INTRODUCTION

Matter can be classified into three categories depending upon its physical state namely solid, liquid and gaseous states. Solids have a definite volume and shape; liquids also have a definite volume but no definite shape; gases have neither a definite volume nor a definite shape.

DISTINCTION BETWEEN THREE STATES OF MATTER

Sl. No. Solids Liquids Gases1. Particles are very Closely

packedParticles are loosely packed

Particles are very loosely packed

2. Voids are extremely small Voids are relatively larger Voids are very large3. Inter particle forces are

largeInter particle forces are intermediate

Intermediate forces are negligible

4. Particle motion is restricted to vibratory motion.

Particle motion is very slow Particle motion is very rapid and also random.

MEASURABLE PROPERTIES OF GASES

Mass, volume, temperature are the important measurable properties of gases. Mass: The mass of the gas is related to the number of moles as

n =

Where n = number of molesw = mass of gas in gramsM = molecular mass of the gas

Volume: Since gases occupy the entire space available to them, therefore the gas volume means the volume of the container in which the gas is enclosed.Units of Volume: Volume is generally expressed in litre or cm3 or dm3 1m3 = 103 litre = 103 dm3 = 106 cm3.

Pressure: The force exerted by the gas per unit area on the walls of the container is equal to its pressure.Units of Pressure: The pressure of a gas is expressed in atm, Pa, Nm–2, bar or, lb/In2 (psi).760 mm = 1 atm = 10132.5 KPa = 101325 Pa = 101325 Nm–2 760 mm of Hg = 1.01325 bar = 1013.25 milli bar = 14.7 lb/2n2 (psi)

Temperature: Temperature is defined as the degree of hotness. The SI unit of temperature is Kelvin. On the Celsius scale water freezes at 0°C and boils at 100°C where as in the Kelvin scale water freezes at 273 K and boils at 373 K.

GAS LAWS

The state of a sample of gas is defined by 4 variables i.e. P, V, n & T. Gas laws are the simple relationships between any two of these variables when the other two are kept constant.

Boyle’s Law: The changes in the volume of a gas by varying pressure at a constant temperature of a fixed amount of gas was quantified by Robert Boyle in 1662. The law was named after his name as Boyle’s law. It states that:

The volume of a given mass of a gas is inversely proportional to its pressure at a constant temperature.Mathematically

P (n, T constant)

V (n, T constant)

i.e. P = (where K is the constant proportionality)

or PV = K (constant)Let V1 be the volume of a given mass of the gas having pressure P1 at temperature T. Now, if the pressure is changed to P2 at the same temperature, let the volume changes to V2. The quantitative relationship between the four variables P1, V1, P2 and V2 is:

P1V1 = P2V2 (temperature and mass constant)

Graphical Representation of Boyle’s Law

Fig. (a) show the plot of V vs P at a particular temperature. It shows that P increases V decreases.

Plot (b) shows the plot of PV vs P at particular temperature. It indicates that PV value remains constant inspite of regular increase in P.

Illustration 1. A sample of gas occupies 100 litres at 1 atm pressure and at 0°C. If the volume of the gas is to be reduced to 5 litres at the same temperature, what additional pressure must be applied?

Solution: Here P1 = 1 atm P2 = ?V1 = 100 litre V2 = 5 litreT1 = 273 K T2 = 273 KAs temperature is constantThen from Boyle’s lawP1V1 = P2V2

or 1 100 = P2 5 P2 = 20 atm Additional pressure that should be applied = P2 – P1 = 20 – 1 = 19 atm

Exercise 1.A sample of hydrogen gas occupies 10 litre at 190 mm presence. What would be the volume of this sample at 1 atm pressure, if the temperature is kept constant?

Exercise 2.5g of ethane is confined in a bulb of one litre capacity. The bulb is so weak that it will burst if the pressure exceeds 10 atm. At what temperature will the pressure of the gas reach the bursting value?

Charle’s Law: The French Scientist, Jacques Charles in 1787 found that for a fixed amount of a gas at constant pressure, the gas expands as temperature increases. The law can be stated as the volume of a given mass of a gas increases or, decrease by 1/273 of its volume at 0°C for each degree rise or, fall of temperature, provided pressure is kept constant.Charles also found that for a given mass of a gas if pressure is kept constant, the volume increases linearly with temperature.

V = V0 (1 + t) or V – V0 = V0 tIf the temperature is measured in the Celsius scale and V0 is the volume at 0°C, it is found that = 1/273. The volume at temperature T is then;

Vt = V0 = V0 VT =

Where T = 273 + t is the temperature on the Kelvin scale, which has – 273°C as its zero point.Let V1 be the volume of a certain mass of a gas at temperature T1 and at pressure P. If temperature is changed to T2 keeping pressure constant, the volume changes to V2. The relationship between for variables V1, T1, V2 and T2 is:

(Pressure and Mass Constant)

Graphical Representation of Charle’s Law

COMBINED GAS EQUATION

The Boyle’s and Charles’ law can be combined to give a relationship between the three variables P, V and T. Let a certain amount of a gas in a vessel have a volume V1, pressure P1 and temperature T1. On changing the temperature and pressure to T2 and P2 respectively, the gas occupies a volume V2.Then we can write

The above relation is called the combined gas law.

Avogadro’s Law: The Avogadro’s law states that at a given temperature and pressure, the volume of a gas is directly proportional to the amount of gas i.e.

V n (P and T constant)or V = constant n

Where n is the amount of the substanceIt was said that 1 mol of any gas at 0°C and under 1 atm pressure occupies 22.4 10–3 m3 or 22.4 litre.Avogadro further generalised the statement that a mole of any substance contains 6.022 1023

particles (molecules, atoms or any other entities).

IDEAL GAS EQUATION

A gas that would obey Boyle’s and Charle’s law under the conditions of temperature and pressure is called an ideal gas.Here, we combine four measurable variables P, V, T and n to give a single equation.V n [P, T constant] Avogadro’s lawV T [n, P constant] Charle’s law

V [n, T constant] Boyle’s law

The combined gas law can be written as or PV nT

PV = nRT this is called ideal gas equationwhere R is the constant of proportionality or universal gas constant The value of R was found out to be R = 8.314 J mol–1 K–1

R = 0.0821 litre atm K–1 mol–1

R = 2 cal K–1 mol–1

Relation between molar mass and density (From ideal gas equation)

Illustration 2. What is the increase in volume when the temperature of 600 ml of air increases from 27°C to 47°C under constant pressure?

Solution: Charle’s law is applicable as the pressure and amount remains constant.

or V1 =

V1 = = 640 ml

Increase in volume of air = 640 – 600 = 40

Illustration 3. At which of the four conditions, the density of nitrogen will be the largest?(A) STP (B) 273 K and 2 atm(C) 546 K and 1 atm (D) 546 K and 2 atm

Solution: Density of a gas is given = . Obviously the choice that has greater would

have greater density.Hence, (B) is correct.

Exercise 3.An open bulb containing air at 19°C was cooled to a certain temperature at which the no. of moles of gaseous molecules increased by 25%. What is the final temperature?

Exercise 4.A certain mixture of helium and argon weighing 5 g occupies a volume of 10 L at and 1 atm pressure. What is the composition of the mixture in mass percentage?

Gay Lussac’s Law (temperature pressure law)

It state that pressure of the given of mass a gas is directly proportional to the Kelvin temperature at constant volume

(n.v. are constant)

PAY-LOAD

When a balloon is buoyant, it can take along some weight into the upper atmosphere. The maximum weight a balloon can carry along is called its pay-load.

Pay load = weight of the air displaced (weight of the balloon + weight of the gas it contains)Illustration 4. A flask of 2 dm3 capacity contains O2 at 101.325 kPa and 300 K. The gas

pressure is reduced to 0.1 Pa. Assuming ideal behaviour, answer the following:(i) What will be the volume of the gas which is left behind?(ii) What amount of O2 and the corresponding number of molecules are left

behind in the flask?(iii) If now 2g of N2 is introduced, what will be the pressure of the flask?

Solution: Given thatV1 = 2dm3 Pt = 101.325 kPaP2 = 0.1 Pa T = 300 KWe have the following results.(i) The volume of O2 left behind will be the same, i.e. 2dm3

(ii) The amount of O2 left behind is given by

n = P2

(iii) 2g of N2 = 1/14 mol

Total amount of gases in flask, 1/ 14 mol + 8.125 10–8 mol 1/14 mol. Thus, the pressure of the flask is given by

P = =

= 89.08 kPa.

Illustration 5. When 2 g of gaseous substance A is introduced into an initially evacuated flask kept at the pressure is found to be 1 atm. The flask is evacuated and 3 g of B is introduced. The pressure is found to be 0.5 atm at Calculate the

ratio (M – Molecular weight).

Solution:

(because V, R and T are constants)

Illustration 6. A balloon of diameter 20 m weights 100 kg. Calculate its pay – load, if it is filled with He at 1 atm and Density of air is 1.2 kg

Solution: Weight of balloon = 100 kg

Volume of balloon

Weight of gas (He) in balloon

Total weight of gas and balloon

Weight of air displaced

Payload = weight of air displaced – (weight of balloon + weight of gas)Pay load

DALTON’S LAW OF PARTIAL PRESSURE

The relation between the pressures of the mixture of nonreacting gases enclosed in a vessel to their individual pressure is described in the law. The law was given by John Dalton in 1807. It states that. At constant temperature, the pressure exerted by a mixture of two or more nonreacting gases enclosed in a definite volume, is equal to the sum of the individual pressures which each gas would exert if present alone in the same volume.The individual pressures of gases are known as partial pressures.

