5 States of Matter STATES OF MATTER : The three important states of the matter are (i) Solid state (ii) Liquid state (iii) Gaseous state, which can exist together at a particular temperature and pressure e.g. water has three states in equilibrium at 4.58 mm and 0.0098ºC.
MEASURABLE PROPERTIES OF GASES : Four measurable properties are. (i) Mass : It is expressed in grams or kg. 1 kg=103 g moles of
PLASMA STATE : It is the gaseous mixture of electrons and positive ions existing at extremely high temperatures (in the interior of stars) or internal electrical fields in discharge tubes.
(ii) Volume : It is equal to the volume of the container and is expressed in terms of litres (L), millilitres (ml), cubic
SINGLE SUPER ATOM STATE : At extermely low temperature, all atoms lose their identity and get condensed into a single entity behaving like a single super atom. SOME CHARACTERISTICS OF THE THREE COMMON FORMS OF MATTER ARE : 1
Property General
2.
Forces
3. 4.
Density Motion
Gaseous state It has definite mass but no definite shape and volume Almost negligible
5.
Packing
Low Molecules have large rotatory, vibratory and translatory motions No proper packing
6.
Energy
Least
7.
Thermal High Expansion Compression High
8.
9. Intermixing Spontaneous 10. Pressure Exert pressure on the walls of container
Liquid State It has definite mass and volume but no definite shape Weaker than those in solids Lower than solids Low values of motions
Less closely packed Higher than solids Higher than solid Slightly higher than solid Spontaneous but slow Negligible
Solid state It has definite mass, volume and shape. Strongest High No translatory or rotatory motion. Possess vibratory motion Molecules closely packed Molecules possess maximum energy Least Least compressibility Least-intermixing Negligible
gas =
Mass in grams m = Molar mass M
centimeters ( cm 3 ) , cubic meters (m3) or cubic decimeters (dm3). 1 l = 1000 ml = 1000 cm 3 = 1 dm 3 1 m 3 = 10 3 dm 3 = 10 6 cm 3 = 10 6 ml = 10 3 l (iii) Pressure : It is equal to force per unit area and expressed in the units such as atmosphere, millimetres (mm), centimetres (cm), torr, bar etc. SI unit of pressure is pascal (Pa) or kilopascal (kPa) 1atm = 76cm of Hg = 760 mmof Hg = 760 torr 1atm = 101.325 kPa = 101325Pa = 101.325 Nm -2 = 1.01325 bar = 14.7lb m -2 (psi) 1bar = 105Pa. Pressure is measured with manometer
(iv) Temperature : It is measured in celcius scale (°C)or in Kelvin scale (K). SI unit of temperature is Kelvin (K) or absolute degree T (K) = t °C + 273 GAS LAWS : Boyle’s Law - The volume of a given mass of a gas is inversely proportional to its pressure at constant temperature. 1 or VP = k , a constant P Value of k depends on mass, temperature and nature of gas. V µ
When mass and temperature are the same we have P1V1 = P2 V2
2
Physi cs
ISOTHERMS : Graphs of V vs P or PV vs P at constant temperature are known as Isotherms. GRAPHICAL REPERSENTATION OF BOYLES LAW :
P
T2 T1
PV
P V
1/V
1 of its volume at 0°C for each degree rise or fall of 273.15 temperature at constant pressure.
t ö æ V t = V 0 ç1+ at constant.P and n è 273.15 ÷ø or The volume of a given mass of a gas is directly proportional to the absolute temperature at constant pressure.
P1 P2 T
P2 > P1
Isobars
GAY LUSSAC’S LAW : (AMANTON’S LAW) : The pressure of a given mass, of gas increases or decreases by 1 of its pressure at 0°C for each degree rise or fall of 273.15 temperature at constant volume.
t ö æ P t = P 0 ç1 + ÷ at constant V and n è 273.15 ø or The pressure of a given mass of a gas at constant volume is directly proportional to absolute temperature P
P
µT or P =kT or TP =k at constant V and n P P or = T T 1
2
1
2
V2 > V1 V2
T
AVOGADRO’S LAW : The volume of a gas is directly proportional to number of molecules or moles at constant temperature and pressure. V µ n at constant T and P V V1 V2 or = K or = at constant T and P n n1 n 2
GRAPHICAL REPRESENTATION OF AVOGADRO’S LAW : at constant P & T
V
V V V = const. or 1 = 2 T T1 T2
ABSOLUTE ZERO : If in the above equation we put the value of t as –273.15 ºC the volume of the gas will be zero. It means gas will not exist, which is not possible. In fact all gases get liquified before this temperature is reached. This hypothetical or imaginary temperature at which the gases occupy zero volume is called absolute zero. ISOBAR : A graph of V vs T at constant pressure is known as Isobar. GRAPHICAL REPRESENTATION OF CHARLE’S LAW:
V
V1
P
CHARLE’S LAW : The volume of the given mass of a gas increases or decreases by
V µ T at constant P and n or
ISOCHORES : A graph of P vs T at constant volume is known as Isochore GRAPHICAL REPRESENTATION OF GAY LUSSAC’S LAW :
n
IDEAL GAS EQUATION : 1 , T and n constant (Boyle 's law ) P V µ T , P and n constant (Charle 's law ) V µ n , P and T constant ( Avogadro's law ) Vµ
nT or PV µ nT or PV = nRT. P This is known as ideal gas equation. R is known as universal gas constant. ÞVµ
IDEAL GAS : The gas which obeys the equation PV = nRT at every temperature and pressure range strictly. REAL OR NON-IDEAL GASES : Since none of the gases present in universe strictly obey the equation PV = nRT , hence they are known as real or non ideal gases. Real gases behave ideally at low P and high T. DENSITY OF A GAS : We have PV = nRT , PV =
w RT ; M
æ ö weight (w) ç Moles(n) = ÷ Molecular weight (M) ø è
Further, d = w = PM V RT The above equation shows that density of a gas depends on P and T.
Laws of Motion
UNIT OF DENSITY OF GAS : It is usually expressed in gm/ litre VAPOUR DENSITY (V.D.) : It is the ratio under similar conditions of P and T V.D. =
Density of gas Density of H 2 which is 0.00009 Mass of 1litre of gas Mass of V litre of gas = Mass of 1litre of H 2 Mass of V litre of H 2
=
Mass of N molecules of gas Mass of N molecules of H 2
=
Mass of 1 molecule of gas Mass of 1 molecule of H 2
=1.987 or 2.0 Cal K -1mol -1 Since 1 Cal = 4.184 ´ 10 7 ergs
(iii) In mks or SI units :
UNIT OF VAPOUR DENSITY : It is a ratio and has no unit.
Unit of R is JouleK–1 mol -1 107 ergs = 1 Joule or R = 8.314 Joule K–1 mol -1
NATURE OF GAS CONSTANT, (R) :
=
[Force /(Length) 2 ] ´ (length)3 Moles ´ Degree / K
NTP OR STP AND SATP
work Force ´length =Moles ´degree =Moles ´degree
Z = PV/nRT
= Work done per degree per mole Deviation from ideal behaviour in terms of compressibility factor Z N2 H2 O2 CH 4 CO2
1.8 1.6 1.4 1.2 1.0 0.8 0.6 0.4
ideal gas
200 400 600 800 1000 P/bar
Deviation from ideal behaviour in terms of PV-P curve CO CH4
H2 He ideal gas
PV
1´ 76 ´13.6 ´ 980 ´ 22400 273.15
= 8.314 ´ 107 ergs K -1mol-1
1 molecular mass of gas 2
PV (Force / Area) ´ Volume = nT Moles ´ Degree / K
1 ´ 22.4 = 0.0821litre atm K -1 mol-1 273.15 (ii) In cgs system : P = 1 atm = 1×76×13.6×980 dyne cm-2; V = 22400 cm3; T = 273.15 K R=
Mass of 1 molecule of gas = 2 ´ mass of 1 atom of H
R=
NUMERICAL VALUE OF R : As the work can be expressed in different systems of units, R will have different values. (i) In litre - atmosphere : At NTP, P = 1atm, V=22.4 litre and T= 273.15K R=
=
=
Normal or standard temperature & pressure means 0°C or 273.15 K and 1 atm pressure. Normal boiling : At 1 atm pressure the boiling temperature is called Normal boiling point. Standard boiling point : At 1 bar pressure the boiling temperature is called standard boiling point. Normal boiling point of water is 100 °C (373.13 K) Standard boiling point of water is 99.6 °C (372.6K) Standard temperature and pressure (STP) 273.15 K(0 °C) temperature 1 bar (105 pascal) pressure Molar volume at STP = 22.71098 L mol–1 Normal temperature and pressure 273.15 K(0 °C) temperature 1 atom (1.01325 pascal) pressure Molar volume at NTP = 22.413996 L mol–1 Standard ambient temperature and pressure (SATP) 298.15 K (25 °C) temperature 1 bar (105 pascal) pressure Molar volume at SATP = 22.784 L mol–1
BOLTZMANN CONSTANT (k) : It is the gas constant per molecule thus k =
R ; Where R=gas N0
constant., N0= Avogadro number. -1 23 -1 -1 Value of k = 8.314 JK mol /6.02 ´ 10 mol
P
3
= 1.38 ´10-23 J K -1
4
Physi cs
DALTON’S LAW OF PARTIAL PRESSURE : At constant temperature the total pressure, exerted by a mixture of non reacting gases, is the sum of partial pressure of each gas P = p1 + p 2 + p 3 + ..... Partial pressure p of a gas = mole fraction of the gas × total pressure. For a gas A, pA = xA × P If n 1 , n 2 and n 3 are moles of non reacting gases filled in a vessel of volume V at temperature T. The total pressure P is given by PV = (n 1 + n 2 + n 3 )RT This is the equation of state of a gaseous mixture. RELATIONSHIP BETWEEN TOTAL PRESSURE AND INDIVIDUAL PRESSURE : On mixing of gases at constant temperature. If a gas A at pressure P1 and volume V1 is mixed with gas B at pressure p2 and volume V2 at same temperature, then Total pressure, P =
p1V1 +p 2V 2 V1 +V 2
AQUEOUS TENSION : It is the pressure exerted by water vapour at a particular temperature. It depends upon temperature. PRESSURE OF A DRY GAS : When a gas is collected over water, its observed pressure is equal to the sum of the pressure of dry gas and the pressure of water vapour (aqueous tension), then Pressure of dry gas = pressure of moist gas – aqueous tension. DIFFUSION : The tendency of every gaseous substance to distribute itself uniformly throughout the available space is known as diffusion. It also takes place through a porous vessel. EFFUSION : The movement of a gas through a small hole when it is subjected to pressure is known as effusion. GRAHAM’S LAW OF DIFFUSION (OR EFFUSION) : At constant temperature and pressure, the rate of diffusion or effusion of a gas is inversly proportional to the square root of its density. Rate of diffusion µ
1 d
If r1 and r2 are rates of diffusion of two gases and d1 and d 2 their respective densities then, r1 = r2
Rate of diffusion = =
d2 = d1
V.D 2 = V.D1
M2 M1
Volume diffused or effused time taken Moles diffused or effused time taken
r1 V /t = 1 1 = r 2 V2 / t 2
d2 = d1
M2 M1
If volume diffused is same
r1 T2 d2 M2 = = = r2 T1 d1 M1
If time of diffusion is same
r1 V1 d2 M2 = = = r 2 V2 d1 M1
Effect of pressure on the rate of diffusion. Rate of diffusion
µpressure, therefore, r
r1 2
=
p1 M2 × p2 M1
APPLICATIONS OF GRAHAM’S LAW OF DIFFUSION: (i) Determination of densities and molecular masses of unknown gases. (ii) Separating the gases having different densities. (iii) Separation of Isotopes ATMOLYSIS : The phenomenon of separation of a mixture of gases due to difference in their rates of diffusion is called atmolysis. KINETIC THEORY OF GASES : Postulates of kinetic theory of gases. (i) A gas consists of large number of tiny particles called molecules. (ii) Volume occupied by gas molecules, is negligible as compared to the total volume of gas. (iii) There is continuous rapid random motion of gas molecules. The molecules collide with each other and against walls of container. (iv) The molecules are perfect elastic bodies and there is no loss of kinetic energy during collisions. (v) There are no attractive forces between the molecules of gas. (vi) The pressure exerted by a gas is due to bombardment of gas molecules against the walls of the container. (vii) The different molecules possess different velocities and hence different energies. The average K.E. is directly proportional to absolute temperature. KINETIC GAS EQUATION : Based upon the postulates of Kinetic theory of gases, the kinetic gas equation is PV =
1 mNU 2 3
where m = mass of a gas molecule, N= number of molecules, U= Root mean square velocity.
