 6 ´ 10^{3} Kg / m^{3} = 13.6 g / cm^{3} ) and g is the gravity,
then pressure is given by,
Pgas = Patm + hdg
 Two sets of conditions are widely used as ‘standard’ values for reporting
Condition  T  P  V_{m} (Molar volume) 
S.T.P./N.T.P.  273.15 K  1 atm  22.414 L 
S.A.T.P*.  298.15 K  1 bar  24.800 L 
* Standard Ambient temperature and pressure.
 In 1662, Robert Boyle discovered the first of several relationships among gas variables (P, T, V).
 It states that, “For a fixed amount of a gas at constant temperature, the gas volume is inversely proportional to the gas ”
Thus, at constant temperature and mass
or P = K
V
or PV = K
(where K is constant)
For two or more gases at constant temperature and mass.
P1 V1 = P2 V2 =……… = K

Boyle’s law can also be given as, æ dP ö
dV
= – K
v 2
è øT
 Graphical representation of Boyle’s law : Graph between P and V at constant temperature is called
isotherm and is an equilateral (or rectangular) hyperbola. By plotting P versus 1 , this hyperbola can be converted
V
to a straight line. Other types of isotherms are also shown below,
Note : ® The isotherms of CO_{2} were first studied by Andrews.
 At constant mass and temperature density of a gas is directly proportional to its pressure and inversely proportional to its
Thus,
é V = mass ù



d úû
or
 At altitudes, as P is low d of air is That is why mountaineers carry oxygen cylinders.
 Air at the sea level is dense because it is compressed by the mass of air above However the density and pressure decreases with increase in altitude. The atmospheric pressure at Mount Everest is only 0.5 atm.
Example : 1 A sample of a given mass of a gas at a constant temperature occupies 95cm^{2} under a pressure of
9.962 ´ 10^{4} Nm^{–}^{2} . At the same temperature, it volume at a pressure of 10.13 ´ 10^{4} Nm^{2} is [Bihar CEE 1992]
(a)
190 cm^{3}
(b)
93 cm ^{3}
(c)
46.5 cm ^{3}
(d)
47.5 cm ^{3}
Solution: (b)
P1 V1 = P2 V2
9.962 ´ 10^{4} ´ 95 = 10.13 ´ 10^{4} ´ V2
V2 = 93 cm^{3}
Example : 2 A gas occupied a volume of 250 ml at 700 mm Hg pressure and 25 ^{o} C . What additional pressure is required to reduce the gas volume to its 4/5^{th} value at the same temperature
(a) 225 mm Hg (b) 175 mm Hg (c) 150 mm Hg (d) 265 mm Hg
Solution: (b)
P1 V1 = P2 V2
700 ´ 250 = P ´ æ 4 ´ 250ö ; P
= 875 mm Hg

_{2} ç ÷ _{2}
è ø
Additional pressure required = 875 – 700 = 175 mm Hg
Example : 3 At constant temperature, if pressure increases by 1%, the percentage decrease of volume is (a) 1% (b) 100/101% (c) 1/101% (d) 1/100%
Solution: (b)
P1 V1 = P2 V2
If P_{1} = 100 mm ,
P_{2} will be 101 mm
Hence 100 ´ V = 101 ´ V_{2} ,
V2 = 100 ´ V ,
101
Decrease in volume = V – 100V
101
= 1
101
of V i.e. 100 %
101
 French chemist, Jacques Charles first studied variation of volume with temperature, in
 It states that, “The volume of a given mass of a gas is directly proportional to the absolute temperature
(= ^{o} C + 273) at constant pressure”.
Thus, V µ T at constant pressure and mass
or V = KT = K(t( o C) + 273.15) , (where k is constant),
V = K T
For two or more gases at constant pressure and mass
V1 = V2 T1 T2
=…… K ,

Charle’s law can also be given as, æ dV ö
dT
= K .
 If t = 0 ^{o} C , then V = V_{0}
è ø P
hence,
V0 = K ´ 273.15
\ K =
V =
V0
273.15
V0
[t + 273.15] = V
é1 + t ù = V [1 + a t]
273.15
^{0} êë 273.15 úû 0 v
where a _{v} is the volume coefficient,
Thus, for every 1o
change in temperature, the volume of a gas changes by
1 æ»
1 ö of the


volume at 0^{o} C .
273.15 ç
273 ÷
 Graphical representation of Charle’s law : Graph between V and T at constant pressure is called
isobar or isoplestics and is always a straight line. A plot of V versus
t( ^{o}C)
at constant pressure is a straight line
cutting the temperature axis at – 273.15^{o} C . It is the lowest possible temperature.
 To lower the temperature of a substance, we reduce the thermal energy. Absolute zero (0K) is the temperature reached when all possible thermal energy has been removed from a substance. Obviously, a substance cannot be cooled any further after all thermal energy has been
 At constant mass and pressure density of a gas is inversely proportional to it absolute
Thus,
é V = mass ù or



d úû
 Use of hot air balloons in sports and meteorological observations is an application of Charle’s
Example : 4 When the temperature of 23 ml of dry CO2
gas is changed from 10 ^{o}
to 30 ^{o} C
at constant pressure of 760
mm, the volume of gas becomes closest to which one of the following [CPMT 1992]
(a) 7.7 ml (b) 25.5 ml (c) 24.6 ml (d) 69 ml
Solution: (c)
V1 = V2
i.e. 23 = V2 ;
V = 24.6 ml
T1 T2 283 K 303 K ^{2}
Example : 5 The volume of a gas is 100 ml at 100 ^{o} C . If pressure remains constant then at what temperature it will be about 200 ml [Roorkee 1993]
(a)
200 ^{o} C
(b)
473 ^{o} C
(c)
746 ^{o} C
(d)
50 ^{o} C
Solution: (b)
V1 = V2 T1 T2
i.e.
100
373 K
= 200
T2
T_{2} = 746 k = 473^{o} C
Example : 6 If 300 ml of a gas at 27 ^{o} C
is cooled to 7 ^{o} C
at constant pressure, its final volume will be [AIIMS 2000]
(a) 135 ml (b) 540 ml (c) 350 ml (d) 280 ml
Solution: (d)
V1 = V2
i.e.
300 = V_{2}
T1 T2
300 K 280K
V_{2} = 280 ml
Example : 7 A flask containing air (open to atmosphere) is heated from 300 K to 500 K. the percentage of air escaped to the atmosphere is nearly [CBSE PMT 1991]
(a) 16.6 (b) 40 (c) 66 (d) 20
Solution: (c)
V1 = V2 T1 T2
i.e.
V1
300
= V2
500
; V2
= 5 V = 1.66 V
3
Volume escaped = 1.66 V – V = 0.66 V = 66% of V
Example : 8 According to Charle’s law, at constant pressure, 100 ml of a given mass of a gas with
10 ^{o} C
rise in