If P is the total pressure of the mixture of nonreacting gases at temperature T and volume V, and P1, P2, P3 …. represent the partial pressures of the gases, thenP = P1 + P2 + P3+ ……… (T, V are constant)

Partial Pressure in terms of Mole Fraction

Mole fraction defines the amount of a substance in a mixture as a fraction of total amount of all substances. If n=1 moles of way substance is present in n moles of the mixture, then mole

fraction of the substance, 1 =

If is the partial pressure of nitrogen the mixture of SO2 and N2. Then

( = no. of moles of N2)

and Pmixture =

dividing we get,

or,

Illustration 7. A 2.5 litre flask contains 0.25 mole each of sulphur dioxide and nitrogen gas at 27°C. Calculate the partial pressure exerted by each gas and also the total pressure.

Solution: Partial pressure of SO2

= 2.49 105 Nm–2 = 2.49 105 Pa

Similarly = 2.49 105 Pa

Following Dalton’s LawPTotal

= +

= 2.49 105 Pa + 2.49 105 Pa = 4.98 105 Pa

Illustration 8. At an under water depth of 250 ft; the pressure is 8.38 atm. What should be the mole percent of oxygen in the diving gas for the partial pressure of oxygen in the mixture to be 0.21 atm, the same as it is in air at 1 atm?

Solution: Since

So, the diving gas should contain 2.5% of

Exercise 5.A gaseous mixture containing 8g of O2 and 224 ml ofN2 at STP is enclosed in a flask of 5 litre capacity of 0°C. Find the partial pressure of each gas and also calculate the total pressure in the vessel.

GRAHAM’S LAW OF DIFFUSION/EFFUSION

The ability of a gas to spread and occupy the whole available volume irrespective of other gases present in the container is called diffusion.Effusion is the process by which a gas escapes from one chamber of a vessel through a small

opening or an orifice.

Thomas Graham in 1831 proposed the law of gaseous diffusion. The law states under similar conditions of temperature and pressure, the rates of diffusion of gases are inversely proportional to the square roots of their densities.

where is the rate of diffusion and d is the density of the gas.

Now, if there are two gases A and B having r1 and r2 as their rates of diffusion and d1 and d2 their densities respectively. Then

and or, (at same T and P)

Here M1 and M2 are the molecular masses of the gases having densities d1 and d2 respectively.Graham’s law of diffusion also holds good for effusion.

Effect of pressure on state of diffusion

The rate of diffusion (r) of a gas at constant temperature is directly preoperational to its pressure

Illustration 9. Which of the two gases ammonia and hydrogen chloride will diffuse faster and by what factor?

Solution:

= = 1.46 or = 1.46 rHCl

Thus ammonia will diffuse 1.46 times faster than hydrogen chloride gas

Illustration 10. A certain gas diffuses from two different vessels A and B. The vessel A has a circular orifice while vessel B has a square orifice of length equal to the radius of the orifice of vessel A. The ratio of the rates of diffusion of the gas from vessel A to vessel B, assuming same temperature and pressure is;(A) (B) 1/(C) 1:1 (D) 2:1

Solution: The rate of diffusion is directly proportional to the area of orifice. dA r2 dB r2

=

Hence, (A) is correct.

Exercise 6.A gaseous mixture of O2 and X containing 20% of X diffused through a small hole in 134 secs while pure O2 takes 124 secs to diffuse through the same hole. Find molecular weight of X?

Exercise 7.A straight glass tube of 200 cm length has two inlets X and Y at the two ends. HCl gas through inlet X and NH3 gas through inlet Y are allowed to enter in the tube at the same time and same pressure. White fumes first appear at a point P inside the tube. Calculate the distance of P from X.

Exercise 8. For 10 minute each, at from two identical holes nitrogen and an unknown gas are leaked into a common vessel of 3 L capacity. The resulting pressure is 4.18 bar and the mixture contains 0.4 mole of nitrogen. What is the molar mass of unknown gas?

KINETIC MOLECULAR THEORY OF GASES

The various gas laws which we have studied so far were based an experimental facts.

The theory that provides an explanation for the various experimental observations about a gas is

based on the Kinetic Molecular model. It is assumed that all gases are made up of identical

molecules, which are in a state of constant random motion.

Postulates of the Model

The following are the postulates of the model:

A gas consists of a large number of identical molecules of mass m. The dimensions of

these molecules are very small compared to the space between them. Hence the

molecules are treated as point masses.

There are practically no attractive forces between the molecules. The molecules

therefore more independently.

The molecules are in a state of ceaseless and random motion, colliding with each other

and with the walls of the container. The direction of their motions changes only on

collision. These collisions are known as elastic collisions in which the energy and

momenta of the molecules are concerned. In nonelastic collisions these quantities are

not conserved.

The pressure of a gas is the result of collision of molecules with the wall of the container.

The average kinetic energy of the colliding molecules is directly proportional to its

temperature.

Velocities of gas molecules

Average Velocity

As per kinetic theory of gases, each molecule is moving with altogether different velocity. Let ‘n’ molecules be present in a given mass of gas, each one moving with velocity v 1, v2, v3… vn. The average velocity or UaV = average of all such velocity terms.

Average velocity = =

Uav =

Root Mean Square Velocity

Maxwell proposed the term Urms as the square root of means of square of all such velocities.

= =

Also Urms

Most probable velocity

It is the velocity which is possessed by maximum no. of molecules.

Vmp =

Furthermore UMP : UAV : Urms : : : :

: :

1 : 1.128 : 1.224Also Uav = Urms 0.9213

Exercise 9.The average speed at T1k and the most probable speed at T2k of CO2 gas is 9 104 cm sec–1. Calculate the value of T1 and T2.

Exercise 10.Calculate the rms speed of ozone kept in a closed vessel at 20°C and 82 cm Hg pressure.

Exercise 11.Calculate total energy of one mole of an ideal monatomic gas at 27C?

Exercise 12.Calculate Vrms of O2 at 273K and 1 105 Pa pressure. The density of O2 under these conditions is 1.42 kg/m3?

Exercise 13.Calculate the average and total kinetic energies of 0.6 mol of an ideal gas at 0°C.

Kinetic Energy of Gas

As per kinetic equation PV =

For 1 mole m n = Molecular Mass (M)

PV = or .

Also = kT

Where k is the Boltzmann constant

Illustration 11. Calculate rms speed of O2 at 273 K and 1 105Pa pressure. The density of O2

under these conditions is 1.42 kg m–3.

Solution: Data are given in SI units

= 459.63 m sec–1

Illustration 12. At what temperature will the r.m.s. velocity of oxygen be one and half times of its value at N.T.P.?

Solution:

Suppose the temperature required is T when the velocity will be C

or, T = 273 = 614.25°K

Kinetic Gas Equation

The kinetic gas equation has been derived on the basis of postulates of the kinetic theory of gases. The equation is:

PV =

Where, P = PressureV = Volumem = Mass of gas moleculesN = No. of molecules in Volume Vu = Root mean square velocity (rms)

Kinetic Energy of Gas Molecules

The kinetic energy of gas molecules can be easily calculated on the basis of kinetic gas equation.

Average K.E. of one molecule = mu2

K.E. of N molecules = mNu2 =

= [ and PV = RT fare 1 mole of gas]

R is constant KE of a gas T

Illustration 13. Calculate K.E. of 4g of N2 at – 13°C.

Solution: EK = =

= 463.21 J mol–1

Exercise 14.At what temperature rms of SO2 is equal to that of CH4 at 27°C

DEVIATION FROM IDEAL GAS BEHAVIOUR

A gas which obeys the gas laws and the gas equation PV = nRT strictly at all temperatures and pressures is said to be an ideal gas. The molecules of ideal gases are assumed to be volume less points with no attractive forces between one another. But no real gas strictly obeys the gas equation at all temperatures and pressures. Deviations from ideal behaviour are observed particularly at high pressures or low temperatures. The deviation from ideal behaviour is expressed by introducing a factor Z known as compressibility factor in the ideal gas equation. Z

may be expressed as

In case of ideal gas, PV = nRT Z = 1 In case of real gas, PV nRt Z 1

Thus in case of real gases Z can be 1 or 1(i) When Z 1, it is a negative deviation. It shows that the gas is more compressible

than expected from ideal behaviour.(ii) When Z 1, it is a positive deviation. It shows that the gas is less compressible than

expected from ideal behaviour.

Causes of deviation from ideal behaviour

The causes of deviations from ideal behaviour may be due to the following two assumptions of kinetic theory of gases. There are

The volume occupied by gas molecules is negligibly small as compared to the volume occupied by the gas.

The forces of attraction between gas molecules are negligible.

The first assumption is valid only at low pressures and high temperature, when the volume occupied by the gas molecules is negligible as compared to the total volume of the gas. But at low temperature or at high pressure, the molecules being in compressible the volumes of molecules are no more negligible as compared to the total volume of the gas.The second assumption is not valid when the pressure is high and temperature is low. But at high pressure or low temperature when the total volume of gas is small, the forces of attraction become appreciable and cannot be ignored.

Van Der Waal’s Equation

The general gas equation PV = nRT is valid for ideal gases only Van der Waal is 1873 modified the gas equation by introducing two correction terms, are for volume and the other for pressure to make the equation applicable to real gases as well.