Laws of Motion
KINETIC ENERGY OF GAS : It can be obtained from kinetic gas equation
(iv) Calulation of velocity when pressure and temperature are given but are not the same as NTP
KE of one molecule = 1 mv 2 2 PV =
U=
P using relation, P1V1 = P 2 V 2 T1 T2
( N = n and m ´ n = M) 3 \ KE = RT for 1 mole of a gas 2 (i) KE of n moles of gas = 3/2 nRT (ii) At absolute zero, KE is zero
MAXWELL’S GENERALISATION : Kinetic Energy of translation of ideal gas is directly proportional to absolute temperature of gas or its pressure and is independent of the nature of gas. THERMAL MOTION :
MAXWELL’S DISTRIBUTION OF VELOCITIES : The molecules present in a given sample of gas move with different velocites in all possible directions. Velocities and directions of molecules keep on changing due to intermolecular collisions. Hence it is impossible to find out the individual velocity of each molecule. It is however possible to predict fraction ( DN / N ) of the total number of molecules having specific velocities at a particular temperature. As shown by the curve, The gases show ideal behavior at low presence/large volume. Since the volume of molecules can be neglected and at high temperature since intermolecular forces decrease.
\ Uµ T
Most probable velocity Fraction (percentage) of molecules
The molecular velocity of a gas is proportional to square root of the absolute temperature. The molecular motion is called thermal motion of molecules.
CALCULATION OF MOLECULAR VELOCITIES OF GASES : (i) Calculation of velocity when temperature alone is given : From Kinetic gas equation, PV =
1 3PV 3RT MU 2 , U 2 = = 3 M M
\ U=
U=
R = 8.314 ´ 10 7 ergs K -1 mol -1
3RT ; M
T 3 ´ 8.314 ´107 ´ T cm / sec = 1.58 ´104 M M
It is given by the formula,
U=
3PV M
3 ´ 76 ´ 13.6 ´ 981´ 22400 M
3PV = M
3P D
T1
T2
T3
( sin ce, D =
Fraction of molecules having very small or very high velocities is very low. No molecule has zero velocity. Fraction of molecules possessing a particular velocity at a particular temperature is constant. The different velocities possessed by gas molecules are: (i) Most probable velocity ( a ) : It is the velocity possessed by maximum fraction of gas molecules at a particular temperature Mathematically, a =
2RT M
(ii) Average velocity ( v ) : This is the average of the different velocities of all the molecules.
AT NTP, P= 76 ´13.6 ´ 981 dynes cm -2, V = 22400 ml (molar volume). (iii) Calculation of velocity when pressure and density are given. It is given by the following formula, U=
Average velocity RMS velocity
Velocity
(ii) Calculation of velocity at NTP
U=
3PV ; In such case molecular volume (22400ml) is M
converted into the volume under given conditions of T and
2 1 2 1 mNU 2 = . mNU 2 = KE = RT 3 2 3 3
KE µ U 2 or U 2 µ T
5
M ) V
v=
n 1 c1 + n 2 c 2 + n 3 c 3 + ..... n 1 + n 2 + n 3 + ......
where c1,c2,c3 etc. are individual velocities of n1,n2,n 3 molecules. Mathematically, v =
8RT pM
6
Physi cs
(iii) Root mean square velocity (U) : It is the square root of the mean of the square of the different velocites of the molecules n 1 c1 + n 2 c 2 + n 3 c 3 + ... ; n 1 + n 2 + n 3 + ....... 2
U=
2
Mathematically, U =
2
3RT M
RELATIONSHIP BETWEEN DIFFERENT TYPES OF VELOCITIES : a:v:U º
2 RT 8RT : : M pM
3RT 8 : 3 = 2: M p
= 1.414 : 1.595 : 1.732 = 1 : 1.128 : 1.224 Calculation of root mean square velocity (Urms) : As already discussed, the Urms can be calculated by any of the following relations: Urms =
3RT 3PV 3P = = M M D
RATIO OF URMS FOR TWO DIFFERENT GASES AT THE SAME TEMPERATURE U1 M2 = U2 M1
RATIO OF URMS FOR THE SAME GAS AT DIFFERENT TEMPERATURE U1 T1 = U2 T2
EFFECT OF TEMPERATURE ON DISTRIBUTION OF MOLECULAR VELOCITIES: Most proabable velocity increases with the rise in temperature but fraction of molecules possessing such velocity decreases. Deviation from ideal behaviour of gas : At high pressure and low temperature the gases deviate considerably from the ideal behaviour. Deviation can be expressed in terms of compressibility factor (Z), expressed as Z=
PV nRT
In case of ideal gas, PV = nRT, Z = 1 In case of real gas, PV ¹ nRT, Z ¹ 1
NEGATIVE DEIVATION: In such case, Z < 1, gas is more compressible. POSITIVE DEVIATION : In such case, Z >1, gas is less compressible. FACTORS AFFECTING THE DEVIATION : The factors affecting the deviation are: (i) Nature of the gas : In general, the most easily liquefiable and highly soluble gases show larger deviation.
(ii) Pressure : The deviation is more at high pressure. CO2 and N2 show negative deviation at low pressure and positive deviation at high pressure. (iii) Temperature : The deviation is more at low temperature. H2 and He always show positive deviations at 0°C
BOYLE’S TEMPERATURE: Temperature at which a real gas exhibits ideal behaviour for considerable range of pressure is called Boyle’s temperature. Mathematically, Tb =
a , where a and b are Vander Waal’s constants bR
(i) Boyle temperature is different for different gases (ii) Below Boyle’s temperature a gas shows negative deviation. (iii) Above Boyle’s temperature a gas shows positive deviation
CAUS E OF D EVIATION FRO M THE ID EAL BEHAVIOUR : It is due to two faulty assumptions of Kinetic theory of gases particularly not valid at high pressure and low temperature. (i) Volume occupied by the gas molecules is negligible as compared to the total volume of gas. (ii) There are no attractive forces between the gas molecules. VAN DER WAAL'S EQUATION : After volume & pressure correction, van derWaals obtained the following equation for n moles of a gas 2 ö æ ç P + n a ÷(V - nb) = nRT , For one mole ç V 2 ÷ø è
a ö æ ÷÷ (V - b) = RT çç P + V2 ø è
Excluded volume or co-volume, (b) : The constant b in van der Waal's equation is known as excluded volume or co-volume. It is nearly four times the actual volume occupied by the gas molecules. Excluded or co-volume (b) = 4 × actual volume of gas molecules.
MAGNITUDE OF ATTRACTIVE FORCES BETWEEN GAS MOLECULES (a) : It is given by van der Waals constant ‘a’. Different gases have different value for ‘a’. (i) The greater the value of ‘a’, the greater the strength of van der Waals forces. (ii) The greater the value of ‘a’, the greater is the ease with which a gas can be liquefied. UNITS FOR VAN DER WAAL'S CONSTANT: Pressure correction, p=
n 2a V
2
\ a=
pV 2 n
2
Volume correction, V = nb
= atm litre 2 mole -2
\ b=
V = litre mole -1 n
Laws of Motion
vAN dER WAAL’S C ONSTANTS FOR SOME COMMON GASES : Gas
a (lit 2atmmol-2 )
b.(lit mol-1)
NH3
4.17
0.0371
CO2
3.59
0.0427
CO
1.49
0.0399
Cl2
6.49
0.0562
H2
0.024
0.0266
HCl
3.67
0.0408
NO
1.34
0.0279
O2
1.36
0.0318
SO2
6.71
0.0564
He
0.034
0.0237
Water
5.46
0.0305
a V
(i)
2
or PV = RT -
a V2
is negligible, then
V2
LIMITATION OF VAN DER WAAL’S EQUATION : There is specific range of temperature and pressure, to apply the equation. It deviates at too high pressure and too low temperature. OTHER EQUATIONS OF STATES : -a
(i)
P(V - b) = RT e RTV
Dieterici equation :
(P +
(ii) Berthelot’s equation : (iii) Clausius equation :
(P +
a TV 2
a T ( V + c) 2
)(V - b) = RT
)(V - b) = RT
Where c is a new constant (iv) Kammerling onne's equation (virial equation):
)(V - b) = RT
)V = RT
a PV a or = (1 ); V RT RTV
Therefore Z is less than 1. Hence at low pressure the gases show negative deviation. (ii) At high pressure and ordinary temperature : At high pressure, volume V will be quite small. The quantity
a
Pb PV ;Z > 1 =1+ RT RT Hence they always show positive deviation.
At low pressure and ordinary temperature : V is very large and b can be neglected then
(P +
EXCEPTIONAL BEHAVIOUR OF HYDROGEN AND HELIUM : Due to their small size, the attractive forces between the molecules are too small,
DISCUSSION OF VAN DER WAAL’S EQUATION : van der Waal’s equation for one mole of gas (P +
a V
2
become negligible but ‘b’ cannot be ignored. P(V - b) = RT PV Pb = 1+ ; Therefore Z is more than 1. RT RT At high pressure the gases show positive deviation Note : At some intermediate range of pressure and ordinary temperature the gas shows the ideal behaviour.
or
PV = A + BP + CP 2 + DP 3 (a) The coefficients A,B,C,D, etc are known as first, second, third, fourth virial coefficients (b) At low pressure only A (which is equal to RT) is important, the others cancel out. (c) A is always positive and increases with rise of temperature. (d) At Boyle temperature B=0 (e) Relation between B and van der Waals constant is B= b-
a RT
CRITICAL PHENOMENON AND LIQUEFACTION OF GASES : Increase of pressure and decrease of temperature tend to cause liquefaction of gases. The effect of temperature is, however more important. CRITICAL TEMPERATURE (Tc) : It may be defined as the temperature above which no gas can be liquefied howsoever high the pressure may be, critical temperature of CO2 is 31.1°C. Critical temperature (Tc) of some gases,
a
He
5.2
CO2
304.1
CH4
190.2
V2
H2
33.2
N2O
309.6
HCl
324.5
and b is negligible. We have
N2
126.0
NH3
405.5
H2S
373.5
PV =1 ; RT Therefore Z = 1. Hence gas shows ideal behaviour.