temperature will become æ 1 = 0.00366ö
ç 273 ÷
(a) 100.0366 (b) 99.9634 (c) 103.66 (d) 100.366
Solution: (a)
Vt = V0 +
V0
273
´ t = 100 +
1
273
´ 10 = 100 + 0.0366 = 100.0366 ml
 In 1802, French chemist Joseph GayLussac studied the variation of pressure with temperature and extende the Charle’s law so, this law is also called Charle’sGay Lussac’s
 It states that, “The pressure of a given mass of a gas is directly proportional to the absolute temperature
(= ^{o} C + 273) at constant volume.”
Thus,
P µ T
at constant volume and mass
or P = KT = K(t(o C) + 273.15)
P = K T
(where K is constant)
For two or more gases at constant volume and mass
P1 = P2 T1 T2
=…….. = K
 If t = 0^{o} C , then P = P_{0}
Hence,
P0 = K ´ 273.15
\ K =
P0
273.15
P = P0
[t + 273.15] = P é1 + t ù = P [1 + at]
273.15
0 êë
273.15 úû 0
where a _{P} is the pressure coefficient,
Thus, for every 1^{o} change in temperature, the pressure of a gas changes by
1 æ» 1 ö
of the


pressure at 0^{o} C .
(4) Graphical representation of GayLussac’s law : A graph between P
and T at constant V is called isochore.
273.15 ç 273 ÷
Note : ® This law fails at low temperatures, because the volume of the gas molecules become significant.
Example : 9 A sealed tube which can withstand a pressure of 3 atmosphere is filled with air at pressure. The temperature above which the tube will burst will be
27 ^{o} C
and 760 mm
(a)
900 ^{o} C
(b)
627^{o} C
(c)
627 ^{o} C
(d)
1173^{o} C
Solution: (b) The tube will burst when the final pressure exceeds 3 atm. at constant volume,
P1 = P2
i.e.
760
= 3 ´ 760
T1 T2 300 K T2
T_{2} = 900 K = 627 ^{o} C
 According to this law, “Equal volumes of any two gases at the same temperature and pressure contain the same number of molecules.”
Thus, V µ n
or V = Kn
(at constant T and P) (where K is constant)
or V1 n1
= V2 n2
=…….. = K
Example,
2H2 (g)+ O2 (g) ¾¾® 2H2O(g)
2moles  1mole  2moles 
2volumes  1volume  2volumes 
2litres  1litre  2litres 
1litre  1 / 2litre  1litre 
1nlitre  1 / 2nlitre  1nlitre 
 One mole of any gas contains the same number of molecules (Avogadro’s number = 02 ´ 10^{23} ) and by
this law must occupy the same volume at a given temperature and pressure. The volume of one mole of a gas is called molar volume, V_{m} which is 22.4 L mol ^{1} at S.T.P. or N.T.P.
 This law can also express as, “The molar gas volume at a given temperature and pressure is a specific
constant independent of the nature of the gas”.
Thus, Vm = specific constant = 22.4 Lmol ^{1} at S.T.P. or N.T.P.
 This law is widely applicable to solve the problems of reactive gaseous
Note : ® Loschmidt number : It is the number of molecules present in 1 c.c. of a gas or vapour at S.T.P.
Its value is 2.687 ´ 1019 per c.c.
 The simple gas laws relating gas volume to pressure, temperature and amount of gas, respectively, are stated below :
Boyle’s law :
Charle’s law :
Avogadro’s law :
P µ 1
V
V µ T
V µ n
or V µ 1
P
(n and T constant)
(n and P constant) (T and P constant)
If all the above law’s combines, then
V µ nT
P
or V = nRT P

or
This is called ideal gas equation. R is called ideal gas constant. This equation is obeyed by isothermal and adiabatic processes.
 Nature and values of R : From the ideal gas equation,
Force ´ Volume
R = PV
nT
= Pressure ´ Volume mole ´ Temperature
= Area = Force ´ Length = Work or energy .
mole ´ Temperature mole ´ Temperature mole ´ Temperature
So, R is expressed in the unit of work or energy mol ^{1} K ^{1} . Different values of R are summarised below :
R = 0.0821 Latm mol ^{1} K ^{1}
= 8.3143 ´ 10^{7} erg mol ^{1} K ^{1}
= 8.3143 joulemol ^{1} K ^{1}
= 8.3143 Nmmol ^{1} K ^{1}
(S.I. unit)
= 8.3143 KPa dm^{3} mol ^{1} K ^{1}
= 8.3143 MPa cm^{3} mol ^{1} K ^{1}
= 8.3143 ´ 10 ^{3} kJ mol ^{1} K ^{1}
= 5.189 ´ 1019 eV mol 1 K 1
= 1.99 cal mol 1 K 1
= 1.987 ´ 10 3 K cal mol 1 K 1
Note : ® Although R can be expressed in different units, but for pressurevolume calculations, R must be taken in the same units of pressure and volume.
 Gas constant, R for a single molecule is called Boltzmann constant (k)
k = R
N
= 8.314 ´ 107
6.023 ´ 10 23
ergs mole 1 degree 1
= 1.38 ´ 10 ^{16} ergs mol ^{1} degree ^{1} or 1.38 ´ 10 ^{23} joulemol ^{1} degree ^{1}
 Calculation of mass, molecular weight and density of the gas by gas equation
m æ mass of the gas (m) ö
PV = nRT = M RT
ç n = Molecular weight of the gas (M) ÷
è ø
\
or dT = M
æ m ö