Volume correction

Let the correction term be v

Ideal volume vi = (V – v)Now v n or v = nb[n = no. of moles of real gas; b = constant of proportionality called Van der Waal’s constant] Vi = V – nbb = 4 volume of a single molecule.

Pressure Correction

Let the correction term be P Ideal pressure Pi = (P + p)

Now, =

Where a is constant of proportionality called another Van der Waal’s constant.Hence ideal pressure

Pi =

Here, n = Number of moles of real gasV = Volume of the gas a = A constant whose value depends upon the nature of the gas

Substituting the values of ideal volume and ideal pressure, the modified equation is obtained as

Illustration 14. 1 mole of SO2 occupies a volume of 350 ml at 300K and 50 atm pressure. Calculate the compressibility factor of the gas.

Solution: P = 50 atm V = 350 ml = 0.350 litren = 1 mole

T = 300L Z =

Z =

Thus SO2 is more compressible than expected from ideal behaviour.

Exercise 15.Out of NH3 and N2 which will have (a) Larger value of a(b) Larger value of b

Exercise 16.2 moles of NH3 occupied a volume of 5 litres at 27°C. Calculate the pressure if the gas obeyed Vander Waals equation. Given a = 4.17 litre2 atm mole–2, b = 0.037 litre/mole.

Exercise 17.Calculate the percentage of free volume available in 1 mole gaseous water at 1 atm and Density of liquid is 0.958g / mL.

Vander Waals equation, different forms At low pressures: ‘V’ is large and ‘b’ is

negligible in comparison with V. The Vander Waals equation reduces to:

; PV + = RT

PV = RT - or PV RT

This accounts for the dip in PV vs P isotherm at low pressures.

Deviation of gases from ideal behaviour with pressure.

At fairly high pressures

may be neglected in comparison with P.

The Vander Waals equation becomes P (V – b) = RTPV – Pb = RTPV = RT + Pb or PV RTThis accounts for the rising parts of the PV vs P isotherm at high pressures.

The plot of Z vs P for N2 gas at different temperature is shown here.

At very low pressures: V becomes so large that both b and become negligible and

the Vander Waals equation reduces to PV = RT. This shows why gases approach ideal behaviour at very low pressures.

Hydrogen and Helium: These are two lightest gases known. Their molecules have very small masses. The attractive forces between such molecules will be extensively small.

So is negligible even at ordinary temperatures. Thus PV RT. Thus Vander Waals

equation explains quantitatively the observed behaviour of real gases and so is an improvement over the ideal gas equation.Vander Waals equation accounts for the behaviour of real gases. At low pressures, the gas equation can be written as,

or Z =

Where Z is known as compressibility factor. Its value at low pressure is less than 1 and it decreases with increase of P. For a given value of Vm, Z has more value at higher temperature.At high pressures, the gas equation can be written as

P (Vm – b) = RT

Z = = 1 +

Here, the compressibility factor increases with increase of pressure at constant temperature and it decreases with increase of temperature at constant pressure. For the gases H2 and He, the above behaviour is observed even at low pressures, since for these gases, the value of ‘a’ is extremely small.

Illustration 15. One litre of a gas at 300 atm and 473 K is compressed to a pressure of 600 atm and 273 K. The compressibility factors found are 1.072 & 1.375 respectively at initial and final states. Calculate the final volume.

Solution: P1V1 = Z1nRT1 and P2V2 = Z2nRT2

or V2 = = = 370.1 ml

Illustration 16. The behaviour of a real gas is usually depicted by plotting compressibility factor Z versus P at a constant temperature. At high temperature and high pressure, Z is usually more than one. This fact can be explained by van der Waal’s equation when (A) the constant ‘a’ is negligible and not ‘b’(B) the constant ‘b’ is negligible and not ‘a’ (C) both constants ‘a’ and ‘b’ are negligible(D) both the constants ‘a’ and ‘b’ are not negligible.

Solution: (V – nb) = nRT

At low pressures, ‘b’ can be ignored as the volume of the gas is very high. At high temperatures ‘a’ can be ignored as the pressure of the gas is high. P (V–b) = RTPV - Pb = RT PV = RT + Pb

Hence, (A) is correct.

Exercise 18. The compressibility factor for CO2 at 273K and 100atm pressure is 0.2005. Calculate the volume occupied by 0.2 mole of CO2 gas at 100 atm assuming ideal behaviour

Some other important definitions

Relative Humidity (RH)

At a given temperature it is given by equation

RH =

Boyle’s Temperature (Tb)

Temperature at which real gas obeys the gas laws over a wide range of pressure is called Boyle’s Temperature. Gases which are easily liquefied have a high Boyle’s temperature [Tb(O2)] = 46 K] whereas the gases which are difficult to liquefy have a low Boyle’s temperature [T b(He) = 26K].

Boyle’s temperature Tb =

where Ti is called Inversion Temperature and a, b are called van der Waals constant.

Critical Constants

Critical Temperature (Tc): It (Tc) is the maximum temperature at which a gas can be liquefied i.e. the temperature above which a gas can’t exist as liquid.

Critical Pressure (Pc): It is the minimum pressure required to cause liquefaction at Tc

Critical Volume: It is the volume occupied by one mol of a gas at Tc and Pc

Vc = 3bMolar heat capacity of ideal gases: Specific heat c, of a substance is defined as the amount of heat required to raise the temperature of is defined as the amount of heat required to raise the temperature of 1 g of substance through 10C, the unit of specific heat is calorie g-1 K-1. (1 cal is defined as the amount of heat required to raise the temperature of 1 g of water through 10C)Molar heat capacity C, is defined as the amount of heat required to raise the temperature of 1 mole of a gas trough 10C. Thus,Molar heat capacity = Sp. Heat molecular wt. Of the gas For gases there are two values of molar heats, i.e., molar heat at constant pressure and molar heat at constant molar heat at constant volume respectively denoted by Cp and Cv. Cp is greater than Cv and Cp-R = 2 cal mol-1 K-1.From the ratio of Cp and Cv, we get the idea of atomicity of gas.For monatomic gas Cp = 5 cal and Cv =3 cal

for diatomic gas Cp = 7 cal and Cv = 5 cal

For polyatomic gas Cp = 8 cal and Cv= cal

also Cp = Cp m, where, Cp and Cv are specific heat and m, is molecular weight.

Illustration 17. Calculate Vander Waals constants for ethylene TC = 282.8 k; PC = 50 atm

Solution: b = = 0.057 litres/mole

a = = 4.47 lit2 atm mole–2

Gas Eudiometry

The relationship amongst gases, when they react with one another, is governed by two laws, namely Gay-Lussac law and Avogadro’s law.Gaseous reactions for investigation purposes are studied in a closed graduated tube open at one end and the other closed end of which is provided with platinum terminals for the passage of electricity through the mixture of gases. Such a tube is known as Eudiometer tube and hence the name Eudiometry also used for Gas analysis.During Gas analysis, the Eudiometer tube filled with mercury is inverted over a trough containing mercury. A known volume of the gas or gaseous mixture to be studied is next introduced, which displaces an equivalent amount of mercury. Next a known excess of oxygen is introduced and the electric spark is passed, whereby the combustible material gets oxidised. The volumes of carbon dioxide, water vapour or other gaseous products of combustion are next determined by absorbing them in suitable reagents. For example, the volume of CO2 is determined by absorption in KOH solution and that of excess of oxygen in an alkaline solution of pyrogallol. Water vapour produced during the reaction can be determined by noting contraction in volume caused due to cooling, as by cooling the steam formed during combustion forms liquid (water) which occupies a negligible volume as compared to the volumes of the gases considered. The excess of oxygen left after the combustion is also determined by difference if other gases formed during combustion have already been determined. From the data thus collected a number of useful conclusions regarding reactions amongst gases can be drawn.

Volume-volume relationship amongst Gases or simple Gaseous reactions. Composition of Gaseous mixtures. Molecular formulae of Gases. Molecular formulae of Gaseous Hydrocarbons.

The various reagents used for absorbing different gases are O3 turpentine oilO2 alkaline pyrogallolNO FeSO4 solutionCO2,SO2 alkali solution (NaOH, KOH, Ca(OH)2, HOCH2CH2NH2, etc.)NH3 acid solution or CuSO4 solutionCl2 waterEquation for combustion of hydrocarbons.

CxHy + O2 xCO2 + H2O

General Assumptions: In all problems, it is assumed that the sparking occurs at room temperature. This implies that water formed would be in liquid state and that nitrogen gas is inert towards oxidation.

Illustration 18. A gaseous hydrocarbon requires 6 times its own volume of O2 for complete oxidation and produces 4 times its volume of CO2. What is its formula?

Solution: The balanced equation for combustion

CxHy + O2 xCO2 + H2O

1 volume volume = 6 (by equation)

or 4x + y = 24 …(1)Again x = 4 since evolved CO2 is 4 times that of hydrocarbon 16 + y = 24 or y = 8 formula of hydrocarbon C4H8

Exercise 19. 50 ml. of a mixture of NH3 and H2 was completely decomposed by sparking into nitrogen and hydrogen. 40 ml of oxygen was then added and the mixture was sparked again. After cooling to the room temperature the mixture was shaken with alkaline pyrogallol and a contraction of 6 ml. was observed. Calculate the % of NH3 in the original mixture. (Assuming that nitrogen does not react with oxygen.)