CO
134.4
Cl2
417.1
Ar
150.7
O2
154.3
SO2
430.3
(iii) At low pressure and high temperature: The effect of
PV = RT or
7
8
Physi cs
CRITICAL PRESSURE (Pc) : At critical temperature the pressure needed to liquefy a gas is known as critical pressure. CRITICAL VOLUME (Vc) : The volume occupied by one mole of a gas at critical temperature and critical pressure is known as critical volume. RELATIONSHIP BETWEEN CRITICAL CONSTANTS AND VAN DER WAAL’S CONSTANT: V c = 3b, T c =
8a a , Pc= 27Rb 27b 2
Relationship between boiling point and critical temperature: Boiling point =
2 critical temperature. 3
JOULE-THOMSON COEFFICIENT ( m J.T. ) : The temperature change produced per atmosphere drop in pressure under constant enthalpy conditions on passing a gas through the porous plug is called Joule-Thomson coefficient ¶T ö The quantity æç ÷ is called Joule -Thomson coefficient. è ¶P ø H
1 æ 2a æ ¶T ö ö - b÷ ç ÷ = ç P C RT ¶ è øH ø p è
(i)
æ ¶T ö ÷ is positive for real gases, Joule Thomson Coefficient ç è ¶P ø H (cooling)
Or Joule Thomson Coefficient will be positive when
Compressibility for 1 mole of gas at critical state: z=
Pc Vc 3 = = 0.375 , almost constant. RTc 8
JOULE THOMSON EFFECT : When a gas under high pressure is allowed to expand adiabatically into a region of extremly low pressure, it suffers change of temperature. The phenomenon is kown as Joule -Thomson effect. CONDITION FOR JOULE-THOMSON EFFECT (INVERSION TEMPERATURE): In Joule -Thomson effect, cooling is observed only if the gas is present below certain temperature known as inversion temperature, Ti. It is characteristic of each gas and related to van der Waal’s constant a and b as, Ti =
2a Rb
At exact inversion temperature there is no Joule-Thomson effect. Above inversion temperature, there is heating during JouleThomson effect. Below inversion temperature there is cooling during Joule-Thomson effect. Inversion temperature for hydrogen = –80°C and for He = –240°C. Joule-Thomson effect is zero in an ideal gas and enthalpy remains constant. When an ideal gas expands in vacuum, it does no work i.e. W=0 ; DE = 0 (Adiabatic condition). Hence internal energy of a given quantity of an ideal gas at constant temperature is independent of its volume.
æ ¶E ö ÷ = 0 (for an ideal gas) ; This quantity is called internal ç è ¶V ø T pressure and is positive for real gases. Thermodynamically an ideal gas may be defined by following the equations. (i) PV = constant, at constant temperature. (ii)
æ ¶E ö ÷ =0 ç è ¶V ø T
æ ¶T ö (ii) When Joule Thomson Coefficient ç ÷ è ¶P ø H (heating)
2a >b RT
is negative,
Or Joule Thomson Coefficient will be negative when
2a
æ ¶T ö (iii) When Joule Thomson Coefficient ç ÷ is zero, (no heating è ¶P ø H or cooling)
2a =b RT Since a, b and R are constants, the sign of Joule-Thomson Coefficient will depend only upon the temperature at which the gas is being allowed to expand. The temperature at which the Joule-Thomson Cofficient changes sign is known as the inversion temperature.
Or Joule Thomson Coefficient will be zero when
2a 2a = b \ Ti = RTi Rb
LAW OF CORRESPONDING STATES : When the values of pressure, volume and temperature are expressed as fractions of the corresponding critical values we have.
P V T = p, = f, =q Pc Vc Tc
where p , f and q are termed, the reduced pressure, the reduced volume and the reduced temperature respectively. If we replace P,V and T by p Pc, f Vc and q Tc in van der Waal’s equation and put the values of Pc, Vc and Tc terms of a, b and R we get, (p +
3 f2
)(3f - 1) = 8q
If the two substances have the same reduced temperature and the same reduced pressure, they will have the same reduced volume. The statement is known as the law of corresponding states.
Laws of Motion
HEAT CAPACITY OF SYSTEM : For gases it can be at constant volume or at constant pressure. (i) Heat capacity at constant volume (CV) is defined as the increase in internal energy of a gas per degree rise of temperature.
(ii) Collision frequency (Z) : The number of collisions experienced by molecules per cc of a gas per second is known as collision frequency of gas, Z =
RELATION BETWEEN Cp AND Cv : Cp– Cv = R R = 1.987 cal or 8.314 Joule R= PDV = Work done by one mole of an ideal gas in expansion at constant pressure when heated through 1°C.
2
At constant pressure, Z µ T 3 At constant temperature, Z µ p2 (iii) Collision number (Z1) : Number of collisions undergone by a molecule with other molecules per second present in 1cm3. Z1 = Collision number = 2 ps 2 vN s = diameter of molecules, v = Average velocity,, N = Number of molecules per unit volume of the gas (iv) Mean free path (l) : The average distance travelled by the molecule between two successive collisions
l=
MOLAR HEAT CAPACITIES FOR POLYATOMIC GASES: 3 5 R and C p = R are for monoatomic gases 2 2 like He, Ar etc. where the energy supplied increases translational kinetic energy only. In polyatomic gases heat supplied is utilised to increase vibrational and rotational energy also. Thus we have 3 R+x 2
and C p =
5 R 5 = 2 = = 1.66 For monoatomic gas, x = 0, g = CV 3 3 R 2 5 R+R Cp 7 = 2 = = 1.46 for diatomic gases, x = R, g = CV 3 5 R+R 2 5 3 R+ R Cp 3 2 2 = 8 = 1.33 = for triatomic gases, x = R , g = 3 CV 3 6 2 R+ R 2 2
COLLISION PROPERTIES : (i) Molecular diameter or collision diameter : The distance between the centers of the molecules at the point of their closest approach.
2 πσ 2 N
At constant temperature, l µ
1 P
VOLUME COEFFICIENT : It is defined as the ratio of the increase in volume of the gas at constant pressure per degree rise of temperature to its volume at 0°C
5 R+x 2
Cp
1
At constant pressure, l µ T
The values C V =
CV =
σ 2 vN 2
v = average velocity N = number of molecules per cm3 s = collision diameter
(ii) Heat capacity at constant pressure (Cp) is defined as increase in enthalpy of a gas per degree rise of temperature.
For one mole of gas, the heat capacities at constant volume and constant pressure are denoted by Cv and Cp and are termed as molar heat capacities.
1 2
3 æ ¶E ö CV = ç R ÷ = è ¶T ø V 2
5 æ ¶H ö Cp = ç ÷ = R è ¶T ø P 2
9
aV =
v t -v 0 v0 ´ t
V0 =Volume of a given mass of a gas at 0°C Vt= Volume of a given mass of a gas at t°C V t = V0 (1 +a V t ) The value of a v is found be 1/273 for all gases (charle's law) Therefore, V t = V0 (1 +
t ) 273
PRESSURE COEFFICIENT (aP) : It is defined as the ratio of the increase in pressure of the gas at constant volume per degree rise of temperature to its pressure at 0°C ap =
P t -P 0 P 0 ´t
P0=Pressure of given mass of gas at 0°C Pt=Pressure of given mass of gas at t °C
P t =P 0 (1 +a p t ) The value found by Ragnault & Gay Lussac was in the vicinity of 1
æ ö 273 for all gases and hence P t = P0 çè1 + 273 ÷ø t
10
Ph ysi cs
AMAGAT LAW OF PARTIAL VOLUME : The total volume of a mixture of non reacting gases at constant temperature and pressure is equal to the sum of the individual partial volumes of the constituents V(total)= V1+V2+V3+.....+Vn =
åV
i
LOSCHMIDT NUMBER : It is the number of molecules present in 1cc of a gas or vapour at STP. Its value is 2.617×1019 per cc. AVERAGE MOLECULAR WEIGHT OF A GASEOUS MIXTURE :
ån M = ån i
Mmix
Liquefaction is further based on the following principles (a) Cooling by freezing mixture : eg NaCl & ice (-22°C), CaCl2 & ice(-55°C), KOH & ice (-65°C) (b) Cooling by adiabatic expansion (Claude’s Method) : The gas in this process suffers a loss in temperature. DE = q + w if
Work is done by the gas at the cost of internal energy and temeprature is lowered. (c) Cooling by Joule-Thomson effect ( Linde’s method) : Expansion of a gas through a small jet under adiabatic conditions results in cooling and liquefaction of gas
LIQUID STATE :
i
l
i
ni= number of moles Mi= molecular weight of each component
l
BAROMETIC FORMULA : The decrease in the atmospheric pressure with increase in altitude given by the following expression is called barometic formula.
l
æ P ln çç è P0
q = 0 then DE = w or - DE = - w
ö Mgh ÷= ÷ RT ø
P0=Pressure at the sea level P = Pressure at height h
M = Average molecular mass of air, 29 ´ 10 -3 kgmol -1 g = acceleration due to gravity, 9.8 ms-2 R = gas constant ( 8.314 JK-1 mol-1) T= Temperature in kelvin
ESCAPE VELOCITY : Velocity required by an object to escape from the gravitational field of a body. It is given by V e = 2gr
AVERAGE KINETIC ENERGY : It is KE of a single molecule, and KE = æ
Boltzmann constant ç k = è
3 3 R kT = . T , where k= 2 2 No
Gas Constant ö R = ÷ Avogadro Number ø No
Total KE =N (average KE), N ( KE ) = 3 2 nRT LIQUEFACTION OF GASES : LIQUEFACTION OF GAS CAN BE ACHIEVED BY 1. Increasing pressure : It increases attraction among molecules. 2. Decreasing temperature : It decrease Kinetic energy of molecules. The temperature of gas must be lower than its critical temperature, TC.
l
Liquid state of the matter is the intermediate state between the gaseous and the solid state. In liquids the molecules of the matter are held together by strong intermolecular forces in comparison to those in gases. On the basis of kinetic molecular model, the liquid state is described as follow: (i) A liquid is composed of small molecules. (ii) The molecules of the liquid are held closer by some kind of intermolecular forces. (iii) The intermolecular forces are not very strong and thus the molecules are always in constant random motion. (iv) The average kinetic energy of molecules of a liquid is directly proportional to their absolute temperature. Properties of the liquid : Most of the physical properties of liquids are controlled by the strength of intermolecular attractive forces existing between molecules of a liquid. These intermolecular forces arranged in order of their increasing strength are London forces/induced diplole
VAPOUR PRESSURE : At equilibrium state the pressure exerted by vapour phase is called vapour pressure at a specific temperature.
Laws of Motion l
The magnitude of vapour pressure depends upon the following two factors. (i) The liquids having weak intermolecular attraction have greater tendency of escaping from the liquid surface in comparison to liquids having stronger forces of intermolecular attraction. For example, the vapour pressure of ether and acetone is more than that of water or acetic acid at any specific temperature. (ii) The vapour pressure of a liquid increases with the increase in temperature.
l
The vapour pressures of a given liquid at two different temperatures may be compared using Clausius Clapeyron equation log
P2 DH é T2 - T1 ù = ê ú P1 2.303R ë T 1T2 û
where P1and P2 are the vapour pressures at temperature T 1 and T2 D H is the heat of vapourisation and R is molar gas constant. l
The temperature at which the vapour pressure of liquid becomes equal to atmospheric pressure (or the external pressure) is termed as the boiling point of liquid. For example at 1 atmospheric pressure acetone boils at 56°C. benzene at 80°C, ethyl alcohol at 78.4°C and water at 100°C.
l
The temperature at which the solid state and the liquid form of a substance are in equilibrium at one atmospheric pressure is known as freezing point
1.
The density of a gas at 30°C and 1.3 atmosphere pressure is 0.027 g/cc. Calculate the molecular weight of the gas.