ç
è ø
P R
Since M and R are constant for a particular gas,
Thus,
dT = constant
P
Thus, at two different temperature and pressure
d1T1 = d2 T2  
P1  P2 
 Gas densities differ from those of solids and liquids as,
 Gas densities are generally stated in g/L instead of
g / cm^{3} .
pressure.
 Gas densities are strongly dependent on pressure and temperature as,
d µ P
d µ 1
T
Densities of liquids and solids, do depend somewhat on temperature, but they are far less dependent on
 The density of a gas is directly proportional to its molar No simple relationship exists between the density and molar mass for liquid and solids.
 Density of a gas at STP = molar mass
22.4
d(N 2
) at STP =
28
22.4
= 1.25 g L^{1}
d(O2
) at STP = 32 = 1.43 g L^{1}
22.4
Example : 10 The pressure of 2 moles of an ideal gas at 546 K having volume 44.8 L is [CPMT 1995]
(a) 2 atm (b) 3 atm (c) 4 atm (d) 1 atm
Solution: (a)
PV = nRT, P ´ 44.8 = 2 ´ 0.082 ´ 546
P = 2 atm
Example : 11 The number of moles of H _{2} in 0.224 litre of hydrogen gas at STP (273 K, 1 atm.) is [MLNR 1994]
(a) 1 (b) 0.1 (c) 0.01 (d) 0.001
Solution: (c) PV = nRT , 1 ´ 0.224 = n ´ 0.082 ´ 273 n = 0.01mol
Example : 12 120 g of an ideal gas of molecular weight 40
mole ^{–}^{1}
are confined to a volume of 20 L at 400 K. Using
R = 0.0821 Latm K ^{–}^{1} mole ^{–}^{1} , the pressure of the gas is [Pb. CET 1996]
(a) 4.90 atm (b) 4.92 atm (c) 5.02 atm (d) 4.96 atm
Solution: (b)
120 g = 120 = 3 moles
40
P = nRT = 3 ´ 0.0821 ´ 400 = 4.92 atm.
V 20
Example : 13 The volume of 2.8 g of carbon monoxide at 27 ^{o} C and 0.821 atm pressure is (R = 0.0821lit atm K ^{–}^{1} mol ^{–}^{1} )
[Manipal PMT 2001]
 3 litre (b) 1.5 litre (c) 3 litre (d) 30 litre
Solution: (c) 2.8 g CO = 2.8 = mol = 0.1mol
28
PV = nRT
or V = nRT = 0.1 ´ 0.0821 ´ 300 = 3 litre
P 0.821
Example: 14 3.2 g of oxygen (At. wt. = 16) and 0.2 g of hydrogen (At. wt. = 1) are placed in a 1.12 litre flask at 0 ^{o} C . The total pressure of the gas mixture will be [CBSE PMT 1992]
(a) 1 atm (b) 4 atm (c) 3 atm (d) 2 atm
Solution: (b) 3.2 g O_{2} = 0.1mol , 0.2 g H _{2} = 0.1mol,
Total n = 0.2 mol ,
P = nRT = 0.2 ´ 0.082 ´ 273 = 4atm
V 1.12
Example : 15 The density of methane at 2.0 atmosphere pressure and 27 ^{o} C
is [BHU 1994]
(a) 0.13 g L^{–}^{1}
(b) 0.26 g L^{–}^{1}
(c) 1.30 g L^{–}^{1}
(d) 2.60 g L^{–}^{1}
Solution: (c)
d = PM
RT
= 2 ´ 16
0.082 ´ 300
= 1.30 g L^{–}^{1}
Example : 16 The volume of 0.0168 mol of O_{2}
obtained by decomposition of
KClO3
and collected by displacement of
water is 428 ml at a pressure of 754 mm Hg at 25 ^{o} C . The pressure of water vapour at 25 ^{o} C
is[UPSEAT 1996]
(a) 18 mm Hg (b) 20 mm Hg (c) 22 mm Hg (d) 24 mm Hg
Solution: (d) Volume of 0.0168 mol of O_{2} at STP = 0.0168 ´ 22400 cc = 376.3 cc
V1 = 376.3 cc , P1 = 760 mm , T1 = 273 K
V2 = 428 cc , P2 = ? , T2 = 298 K
P1 V1 T1
= P2 V2 T2
gives P_{2}
= 730 mm (approx.)
\ Pressure of water vapour = 754 – 730 = 24 mm Hg
 According to this law, “When two or more gases, which do not react chemically are kept in a closed vessel, the total pressure exerted by the mixture is equal to the sum of the partial pressures of individual gases.”
Thus,
Ptotal
= P_{1} + P_{2} + P_{3} + ………
Where
P1 , P2 , P3 ,…… are partial pressures of gas number 1, 2, 3 ………
 Partial pressure is the pressure exerted by a gas when it is present alone in the same container and at the same
Partial pressure of a gas (P1 ) =
Number of moles of the gas (n1 ) ´ PTotal
Total number of moles (n) in the mixture
= Mole fraction (X_{1}
) ´ PTotal
 If a number of gases having volume container of volume V, then,
V1 , V2 , V3 ……
at pressure
P_{1}, P_{2} , P_{3} ……..
are mixed together in
PTotal
= P1 V1 + P2 V2 + P3 V3 …..
V
or = (n1
 n2
 n3
…..) RT
V
( PV = nRT)
or = n RT
V
( n = n1
 n2
 n3
…..)
 Applications : This law is used in the calculation of following relationships,
 Mole fraction of a gas (X_{1}
) in a mixture of gas = Partial pressure of a gas (P1 )
PTotal
 % of a gas in mixture = Partial pressure of a gas (P1 ) ´ 100
PTotal
 Pressure of dry gas collected over water : When a gas is collected over water, it becomes moist due to water vapour which exerts its own partial pressure at the same temperature of the gas. This partial perssure of water vapours is called aqueous Thus,
Pdry gas = Pmoist gas or PTotal – Pwater vapour
or Pdry gas = Pmoist gas – Aqueous tension (Aqueous tension is directly proportional to absolute temperature)
 Relative humidity (RH) at a given temperature is given by :
RH = Partial pressure of water in air .
Vapour pressure of water
 Limitations : This law is applicable only when the component gases in the mixture do not react with
each other. For example,
N _{2} and O_{2} , CO and
CO2 ,
N _{2} and Cl_{2} , CO and
N _{2} etc. But this law is not applicable
to gases which combine chemically. For example, H _{2}
etc.
and
Cl_{2} , CO and
Cl2 ,
NH 3 , HBr and HCl, NO and O2
Note : ®
N _{2} (80%) has the highest partial pressure in atmosphere.
 Another law, which is really equivalent to the law of partial pressures and related to the partial volumes of gases is known as Law of partial volumes given by Amagat. According to this law, “When two or more gases, which do not react chemically are kept in a closed vessel, the total volume exerted by the mixture is equal to the sum of the partial volumes of individual gases.”
Thus, VTotal = V1 + V2 + V3 + ……
Where V1 , V2 , V3 ,…… are partial volumes of gas number 1, 2, 3…..
Example: 17 What will be the partial pressure of H _{2} in a flask containing 2 g of H _{2} , 14 g of
N _{2} and 16 g of O_{2}
[Assam JET 1992]
(a) 1/2 the total pressure (b) 1/3 the total pressure(c) 1/4 the total pressure (d) 1/16 the total pressure
Solution: (a)
n(H 2 ) = 2 = 1, n(N 2 ) = 14 = 0.5 , n(O2 ) = 16 = 0.5, p(H 2 ) = 1 p = 1 p
2 28 32 1 + 0.5 + 0.5 2
Example : 18 Equal weights of methane and oxygen are mixed in an empty container at 25 ^{o} C . The fraction of the total pressure exerted by oxygen is [IIT 1981]
(a) 1/3 (b) 1/2 (c) 2/3 (d) 1 / 3 ´ 273 / 298
Solution: (a)
n(CH 4 ) = w
16
= 1, n(O2 ) = w
32
p(O2 ) = w / 32 = 1 = 1
w / 16 + w / 32 2 + 1 3
Example : 19 In a flask of volume V litres, 0.2 mol of oxygen, 0.4 mol of nitrogen, 0.1 mol of ammonia and 0.3 mol of
helium are enclosed at
27 ^{o} C . If the total pressure exerted by these nonreating gases is one atmosphere, the
partial pressure exerted by nitrogen is [Kerala MEE 2001]
(a) 1 atm (b) 0.1 atm (c) 0.2 atm (d) 0.4 atm
Solution: (d)
P = Mol. fraction of N ´ Total perssure = 0.4 ´ 1 atm = 0.4 atm .