Exercise 20.When a certain quantity of oxygen was ozonised in suitable apparatus, the volume decreases by 4 ml. On addition of turpentine oil, the volume decreases further by 8 ml. Find the formula of ozone, assuming all measurements are made at same temperature and pressure.

Liquid State

The liquid molecules are relatively close together. The intermolecular forces of attraction in case of liquids are much larger than in gases. Unlike gases, liquids have a definite volume although no definite shape (similarity with

gases). The molecules are in constant random motion. The average kinetic energy of molecules in a given sample is proportional to the absolute

temperature. Guldberg’s rule: Normal boiling point of the liquid is nearly two – third of its critical

temperature when both are expressed on the absolute scale.

Trouton’s rule : The molar heat of vapourisation of a liquid expressed in Joules divided by normal b.p. of the liquid on the absolute scale is approximately equal to 88.

Evaporation

Evaporation is the spontaneous change in which a liquid changes into vapours at the surface of liquid. Evaporation occurs at all temperatures. Evaporation increases with increase in surface area, increase in temperature and decrease in intermolecular attractive forces. In contrast to evaporation, boiling takes place at a definite temperature and it involves bubble formation below the surface. Evaporation produces cooling.

Vapour Pressure

Vapour pressure of a liquid at any temperature may be defined as the pressure exerted by the vapour present above the liquid in equilibrium with the liquid at that temperature. The magnitude of vapour pressure depends upon the nature of liquid and temperature.

Non – polar or less polar liquids have fairly high vapour pressure due to weak forces of attraction (e.g. ether etc). Polar liquids (e.g. water, alcohols, etc.) have lower vapour pressure because of strong dipole – dipole interaction between their molecules.

Vapour pressure of a liquid is constant at a given temperature. Further the vapour pressure of a liquid increases with increase in temperature. When the vapour pressure of the liquid is equal to the external pressure (normal pressure or 1 atm pressure) acting upon the surface of the liquid, the bubbles increase in size and escape freely; the temperature at which this happens is called the boiling point of the liquid. In case the external pressure is more than the atmospheric pressure, more heat will be required to make the vapour pressure equal to the external pressure and hence higher will be the boiling point. In case, the external pressure is low (as on the top of a mountain), the boiling point of the liquid decreases. This explains why a liquid boils at a lower temperature on the top of a mountain. (where pressure in low) than on the sea shore.

Substances having high vapour pressure (e.g. petrol) evaporate more quickly than substances of low vapour pressure. (e.g. motor oil)

SURFACE TENSION

Surface Tension of a liquid is defined as the force acting at right angles to the surface along one centimeter length of the surface. It is represented by the greek letter gamma, Due to surface tension molecules tend to leave the surface, i.e. the surface of the liquid tends to contract to the smallest possible area for a given volume of the liquid. Further for a given volume of the liquid, sphere has the minimum surface area. This explains why the drops of a liquid are spherical.Thus it is apparent that in order to increase its surface area, force must be exerted to overcome the surface tension. In other words, work will have to be done to increase the surface area. Thus the surface tension of a liquid is defined as the work (energy) required to expand the surface of a liquid by unit area. Mathematically,

Thus surface tension of a liquid may also be defined as the force in dynes necessary to rupture its surface along one centimeter length. In SI units it is defined as the force in newtons

required to rupture 1 meter length of the surface of a liquid.Thus the units of surface tension are dynes per cm (or Newtons per metre, in SI system).Surface tension of a liquid is measured with the help of apparatus called stalgmometer.The surface tension of a liquid decreases with increase of temperature and becomes zero at its critical temperature (where the surface of separation between liquid and its vapour disappears). The decrease in surface tension with increase of temperature is due to the fact that with increase of temperature, the kinetic energy of the molecule (and hence the speed of molecules) increases and hence the intermolecular forces of attraction decreases.

Surface tension in everyday life

Cleansing action of soap and detergents is due to their property of lowering the interfacial tension between water and greasy substances. Thus soap solution due to its lower surface tension can penetrate into the fibre to surround the greasy substances and wash them away.

Efficacy of tooth pastes, mouth washes and nasal jellies is partly due to the presence of substances having lower surface tension. Lowering of interfacial tension helps these preparations to spread evenly over the surface with which they come in contact thereby increasing the efficiency of their antiseptic action.

VISCOSITY

It is well known that all liquids do not flow with the same speed. Some liquids like water, alcohol, ether, etc. flow very rapidly while some one like glycerine, honey, castor oil, etc. flow slowly. This indicates that every liquid has some internal resistance to flow. This internal resistance to flow possessed by a liquid is called its viscosity. The liquids which flow rapidly have low internal resistance and hence are said to be low viscous, i.e. their viscosity is less. On the other hand, the liquids which flow slowly have high internal resistance and hence are said to be more viscous, i.e. their viscosity is high. This force of friction which one part of the liquid offers to another part of the liquid is called Viscosity.Coefficient of viscosity may be defined as the force per unit area required to maintain unit difference of velocity between two parallel layers in the liquid, one unit apart. In C.G.S. units, it is expressed in dynes per square centimeter . This unit is called a poise after the

name of Poiseulle who pioneered the study of viscosity. Low values of viscosity are expressed in centipoise and millipoise . In S.I. units, viscosity is expressed in

or Pas (pascal second).

The viscosity of a liquid generally decreases with rise in temperature. With increase of temperature, the kinetic energy of the molecules of the liquid increases and hence the liquid starts flowing faster, i.e. the viscosity decreases. The decrease in viscosity is found to be about 3% per degree rise of temperature.

Viscosity in everyday life

Lubricating oils are graded according to their viscosity. A good quality or ‘all – weather’ lubricating is one whose viscosity and hence lubricating property does not change much with increase of temperature. Such oils are obtained by adding long chain coiling polymers to the oil. As temperature rises, the polymer particles tend to uncoil and thus increase the viscosity of oil thereby compensating for the decrease of viscosity of the oil with rise of temperature.

The condition of high blood pressure and thus strain on heart may also be explained on the property of viscosity. In arteriosclerosis (hardening of arteries), arterial walls contract resulting in decrease of diameter of capillaries.Narrow capillaries offer resistance to the flow of blood due to viscosity with the result greater force is needed to make blood flow through capillaries. This results in a condition of high blood pressure and strain on the heart.

The increased blood circulation required during fever is supplied by its temperature dependence properly (recall that rise of every degree centigrade temperature decreases viscosity of blood by about 3%). Thus lowering of viscosity results in a more rapid flow of blood without any extra strain on the heart.

In case of asphyxia, concentration of CO in blood increases resulting in swelling of corpuscles which then increases viscosity of blood.

LIQUID CRYSTALS

In a temperature range just above the melting point, crystals of certain substances can exist in a definite pattern as in solid but can flow like a liquid. Such crystals are called liquid crystals. When white light falls on a liquid crystal, it reflects only one colour, and as the temperature is changed it reflects different light. Thus, liquid crystals can be used to detect even small temperature changes. There are two important types of liquid crystals namely nematic liquid crystals (needle like) and smectic (soap like) liquid crystals.

THE SOLID STATE

Solids are rigid and have definite shapes. They also possess definite volume which is independent of the volume of the vessel.Solids possess higher densities and are almost incompressible. All these characteristic properties are due to strong inter particle forces, smaller inter particle spares and restricted motion of the particles in the solid state.The temperature at which solid changes into liquid state at normal pressure is called melting point of the substance.

Classification of Solids

Solids are broadly classified into two types: Crystalline solids and amorphous solids.

Distinction between Crystalline and Amorphous Solids

Sl. No. Crystalline Solids Amorphous Solids1. The internal arrangement of particles is

regularThe internal arrangement of particles is not regular

2. They have sharp melting points. They do no have sharp melting points.3. Crystalline solids are regarded as true

solidAmorphous solids are regarded as super cooled liquids or pseudo solids

4. Crystalline solids give a regular cut when cut with sharp edged knife.

Amorphous solids give irregular cut.

5. Crystalline solids are anisotropic. Amorphous solids are isotropic.

Classification of Solids based on different binding forces

Crystalline solids can be classified into different categories depending upon the type of constituent particles and the nature of attractive forces operating between them. Various categories are:

Atomic solids Molecular solids Ionic solids Covalent solids Metallic solids

Atomic Solids

In these solids the constituent particles are atoms. These closely packed atoms are held up by London dispersion forces. Some examples are crystals of noble gases. Such solids are very soft, possess very low melting points and poor conductors of heat and electricity.

Molecular Solids

In these solids, the constituent particles which pack up together are molecules of the substance. These molecules may be non – polar (dipole moment = 0) such as etc. or they may be polar (dipole moment > 0) like etc.

In case of non – polar molecules, the attractive forces operating between the molecules are Vander Waal forces (also called dispersion forces). The example of such solids are : dry ice (Solid , iodine (crystals).

In case of polar molecules, the attractive forces operating between the molecules in solid state are dipole – dipole forces. The examples of such solids are : solid , solid HCl.

However, in some solids with polar molecules, the interparticle forces are hydrogen bonds. The examples of such solids are ice ; solid hydrogen fluoride (HF); solid ammonia , etc.

Characteristics of Molecular Solids

Some of the general characteristics of molecular solids are : They are generally soft. Their melting points are low to moderately high. The melting points of solids with non –

polar molecules are relatively low whereas solids with polar molecules have moderately high melting points.

They are generally bad conductors of heat and electricity. They have generally low density.