RTd Sol. M = ; P
R = 0.0821 atm lit K -1 mol-1;
d = .027 g / cc = 27 g / l;
T = 273 + 30 = 303K; P =1.3 atm M=
27 ´ .0821 ´ 303 = 516 .66 g / mol 1. 3
11
SURFACE TENSION : l The surface tension is defined as the force per cm acting perpendicular to the tangential line on the surface of the liquid which tend to compress the surface area. The units of surface tension are force per unit length i.e. dynes cm–1 or Nm–1 (in SI units) l Stronger the forces of intermolecular attraction (cohesive forces) greater is the surface tension. l The surface tension decreases with rise in temperature or surface tension in inversely proportional to temperature. Surface tension µ 1/Temperature Measurement of surface tension of a liquid by the drop number method is the most convenient method. VISCOSITY: l Viscosity may be defined as the force of friction between two layers of a liquid moving past one another with different velocities. l The viscosities of liquids are compared in terms of coefficient of viscosity which is defined as the force per unit area needed to maintain a unit difference in velocities between two consecutive parallel layers which are one cm. apart. l The units of viscosity are poise (P) where 1P = g cm–1 sec–1. In SI units of viscosity 1P = 0.1 N sec m –2. l The liquid having stronger forces of attraction has a higher viscosity. l With the rise in temperature, viscosity of a liquid decreases because the intermolecular attractive force between consecutive layers decrease as temperature increases.
2.
Density of ammonia is 0.77 g/l. Calculate its vapour density.
Sol. V.D. = 3.
0.77 Densityof gas = =8.55 Densityof H 2 0.09g / l 0.09
A gas cylinder containing cooking gas can withstand a pressure of 14.9 atmosphere. The pressure gauge of cylinder indicates 12 atmosphere at 27°C. Due to sudden fire in the building the temperature starts rising. At what temeprature, cylinder will explode.
Sol. P1 = P2 Þ T1 T2
14.9 12 = Þ T1 = 372.5 K T1 300
12
Ph ysi cs
The total pressure exerted by a mixture of gases containing 0.4g H2, 2.2g of CO2, 1.4 g N2 and 3.2g SO2 is 2.5 atmosphere . What are the partial pressures of each gas under the same conditions? Sol. Partial pressure of a gas = mole fraction × total pressure 4.
Total moles = pH 2 =
12 ´ .0821(t + 273 + 10) 120
1 t + 273 = or t = - 173°C 1.1 t + 283
Dividing
= 273 - 173° C = 100 K
0.4 2.2 1.4 3.2 + + + = 0.35 2 44 28 64
12 ´ 0.0821(100) = 0.821 litre 120 Calculate the molecular mass of a gas if its specific heat at constant pressure is 0.125 and at constant volume is 0.075 Þ V=
0.2 ´ 2.5 = 1.428 atm; 0.35
pCO2 =
Second equation 1.1 ´ V =
8.
0.05 ´ 2.5 = 0.357 atm 0.35
Sol.
0.125 5 g=CC =0.075 = 3 =1.66 hence gas is monoatomic p v
=3 2 R =3 2 ´2 =3 cal. Also, specific heat ´molar mass =C 3 \M =0.075 =40 g / mol 0.075 ´M =3
0.05 ´ 2.5 = 0.357 atm; pN 2 = 0.35 pSO2 =
5.
Molar heat at constant volume C v
0.05 ´ 2.5 = 0.357 atm 0.35
v
Calculate the total pressure in a 10 litre cylinder which contains 0.4 g He, 1.6g O2 and 1.4 g of N2 at 27°C. Also calculate the partial pressure of the He gas in the cylinder. Assume ideal behaviour of gases, R = 0.0821 atm lit K-1 mol-1.
Sol. PV = (n1 + n 2 + n 3 ) RT æ 0.4 1.6 1.4 ö P ´10 = ç + + ÷ ´ 0.821 ´ 300 ; P= 0.4926 atm 32 28 ø è 4 Total moles = 0.2
pHe
0.1 =0.2 ´0.4926 =0.2463atm
\ n1T1 = n 2 T2
2 NO2 + 0.5 O2 .Calculate For the reaction N2O5 (g) the mole fraction of N2O5(g) decomposed at constant volume and temperature, if initial pressure is 600 mm Hg and pressure at any time is 960 mm Hg. Assume ideal gas behaviour. Sol. At constant V and T, Pressure is directly proportional to number of moles.
6.
N 2 O 5 (g ) Initial Moles At time t moles Total moles
1 1– x
2 NO 2 + 0.5O 2 0 0 2x 0.5x = 1+1.5x
Pressure of (1 + 1.5x) moles = 960 mm
960 =600 (1 + 1.5x) Mole fraction decomposed, x = 0.4 7. The pressure exerted by 12 g of an ideal gas at temperature tºC in a vessel of volume V litres is one atmosphere. When the temperature is increased by 10 degrees at the same volume, the pressure increases by 10%. Calculate t and V (molecular weight of gas = 120). Sol. PV = nRT Pr essure of 1mole =
First equation
An open vessel at 27°C is heated until 3 5 th of the air in it has been expelled. Assuming that the volume of the vessel remains constant find, (a) Temperature at which vessel is heated (b) The air escaped out of it if vessel is heated to 900 K (c) The temperature at which the half of air escapes out Sol. An open vessel has constant V and P. 9.
12 ´ .0821(t + 273) 1´ V = 120
(a) 1 ´ 300 = 2 T2 5
(
\T2 = 750K = 477 ° C 3 expelled, 2 left 5 5
)
(b) 1 ´ 300 = n 2 ´ 900 \ n 2 = 1 (1/ 3left, 2 expelled) 3 3 (c) 1´ 300 = 1 ´ T 2
æ1 ö = 600K = 327° C ç left, 1 2 expelled ÷ è2 ø 10. A balloon of diameter 20m weights 100 kg. Calculate its pay load if it is filled with helium at 1.0 atm and 27°C. Density of air = 1.2 kg/m3, R= 0.0821 dm3 atm K–1 mol–1.
=4 3 pr =34 ´227 (10) =4190.47m Mass of the air displaced =4190.47 ´ 1.2 =5028.56 kg 1 ´4190.47 ´ 10 Moles of helium in the balloon = =170344 0.082 ´300
Sol. Volume of balloon
3
3
3
3
=´
= Mass of filled balloon =681.376 +100 =781.376 kg Pay load =Mass of air displaced -Mass of filled balloon =5028.56 -781.376 =4247.184 kg Mass of helium 4 170344 g 681.376 kg
11. An evacuated glass vessel weighs 50.0g when empty, 148.0g when filled with a liquid of density 0.98g/ml and 50.5g when filled with an ideal gas at 760 mm Hg and 300K. Determine the molecular weight of gas.
= -= 98 =100 ml =volumeof vessel Volume of liquid = 0.98 weight of gas =50.5 -50.0 =0.5g 760 100 w Gas equation PV = ´RT or ´ M 760 1000
Sol. Weight of liquid 148 50 98g
=
= ´ ´
0.5 .0821 300 M
or M 123 12. 20 dm3 SO2 diffuse through a porous partition in 60 seconds What volume of O2 will diffuse under similar conditions in 30 seconds. Sol.
V / 30 64 = or V =14.1 dm 3 20 / 60 32
13. At 27°C hydrogen is leaked through a tiny hole into a vessel for 20 minutes. Another unknown gas at the same temperature and pressure as that of hydrogen leaked through the same hole for 20 minutes. After the effusion of the gases the mixture exerts a pressure of 6 atm. The hydrogen content of the mixture is 0.7 mole. If the volume of the container is 3 litre, what is the molecular mass of the unknown gas? Sol. The partial pressure of hydrogen and unknown gas (ug) can be obtained as p H2 pug pH 2
=0.73 ´0.0821 ´300 =n3 ´0.0821 ´300
n
Laws of Motion Sol. r1 = P1 ´ M 2 or r2 P2 M1
The compound is XeF6 = 131 + 19 ´ 6 = 245
Where atomic weight of Xe is 131 and atomic weight of F is 19. If two Xe atoms were there, weight would have exceeded 252 which is not possible. 15. A 4:1 molar mixture of He and CH4 is contained in a vessel at 20 bar pressure. Due to hole in the vessel, the gas mixture leaks out what is the composition of the mixture effusing out initially? (IIT 1994). Sol. Molar ratio of He and CH4 is 4:1 Partial pressure ratio of He and CH 4 = 16 : 4 (QTotal Pressure = 20bar) r1 r2
=pp
1
2
or m =10.33 14. One mole of nitrogen gas at 0.8 atm takes 38 seconds to diffuse through a pinhole, whereas one mole of an unknown compound of xenon with fluorine at 1.6 atm. takes 57 seconds to diffuse through the same hole. Calculate the molecular formula of the compound.
=164 ´ 164 =8 :1
each weighing 10-22 g. The root mean square velocity is
1.5 ´105 cms -1 Calculate (i) Total pressure (ii) Total kinetic energy (iii) Temperature of the gas Kinetic gas equation is PV
Sol. (i)
=13 mNU
2
=13 ´10 ´102 ´10´(1.5 ´10 ) =3.75 ´10 dynes / cm 22
P
24
5 2
8
3
æ1 2ö (ii) Total Kinetic energy = N ´ ç mU ÷ ø è2
=
1024 ´10 -22 (1.5 ´ 105 ) 2 = 1.125 ´1012 ergs 2
(iii) The number of moles =
1´10 24 6.02 ´10 23
=1.66 moles
KE = 3 nRT 2
or n = 0.0308 mole 0.7 / 20 m = , 0.308 / 20 2
r M2 ; or He M1 rCH4
Composition of the mixture effusing initially for He and CH4 is 8:1. 16. A container of capacity 2 litre contains 1024 gas molecules
ug
Applying law of diffusion ,
n1 t 2 P1 M2 ´ = ´ n 2 t 1 P2 M1
1 57 0.8 M 2 ´ = 38 1 1.6 28 M 2 = 252
=number of moles
+p =6 =13 ´0.0821 ´300(0.7 +n)
13
2KE 2 ´ 1.125 ´ 1012 = = 5434.3K 3nR 3 ´ 1.66 ´ 8.314 ´ 10 7 17. Calculate translational KE of 2 moles of a gas at 27°C \T =
Sol. KE=
3 3 7 nRT = ´ 2 ´ 8.314 ´10 ´ 300 ergs 2 2
7482.5 ´107 ergs = 7482.5 Joules
2
14
Ph ysi cs
18. Calculate the average and total kinetic energy of 0.5 mole of an ideal gas at 0°C
\
U rms =
8.314 ´ 107 =3 ´ ´ 273 =5.65 ´ 10-14 ergs 2 6.023 ´ 1023 Joules = 3 nRT = 3 ´ 0.5 ´ 8.314 ´ 10 ´ 373 2 2
M kr 84 Þ M mix = 62.425 or 1.16 = M mix M
Dissociation:
Cl2 1-a Total number of moles = 1+a Normal M.wt Experimental M. wt a = 0.137 = 13.7%
= 1+ a ;
2Cl 2a 71 =1+ a 62.425
20. The average speed at T1 K and the most probable speed at T2 K of CO2 gas is 9×104 cm sec-1. Calculate the value of T1and T2. Sol. Average speed at T1K =
8RT1 = pM
2RT 2 M 2RT2 = 9 ´10 4 M
8 ´ 8.314 ´10 7 ´ T1 2 ´ 8.314 ´10 7 ´ T2 = = 9 ´ 10 4 \ 3.14 ´ 44 44 On calculating, we get T2 = 2143.37K and T1 = 1684.0 K
21. A glass bulb of 1 litre capacity contains 2 ´ 1021 molecules of N2 exerting pressure 7.57×103 Nm–2. Calculate the RMS speed and temperature of gas molecules. If the ratio of Ump to Urms is 0.82 calculate the Vmp for these molecules at this temperature. Sol. Given P = 7.57 ´103 Nm-2 V = 1litre = 10 -3 m 3 , R =8.3145 J K -1 mol-1
n=
28 ´ 10-3
(M.W.in Kg)
2 ´ 10 21 6.023 ´ 10 23 mol
= 0.82 ; U mp = 494.22 ´ 0.82 = 405.26 ms -1
22. A gaseous mixture of helium and oxygen is found to a density of 0.518 g dm-3 at 25°C and 720 torr. What is the percent by mass of helium in the mixture? Sol. M av = dRT = 0.518 ´ 0.0821´ 298 = 13.36 g mol -1 P 720 / 760 Let x be the mole fraction of helium in the mixture, then M av = xM He + (1 - x ) M O 2 13.36 = x ´ 4 + (1 - x )32
\ x = 0.666
0.666 ´ 4 ´ 100 (0.666 ´ 4 + 0.334 ´ 32) = 19.95
Mass percent of helium =
23. Using Van der Waal's equation calculate the constant ‘a’ when the moles of a gas confined in a 4 l flask exerts a pressure of 11.0 atm at a temperature of 30 K. The value of ‘b’ is 0.05 lit mol-1.