2 0.2 + 0.4 + 0.1 + 0.3
Example : 20 Equal weights of ethane and hydrogen are mixed in an empty container at
25 ^{o} C . The fraction of the total
Solution: (d)
pressure exerted by hydrogen is [IIT 1993]
(a) 1 : 2 (b) 1 : 1 (c) 1 : 16 (d) 15 : 16
n(C2 H6 ) = w , n(H 2 ) = w
30 2
p(H
) = w / 2 = 1 = 15
^{2} w / 2 + w / 30
1 + 1 16
15
Example : 21 A gaseous mixture contains 56 g of
N _{2} , 44 g of
CO2
and 16 g of
CH _{4} . The total pressure of the mixture is
720 mm Hg. The partial pressure of CH _{4}
is [IIT 1993]
Solution: (a)
(a) 180 mm (b) 360 mm (c) 540 mm (d) 720 mm
p(CH 4 ) = 16 / 16 ´ 720 = 1 ´ 720 = 1 ´ 720 = 180 mm .
5 / 28 + 44 / 44 + 16 / 16 2 + 1 + 1 4
 Diffusion is the process of spontaneous spreading and intermixing of gases to form homogenous mixture irrespective of force of While Effusion is the escape of gas molecules through a tiny hole such as pinhole in a balloon.
 All gases spontaneously diffuse into one another when they are brought into
 Diffusion into a vacuum will take place much more rapidly than diffusion into another
 Both the rate of diffusion of a gas and its rate of effusion depend on its molar mass. Lighter gases diffuses faster than heavier The gas with highest rate of diffusion is hydrogen.
 According to this law, “At constant pressure and temperature, the rate of diffusion or effusion of a gas is inversely proportional to the square root of its vapour density.”
Thus, rate of diffusion (r) µ 1
(T and P constant)
For two or more gases at constant pressure and temperature,
r1 =
r2
Note : ®
Always remember that vapour density is different from absolute density. The farmer is
independent of temperature and has no unit while the latter depends upon temperature and expressed in
 Graham’s law can be modified in a number of ways as,
 Since, 2 ´ vapour density (V.D.) = Molecular weight
gm^{1} .
then, r1 = = =
r2
where,
M1 and M 2
are the molecular weights of the two gases.
 Since, rate of diffusion (r) = Volume of a gas diffused
Time taken for diffusion
then,
r1 =
r2
V1 / t1 =
V2 / t 2
 When equal volume of the two gases diffuse, e. V_{1} = V_{2}
then,
r1 = t 2 =
r2 t1
 When volumes of the two gases diffuse in the same time , e. t_{1} = t _{2}
then,
r1 = V1 =
r2 V2
 Since, r µ p (when p is not constant)
then,
r1 = P1 =
r2 P2
æ ö