Ionic Solids

In ionic solids, the constituent particles are ions of opposite charges. Each ion is surrounded by a definite number of ions of opposite charge. The number of ions that surround a particular ion of opposite charge its called co – ordination number of the ion. For example, in sodium chloride crystal each sodium ion is surrounded by six chloride ions. Hence coordination number of

is 6. At the same time each chloride ion is surrounded by six ions. Therefore the co – ordination number of ion is also 6. However, in calcium fluoride crystal each ion is

surrounded by eight fluoride ions and each ion is surrounded by four ions. Thus, in

crystal co – ordination numbers of and ions are respectively 4 and 8. The

interparticle forces in ionic solids are ionic bonds operating between the ions of opposite charges some examples of ionic solids are : sodium chloride (NaCl) ; ceasium chloride (CsCl), zinc sulphide (ZnS), calcium fluoride , etc.

Characteristics of Ionic Solids

Some common characteristics of ionic solids are as follows: They are hard, brittle and have low volatility. They have high melting points. They are poor conductors of electricity in solid state, however they become good

conductors of electricity in molten state or in dissolved state. They are generally soluble in polar solvents like water.

Covalent Solids

In these types of solids the constituent particles are atoms of same or different elements connected to each other by covalent bond network. For example, in diamond only carbon atoms constitute the covalent network while carborundum covalent bond network is constituted by silicon and carbon atoms. Obviously, the interparticle forces operating in these solids are covalent bonds. These solids are also called network solids because the covalent bonds extend in three dimensions forming a giant interlocking structure. Some examples of covalent solids are :Diamond, silicon carbide, aluminium nitrite etc.

Characteristics of Covalent Solids

Some common characteristics of covalent solids are : They are very hard. Diamond is the hardest naturally occurring substance. They have very high melting points. They are poor conductors of heat and electricity. They have high heats of fusion.

Metallic Solids

In these type of solids, the constituent particles are metal atoms. The interparticle forces in these solids are metallic bonds. In the metallic crystals the metal atoms occupy the fixed positions but their valence electrons are mobile. The close packed assembly of metal kernels (part of metal atom without valence electrons) remain immersed in the sea of mobile valence electrons. The attractive force between the kernels and mobile valence electrons is termed as metallic bond.

Characteristics of Metallic SolidsThe common characteristics of metallic solids are as follows:

They generally range from soft to very hard. They are malleable and ductile. They are good conductors of heat and electricity. They possess bright lustre. They have high melting and boiling points. They have moderate heats of fusion.

The summary of classification of solids on the basis of interparticle forces is given in

Classification of Solids on the Basis of Binding Forces

Crystal Classification

Unit Particles Binding Forces Properties Examples

Atomic Atoms London dispersion forces

Soft, very low melting, poor thermal and electrical conductors

Noble gases

Molecular Polar or non – polar molecules

Vander Waal’s forces (London dispersion, dipole – dipole forces hydrogen bonds)

Fairly soft, low to moderately high melting points, poor thermal and electrical conductors

Dry ice (solid,

methane

Ionic Positive and negative ions

Ionic bonds Hard and brittle, high melting points, high heats of fusion, poor thermal and

NaCl, ZnS

electrical conductorsCovalent Atoms that are

connected in covalent bond network

Covalent bonds Very hard, very high melting points, poor thermal and electrical conductors

Diamond, quartz, silicon

Metallic Cations in electron cloud

Metallic bonds Soft to very hard, low to very high melting points, excellent thermal and electrical conductors, malleable and ductile

All metallic elements, for example, Cu, Fe, Zn

Intermolecular forces or Vander Waal’s forces

Intermolecular forces or vander Waals’ forces originate from the following three types of interactions.

Dipole – Dipole interactions: In case of polar molecules, the vander waals’ forces are mainly due to electrical interaction between oppositively charged ends of molecules (Fig. 1. a) called dipole – dipole interactions. For example, gases such as etc.have permanent dipole moments as a result of which there is appreciable dipole – dipole interactions between the molecules of these gases. The magnitude of this type of interaction depends upon the dipole moment of the molecule concerned. Evidently, greater the dipole moment, stronger is the dipole – dipole interactions. Because of these attractive forces, these gases can be easily liquefied.

Dipole – Induced dipole Interactions: A polar molecule may sometimes polarize a non – polar molecule which lies in its vicinity and thus induces polarity in that molecule just as a magnet induces magnetic polarity in a neutral piece of iron lying close by. The induced dipole then interacts with the dipole moment of the first molecule and thereby the two molecules are attracted together (Fig. 1. b). The magnitude of this interaction, evidently depends upon the magnitude of the dipole moment of the polar molecule and the polarizability of the non – polar molecule. The solubility of inert gases in increases

from He to Rn due to a corresponding increase in magnitude of the dipole – induced dipole interactions as the polarizability of the noble gas increases with increase in size from He to Rn.

Momentary dipole – induced dipole interactions: The electrons of neutral molecules keep on oscillating w.r.t. the nuclei of atoms. As a result, at a given instant, one side of the molecule may have a slight excess of electrons relative to the opposite side. Thus a

non – polar molecule may become momentarily self – polarized. This polarized molecule may induce a dipole moment in the neighbouring molecule. These two induced dipoles then attract each other (Fig. 1. c). These momentary dipole – induced dipole attractions are also called London forces or dispersive forces. The magnitude of these forces depends upon the following:

(i) Size or molecular mass: The melting points and boiling points of non – polar molecules increase as the size or molecular mass of the molecule increases. For example, the m.p. and b.p. of alkanes, halogens, noble gases etc. increase as the molecular mass of the molecule increases.

(ii) Geometry / Shape : For example, isomer n – pentane has higher boiling point than neo – pentane because the former is zig – zag chain with larger sites of contact and hence large intermolecular forces whereas the latter is nearly spherical and hence has less contact and weaker forces.

Illustration19. Why liquid have a definite volume but no definite shape?

Solution: This is because the intermolecular forces are strong enough to hold the molecule together but not so strong as to fix them into definite positions (as in solids) instead, they possess fluidity and hence no definite shape.

Illustration 20. Why diethyl ether has higher vapour pressure than ethyl alcohol at a particular temperature.

Solution: This is because the intermolecular forces of attraction in ethyl alcohol are stronger than those present in diethyl ether.

Illustration 21. Why falling liquid drops are spherical?

Solution: This is due to property of surface tension possessed by the liquids. This makes the surface area minimum. For a given volume, sphere has minimum surface area.

Illustration 22. Arrange the following in the increasing order of melting point of different types of crystalline solid (i) Covalent solid (ii) Metallic solid (iii) Molecular (iv) Ionic

Solution: Covalent > Ionic > Metallic > Molecular

Illustration 23. Explain (a) Aerated water bottles are kept under water during summer.(b) A bottle of liquid ammonia is cooled before opening the seal.

Solution: (a) Aerated water contains dissolved in water. The solubility of gas decreases with rise in temperature. Thus, amount of free gas in bottle can increase in summer, which may result into increase in pressure and ultimately explosion. To avoid this they are kept under cold water.

(b) If the seal is opened without cooling, the liquid ammonia suddenly vaporises and gushes out of the bottle with force. This may lead to serious accidents.

ANSWER TO EXERCISEExercise 1:

2.5 L

Exercise 2:730.8 K

Exercise 3:39.4C

Exercise 4:He = 25.24%, Ar = 74.76%

Exercise 5: = 1.12 atm , = 0.448 atm, Ptotal = 1.568 atm

Exercise 6:58.84

Exercise 7: 81.1 cm

Exercise 8:428

Exercise 9:T1 = 168K, T2 = 2143 K

Exercise 10:3.9 104 cm/sec

Exercise 11:900 cal

Exercise 12:459.63 m / s

Exercise 13., 2.043 KJ

Exercise 14.927C

Exercise 15:(a) NH3

(b) N2

Exercise 16:9.33 atm

Exercise 17:99.9386%

Exercise 18. 0.448 litre

Exercise 19:72%

Exercise 20:O3

MISCELLANEOUS EXERCISES

Exercise 1: Can we apply Dalton’s law of partial pressure to a mixture of carbon monoxide and oxygen?

Exercise 2: Out of dry air and wet air, which is heavier?

Exercise 3: Carbon dioxide is heavier than oxygen and nitrogen but is does not form the lower layer of the atmosphere. Explain.

Exercise 4: The molecular speeds of gaseous molecules are analogous to those of rifle bullets; why is then odour of the gaseous molecule not detected so fast?

Exercise 5: Liquid ammonia bottle is cooled before opening the seal. Explain.

Exercise 6: 100 mL of O2 is collected over water at 23oC and 800 torr. Compute the standard volume of the dry oxygen. Vapour pressure of water at 23oC is 21.1 torr.

Exercise 7: A mixture of Na2CO3 and NaHCO3 has a mass of 22 g. Treatment with excess HCl solution liberates 6 L of CO2 at 25oC and 0.947 atm pressure. Determine the percent Na2CO3 in the mixture.

Exercise 8: A jar contains a gas and a few drops of water. The pressure in the jar is 830 mm of Hg. The temperature of the jar is reduced by 1%. The vapour pressure of water at two temperatures are 30 and 25 mm of Hg. Calculate the new pressure in the jar.

Exercise 9: The radius of an Xe atom is 1.3 10-8 cm. A 100 cm3 container is filled with Xe at a pressure of 1 atm and a temperature of 273 K. Calculate the fraction of the volume that is occupied by Xe atoms.