æ n 2a ö Sol. çç P + 2 ÷÷ (V - nb) = nRT V ø è
8RT1 pM
Most probable speed at T2 K = According to Question ,
U mp U rms
19. The composition of the equilbrium mixture, Cl2 2Cl, which is attained at 200º C, is determined by measuring the rate of effusion through a pin hole. It is observed that at 1.80 mm Hg pressure, the mixture effuses 1.16 times as fast as krypton effuses under the same condition. Calculate the fraction of the chlorine molecules dissociated into atmosphere. (Atomic wt. of krypton 84) r Kr
3 ´ 8.314 ´ 274.2
7
=1.702 ´ 1010 ergs =1.702kJ
Sol. r mix =
3RT = M
= 494.22 ms -1
Total Kinetic energy of 0.5moles of a gas = 5.65 ´ 10
PV 7.57 ´ 103 ´ 10-3 = ´ 6.023 ´ 10 23 nR 2 ´ 10 21 ´ 8.314
T = 274.2 K
3 R T 2 N0
Sol. Average KE per molecule of the gas = .
-21
T=
2 ö æ ç11 + 2 a ÷ ( 4 - 2 ´ 0.05) = 2 ´ 0.0821´ 300 ç 4 2 ÷ø è
a = 6.46 atm lit 2 mol - 2 24. (a) Calculate the pressure exerted by 5 mol of CO2 in one litre vessel at 47°C using van der Waal's equation. Also report the pressure of gas if it behaves ideally in nature. Given that a = 3.592 atm ltr2 mol-2 . b = 0.042 litre mol-1 (b) If volume occupied by CO2 molecules is negligible, then calculate the pressure exerted by one mol of CO2 gas at 273K . 2 ö æ Sol. (a) ç P + n a ÷ ( V - nb) = nRT ç V 2 ÷ø è 2 æ ö ç P + 5 ´ 3.592 ÷ (1 - 5 ´ 0.0427) = 5 ´ .0821´ 320 ç ÷ 1 è ø P = 77.218 atm when the gas behaves ideally then PV = nRT P ´1 = 5 ´ .0821´ 320 . Þ P = 131.36 atm
Laws of Motion é
a ù ú [V - b] = RT V2 û ë RT a .0821´ 273 3.592 = If b is negligible P = 2 V V 22.4 (22.4) 2
(b) For 1 mole êP +
P =0.9922 atm
25. The compression factor's (compressibility factor) for 1 mol of a van der Waal's gas at 0°C and 100 atmospheric pressure is found to be 0.5. Assuming that the volume of gas molecules is negligible, calculate the van der Waals constant ‘a’. PV 100 ´ V \V = 0.112litre. , 0.5 = 1 ´ .0821 ´ 273 nRT van der Waal 's equation when ' b ' is negligible
Sol. Z =
é ù é a ù a = 0.0821 ´ 273 ê P + 2 ú V = RT or ê100 + 2ú ë (0.112) ûú V û êë a = 1.253atm litre2 mol-2
26. Using van der Waal’s equation calculate the constant ‘a’ when two moles of a gas is confined in a four litre flask exerts a pressure of 11.0 atm at a temperature of 300 K. The value of ‘b’ is 0.5 litre mol -1 .
28. It is found that at 1470 K the mixture ( Cl 2
Sol. Cl 2 2Cl 1–x 2x Where x is degree of dissociation. The molecular weight of mixture of Cl 2 and Cl at equilibrium is given by M mix =
27. At room temperature ammonia gas at 1 atm pressure and hydrogen chloride at p atmospheric pressure are allowed to diffuse through identical pin holes from opposite ends of a glass tube of 1 meter length and uniform cross section. Ammonium is first formed at a distance of 60 cm from the through which HCl gas is sent in what is the value of p? Sol.
rNH3 rHCl
=
PNH3 M HCl 40 = × M NH3 60 PHCl
40 1 36.5 × = 60 P 17 HCl
PHCl = 2.198 atm
83.5 = 1.16 71 /(1 + x )
29. What is the molar volume of N 2 at 500 K and 600 atm according to (a) ideal gas law (b) virial equation. The virial coefficient. B of N 2 (g) at 500 K = 0.0169 lit mol -1 . How do you interpret the result. Sol. For 1 mole of ideal gas PV = 1RT 0.0821´ 500 = 6.84 × 10 - 2 lit mol -1 600 (b) using virial equation, neglecting high powers
V=
Z=
B PV BP =1+ =1+ V RT RT
0.0169 ´ 600 = 1.247 0.082 ´ 500 For 1 mol of a real gas
Z= 1 +
Z=
PV ZRT 1.247 ´ 0.0821´ 500 = ; V= RT P 600
= 8.53 × 10 - 2 lit mol -1 The molar volume of the real gas is greater because of finite volume of the gas molecules. 30. An evacuated glass vessel weighs 50 g when empty 148 g when filled with a liquid of density 0.98 g/ml and 50.5 g when filled with ideal gas at 760 mm Hg at 300 K. Determine the molecular wt of the gas. Sol. Weight of liquid = 148 – 50 = 98 g Volume of liquid =
w 98 = = 100 ml d 0.98
wt of gas = 50.5 – 50 = 0.5g PV =
or
M Kr = M mix
\ x = 0.14
é 22 a ù ê11 + 2 ú [4.2 × 0.05] = 2 × 0.0821 × 300 4 úû êë a = 6.46 atm litre 2 mol - 2
71 (1 - x ) ´ 71 + 2x ´ 35.5 = 1+ x (1 - x) + 2 x
R mix = R Kr
2
Given T = 300 K, V = 4 litre, P = 11.0 atm, n = 2 We have
2Cl) diffuses
1.16 times as fast as Krypton (83.8) diffuses under identical conditions. Find the degree of dissociation at equilibrium.
Sol. van der Waals equation for n moles é n aù T êP + 2 ú (V – nb) = nRT v ûú ëê
15
W RT T M
0.5 760 ´ 100 × 0.821 × 300 = m 760 ´1000 or m = 123.15
or
16
Ph ysi cs
Very Short/Short Answer Questions 1. 2. 3. 4. 5. 6. 7.
8.
9.
10.
11.
When do real gases behave as ideal gas? The size of weather balloon becomes larger and larger as it ascends up into higher altitudes. Why? What do you mean by Boyle temperature ? Give its expression and its relation with inversion temperature. Liquid is transferred from a large beaker to a small beaker, what will be the effect on its vapour pressure ? Explain, why the bubbles of a gas in a boiling liquid generally increase in volume as they approach the upper surface ? What is the effect of temperature on the vapour pressure of a liquid ? Two different gases ‘A’ and ‘B’ are filled in separate containers of equal capacity under the same conditions of temperature and pressure. On increasing the pressure slightly the gas ‘A’ liquefies but gas B does not liquify even on applying high pressure until it is cooled. Explain this phenomenon. A gas occupying a volume of 100 litres is at 20°C under a pressure of 2 bar. What temperature will it have when it is placed in an evacuated chamber of volume 175 litres ? The pressure of the gas in the chamber is one-third of its initial pressure. The values of the van der Waal’s constants for a gas are a = 4.10 dm6 bar mol–2 and b = 0.035 dm3 mol–1. Calculate the values of the critical temperature and critical pressure for the gas. Compressibility factor, Z, of a gas is given as Z
pV =nRT
(i) What is the value of Z for an ideal gas ? (ii) For real gas what will be the effect on value of Z above Boyle’s temperature ? For real gases the relation between p, V and T is given by van der Waals equation :
æ ö çèp +anV ÷ø(V -nb)=nRT 2
2
where ‘a’ and ‘b’ are van der Waals constants, ‘nb’ is approximately equal to the total volume of the molecules of a gas. ‘a’ is the measure of magnitude of intermolecular attraction. (i) Arrange the following gases in the increasing order of ‘b’. Give reason. O2, CO2, H2, He (ii) Arrange the following gases in the decreasing order of magnitude of ‘a’. Give reason. CH4, O2, H2
12. A bacterial culture isolated from sewage produced 41.3 ml of methane, CH4 at 31ºC and 753 mm Hg. What is the volume of this methane at STP?
Long Answer Questions 13. (i)
A quantity of hydrogen is confined in a chamber of constant volume. When the chamber is immersed in a bath of melting ice, the pressure of the gas is 1000atm. (a) What is the Celsius temperature when the pressure manometer indicates an absolute pressure of 400 atm? (b) What pressure will be indicated when the chamber is brought to 100 ºC?
(ii) A steel tank containing air at 15 atm pressure at 15° C is provided with a safety valve that will yield at a pressure of 30 atm. To what minimum temperature must the air be heated to blow the safety valve? 14. (i)
One mole of the CO2 occupies 1.5 L at 25ºC. Calculate the pressure exerted by the gas using:
(a) an ideal gas equation (b) van der Waal’s equation if: a = 3.012 atm mol–2 and b = 0.04 L mol–1. (ii) The critical temperature and pressure for NO gas are 177 K and 64.5 atm. respectively. Calculate the van der Waal’s constants ‘a’ and ‘b’. 15. (i)
A spherical balloon of 21 cm diameter is to be filled with hydrogen at STP from a cylinder containing the gas at 20 atm at 27ºC. If the cylinder can hold 2.82 L of water, calculate the number of balloons that can be filled up.
(ii) Two van der Waals gases A and B have volumes 0.112 and 0.111 L mol –1 r espectively. Calculate the compressibility factors for one mole of each at 273 K and 200 atm and hence state which gas is more compressible.
Multiple Choice Questions 16. Dipole-dipole forces act between the molecules possessing permanent dipole. Ends of dipoles possess ‘partial charges’. The partial charge is (a) more than unit electronic charge (b) equal to unit electronic charge (c) less than unit electronic charge (d) double the unit electronic charge
17
Laws of Motion
p1 > p1 = p1 < p1 <
p2 p2 p2 p2
> = < =
p3 p3 p3 p3
> = < <
p4 p4 p4 p4
p1 p2
Volume (mL)
(a) (b) (c) (d)
p3 p4
Temperature (K)
18. The pressure of a 1 : 4 mixture of dihydrogen and dioxygen enclosed in a vessel is one atmosphere. What would be the partial pressure of dioxygen ? (a) 0.8 × 105 atm (b) 0.008 Nm–2 4 –2 (c) 8 × 10 Nm (d) 0.25 atm 19. The ratio of Boyle’s temperature and critical temperature for a gas is :
20.
1.
2.
3.