ç r µ ÷
è ø
Note : ® It should be noted that this law is true only for gases diffusing under low pressure gradient.
 CO2 > SO2 > SO3 > PCl3 is order of rates of
 Rate of diffusion and effusion can be determined as,
 Rate of diffusion is equal to distance travelled by gas per unit time through a tube of uniform cross
section.
 Number of moles effusing per unit time is also called rate of diffusion.
 Decrease in pressure of a cylinder per unit time is called rate of effusion of
 The volume of gas effused through a given surface per unit time is also called rate of
 Applications : Graham’s law has been used as follows :
 To determine vapour densities and molecular weights of
 To prepare Ausell’s marsh gas indicator, used in
 Atmolysis : The process of separation of two gases on the basis of their different rates of diffusion due to difference in their densities is called It has been applied with success for the separation of
isotopes and other gaseous mixtures. Example, this process was used for the largescale separation of
gaseous
235 UF6
and
238 UF6
during the second world war.
Example : 22 The time taken for a certain volume of a gas ‘X’ to diffuse through a small hole is 2 minutes. It takes 5.65 minutes for oxygen to diffuse under the similar conditions. The molecular weight of ‘X’ is [NCERT 1990] (a) 8 (b) 4 (c) 16 (d) 32
Solution: (b)
rX =
rO2
v / 2 =
32 , 5.65 =
32 , M X = 4
v / 5.65
M X 2 M X
Example : 23 The rate of diffusion of methane at a given temperature is twice that of gas X. The molecular weight of X is
[IIT 1990]
(a) 64.0 (b) 32.0 (c) 4.0 (d) 8.0
Solution: (a)
rCH4 =
rX
MX Þ 2 =
MCH4
MX Þ MX
16
= 64 .
Example : 24 Density ratio of O_{2} and H _{2} is 16 : 1. The ratio of their r.m.s. velocities will be [AIIMS 2000]
(a) 4 : 1 (b) 1 : 16 (c) 1 : 4 (d) 16 : 1
Solution: (c)
r1 = v1 =
r2 v2
= = 1 : 4 .
Example : 25 The rate of diffusion of a gas having molecular weight just double of nitrogen gas is 56 ml s ^{–}^{1} . The rate of diffusion of nitrogen will be [CPMT 2000]
(a) 79.19 ml s ^{–}^{1}
(b) 112.0 ml s ^{–}^{1}
 56 ml s ^{–}^{1}
(d) 90.0 ml s ^{–}^{1}
Solution: (a)
rX =
rN2
= = ;
56
rN2
= 1
or rN 2
= 56
= 79.19 ms ^{–}^{1}
Example : 26 50 ml of gas A diffuse through a membrane in the same time as 40 ml of a gas B under identical pressure temperature conditions. If the molecular weight of A is 64, that of B would be [CBSE PMT 1992]
(a) 100 (b) 250 (c) 200 (d) 80
Solution: (a)
r1 =
r2
Þ 50 / t =
40 / t
50 =
or 5 =
or 25 = M _{2}
or M
= 100
40 4
16 64 ^{2}
 For gaseous systems, gravitational force is negligible but this is not true for the gases of high molecular masses such as polymer. In this case, the pressure will be different in vertical positions in a container. The variation of pressure with altitude is given by the socalled Barometric formula.
where, P^{o} and P are the pressure of the gas at the ground level and at a height ‘h‘ from the ground respectively.
M is molecular mass of the gas, g is acceleration due to gravity, R is gas constant and T is temperature in kelvin.
 Since number of moles of gas ‘n‘ and density of the gas ‘d‘ are directly proportional to pressure hence the above equation may be expressed as, d = d^{o}e ^{–}^{Mgh} ^{/} ^{RT} and n = n^{o}e ^{–}^{Mgh} ^{/} ^{RT} .
 The above equations may be expressed as,
 Kinetic theory was developed by Bernoulli, Joule, Clausius, Maxwell and Boltzmann and represents dynamic particle or microscopic model for different gases since it throws light on the behaviour of the particles (atoms and molecules) which constitute the gases and cannot be seen. Properties of gases which we studied earlier are part of macroscopic model.
 Postulates
 Every gas consists of a large number of small particles called molecules moving with very high velocities in all possible
 The volume of the individual molecule is negligible as compared to the total volume of the
 Gaseous molecules are perfectly elastic so that there is no net loss of kinetic energy due to their
 The effect of gravity on the motion of the molecules is
 Gaseous molecules are considered as point masses because they do not posses potential So the attractive and repulsive forces between the gas molecules are negligible.
 The pressure of a gas is due to the continuous bombardment on the walls of the containing
 At constant temperature the average E. of all gases is same.
 The average E. of the gas molecules is directly proportional to the absolute temperature.
 Kinetic gas equation : On the basis of above postulates, the following gas equation was derived,
where, P = pressure exerted by the gas, V = volume of the gas, m = average mass of each molecule,
n = number of molecules, u = root mean square (RMS) velocity of the gas.
 Calculation of kinetic energy
We know that,
K.E. of one molecule = 1 mu^{2}
2
K.E. of n molecules = 1 mnu2 = 3 PV ( PV = 1 mnu2 )
2 2 3
n = 1, Then K.E. of 1 mole gas = 3 RT
2
( PV = RT)
= 3 ´ 8.314 ´ T = 12.47 T Joules .
2
= Average K.E. per mole = 3 RT
= 3 KT
æ K =
R = Boltzmann constantö
N(Avogadro number) 2 N 2 ç ÷



This equation shows that K.E. of translation of a gas depends only on the absolute temperature. This is
known as Maxwell generalisation. Thus average K.E. µ T.
If T = 0K (i.e., – 273.15^{o} C) then, average K.E. = 0. Thus, absolute zero (0K) is the temperature at which
molecular motion ceases.
 Kinetic gas equation can be used to establish gas
Example : 27 The kinetic energy for 14 grams of nitrogen gas at 127 ^{o} C
constant = 8.31 JK ^{–}^{1} mol ^{–}^{1} )
is nearly (mol. mass of nitrogen = 28 and gas
[CBSE PMT 1999]
(a) 1.0 J (b) 4.15 J (c) 2493 J (d) 3.3 J
Solution: (c) K.E. = 3 RT mol ^{–}^{1}
2
or K.E. = 3 nRT = 3 ´ 14 ´ 8.31 ´ 400 J = 2493 J
2 2 28
 The closest distance between the centres of two molecules taking part in a collision is called molecular or collision diameter (s). The molecular diameter of all the gases is nearly same lying in the
order of 10 ^{8} cm .
 The number of collisions taking place in unit time per unit volume, called collision frequency (z).
 The number of collision made by a single molecule with other molecules per unit time are given by,
ZA = 2ps 2uav.n
where n is the number of molecules per unit molar volume,
n = Avogadro number(N_{0} ) = 6.02 ´ 10^{23} m3
Vm 0.0224
 The total number of bimolecular collision per unit time are given by,
ZAA
= 1 ps ^{2}u
av.n2
 If the collisions involve two unlike molecules, the number of bimolecular collision are given by,
é (M + M )ù1 / 2
ZAB = s ^{2} ê8pRT ^{A} ^{B} ú
AB ë M
A MB û
where, s _{AB}
= s _{A} + s _{B}
2
and
MA ,
M B are molecular weights (M = mN0 )
 (a) At particular temperature; Z µ p ^{2}
 At particular pressure;
 At particular volume;
Z µ T 3 / 2
Z µ T1 / 2
 During molecular collisions a molecule covers a small distance before it gets The average distance travelled by the gas molecules between two successive collision is called mean free path (l).
l = Average distance travelled per unit time(u_{av} ) = u_{av} = 1 .
No. of collisions made by single molecule per unit time (ZA )
 Based on kinetic theory of gases mean free path, l µ T . Thus,
P
2ps 2uavr.n
 Larger the size of the molecules, smaller the mean free path, e., l µ
1
(radius)^{2}
 Greater the number of molecules per unit volume, smaller the mean free
 Larger the temperature, larger the mean free
 Larger the pressure, smaller the mean free
 Relation between collision frequency (Z) and mean free path (l) is given by,
Z =  urms 
l 
 At any particular time, in the given sample of gas all the molecules do not possess same speed, due to the frequent molecular collisions with the walls of the container and also with one another, the molecules move with ever changing speeds and also with ever changing direction of
 According to Maxwell, at a particular temperature the distribution of speeds remains constant and this distribution is referred to as the MaxwellBoltzmann distribution and given by the following expression,
3 / 2
dn0 n
= 4p æ M ö