Exercise 10: A container with a volume of 3.0 L holds N2 (g) and H2O (I) at 29° C. The pressure is found to be 1 atm. The water is then split into hydrogen and oxygen by electrolysis, according to the reaction, H2O (l) H2 (g) +1/2 O2 (g). After the reaction is complete, the pressure is 1.86 atm. What mass of water was present in the container? The aqueous tension of water at 29° C is 0.04 atm.

ANSWERS TO MISCELLANEOUS EXERCISES

Exercise 1: No, the law cannot be applied because these gases readily combine to form carbon dioxide. The law can be applied only to the non reacting gases.

Exercise 2: Dry air mainly contains gases like nitrogen and oxygen along with small proportions of the other gases. The wet air on the other hand, also contains a small amount of water vapours. Now, vapour density of water vapours is less than that of nitrogen and oxygen because molar mass of H2O (18) is less than that of N2 (28) and O2 (32). This means that heavier molecules or nitrogen and oxygen have been replaced by lighter molecules of water vapours when the dry air changes to we air. Therefore, dry air is heavier than wet air.

Exercise 3: The diffusion of the gases is quite independent of the gravitational pull by the earth. The molecules of carbon dioxide remain distributed throughout the atmosphere. Therefore, carbon dioxide does not form the lower layer of the atmosphere.

Exercise 4: No, doubt molecules in the gaseous state move almost with the speed of rifle bullets. But these molecules donot follow a straight path. Since they undergo collisions with one another at a very fast rate, the path followed is zig-zag. Thus, the odour cannot be detected at the same speed at which the molecules move.

Exercise 5: Liquid ammonia bottle contains the gas under very high pressure. If the bottle is opened as such, then the sudden decrease in pressure will lead to a large increase in the volume of the gas. As a result, the gas will come out of the bottle all of a sudden with force. This will lead to the breakage of the bottle and also some serious accident. However, if the bottle is cooled under tap water for sometime, this will lead to a decrease in the volume of the gas to a large extent. If the seal is opened now, the gas will come out of the bottle at a slow rate, thus, reducing the chances of accident.

Exercise 6: The gas collected is actually a mixture of oxygen and water vapour. The partial pressure of water in the mixture at 23oC is 21.1 torr. HencePressure of dry oxygen = (total pressure) (vapour pressure of water) = 799 torrThus, for the dry oxygen V1 = 100 mL, T1 = 23 + 273 = 296 KP1 = 799 torr converting to STP

Exercise 7:

The number of mol of CO2 produced is equal to the number of mol of Na2CO3

plus number of mol of NaHCO3 initially present

Let x = number of g Na2CO3, then 22 x = number of g NaHCO3. The number of mol of each reactant is equal to the mass divided by the formula weight.

x = 12.1 g22 x = 9.9 g

Exercise 8: At T KPgas = Pdry gas + Pmoisture

Pdry gas = 830 – 30 = 800 mm

Now at new temperature

Since, V1 = V2; or

Pgas = Pdry gas + Pmoisture = 792 + 25 = 817 mm of Hg

Exercise 9: We have

Now, the number of Xe atoms is N = nNA = (4.467 10-3 mol)( 6.023 1023 mol-1) = 2.69 1021

Volume occupied by atoms = {4/3 r3} N = 4/3 3.14 (1.3 10-8 cm)3 (2.69 1021) = 2.47 10-2 cm3

Percentage of volume occupied by Xe atoms

Exercise 10:

Total number of moles after water has split up

Total number of moles of that have come into vapour state= 0.225 – 0.120834 = 0.104218

So, number of moles water initially

Mass of

SOLVED PROBLEM

Subjective:

Board Type Questions

Prob 1. A gas occupies one litre under atmospheric pressure. What will be the volume of the same amount of gas under 750 mm of Hg at the same temperature?

Sol. Given V1 = 1 litre P1 = 1 atm

V2 = ? P2 = atm

P1V1 = P2V2

1 1 =

V2 = 1.0133 litre = 1013.3 ml

Prob 2. How large a balloon could you fill with 4g of He gas at 22°C and 720 mm of Hg?

Sol. Given, P = atm, T = 295K, w = 4g

and m = 4 for He

PV =

=

V = 25.565 litre

Prob 3. Calculate the temperature at which 28g N2 occupies a volume of 10 litre at 2.46 atm

Sol. w = 28g, P = 2.46 atm, V = 10 litre, m = 28

Now, PV = (R = 0.0821 litre atm K–1 mole–1)

T = 299.6 K

Prob 4. A gas occupies 300 ml at 27°C and 730 mm pressure what would be its volume at STP

Sol. V2 = litre, P2 = atm, T2 = 300K

At STP, V1= ? P1= 1 atm, T1= 273K

or V1 = 0.2622 litre

Volume at STP = 262.2 ml

Prob 5. In Victor Meyer’s experiment, 0.23g of a volatile solute displaced air which measures 112 ml at NTP. Calculate the vapour density and molecular weight of the substance.

Sol. Volume occupied by solute at NTP = Volume of air displaced at NTP= 112 ml

For volatile solute PV =

at NTP, P = 1 atm, T = 273 K

=

m = 46.02 and V.D. = 23.01

IIT Level Questions

Prob 6. A cylindrical balloon of 21 cm diameter is to be filled up with H2 at NTP from a cylinder containing the gas at 20 atm at 27°C. The cylinder can hold 2.82 litre of water at NTP. Calculate the number of balloons that can be filled up.

Sol. Volume of 1 balloon which has to be filled = = 4851 ml

= 4.851 litreLet n balloons be filled, then volume of H2 occupied by balloons = 4.851 nAlso, cylinder will not be empty and it will occupy volume of H2 = 2.82 litre. Total volume occupied by H2 at NTP = 4.851 n + 2.82 litre At STP P2 = 1 atm Available H2

V1= 4.851 n + 2.82 P2 = 20 atmT1 = 273 K T2 = 300K

= V2= 2.82 litre

or n = 10

Prob 7. A 20g chunk of dry ice is placed in an empty 0.75 litre wire bottle tightly closed what would be the final pressure in the bottle after all CO2 has been evaporated and temperature reaches to 25°C?

Sol. w = 20g dry CO2 which will evaporate to develop pressurem = 44, V = 0.75 litre, P = ? T = 298K

PV =

P 0.75 = 0.0821 298

P = 14.228 atmPressure inside the bottle = P + atm pressure = 14.828 + 1 = 15.828 atm

Prob 8. The pressure of the atmosphere is 2 10–6 mm at about 100 mile from the earth and temperature is – 180°C. How many moles are there in 1 ml gas at this attitude?

Sol. Given, P = atm

T = – 180 + 273 = 93 K

V = 1 ml = litre

PV = nRT

= n 0.0821 93

n = 3.45 10–13 mole

Prob 9. 50 litre of dry N2 is passed through 36g of H2O at 27°C. After passage of gas, there is a loss of 1.20g in water. Calculate vapour pressure of water.

Sol. The water vapours occupies the volume of N2 gas i.e. 50 litre For H2O vapour V = 50 litre, w = 1.20g, T = 300K, m = 18

or P 50 =

P = 0.03284 atm= 24.95 mm

Prob 10. A mixture of CO and CO2 is found to have a density of 1.50g/litre at 30°C and 730 mm. What is the composition of the mixture?

Sol. For a mixture of CO and CO2, d = 1.50 g/litre

P = T = 303K

PV = ;

= 38.85

i.e. molecular weight of mixture of CO and CO2 = 38.85Let % of mole of CO be a in mixture then

Average molecular weight =

38.85 =

a = 32.19Mole % of CO = 32.19Mole % of CO2 = 67.81

Prob 11. The average speed of an ideal gas molecule at 27°C is 0.3 m sec–1. Calculate average speed at 927°C.

Sol. uav=

Given uav = 0.3 m sec–1 at 300K

u1= 0.3 =

at T = 273 + 927 = 1200K

u2 =

u2 = 0.6 m sec–1

Prob 12. Pure O2 diffuses through an aperture in 224 seconds, whereas mixture of O2 and another gas containing 80% O2 diffuses from the same in 234 sec. What is the molecular weight of gas?

Sol. For gaseous mixture 80% O2, 20% gas

Average molecular weight Mm =

Now, for diffusion of gaseous mixture and pure O2

or

Mm= 34.92 = 34.92

m = 46.6

Prob 13. Calculate the pressure exerted by 10–23 gas molecules, each of mass 10–22 g in a container of volume one litre. The rms speed is 105 cm sec–1.

Sol. Given, n = 1023 m = 10–22g V = 1 litre = 103ccurms = 105cm sec–1

PV =

P 103 =

P = 3.3 107dyne cm–2

Prob 14. Calculate the temperature at which CO2 has the same rms speed to that of O2 at STP.

Sol. rrms of O2 = at STP, urms of O2 =

For CO2 urms CO2 =

Given both are same;

T =375.38 K = 102.38°C

Prob 15. Calculate the compressibility factor for CO2, if one mole of it occupies 0.4 litre at 300K and 40 atm. Comment on the result.

Sol. Compressibility factor (Z) =

Z = = 0.65

Z value is lesser than 1 and thus, nRT PV. In order to have Z = 1, volume of CO2

must have been more at same P and T or CO2 is more compressible than ideal gas.