8 27 1 2 (a) (b) (c) (d) 27 8 2 1 Gases possess characteristic critical temperature which depends upon the magnitude of intermolecular forces between the particles. Following are the critical temperature of some gases. Gases H2 He O2 N2 Critical temperature 33.2 5.3 154.3 126 in Kelvin
Which one of the following statements is wrong for gases? (a) Gases do not have a definite shape and volume (b) Volume of the gas is equal to the volume of the container confining the gas (c) Confined gas exerts uniform pressure on the walls of its container in all directions (d) Mass of the gas cannot be determined by weighting a container in which it is enclosed. Non reacting gases have a tendency to mix with each other. This property is known as (a) diffusion (b) fusion (c) mixing (d) None of these Which of the following mixtures of gases does not obey Dalton’s law of partial pressure ? (a) O2 and CO2 (b) N2 and O2 (c) Cl2 and O2 (d) NH3 and HCl
From the above data what would be the order of liquefaction of these gases ? Start writing the order from the gas liquefying first (a) H2, He, O2, N2 (b) He, O2, H2, N2 (c) N2, O2, He, H2 (d) O2, N2, H2, He 21. Atmospheric pressures recorded in different cities are as follows : Cities p in N/m2 Shimla 1.01 × 105 Bangalore 1.2 × 105 Delhi 1.02 × 105 Mumbai 1.21 × 105 Consider the above data and mark the place at which liquid will boil first. (a) Shimla (b) Bangalore (c) Delhi (d) Mumbai 22. Which curve in figure represents the curve of ideal gas ? (a) B only FE (b) C and D only (c) E and F only D (d) A and B only C
A B
pV
17. A plot of volume (V) versus temperature (T) for a gas at constant pressure is a straight line passing through the origin. The plots at different values of pressure are shown in figure. Which of the following order pressure is correct for this gas ?
0
p
23. Increase in kinetic energy can overcome intermolecular forces of attraction. How will the viscosity of liquid be affected by the increase in temperature ? (a) Increase (b) No effect (c) Decrease (d) No regular pattern will be followed
4.
5.
6.
Densities of two gases are in the ratio 1:2 and their temperatures are in the ratio 2:1 then the ratio of their respective pressures is (a) 1:1 (b) 1:2 (c) 2:1 (d) 4:1 Gas equation PV = nRT is obeyed by (a) only isothermal process (b) only adiabatic process (c) both (a) and (b) (d) None of these The following graph illustrates V (a) Dalton’s law (b) Charle’s law (c) Boyle’s law (d) Gay-Lussac’s law Temp. (ºC)
18 7.
8.
Ph ysi cs 4.4 g of a gas at STP occupies a volume of 2.24 L, the gas can be (a) O2 (b) CO (b) NO2 (d) CO2 Which of the following volume (V) - temperature (T) plots represents the behaviour of one mole of an ideal gas at one atmospheric pressure ?
(22.4 L 273K)
(36.8 L 373 K)
(22.4 L 273K)
é dE ù (a) ê dV ú = 0 ë ûT
é dE ù (b) ê dT ú = 0 ë ûP
é dE ù (c) ê dT ú = 0 ë ûV
(d) All of these
16. The value of R in SI units is :
V(L)
V(L)
15. For an ideal gas, correct relation is-
(26.8 L 373 K)
(a) 8.315 ´10 7 erg K -1 mol -1 (b) 8.315 JK -1 mol-1 (c) 0.0815 litre atm K -1 mol -1
(a)
T(K)
V(L)
(22.4 L 273K)
(b)
T(K)
V(L)
(30.6 L 373 K)
(22.4 L 273K) (14.2 L 373 K)
(c)
9.
T(K)
(d)
T(K)
If the four tubes of a car are filled to the same pressure with N2, O2, H2 and Ne separately, then which one will be filled first ? (a) N 2
(b)
O2
(c) H 2
(d) Ne
10. When the product of pressure and volume is plotted against pressure for a given amount of the gas, the line obtained is (a) parallel to X-axis (b) parallel to Y-axis (c) linear with positive slope (d) linear with negative slope 11. Air at sea level is dense. This is a practical application of (a) Boyle’s law (b) Charle’s law (c) Kelvin’s law (d) Brown’s law 12. Use of hot air balloons in sports and meteorological observations is an application of (a) Boyle’s law (b) Charle’s law (b) Kelvin’s law (d) Gay-Lussac’s law 13. “Equal volumes of all gases at the same temperature and pressure contain equal number of particles.” This statement is a direct consequece of (a) Perfect gas law (b) Avogadro’s law (c) Charle’s law (d) Boyle’s law 0 14. If 300 ml of a gas at 27 C is cooled to 70 C at constant pressure, its final volume will be (a) 135 ml (b) 540 ml (c) 350 ml (d) 280 ml
(d) 2 cal K -1 mol-1 17. The compressibility factor for H2 and He is usually : (a) >1 (b) =1 (c) <1 (d) Either of these 18. The densities of two gasses are in the ratio of 1: 16. The ratio of their rates of diffusion is (a) 16:1 (b) 4:1 (c) 1:4 (d) 1:16 19. At what temperature, the rate of effusion of N2 would be o 1.625 times than that of SO2 at 50 C ? (a) 110 K (b) 173 K (c) 373 K (d) 273 K 20. The rate of diffusion of methane at a given temperature is twice that of X. The molecular weight of X is (a) 64.0 (b) 32.0 (c) 40.0 (d) 80.0 21. X ml of H2 gas effuse through a hole in a container in 5 seconds. The time taken for the effusion of the same volme of the gas specified below under identical conditions is (a) 10 seconds : He
(b) 20 seconds : O2
(c) 25 seconds : CO
(d) 55 seconds : CO 2
22. 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 (a) 79.19 ml s–1 (b) 112.0 ml s–1 –1 (c) 56.0 ml s (d) 90.0 ml s–1 23. Which pair of the gaseous species diffuse through a small jet with the same rate of diffusion at same P and T (a) NO, CO (b) NO, CO2 (c) NH3, PH3 (d) NO, C2H6 24. The rate of diffusion of SO2, CO2 , PCl3 and SO3 are in the following order (a) PCl3 > SO3 > SO2 >CO2 (b) CO2 > SO2 > PCl3 > SO3 (c) SO2 > SO3 > PCl3 > CO2 (d) CO2 > SO2 > SO3 > PCl3
Laws of Motion 25. Which of the following expression correctly represents the relationship between the average molar kinetic energy KE of CO and N2 molecules at the same temperature (a) KE CO = KE N 2 (b) KE CO > KE N 2 (c) KE CO < KE N 2 (d) Can not be predicted unless the volumes of the gases are not given 26. Kinetic theory of gases presumes that the collisions between the molecules to be perfectly elastic because (a) the gas molecules are tiny particles and not rigid in nature (b) the temperature remains constant irrespective of collision (c) collision will not split the molecules (d) the molecules are large particleand rigid in nature 27. The average kinetic energy of an ideal gas per molecule in SI unit at 25° C will be (a) 6.17 × 10-21 kJ (b) 6.17 × 10–21 J (c) 6.17 × 10–20 J (d) 7.16 × 10–20 J 28. Boyle’s law, according to kinetic equation can be expressed as (a) PV = KT (c) PV =
3 kT 2
(b)
PV = RT
(d)
PV =
2 kT 3
29. The ratio between the root mean square velocity of H2 at 50 K and that of O2 at 800 K is (a) 4 (b) 2 (c) 1 (d) 1/4 30. The r.m.s velocity of hydrogen is 7 times the r.m.s velocity of nitrogen. If T is the temperature of the gas , than (a) T(H 2 ) = T ( N 2 )
(b)
T (H 2 ) > T ( N 2 )
(c) T(H 2 ) < T ( N 2 )
(d)
T (H 2 ) = 7 T ( N 2 )
31. The r.m.s velocity of CO2 at temperature T (in kelvin) is x cms -1 . At what temperature (in kelvin) the r.m.s. velocity
be 4x cms–1 ?
of nitrous oxide would (a) 16 T (b) 2 T (c) 4 T (d) 32 T 32. Which of the following has maximum root mean square velocity at the same temperature? (a) SO2 (b) CO2 (c) O2 (d) H2 33. Density ratio of O2 and H2 is 16:1. The ratio of their r.m.s. velocities will be (a) 4 : 1 (b) 1 : 16 (c) 1 : 4 (d) 16 : 1
19
34. At what temperature the RMS velocity of SO2 be same as that of O2 at 303 K ? (a) 273 K (b) 606 K (c) 303 K (d) 403 K 35. The temperature of an ideal gas is reduced from 9270C to 270C. the r.m.s. velocity of the molecules becomes. (a) double the inital value (b) half of the initial value (c) four times the initial value (d) ten times the initial value 36. If the average velocity of N2 molecues is 0.3 m/s at 27ºC, then the velocity of 0.6 m/s will take place at (a) 273 K (b) 927 K (c) 1000 K (d) 1200 K 37. Gas deviates from ideal gas nature because molecules (a) are colouress (b) attaract each other (c) contain covalent bond (d) show Brownian movement. 38. The compressibility factor for an ideal gas is (a) 1.5 (b) 1.0 (c) 2.0 (d) ¥ 39. A gas will approach ideal behaviour at (a) low temperature and low pressure (b) low temperature and high pressure (c) high temperature and low pressure (d) high temperature and high pressure 40. The compressibility of a gas is less than unity at STP. Therefore (a) Vm > 22.4 litres
(b) Vm
(c) Vm
(d) Vm
=22.4 litres
<22.4 litres =44.8 litres
41. An ideal gas will have maximum density when (a) P = 0.5 atm, T = 600 K (b) P = 2 atm, T = 150 K (c) P = 1 atm, T = 300 K (d) P = 1 atm, T = 500 K 42. 120 g of an ideal gas of molecular weight 40 g mol -1 are confined to a volume of 20 L at 400 K. Using R=0.0821 L atm -1
K-1 mol , the pressure of the gas is (a) 4.90 atm (b) 4.92 atm (c) 5.02 atm (d) 4.96 atm 43. 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 250 C. The pressure of water vapour at 250 C is (a) 18 mm Hg (b) 20 mm Hg (c) 22 mm Hg (d) 24 mm Hg. 44. Pressure of a mixture of 4 g of O2 and 2 g of H2 confined in a bulb of 1 litre at 00 C is (a) 25.215 atm (b) 31.205 atm (c) 45.215 atm (d) 15.210 atm
20
Ph ysi cs
45. Same mass of CH4 and H2 is taken in container, The partial pressure caused by H2 is (a)
8 9
(b)
1 9
(c)
1 2
(d)
1
46. There are 6.02 ´ 1022 molecules each of N2, O2 and H2
47.
48.
49.
50.
51.
52.