2pRT
.e –Mu^{2} / 2RT .u2 dc
where,
è
dn0 =
ø
Number of molecules out of total number of
molecules n, having velocities between c and
c + dc ,
dn0 / n = Fraction of
the total number of molecules, M = molecular weight, T = absolute
temperature. The exponential factor e ^{–} ^{M}^{u}2 / 2RT
is called Boltzmann factor.
 Maxwell gave distribution curves of molecular speeds for
CO2 at
different temperatures. Special features of the curve are :
 Fraction of molecules with two high or two low speeds is very
 No molecules has zero
 Initially the fraction of molecules increases in velocity till the peak of the curve which pertains to most probable velocity and thereafter it falls with increase in
(4) Types of molecular speeds or Velocities :
 Root mean square velocity (u_{rms}) : It is the square root of the mean of the squares of the velocity of a large number of molecules of the same
urms =
urms = =
= = =
where k = Boltzmann constant = R
N0
 For the same gas at two different temperatures, the ratio of RMS velocities will be, u1=
u2
 For two different gases at the same temperature, the ratio of RMS velocities will be, u1=
u2
 RMS velocity at any temperature t ^{o} C
may be related to its value at S.T.P. as, ut = .
Note : ® RMS velocity explained the nonexistence of gases in the atmosphere of moon.
 When temperature alone is given then, u
rms
= 1.58 ´
´ 10^{4} cm / sec .
 If P and T both are given, use equation in terms of temperature, e., use u
rms
= and not
 Average velocity (vav ) : It is the average of the various velocities possessed by the
vav
= v1 + v2 + v3 +…… vn
n
vav = =
(iii) Most probable velocity
gas at a given temperature.
(a _{mp} ) : It is the velocity possessed by maximum number of molecules of a
a _{mp} = = =
(5) Relation between molecular speeds or velocities,
 Relation between urms
 Relation between a _{mp}
 Relation between a _{mp}
and vav : vav = 0.9213 ´ urms and urms : a _{mp} = 0.816 ´ urms and vav : vav = 1.128 ´ a _{mp}
or urms
or urms
= 1.085 ´ vav
= 1.224 ´ a _{mp}
 Relation between a _{mp} , vavand urms :
a mp
: vav
:
:
: urms
:
:
1.414 : 1.595 : 1.732
1 : 1.128 : 1.224 i.e., a _{mp} < vav < urms
Example : 28 The rms velocity of CO_{2} at a temperature T (in kelvin) is x cm s ^{–}^{1} . At what temperature (in kelvin) the rms
Solution: (a)
velocity of nitrous oxide would be 4x cm s ^{–}^{1}
(a) 16 T (b) 2 T (c) 4 T (d) 32 T
u =
[EAMCET 2001]
\ uCO2 =
u N2O
i.e., x =
4 x
or 1 =
4
or TN2O = 16 T
Example : 29 The rms velocity of an ideal gas at 27^{o} C
is 0.3 ms^{–}^{1} . Its rms velocity at 927^{o} C (in ms^{–}^{1} ) is
[IIT 1996; EAMCET 1991]
Solution: (d)
(a) 3.0 (b) 2.4 (c) 0.9 (d) 0.6
u =
For the same gas at two different temperatures,
u1 =
u2
; 0.3 =
u2
= 1 , u2 = 0.6 ms^{–}^{1}
2
Example : 30 The rms velocity of hydrogen is
times the rms velocity of nitrogen. If T is the temperature of the gas
[IIT 2000]
(a)
T(H_{2}) = T(N_{2})
(b)
T(H_{2}) > T(N_{2})
(c)
T(H_{2}) < T(N_{2})
T(H2 ) =
7T(N2 )
Solution: (c) u =
; \ u (H 2 ) =
T(H 2 ) ´ M(N 2 ) or =
or 7 = T(H 2 ) ´ 14 or
T(H _{2} ) = 1
u (N 2 )
M(H _{2} ) T(N _{2} )
T(N 2 )
T(N 2 ) 2
or T(N 2 ) = 2 ´ T(H 2 ) i.e., T(N 2 ) > T(H 2 )
Example : 31 If the average velocity of
N 2 molecules is 0.3 m/s at 27 ^{o} C , then the velocity of 0.6 m/s will take place at
[Manipal PMT 2001]
Solution: (d)
(a) 273 K (b) 927 K (c) 1000 K (d) 1200 K
v = 0.921u
\ v1
v2
= u1 =
u2
\ 0.3 =
0.6
or 1 =
2
or T_{2} = 300 ´ 4 = 1200 K
Example : 32 The temperature of an ideal gas is reduced from
927 ^{o} C
to 27 ^{o} C . The rms velocity of the molecules
Solution: (b)
becomes [Kerala CEE 2001]
(a) Double the initial value (b) Half of the initial value
 Four times the initial value (d) Ten times the initial value
u =
\ u1 = =
u2
= = = 2
\ u2 = 1 u1
2
 Gases which obey gas laws or ideal gas equation (PV = nRT) at all temperatures and pressures are called ideal or perfect Almost all gases deviate from the ideal behaviour i.e., no gas is perfect and the concept of perfect gas is only theoretical.
 Gases tend to show ideal behaviour more and more as the temperature rises above the boiling point of their liquefied forms and the pressure is Such gases are known as real or non ideal gases. Thus, a “real gas is that which obeys the gas laws under low pressure or high temperature”.
 The deviations can be displayed, by plotting the PV isotherms of real gas and ideal
 It is difficult to determine quantitatively the deviation of a real gas from ideal gas behaviour from the PV
isotherm curve as shown above. Compressibility factor Z defined by the equation,
PV = ZnRT
or Z = PV / nRT = PVm / RT
is more suitable for a quantitative description of the deviation from ideal gas behaviour.
 Greater is the departure of Z from unity, more is the deviation from ideal Thus, when
(i)
Z = 1, the gas is ideal at all temperatures and pressures. In case of
N _{2} , the value of Z is close to 1 at
50^{o} C . This temperature at which a real gas exhibits ideal behaviour, for considerable range of pressure, is known as Boyle’s temperature or Boyle’s point (TB ) .
 Z > 1 , the gas is less compressible than expected from ideal behaviour and shows positive deviation, usual
at high P i.e. PV > RT .
 Z < 1 , the gas is more compressible than expected from ideal behaviour and shows negative deviation,
usually at low P i.e. PV < RT .
Z > 1 for H _{2}
and He at all pressure i.e., always shows positive deviation.
 The most easily liquefiable and highly soluble gases
(NH 3 , SO2 )
show larger deviations from ideal
behaviour i.e. Z << 1 .
 Some gases like CO_{2} show both negative and positive
 Causes of deviations of real gases from ideal behaviour : The ideal gas laws can be derived from the kinetic theory of gases which is based on the following two important assumptions,
 The volume occupied by the molecules is negligible in comparison to the total volume of
 Causes of deviations of real gases from ideal behaviour : The ideal gas laws can be derived from the kinetic theory of gases which is based on the following two important assumptions,
 The molecules exert no forces of attraction upon one It is because neither of these assumptions can be regarded as applicable to real gases that the latter show departure from the ideal behaviour.
 To rectify the errors caused by ignoring the intermolecular forces of attraction and the volume occupied by molecules, Vander Waal (in 1873) modified the ideal gas equation by introducing two corrections,
 Volume correction
 Pressure correction
 Vander Waal’s equation is obeyed by the real gases over wide range of temperatures and pressures, hence it is called equation of state for the real
 The Vander Waal’s equation for n moles of the gas is,
 To rectify the errors caused by ignoring the intermolecular forces of attraction and the volume occupied by molecules, Vander Waal (in 1873) modified the ideal gas equation by introducing two corrections,
a and b are Vander Waal’s constants whose values depend on the nature of the gas. Normally for a gas a >> b .
 Constant a : It is a indirect measure of magnitude of attractive forces between the Greater is the value of a, more easily the gas can be liquefied. Thus the easily liquefiable gases (like
SO2 > NH 3 > H 2 S > CO2 ) have high values than the permanent gases (like N 2 > O2 > H 2 > He) .
Units of ‘a‘ are : atm. L^{2} mol ^{2}
or atm. m6mol 2 or
N m^{4} mol ^{2} (S.I. unit).
 Constant b : Also called covolume or excluded volume,
It’s value gives an idea about the effective size of gas molecules. Greater is the value of b, larger is the size and smaller is the compressible volume. As b is the effective volume of the gas molecules, the constant value of b for any gas over a wide range of temperature and pressure indicates that the gas molecules are incompressible.
Units of ‘b‘ are :
Lmol ^{1}
or m^{3} mol ^{1} (S.I. unit)
 Vander Waal’s constant for some gases are,
Name of gas  a  b  
atm litre^{2}mol  2  Nm4mol2  litre mol 1  m3mol1  
Hydrogen  0.245  0.0266  0.0266  0.0266  
Oxygen  1.360  0.1378  0.0318  0.0318  
Nitrogen  1.390  0.1408  0.039  0.0391  
Chlorine  6.493  0.6577  0.0562  0.0562  
Carbon dioxide  3.590  0.3637  0.0428  0.0428  
Ammonia  4.170  0.4210  0.0371  0.0371  
Sulphur dioxide  6.170  0.678  0.0564  0.0564  
Methane  2.253  0.0428 
 The two Vander Waal’s constants and Boyle’s temperature (TB
) are related as,
(4) Vander Waal’s equation at different temperature and pressures :
 When pressure is extremely low : For one mole of gas,