Objective:

Prob 1. An LPG cylinder contains 15 kg of butane gas at 27°C and 10 atmospheric pressure. It was leaking and its pressure fell down to 8 atmospheric pressure after one day. The gas leaked is:(A)1kg (B) 2 kg(C) 3 kg (D) 4 kg

Sol. For the cylinder V = constantHence P1V = n1RT and P2V = n2RT

Hence w2 = 12 kg

Gas leaked out = 15 – 12 = 3 kg (C)

Prob 2. How much should the pressure be increased in order to decrease the volume of the gas by 5% at constant temperature?(A) 5% (B) 5.26%(C) 10% (D) 4.26%

Sol. PV= constant at constant temperatureNew volume = 0.95VP1V = P2 0.95V

P2 = = 1.0525P1

Increase in pressure = 1.0526 P1 – P1 = 0.0526 P1 = 5.26% of P1

(B)

Prob 3. In what ratio by mass carbon monoxide and nitrogen should be mixed so that partial pressure exerted by each is same?(A)1:1 (B) 1:2(C) 2:1 (D) 3:4

Sol. CO = w1g = mol, N2 = w2g = mol

as P =

Hence w1 = w2

w1 : w2= 1:1 (A)

Prob 4. A box of 1 litre capacity is divided into two equal compartments by a thin partition which are filled with 6g H2 and 16g CH4 respectively. The pressure in each compartment is recorded as P atm. The total pressure when partition is removed will be(A) P (B) 2P(C) P/2 (D) P/4

Sol. As pressure in each compartment is P atm, after the partition is removed the total pressure will be P atm. (A)

Prob 5. At 27°C, a gas is compressed to half of its volume. To what temperature it must now be heated so that gas occupies just its original volume?(A) 54°C (B) 600°C(C) 327°C (D) 327K

Sol. PV =

P2 = 2P now; initial volume = V2

Final volume = V

Keeping pressure constant

, T2 = 600K = 327°C

(C)

Prob 6. The vapour density of a gas is 11.2. The volume occupied by 11.2g of this gas at N.T.P. is:(A) 1 litre (B) 11.2 litre(C) 22.4 litre (D) 20 litre

Sol. Vapour Density = 11.2Molecular weight = 22.4Hence 22.4g = 22.4 litre 11.2g = 11.2 litre (B)

Prob 7. A gas is found to have a density of 1.80g litre–1 at 760 mm of pressure and 27°C. The gas will be (A) O2 (B) CO2

(C) NH3 (D) SO2

Sol. M = = 44

Hence the gas is CO2

(B)

Prob 8. A mixture of He and Ar contains 8 mole of He for every 2 mole of Ar. The partial pressure of Ar is: (A)2/3 the total pressure (B) 1/3 total pressure(C) 2/5 the total pressure (D) 1/5 the total pressure

Sol. PArgon=

(D)

Prob 9. The vapour densities of CH4, O2 are in the ratio 1:2. The ratio of diffusions of O2 and CH4 at same P and T is:(A) 1:2 (B) 2:1(C) 1:1.424 (D) 1.414 : 1

Sol.

(C)

Prob 10. A per weighed vessel was filled with oxygen at NTP and weighed. It was then evacuated, filled with SO2 at the same temperature and pressure and again weighed. The weight of oxygen is:(A)The same as that of SO2 (B) 1/2 that of SO2

(C) Twice that of SO2 (D) 1/4 that of SO2

Sol. PV = nRT n1 = n2

or

w1 =

(B)

Prob 11. The gases are at absolute temperature 300K and 350 K respectively. The ratio of average kinetic energy of their molecules is:(A) 7:6 (B) 6:7(C) 36:49 (D) 49:36

Sol.

(B)

Prob 12. 5g of each of the following gases at 87°C and 750 mm pressure are taken. Which of them will have the least volume?(A)HF (B) HCl(C) HBr(D) Hi

Sol. PV = . If other factors are same,

Volume will be least in case of Hi because HI low the maximum molecular weight.(D)

Prob 13. The rms speed of hydrogen is times the rms speed of nitrogen. If T is the temperature of the gas then(A) (B)

(C) (D)

Sol. urms (H2) =

or, =

or (B)

Prob 14. A gaseous mixture contains 1g of H2, 4g of He, 7g of N2 and 8g O2. The gas having the highest partial pressure is:(A)H2 (B) O2

(C) He (D) N2

Sol. P = mole fraction Pm

The gas having highest mole fraction will have highest partial pressure Helium (C)

Prob 15. Which sample contains least number of molecules?(A)1g of CO2 (B) 1g of N2

(C) 1g of H2 (D) 1g of CH4

Sol. 1g of CO2 = = least number of moles

= least number of molecules (A)

True and False

Prob 16. In the van der Waals’ equation

The constant ‘a’ represents the actual volume of the gas molecules.

Sol. False

Prob 17. A mixture of ideal gases is cooled upto liquid helium temperature (4.22 K) to form an ideal solution.

Sol. False

Fill in the Blanks

Prob 18. A graph between volume (in L) vs. temperature (in oC) at constant pressure on extrapolation cuts the temperature axis at………………..

Sol. 273oC

Prob 19. Aqueous tension is the vapour pressure of …………. and depends only upon ………

Sol. water vapours, temperature

Prob 20. Boiling point is the temperature at which the vapour pressure becomes equal to ……………………………

Sol. external pressure (atmospheric pressure)

ASSIGNMENT PROBLEMS

Subjective:

Level - O

1. Account for the following properties of gases on the basis of kinetic molecular theory(i) High compressibility(ii) Gases occupy whole of the volume available to them.

2. What would have been the effect of gas pressure if the collisions between gas molecules were not elastic?

3. A sample of a mixture of gases contains 2 moles of N2, 3 moles of H2 and 5 moles of O2. What is the mole fraction of each gas in the mixture?

4. A gas diffuses 1.27 times as rapidly as Cl2 under the same conditions of temperature and pressure. Answer the following:(i) Is the gas heavier or lighter than Cl2.(ii) What is the ratio of the molecular mass of the gas to that of Cl2?

5. 1 mole of SO2 occupies a volume of 350 ml at 300K and 50 atm pressure. Calculate the compressibility factor of the gas.

6. Calculate the pressure exerted by 8.5g of ammonia (NH3) contained in 0.5 litre vessel of 300K. For ammonia a = 4 atm L2 mol–2, b = 0.036 L mol–1.

7. Explain following observationAerated water bottles are kept under water during summer.

8. Draw the plot of log P vs log V for Boyle’s law.

9. Is it possible to cool the gas to 0 Kelvin.

10. How is partial pressure of a gas related to the total pressure in a mixture of gases?

11. A gas at – 20°C occupies the volume 140 ml. Calculate the temperature at which the volume of gas becomes 65 ml. Assume P as constant.

12. Calculate the volume occupied by 7g N2 at 27°C and 750 mm of Hg.

13. O2 is present in 1 litre flask at a pressure of 7.6 10–10 mm of Hg. Calculate no. of molecules at 0°C.

14. Find the volume of vessel containing 16 g O2 gas at NTP.

15. A gas at 0°C and 1 atm pressure occupies 2.5 litres. What change in temperature would be necessary if the pressure is to be adjusted to 1.5 atm and the gas has been transferred to a 2.0 litre container?

Level – I

1. O2 is collected over water at 20°C. The pressure inside shown by the gas is 750 mm of Hg. What is the pressure due to O2 if vapour pressure of H2O is 18 mm at 20°C?

2. The volume of the average adult lung when expanded is about 6 litre at 98.4°F if the pressure of oxygen in inhaled air is 168 mm of Hg, calculate the mass of O2 required to occupy the lung at 98.4°F.

3. Calculate the total pressure in a mixture of 4g of O2 and 2g of H2 confined in a bulb of 1 litre at 0°C

4. A gaseous mixture of O2 and N2 are in the ratio of 1:4 by weight. Calculate the ratio of molecules.

5. The best vacuum so far attained in laboratory in laboratory is 10–10 mm of Hg. How many molecules of gas remain per cc at 20°C in this vacuum?

6. Calculate root mean square speed of O3 at 20°C and 82 cm Hg

7. The mass of molecule A is twice the mass of molecule B. The rms speed of A is twice the rms speed of B. If the samples of A and B contain the same number of molecules, what will be the ratio of pressure of two samples in separate containers of equal volume?

8. Compressibility factor (Z) for N2 at – 50°C and 800 atm pressure is 1.95. Calculate mole of N2

gas required to fill a gas cylinder of 100 litre capacity under the given conditions.

9. Calculate the temperature of gas if it obeys Van der Waal’s equation from the following data. A flask of 25 litre contains 10 mole of agas under 50 atm. Given: a = 5.46 atm lit2 mol–2 and b = 0.031 lit mol–1

10. The kinetic molecular theory attributes an average translational kinetic energy of to

each particle. What rms speed would a mist particle of mass 10–12g have at room temperature (27°C) according to kinetic theory of gases.

11. (a) Two flasks of equal volume connected by a narrow tube (of negligible volume) are at 27°C and contain 0.7 moles of H2 at 0.5 atm. One of the flask is then immersed into a bath kept at 127°C, while the other remains at 27°C. Calculate the final pressure and the number of moles of H2 is each flask.

(b) 1 gm of an alloy of Al and Mg reacts with excess HCl to form AlCl3, MgCl2 and H2. The evolved H2 collected over mercury at 0°C occupied 1200 ml at 699 mm Hg. What is the composition of alloy?

12. Ladenburg found that a sample of ozonised oxygen containing 86.16% of O3 by weight required 430 secs to diffuse under conditions where pure O2 required 369.5 secs. Determine the vapour density of O3.