53.
which are mixed together at 760 mm and 273 K. The mass of the mixture in grams is (a) 6.02 (b) 4.12 (c) 3.09 (d) 6.2 The density of air is 0.00130 g/ml. The vapour density of air will be (a) 0.00065 (b) 0.65 (c) 14.4816 (c) 14.56 An ideal gas is one which obeys the gas laws under (a) a few selected experimental conditions (b) all experimental conditions (c) low pressure alone (d) high temperature alone The temperature at which real gases obey the ideal gas laws over a wide range of pressure is called (a) Critical temperature (b) Boyle’s Temperature (c) Inversion temperature (d) Reduced temperature Which of the following exhibits the weakest inter- molecular forces (a) NH3 (b) HCl (c) He (d) H2O The van der Waal’s equation reduces itself to the ideal gas equation at (a) high pressure and low temperature (b) low pressure and low temperature (c) low pressure and high temperature (d) high pressure alone An ideal gas obeying kinetic theory of gases can be liquefied if : (a) its temperature is more than critical temperature TC (b) its pressure is more than critical pressure PC (c) its pressure is more than PC at a temperature less than TC (d) it cannot be liquefied at any value of P and T The ratio of Boyle’s temperature and critical temperature for a gas is : (a)
8 27
(b)
27 8
(c)
1 2
(d)
2 1
54. The value of van der waals constant ‘a’ for gases O2, N2 , NH3 and CH4 are 1.360, 1.390, 4.170 and 2.253 liter 2 atm mol–2 respectively. The gas which can most easily be liquefied is : (a) O2 (b) N2 (c) NH3 (d) CH4
æ dT ö ÷ for an ideal gas is è dP ø H
55. Joule- Thomson coefficient ç
(a) zero (b) positive (c) negative (d) None of these 56. The inversion temperature (T i) for a gas is given by (a) a / Rb (b) 2a / Rb (c)
Rb a
(d)
2Rb a
57. The Joule-Thomson coefficient for a gas is zero at (a) Inversion temperature (b) Critical temperature (c) Absolute temperature (d) Below 0ºC 58. Longest mean free path under similar conditions of P and T stands for. (a) N2 (b) O2 (c) H2 (d) Cl2 59. The mean free path ( l ) of a gas sample is given by
l=
1
(a) l = 2ps 2 N
(b)
(c) l = 2pus 2 N
(d) None of these
2.ps2 N
60. Which of the following statements is false (a) Avogadro Number = 6.02 × 1021 (b) The relationship between average velocity (n ) and root mean square velocity (u) is (n ) =.9213u (c) The mean kinetic energy of an ideal gas is independent of the pressure of the gas (d) The root mean square velocity of the gas can be calculated by the formula (3RT / M)½ 61. At higher altitude the boiling point of water lowers because (a) atmospheric pressure is low (b) temperature is low (c) atmospheric pressure is high (d) None of these 62. Which of the following liquid will exhibit highest vapour pressure? (a) C2H5OH(l) (b) NH3 (l) (c) HF (l) (d) H2O(l)
Laws of Motion
21
63. A manifestation of surface tension is : (a) rise of liquid in a capillary tube (b) spherical shape of liquid drops (c) upward movement of water in soils (d) All the above 64. Generally , liquid drops assume spherical shape because: (a) a sphere has maximum surface area (b) a sphere has minimum surface area (c) sphere is symmetrical in shape (d) none of these 65. On heating a liquid, its surface tension (a) increases (b) decreases (c) remains same (d) is reduced to zero
66. Water drops stick to a glass surface due to: (a) cohesion (b) adhesion (c) flocculation (d) None of these 67. The internal resistance to flow in liquid is called (a) Fluidity (b) Specific resistance (c) Viscosity (d) Surface tension. 68. With rise in temperature, viscosity of a liquid (a) increase (b) decrease (c) remains constant (d) may increase or decrease. 69. With the increasing molecular weight of a liquid, the viscosity (a) decreases (b) increases (c) remain constant (d) All are wrong
1.
7.
2.
3.
4.
5.
6.
A weather ballon filled with hydrogen at 1 atm and 27°C has volume equal to 12000 litres. On ascending it reaches a place where the temperature is –23°C and pressure is 0.5 atm. The volume of the balloon is [CBSE-PMT 1991] (a) 24000 litres (b) 20000 litres (c) 10000 litres (d) 12000 litres If a gas expands at constant temperature, it indicates that : [CBSE-PMT 2008] (a) kinetic energy of molecules decreases (b) pressure of the gas increases (c) kinetic energy of molecules remains the same (d) number of the molecules of gas increases The pressure exerted by 6.0g of methane gas in a 0.03 m3 vessel at 129°C is (Atomic masses : C = 12.01, H = 1.01 and R = 8.314 kpa dm3K–1 mol –1) [CBSE-PMT 2010] (a) 31684 Pa (b) 215216 Pa (c) 13409 Pa (d) 41777 Pa By what factor does the average velocity of a gaseous molecule increase when the temperature (in Kelvin) is doubled ? [CBSE-PMT 2011] (a) 2.0 (b) 2.8 (c) 4.0 (d) 1.4 Two gases A and B having the same volume diffuse through a porous partition in 20 and 10 seconds respectively. The molecular mass of A is 49 u. Molecular mass of B will be : [CBSE-PMT 2011] (a) 50.00 u (b) 12.25 u (c) 6.50 u (d) 25.00 u A gaseous mixture was prepared by taking equal mole of CO and N2. If the total pressure of the mixture was found 1 atmosphere, the partial pressure of the nitrogen (N2) in the mixture is : [CBSE-PMT 2011] (a) 0.5 atm (b) 0.8 atm (c) 0.9 atm (d) 1 atm
8.
9.
10.
11.
12.
13.
A bubble of air is underwater at temperature 15°C and the pressure 1.5 bar. If the bubble rises to the surface where the temperature is 25°C and the pressure is 1.0 bar, what will happen to the volume of the bubble ? [CBSE-PMT 2011M] (a) Volume will become greater by a factor of 1.6. (b) Volume will become greater by a factor of 1.1. (c ) Volume will become smaller by a factor of 0.70. (d) Volume will become greater by a factor of 2.5. 50 mL of each gas A and of gas B takes 150 and 200 seconds respectively for effusing through a pin hole under the similar condition. If molecular mass of gas B is 36, the molecular mass of gas A will be : [CBSE-PMT 2012] (a) 96 (b) 128 (c) 20.25 (d) 64 A certain gas takes three times as long to effuse out as helium. Its molecular mass will be : [CBSE-PMT 2012] (a) 27 u (b) 36 u (c) 64 u (d) 9 u Maximum deviation from ideal gas is expected from : (a) N2(g) (b) CH4(g) [NEET 2013] (c) NH3 (g) (d) H2(g) For an ideal gas, number of moles per litre in terms of its presure P, gas constant R and temperature T is [AIEEE 2002] (a) PT/R (b) PRT (c) P/RT (d) RT/P Value of gas constant R is [AIEEE 2002] (a) 0.082 litre atm (b) 0.987 cal mol–1 K–1 (c) 8.3 J mol–1 K–1 (d) 83 erg mol–1 K–1. Kinetic theory of gases proves [AIEEE 2002] (a) only Boyle’s law (b) only Charles’ law (c) only Avogadro’s law (d) All of these.
22
Ph ysi cs
14. The heat required to raise the temperature of body by 1 K is called [AIEEE 2002] (a) specific heat (b) thermal capacity (c) water equivalent (d) none of these. 15. What volume of hydrogen gas, at 273 K and 1 atm pressure will be consumed in obtaining 21.6 g of elemental boron (atomic mass = 10.8) from the reduction of boron trichloride by hydrogen ? [AIEEE 2003] (a) 67.2 L (b) 44.8 L (c) 22.4 L (d) 89.6 L 16. According to the kinetic theory of gases, in an ideal gas, between two successive collisions a gas molecule travels (a) in a wavy path [AIEEE 2003] (b) in a straight line path (c) with an accelerated velocity (d) in a circular path 17. As the temperature is raised from 20ºC to 40ºC, the average kinetic energy of neon atoms changes by a factor of which of the following ? [AIEEE 2004] (a) 313 293
(b)
22.
23.
diffusion (rA / rB ) of two gases A and B, is given as : [AIEEE 2011 RS]
24.
(313 / 293)
(d) 2 (c) 1 2 18. In van der Waals equation of state of the gas law, the constant ‘b’ is a measure of [AIEEE 2004] (a) volume occupied by the molecules (b) intermolecular attraction (c) intermolecular repulsions (d) intermolecular collisions per unit volume 19. Which one of the following statements is NOT true about the effect of an increase in temperature on the distribution of molecular speeds in a gas? [AIEEE 2005] (a) The area under the distribution curve remains the same as under the lower temperature (b) The distribution becomes broader (c) The fraction of the molecules with the most probable speed increases (d) The most probable speed increases
If 10–4 dm3 of water is introduced into a 1.0 dm3 flask at 300 K, how many moles of water are in the vapour phase when equilibrium is established ? [AIEEE 2010] (Given : Vapour pressure of H2O at 300 K is 3170 Pa; R = 8.314 J K–1 mol–1) (a) 5.56× 10–3 mol (b) 1.53 × 10–2 mol (c) 4.46 × 10–2 mol (d) 1.27 × 10–3 mol When r, P and M represent rate of diffusion, pressure and molecular mass, respectively, then the ratio of the rates of
25.
(a)
( PA / PB ) ( M B / M A )1/ 2 (b) ( PA / PB )1/ 2 ( M B / M A )
(c)
( PA / PB ) ( M A / M B )1/ 2 (d) ( PA / PB )1/ 2 ( M A / M B )
The molecular velocity of any gas is [AIEEE 2011 RS] (a) inversely proportional to absolute temperature. (b) directly proportional to square of temperature. (c) directly proportional to square root of temperature. (d) inver sely proportion al to the squar e root of temperature. The compressibility factor for a real gas at high pressure is : [AIEEE 2012] (a) 1 +
(b)
21.
(c)
1 273 ´ 3 298
(d) 1/3.
(d) 1 –
pb RT
At constant volume and temperature conditions, the rate of diffusion DA and DB of gases A and B having densities rA and rB are related by the expression. [IIT-JEE 1993]
ùú =éêëD r rû ö =D æçèr r ÷ø
(a)
DA
B
(b)
DA
12
12
DA
B
A
28.
B
B
A
B
(c)
r ùú =éêëD r û rö =D æçèr ÷ø
12
12
A
B
VC ~ (d) -1 VS Equal masses of methane and oxygen are mixed in an empty container at 25°C. The fraction of the total pressure exerted by oxygen is [AIEEE 2007] (a) 1/2 (b) 2/3
pb RT
27.
VC ~ - 3 - 10 VS
VC ~ 23 - 10 (c) VS
(c) 1 +
For gaseous state, if most probable speed is denoted by C*, average speed by C and mean square speed by C, then for a large number of molecules the ratios of these speeds are : (a) C* : C : C = 1.225 : 1.128 : 1 [JEE Main 2013] (b) C* : C : C = 1.128 : 1.225 : 1 (c) C* : C : C = 1 : 1.128 : 1.225 (d) C* : C : C = 1 : 1.225 : 1.128
volume of a solute particle in a true solution VS , could be [AIEEE 2005] VC ~ 3 - 10 VS
(b) 1
26.
20. The volume of a colloidal particle, VC as compared to the
(a)
RT pb
(d)
DA
B
B
A
The compression factor (compressibility factor) for 1 mole of a van der Waal’s gas at 0°C and 100 atm pressure if found to be 0.5. Assuming that the volume of gas molecules is negligible, calculate the van der Waal’s constant 'a'. [IIT-JEE 2001] (a) 0.253 L2 mol–2 atm (b) 0.53 L2 mol–2 atm (c) 1.853 L2 mol–2 atm (d) 1.253 L2 mol–2 atm
Laws of Motion 29. The given graph represents the variation of Z PV ) versus P, for three real gases (compressibility factor = nRT A, B and C. Identify the only incorrect statement [IIT-JEE 2006]
C
A
[IIT-JEE 2009] (a) nb
PV(liter atm mol )
Z 2.
–1
B P (atm)
N2O4 is 20 % dissociated at 27°C and 760 torr. The density of the equilibrium mixture is (a) 3.1 g/l (b) 6.2 g/l (c) 12.4g/l (d) 18.6 g/l Helium atom is two times heavier than a hydrogen molecule at. 298K. The average KE of helium is (a) 2 times of H2 molecule (b) same as that of H2 molecule (c) 4 times that of hydrogen molecule (d)
3.
4.
an2
(c)
–
an2
24.6
(a) For the gas A, a = 0 and its dependence on P is linear at all pressure. (b) For the gas B, b = 0 and its dependence on P is linear at all pressure (c) For the gas C, which is typical real gas for which neither a nor b = 0. By knowing the minima and the point of intersection, with Z = 1, a and b can be calculated (d) At high pressure, the slope is positive for all real gases 30. The term that corrects for the attractive forces present in a real gas in the van der Waals equation is
1.
(b)
(d) – nb V2 V2 31. For one mole of a van der Waals gas when b = 0 and T = 300 K, the PV vs, 1/V plot is shown below. The value of the van der Waals constant a (atm. liter2 mol–2) is : [IIT-JEE 2012]
1
0
23
1 that of H2 molecule 2
The units of ‘a’ in van der Waals equation of state is (a) atm. litre mol–1 (b) atm. litre2 mol–2 2 2 (c) atm litre mol (d) atm. litre mol–2 A container contains certain gas of mass ‘m’ of high pressure. Some of the gas has been allowed to escape from the container and after some time the pressure of the gas becomes half and its absolute temperature 2/3 rd. The amount of the gas escaped is (a) 2/3 m (b) 1/2 m (c) 1/4 m (d) 1/6 m
23.1 21.6 20.1
0
(a) 1.0
5.
2.0 3.0 –1 1/V(mol liter )
(b) 4.5
(c) 1.5
(d) 3.0
The partial pressure of hydrogen in a flask containing 2.016 g of H2 and 96.0 g of O2 is (a) 1/8 of the total pressure (b) 1/6 of the total pressure (c) 1/4 of the total pressure (d) 2/3 of the total pressure
6.
At 27°C a gas was compressed to half of its volume. To what temperature it must be now heated so that it occupies just its original volume. The pressure remains constant (a) 54°C (c) 600°C
7.
Equal volumes of the gases which do not react together are confined in separate vessels. The pressure is 200 mm and 400 mm of Hg respectively. If the two gases are mixed together what will be the pressure of the resulting mixture (temperature remaining constant) (a) 400 mm
8.
(b) 327°C (d) 327 °C.
(b)
400 mm
(c) 300 mm (d) 200 mm A flask containing air (open to the atmosphere) is heated from 300 K to 500 K. The percentage of the air escaped into the atmosphere is (a) 16.6 (b) 40 (c) 60 (d) 20
24 9.
10.
11.
12.
13.
Ph ysi cs A sample of O2 gas is collected over water at 23°C at a barometer pressure of water at 751 mm Hg (Vapour pressure of water at 23°C is 21 mm Hg). The partial pressure of O2 gas in the sample collected is (a) 21 mm Hg (b) 751 mm Hg (c) 0.96 atm. (d) 1.02 atm. Which of the following is a false statement ? (a) Gases having same molecular masses diffuse at the same rate (b) 0.5 litre of nitrogen and 1 litre of helium will have the same number of the atoms at the same temperature and pressure (c) The value of molar gas constant does not vary with the nature of the gas (d) None is false The density of SO2 at STP is 2.06 kg m-3. Its density at 819°C and 2 atmosphere is (a) 2.86 kg m-3 (b) 1.43 kg m-3 -3 (c) 0.715 kg m (d) 4.2686 kg m-3 When helium is allowed to expand into vacuum, heating effect is observed. Its reason is that (a) helium is an ideal gas (b) helium is an inert gas (c) inversion temperature of helium is low (d) the boiling point of helium is the lowest amongst the element. What would be the vapour density of a gas 260 cm3 of which at 290 K and 100.40 KPa pressure weights 0.160g (a) 17.9 gL-1
(b) 14.2 gL-1
(c) 7.4 g cm -3
(d) None of these
14. Reducing the pressure from 1.0 atm to 0.5 atm would change the number of molecules in one mole of ammonia to (a) 25% of its initial value (b) 50% of its initial value (c) 75% of its initial value (d) None of the above 15. The ratio of diffusion of nitrogen at 25°C would be ......... times that of carbon dioxide at 75°C (a) 0.90 (b) 1.16 (c) 1.41 (d) 1.76 16. The average kinetic energy of 28 g CO at, 300 K is E kcal. The average kinetic energy of 2 g H2 at the same temperature would be..... kcal. (a) E (b) 14 E (c) 1/14 E (c) 28 E 17. The critical temperature of water is higher than that of O2 because the H2O molecule has (a) fewer electrons than O2 (b) two covalent bonds (c) V-shape (d) dipole moment.
18. A bubble of the gas released at the bottom of a lake increases to eight times the original volume when it reaches at the surface. Assuming that the atmospheric pressure is equivalent to pressure exerted by a column of water 10 m high, what is the depth of the lake (a) 80 m (b) 90 m (c) 10 m (d) 70 m 19. SO 2 and He are kept in a container at partial pressure P1 and P2 . A thin perforation is made in the wall of the container and it is observed that gases effuse at the same rate. The ratio of P1 and P2 will be (a) 4 : 1 (c) 1 : 16
(b) 1 : 4 (d) 16 : 1
20. Helium has the van der Waals constant b = 24 ml mol -1 . The molecular diameter of helium will be (a) 267 pm (b) 133.5 pm (c) 26.7 pm (d) Data not sufficient for calculation the diameter. 21. At 100°C and 1 atm, if the density of liquid water is 1.0 g cm–3 and that of water vapour is 0.0006 g cm–3, then the volume occupied by water molecules in 1 litre of steam at that temperature is (a) 6 cm3 (b) 60 cm3 3 (c) 0.6 cm (d) 0.06 cm3 22. Which of the following graphs is not a straight line for an ideal gas? (a) n ® T (b) T ® p (c) n ®
1 T
(d)
n®
1 p
23. A volume V of a gas at temperature T1 and a pressure p is enclosed in a sphere. It is connected to another sphere of volume V/2 by a tube and stopcock. The second sphere is initially evacuated and the stopock is closed. If the stopcock is opened the temperature of the gas in the second sphere becomes T2. The first sphere is maintained at a temperature T1. What is the final pressure p1 within the apparatus ? (a)
2pT2 2T2 + T1
(b)
2pT2 T2 + 2T1
(c)
pT2 2T2 + T1
(d)
2pT2 T1 + T2
24. The molecular velocities of two gases at the same temperature are u1 and u2 and their masses are m1 and m2 respectively. Which of the following expressions are correct ? (a)
(c)
m1
=
m2 u 22
(b)
m1u1 = m 2 u 2
m1 m 2 = u1 u2
(d)
m1u12 = m 2 u 22
u12
Laws of Motion 25. At low pressure, the van der Waal's equation is reduced to (a) Z =
pVm aP = 1RT RT
(b)
Z=
pVm b p = 1+ RT RT
pVm a = 1RT RT 26. The van der Waal's equation for n = 1 mol may be expressed
(c) pVm = RT
(d)
Z=
æ RT ö 2 aV ab as V 3 - ç b + ÷V + =0 ç p ÷ø p p è
Where V is the molar volume of the gas. Which of the following is correct? (a) For a temperature less than Tc, V has three real roots (b) For a temperature more than Tc, V has one real and two imaginary roots (c) For a temperature equal to Tc all three roots of V are real and identical (d) All of these 27. According to kinetic theory of gases, for a diatomic molecule (a) the pressure exerted by the gas is proportional to the mean velocity of the molecule. (b) the root mean square velocity of the molecule is inversely proportional to the temperature. (c) the pressure exerted by the gas is proportional to the rms velocity of the molecule. (d) the mean translational kinetic energy of the molecule is proportional to the absolute temperature. 28.
A perfect gas is found to obey the relation PV 3 / 2 = constant. If the gas is compressed to half of its volume at temperature T adiabatically, the final temperature of the gas will be (a)
2T 2
32.
33.
34.
35.
36.
(b) 4T
27°C. The cylinder can hold 2.82 L of water at NTP. The number of balloons that can be filled up is (a) 15 (b) 10 (c) 20 (d) 25 The mass of N 2 in a 15 L gaseous mixture at 20°C and 740 mm pressure of the composition of the mixture by volume is H 2 = 10%, O 2 = 20% and N 2 = 70% (a) 11.91 g (b) 16.2 g (c) 21.91 g (d) 28.00 g What will be the mole fraction of N 2 in a mixture of N 2 and O 2 if partial pressure of O 2 is 63 cm and total pressure of the mixture is 90 cm (a) 3.0 (b) 0.3 (c) 0.7 (d) 0.5 The bottles of NH 3 and HBr gases are connected through a tube of 1 metre length. The distance of white solid formed in the tube from the end of NH 3 (a) 68.56 cm from NH 3 bottle (b) 68.56 cm from HBr bottle (c) At the centre of the tube (d) None is correct. At a temperature TK the pressure of 4 g of argon in a bulb is p. The bulb is put in a bath having temperature higher by 50 K than the first one. 0.8 g of argon has to be removed to maintain original pressure. The temperature T is equal to (a) 510 K (b) 200 K (c) 73 K (d) 100 K Let the most probable velocity of hydrogen molecules at a temperature of t° C be V0 . When the temperature is raised to (2t + 273)°C the new rms velocity is (suppose all the molecules dissociate into atoms at latter temperature) (a)
(c) 29.
30.
(d) 2T T 2 A vessel containing gas at pressure 60 cm of Hg was connected to an arm A of open end manometer. The atmospheric pressure was recorded as 74 cm of Hg. If mercury in arm A stands at 84.5 cm height, the mercury in arm B will stand at (a) 24.5 cm (b) 70.5 cm (c) 88 cm (d) 74 cm
V1 + V2 + V3
(b)
V1 + V2 + V3 3
V1 + V2 + V3 (d) V1 ´ V2 ´ V3 P A spherical balloon of 21 cm diameter is to be filled with
(c)
H 2 at NTP from a cylinder containing the gas at 20 atm at
2 3V0 273 ö æ 3ç 2 + ÷V0 t ø è
38.
(b)
6V0
(d)
2 V0 3
A vessel is filled with a mixture of O 2 and N 2 . At what ratio of partial pressures will be the mass of gases be identical (a) p( O ) = 8.75 p( N ) 2
Three gases A, B, C have volumes V1 , V2 , V3 at pressure P and temperature T are mixed keeping the temperature and pressure constant. The final volume of the gaseous mixture will be (a)
31.
(c) 37.
25
2
(b)
p( O ) = 0.78 p( N ) 2 2
(c)
p( O ) = 0.875 p ( N ) 2 2
(d) p( O 2 ) = 11.4 p( N 2 ) A compound exists in the gaseous state both as monomer A and dimer A 2 . The M. wt. of monomer is 48. In an experiment 96 g of the compound was confined in vessel of 33.6 L and heated to 273° C. Calculate the pressure developed, if the compound exists as a dimer to the extent of 50% by weight under the conditions (a) 0.9 atm (b) 4.0 atm (c) 2.0 atm (d) 1.0 atm
26 39.
Ph ysi cs A mixture of 1 mol of H 2 and 1 mole of Cl 2 with little charcoal in a 10 L evacuated flask was irriadiated with light until the reaction was completed. Subsequently 5L of water was introduced into the flask and the flask was cooled to 27°C. The pressure exerted by the system is approximately equal to (a) (b) (c)
2 ´ 0.0821 atm 10
4 ´ 0.0821´ 300 atm 5 The relative humidity of air is 80% at 27°C. If the aqueous tension at the same temperature is 27 mm Hg. The partial pressure of water vapour in the air will be (a) 21.60 mm Hg (b) 27 mm Hg (c) 25 mm Hg (d) 23 mm Hg
(d)
40.
2 ´ 0.0821´ 300 atm 5 » 26 mm Hg
41.
A closed vessel contains He and Ozone at pressure of P atm. The ratio of He and oxygen atoms is 1 : 1. If He is removed from the vessel, the pressure of the system will reduce to (a) 0.25 P (b) 0.5 P (c) 0.75 P (d) 0.33 P 42. The factor that has the largest effect on vapour pressure of a liquid is (a) liquid surface area (b) molecular dipole moment (c) presence of H-bonding (d) molecular mass of liquid 43. The molecular mass of a compound does not effected by (a) vapour pressure of a liquid (b) vapour density (c) vapour pressure of solid (d) molar volume of vapour 44. Which of the following liquids has the highest viscosity? (a) Benzene (b) Carbon disulphide (c) Acetone (d) Ethanol