æ P + a ö(V – b) = RT or PV = RT – a + Pb + ab
ç V 2 ÷
V V ^{2}
 When pressure is extremely high : For one mole of gas,
PV = RT + Pb ;
PV = 1 + Pb RT RT
or Z = 1 + Pb
RT
where Z is compressibility factor.
 When temperature is extremely high : For one mole of gas,
PV = RT .
 When pressure is low : For one mole of gas,

æ P +
a ö(V – b) = RT
or PV = RT + Pb – a + ab
or PV
= 1 – a
or Z = 1 – a
ç V 2 ÷
V V ^{2} RT
VRT
VRT

 For hydrogen : Molecular mass of hydrogen is small hence value of ‘a‘ will be small owing to smaller
intermolecular force. Thus the terms
a and
V
ab may be ignored. Then Vander Waal’s equation becomes,
V 2
PV = RT + Pb
or PV RT
= 1 + Pb
RT
or Z = 1 + Pb
RT
In case of hydrogen, compressibility factor is always greater than one.
(5) Merits of Vander Waal’s equation :
 The Vander Waal’s equation holds good for real gases upto moderately high
 The equation represents the trend of the isotherms representing the variation of PV with P for various
 From the Vander Waal’s equation it is possible to obtain expressions of Boyle’s temperature, critical constants and inversion temperature in terms of the Vander Waal’s constants ‘a‘ and ‘b‘.
 Vander Waal’s equation is useful in obtaining a ‘reduced equation of state’ which being a general equation of state has the advantage that a single curve can be obtained for all gases when the equation if graphically represented by plotting the
(6) Limitations of Vander Waal’s equation :
 This equation shows appreciable deviations at too low temperatures or too high
 The values of Vander Waal’s constants a and b do not remain constant over the entire ranges of T and
P, hence this equation is valid only over specific range of T and P.
 Other equations of state : In addition to Vander Waal’s equation, there are also equations of state which have been used to explain real behaviour of gases are,
(i) Clausius equation :
éP + a ù (V – b) = RT . Here ‘c‘ is another constant besides a, b and R.


ê T(V + c)^{2} ú
(ii)Berthelot equation :
æ P + a ö(V – b) = RT .
(iii) Wohl equation :
ç

è
P = RT
TV ^{2} ÷
 a + c
(V – b) V(V – b) V ^{2}
(iv) Dieterici equation :
P = RT
V – b
.e ^{–}^{a} ^{/} ^{RTV} . The expression is derived on the basis of the concept that
molecules near the wall will have higher potential energy than those in the bulk.
 Kammerlingh Onnes equation : It is the most general or satisfactory expression as equation expresses PV as a power series of P at a given
PV = A + BP + CP ^{2} + DP ^{3} + ……
Here coefficients A, B, C etc. are known as first, second and third etc. virial coefficients.
 Virial coefficients are different for different
 At very low pressure, first virial coefficient, A = RT.
 At high pressure, other virial coefficients also become important and must be
 A state for every substance at which the vapour and liquid states are indistinguishable is known as critical state. It is defined by critical temperature and critical
 Critical temperature (T_{c}) of a gas is that temperature above which the gas cannot be liquified however large pressure is It is given by,
 Critical pressure (P_{c}) is the minimum pressure which must be applied to a gas to liquify it at its critical It is given by,