13. Assuming O2 molecule spherical in shape and occupying the radius 150 pm. Calculate

(i) the volume of single molecule of gas(ii) the percentage of empty space in one mole of O2 at NTP

14. 9 volumes of gaseous mixture consisting of gaseous organic compound A and just sufficient amount of O2 required for complete combustion yielded on burning 4 volumes of CO2, 6 volumes of water vapour and 2 volumes of N2 all volumes measured at the same temperature and pressure. If the compound A contained only C, H and N (i) how many volumes of O2 required for complete combustion?(ii) What is the molecular formula of the compound A?

15. Two gases in adjoining vessels are brought into contact by opening a stop cock between them. The one vessel measured 0.25 lit and contained NO at 800 torr and 220 K, the other measured 0.1 lit and contained O2 at 600 torr and 220 K. The reaction to form N2O4 (solid) exhausts the limiting reactant completely.(a) Neglecting the vapour pressure of N2O4 what is the pressure of the gas remaining at 220

K after completion of the reaction?(b) What weight of N2O4 is formed?

Level – II

1. (a) How large a balloon could you fill with 4.0 g of He gas at 22oC and 720 mm of Hg?

(b) Calculate the density of CO2 at 100 0C and 800 mm Hg pressure.

2. (a) A container has 3.2 g of a certain gas at NTP. What would be the mass of the same gas contained in the same vessel at 200C and 16 atm. Pressure?

(b) A certain quantity of a gas measured 500 mL at a temperature of 15°C and 750 mm Hg. What pressure is required to compress this quantity of gas into a 400 mL vessel at a temperature of 50°C?

3. (a) Three footballs are respectively filled with N2, H2 and He to the same pressure. In what order are these footballs to be reinflated?

(b) The ratio of rates of diffusion of two gases A and B under same pressure is 1:4. If the ratio of their masses present in the mixture is 2:3. What is the ratio of their mole fraction in mixture?

4. (a) Calculate the rms speed in cm/sec at 25°C of a free electron and a molecule of UF6.(b) The flask (A) and (B) have equal volumes. Flask (A) contains H2 gas at 300K, while (B)

contains equal mass of CH4 at 900K. Calculate the ratio of average speed of molecules in flask (A) and (B)/

5. (a) Calculate total energy of one mole of an ideal monatomic gas at 27C?(b) How much thermal energy should be added to 3.45 g Ne in a 10 litre flask to raise the

temperature from 0°C to 100°C. (Atomic weight Ne = 20.18)?

6. The critical temperature and pressure of CO2 gas are 304.2K and 72.9 atm respectively. What is the radius of O2 molecule assuming it to behave as Vander Waal’s gas?

7. Argon has TC= –122°C PC = 48 atm. What is the radius of the argon atm?

8. 0.22g of a sample of a volatile compound containing C, H and Cl only yielded on combustion in O2 0.195g of CO2 and 0.0804g of H2O. 0.12g of the compound occupied a volume of 37.24 ml at 105°C and 768 mm Hg pressure. Calculate the molecular formula of the compound.

9. 10 ml of mixture of CH4, C2H4 and CO2 are exploded with excess O2. A contraction of 17 ml was observed after explosion. After the treatment with KOH solution there was a 2 nd

contraction of 14 ml. Find the % of gases in mixture?

10. 25cc of a gaseous mixture containing N2 and N2O is passed through a hot Cu tube. The resulting gas has a volume of 20 ml. find the percentage of N2 and N2O in the original gas mixture (The volumes have been measured under same conditions of temperature & pressure)?

11. Calculate the pressure exerted by 1 mole of methane (CH4) in a 250 ml container at 300 K using van der Waal’s equation. What pressure will be predicted by ideal gas equation?

a = 2.253L2 atm. mol-2,

b = 0.04281 lit mol-1R = 0.0821 L atm mol-1K.

12. To what minimum degree centigrade the temperature of earth will have to be raised to get atmosphere free earth like moon?Given Vesp = When Re = 6.37 106 mAssume: atmospheric air chiefly contains N2 and O2 only.

13. (a) A 10 litre box contained 41.4 gm of a mixture of CxH8 and CxH12. At 44C the total pressure is 1.56 atm. Analysis of the gas mixture shows 86% of C & 14% of hydrogen by weight of hydrocarbon sample. What gases are in the box?How many moles of each gas are in the box?

(b) 24 ml. of water gas containing only hydrogen and carbon monoxide in equal proportions by volume, are exploded with 80 ml. of air, 20% by volume of which is oxygen. If all gaseous volumes are measured at room temperature and pressure, calculate the composition by volume of the resulting gaseous mixture.

14. An open vessel at 27C is heated until three-fifths of the air in it has been expelled. Assuming the volume of the vessel remains constant, find the temperature to which the vessel has to be heated.

15. At 60°C the density of N2O4 gas was found to be 30.2. Calculate the percentage of NO2

molecules by weight and by volume.

Objective:

Level – I

1. According to the kinetic theory of gases(A) There are intermolecular attractions(B) Molecules have considerable volume(C) No intermolecular attractions(D) The velocity of molecules decreases after each collision

2. The Van der Waal’s equation explains the behaviour of (A) Ideal gases (B) Real gases(C) Vapour (D) Nonreal gases

3. Which of the following is not a crystalline solid?(A) KCl (B) CsCl(C) Glass (D) Rhombic sulphur

4. The density of neon will be highest at(A) STP (B) 0°C, 2 atm(C) 273°C, 1 atm (D) 273°C, 2 atm

5. Absolute zero is defined as the temperature(A) at which all molecular motion ceases (B) at which liquid helium boils(C) at which ether boils (D) all of the above

6. At constant temperature, in a given mass of an ideal gas(A) The ratio of pressure and volume always remains constant(B) Volume always remains constant(C) Pressure always remains constant(D) The product of pressure and volume always remains constant

7. At the same temperature and pressure, which of the following gases will have the highest kinetic energy per mole?(A) Hydrogen (B) Oxygen(C) Methane (D) All are same

8. Which of the following mixture 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

9. The temperature at which real gases obey the ideal gas laws over a wide range of pressure is called(A) Critical temperature (B) Boyle temperature(C) Inversion temperature (D) Reduced temperature

10. The compressibility factor for an ideal gas is:(A) 1.5 (B) 1.0(C) 2.0 (D)

11. According to Graham’s law at a given temperature, the ratio of the rates of diffusion rA/rB of gases A and B is given by

(A) (B)

(C) (D)

12. 4.4 gm of a gas at STP occupies a volume of 2.24 litre, the gas can be(A) O2 (B) CO(C) NO2 (D) CO2

13. A certain sample of gas has a volume of 0.2 litre measured to 1 atm. Pressure and 0°C. At the same pressure but at 273°C, its volume will be(A) 0.4 litre (B) 0.8 litre(C) 27.8 litre (D) 55.6 litre

14. If the four tubes of a car are fitted to the same pressure with N2, O2, H2 and Ne separately, then which one will be filled first?(A) N2 (B) O2

(C) H2 (D) Ne

15. If pressure becomes double at the same absolute temperature of 2 litre CO2, then the volume of CO2 becomes(A) 2 litre (B) 4 litre(C) 5 litre (d) 1 litre

16. A gaseous mixture contains 56g of N2, 44g of CO2 and 16g of CH4. The total pressure of the mixture is 720 mm Hg. The partial pressure of CH4 is:(A) 180 mm (B) 360 mm(C) 540 mm (D) 720 mm

17. Same mass of CH4 and H2 is taken in container. The partial pressure caused by H2 is

(A) (B)

(C) (D) 1

18. The major binding force of diamond, silicon and quartz is(A) Electrostatic force (B) Electrical attraction(C) Covalent bond force (D) Van der Waal’s force

19. The density of air is 0.00130 g/ml. The vapour density of air will be(A) 0.0065 (B) 0.65(C) 14.4816 (D) 14.56

20. Molecular weight of a gas that diffuses twice as rapidly as the gas will molecular weight 64 is:(A) 16 (B) 8(C) 64 (D) 6.4

Level –II

1. If P, V, T represent pressure, volume and temperature of a gas, the correct representation of Boyle’s law is

(A) (at constant P) (B) PV = RT

(C) (at constant T) (D) PV = nRT

2. Five grams each of the following gases at 87°C and 750mm pressure are taken. Which of them will have the least volume?(A) HF (B) HCl(C) HBr (D) HI

3. A real gas most closely approaches the behaviour of an ideal gas at(A) 15 atm and 200K (B) 1 atm and 273K(C) 0.5 atm and 500K (D) 15 atm and 500K

4. Which one of the following statements is true?(A) The gas equation is not valid at high pressure and low temperature(B) The product of pressure and volume of a fixed amount of gas is independent of

temperature(C) Molecules of different gases have the same kinetic energy at a give temperature(D) The gas constant per molecule is called as Boltzmann’s constant

5. Two separate bulbs contain ideal gas A and B. The density of gas A is twice that of B. The molecular mass of A is half that of B. The two gases are at the same temperature. The ratio of pressure A to that of gas B is:

(A) 2 (B)

(C) 4 (D)

6. Use of hot air balloons in sports and metrological observations is an application of(A) Boyle’s law (B) Kelvin’s law(C) Charle’s law (D) Brown’s law

7. For an ideal gas, the number of moles per litre in terms of its pressure P, gas constant R and temperature T is:

(A) (B) PRT

(C) (D)

8. 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

9. Air at sea level is dense. This is a practical application of(A) