# 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

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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 (cm3 ) or cubic metres(m3 ) .

(iii) 1L = 1000 mL ; 1 mL = 10 -3 L

1 L = 1dm3 = 103 cm3

1m3 = 103 dm3 = 106 cm3 = 106 mL = 103 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 = 103 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 (Fo).

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 mv2 )

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 = 1Nm-2 = 1kg m-1 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 = 105 Pa = 105 Nm-2 = 100KNm-2 = 100KPa

• Several other units used for pressure are,

 Name Symbol Value bar bar 1bar = 105 Pa atmosphere atm 1atm = 1.01325 ´ 105 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 ´ 103 Kg / m3 = 13.6 g / cm3 ) 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 Vm (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.

PV1 = P2 V2 =……… = K

 ç      ÷

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 CO2 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

 ê
 ë
 d1 = P1 = V2 = ……. = K d2 P2 V1

         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 95cm2 under a pressure of

9.962 ´ 104 Nm2 . At the same temperature, it volume at a pressure of 10.13 ´ 104 Nm-2 is [Bihar CEE 1992]

(a)

190 cm3

(b)

93 cm 3

(c)

46.5 cm 3

(d)

47.5 cm 3

Solution: (b)

P1 V1 = P2 V2

9.962 ´ 104 ´ 95 = 10.13 ´ 104 ´ V2

V2 = 93 cm3

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 4/5th 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

 5

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 P1 = 100 mm ,

P2 will be 101 mm

Hence 100 ´ V = 101 ´ V2 ,

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 = 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 ,

 ç      ÷

Charle’s law can also be given as, æ dV ö

dT

= K .

• If t = 0 o C , then V = V0

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

 è
 ø

volume at 0o 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( oC)

at constant pressure is a straight line

cutting the temperature axis at – 273.15o 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

 ê
 ë
 d1 = T2 = V2 = …… = K d2 T1 V1

         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

T2 = 746 k = 473o 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   =     V2

T1       T2

300 K       280K

V2 = 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

 è
 ø

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 Gay-Lussac studied the variation of pressure with temperature and extende the Charle’s law so, this law is also called Charle’s-Gay 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 = 0o C , then P = P0

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 1o change in temperature, the pressure of a gas changes by

1     æ»  1 ö

of the

 è
 ø

pressure at 0o C .

(4) Graphical representation of Gay-Lussac’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)

627o C

(c)

627 o C

(d)

1173o 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

T2 = 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 ´ 1023 ) 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, Vm 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 :

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

 PV = nRT

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 ´ 107 erg mol -1 K -1

= 8.3143 joulemol -1 K -1

= 8.3143 Nmmol -1 K -1

(S.I. unit)

= 8.3143 KPa dm3 mol -1 K -1

= 8.3143 MPa cm3 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 pressure-volume 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

çn = Molecular weight of the gas (M) ÷

è                                                           ø

\

or dT M

æ          m ö

 V ÷
 d
 =

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 / cm3 .

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 O2 = 0.1mol , 0.2 g H 2 = 0.1mol,

Total n = 0.2 mol ,

P nRT0.2 ´ 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 L1

(b) 0.26 g L1

(c) 1.30 g L1

(d) 2.60 g L1

Solution: (c)

d PM

RT

=        2 ´ 16

0.082 ´ 300

= 1.30 g L1

Example : 16 The volume of 0.0168 mol of O2

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 O2 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 VT1

P2 V2 T2

gives P2

= 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

= P1 + P2 + P3 + ………

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 (X1

) ´ PTotal

• If a number of gases having volume container of volume V, then,

V1 , V2 , V3 ……

at pressure

P1, P2 , P3 ……..

are mixed together in

PTotal

PV1 + 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 (X1

) 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 O2 , CO and

CO2 ,

N 2 and Cl2 , CO and

N 2 etc. But this law is not applicable

to gases which combine chemically. For example, H 2

etc.

and

Cl2 , 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 O2

[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

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 non-reating 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
 N

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. V1 = V2

then,

r1 = t 2 =

r2       t1

• When volumes of the two gases diffuse in the same time , e. t1 = t 2

then,

r1 = V1 =

r2       V2

• Since, r µ p (when p is not constant)

then,

r1 = P1 =

r2       P2

æ                 ö

 1

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 large-scale 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   =

325.65 =

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 O2 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 25M 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 so-called Barometric formula.

where, Po 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 = doe Mgh / RT and n = noe 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 mu2

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                                              ÷

 N
 è
 ø

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., – 273.15o 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(N0 ) = 6.02 ´ 1023 m-3

Vm                                                0.0224

• The total number of bimolecular collision per unit time are given by,

ZAA

= 1 ps 2u

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(uav )               =            uav        =          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

• 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 Maxwell-Boltzmann distribution and given by the following expression,

3 / 2

dn0 n

= 4p æ   M    ö

 ç            ÷

2pRT

.e Mu2 / 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 Mu2 / 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 (urms) : 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 non-existence of gases in the atmosphere of moon.

• When temperature alone is given then, u

rms

= 1.58 ´

´ 104 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 CO2 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 27o C

is 0.3 ms1 . Its rms velocity at 927o C (in ms1 ) 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 ms1

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(H2) = T(N2)

(b)

T(H2) > T(N2)

(c)

T(H2) < T(N2)

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 T2 = 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 P-V isotherms of real gas and ideal
• It is difficult to determine quantitatively the deviation of a real gas from ideal gas behaviour from the P-V

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

50o 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 CO2 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
• 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,

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. L2 mol -2

or atm. m6mol -2  or

N m4 mol -2 (S.I. unit).

• Constant b : Also called co-volume 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 m3 mol -1 (S.I. unit)

• Vander Waal’s constant for some gases are,

 Name of gas a b atm litre2mol -2 Nm4mol-2 litre mol -1 m3mol-1 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,
 ø
 è

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,

 è

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

 ø
• 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 (b) = RT . Here ‘c‘ is another constant besides a, b and R.

 û
 ë

ê       T(V + c)2 ú

# (ii)Berthelot equation :

æ P +    a    ö(V b) = RT .

# (iii)            Wohl equation :

ç

 ø

è

P =    RT

TV 2 ÷

• a c

(V b)    V(V b)    V 2

# (iv) Dieterici equation :

P =   RT

Vb

.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 (Tc) of a gas is that temperature above which the gas cannot be liquified however large pressure is It is given by,
• Critical pressure (Pc) is the minimum pressure which must be applied to a gas to liquify it at its critical It is given by,
 Vc = 3b
• Critical volume (Vc) is the volume occupied by one mole of the substance at its critical temperature and critical It is given by,

• Critical compressibility factor (Zc) 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 non-linear 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 1o 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 1o 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.,

1o 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 Cp and Cv

• 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 Cp Cv Cp– Cv Cp/Cv= 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

O2 , 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 Joule-Thomson 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
• 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 jet-propelled planes and
• Compressed oxygen is used for welding
• Compressed helium is used in
• Joule-Thomson 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 Joule-Thomson effect.

During Joule-Thomson effect, enthalpy of the system remains constant.

Joule-Thomson 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., Joule-Thomson 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.