 Critical volume (V_{c}) is the volume occupied by one mole of the substance at its critical temperature and critical It is given by,
 Critical compressibility factor (Z_{c}) is given by,
A gas behaves as a Vander Waal’s gas if its critical compressibility factor (Zc ) is equal to 0.375.
Note : ® A substance in the gaseous state below Tc
is called vapour and above Tc
is called gas.
 In 1881, Vander Waal’s demonstrated that if the pressure, volume and temperature of a gas are expressed
in terms of its be obtained.
Pc , Vc
and
Tc , then an important generalization called the principle of corresponding states would
 According to this principle, “If two substances are at the same reduced temperature (q) and pressure (p) they must have the same reduced volume (f),” e.
where, f = V / Vc
or V = fVc ; p = P / Pc
or P = pPc ; q = T / Tc
or T = qTc
This equation is also called Vander Waal’s reduced equation of state. This equation is applicable to all substances (liquid or gaseous) irrespective of their nature, because it is not involving neither of the characteristic constants.
 This principle has a great significance in the study of the relationship between physical properties and chemical constitution of various
 The motion of atoms and molecules is generally described in terms of the degree of freedom which they
 The degrees of freedom of a molecule are defined as the independent number of parameters required to describe the state of the molecule
 When a gaseous molecule is heated, the energy supplied to it may bring about three kinds of motion in it, these are,
 The translational motion (ii) The rotational motion (iii) The vibrational This is expressed by saying that the molecule possesses translational, rotational and vibrational degrees of freedom.
 For a molecule made up of N atoms, total degrees of freedom = 3N. Further split up of these is as follows :
Translational  Rotational  Vibrational  
For linear molecule :  3  2  3N – 5 
For nonlinear molecule :  3  3  3N – 6 
 Specific heat (or specific heat capacity) of a substance is the quantity of heat (in calories, joules, kcal, or kilo joules) required to raise the temperature of 1g of that substance through 1^{o} C . It can be measured at constant pressure (c p ) and at constant volume (cv ).
 Molar heat capacity of a substance is the quantity of heat required to raise the temperature of 1 mole of the substance by 1^{o} C .
\ Molar heat capacity = Specific heat capacity ´ Molecular weight, i.e.,
Cv = cv ´ M and Cp = c p ´ M .
 Since gases upon heating show considerable tendency towards expansion if heated under constant
pressure conditions, an additional energy has to be supplied for raising its temperature by required under constant volume conditions, i.e.,
1^{o} C
relative to that
Cp > Cv or Cp = Cv + Work done on expanson, PDV(= R)
where, Cp = molar heat capacity at constant pressure; Cv =
molar heat capacity at constant volume.
Note : ®
Cp and Cv
for solids and liquids are practically equal. However, they differ considerable in
case of gas because appreciable change in volume takes place with temperature.
(4) Some useful relations of C_{p} and C_{v}
 Cp – Cv = R = 2 calories = 314 J
C = 3 R (for monoatomic gas) and C
v 2 v
= 3 + x
2
(for di and polyatomic gas), where x varies from gas to gas.
 Cp = g
Cv
(Ratio of molar capacities)
 For monoatomic gas Cv= 3 calories whereas, Cp = Cv + R = 5calories
 For monoatomic gas, (g ) = Cp
5 R
= 2 = 1.66 .
Cv 3 R
2
C 7 R
 For diatomic gas (g ) = ^{p}
Cv
= 2 = 1.40
5 R
2
 For triatomic gas (g ) = Cp
Cv
= 8R = 1.33 6R
Values of Molar heat capacities of some gases,
Gas  C_{p}  C_{v}  C_{p}– C_{v}  C_{p}/C_{v}=  g  Atomicity 
He  5  3.01  1.99  1.661  1  
N 2  6.95  4.96  1.99  1.4  2  
O2  6.82  4.83  1.99  1.4  2  
CO2  8.75  6.71  2.04  1.30  3  
H2 S  8.62  6.53  2.09  1.32  3 
 A gas may be liquefied by cooling or by the application of high pressure or by the combined effect of The first successful attempt for liquefying gases was made by Faraday (1823).
 Gases for which the intermolecular forces of attraction are small such as
H 2 ,
N _{2} , Ar and
O_{2} , have low
values of Tc
and cannot be liquefied by the application of pressure are known as “permanent gases” while the
gases for which the intermolecular forces of attraction are large, such as polar molecules
NH 3 ,
SO2
and
H2O
have high values of Tc
and can be liquefied easily.
 Methods of liquefaction of gases : The modern methods of cooling the gas to or below their Tc
hence of liquefaction of gases are done by Linde’s method and Claude’s method.
and
 Linde’s method : This process is based upon JouleThomson effect which states that “When a gas is allowed to expend adiabatically from a region of high pressure to a region of extremely low pressure, it is accompained by ”
 Claude’s method : This process is based upon the principle that when a gas expands adiabatically against an external pressure (as a piston in an engine), it does some external work. Since work is done by the molecules at the cost of their kinetic energy, the temperature of the gas falls causing
 By adiabatic
 Uses of liquefied gases : Liquefied and gases compressed under a high pressure are of great importance in
 Liquid ammonia and liquid sulphur dioxide are used as refrigerants.
 Liquid carbon dioxide finds use in soda
 Liquid chlorine is used for bleaching and disinfectant
 Liquid air is an important source of oxygen in rockets and jetpropelled planes and
 Compressed oxygen is used for welding
 Compressed helium is used in
 JouleThomson effect : When a real gas is allowed to expand adiabatically through a porous plug or a fine hole into a region of low pressure, it is accompanied by cooling (except for hydrogen and helium which get warmed up).
Cooling takes place because some work is done to overcome the intermolecular forces of attraction. As a result, the internal energy decreases and so does the temperature.
Ideal gases do not show any cooling or heating because there are no intermolecular forces of attraction i.e., they do not show JouleThomson effect.
During JouleThomson effect, enthalpy of the system remains constant.
JouleThomson coefficient.
m = (¶T / ¶P)H . For cooling,
m = +ve
(because dT and dP will be
 ve ) for
heating
m = –ve (because dT = +ve, dP = –ve) . For no heating or cooling
m = 0
(because dT = 0).
 Inversion temperature : It is the temperature at which gas shows neither cooling effect nor heating effect
i.e., JouleThomson coefficient shows heating effect.
m = 0 . Below this temperature, it shows cooling effect and above this temperature, it
Any gas like
H2 , He
etc, whose inversion temperature is low would show heating effect at room
temperature. However, if these gases are just cooled below inversion temperature and then subjected to Joule Thomson effect, they will also undergo cooling.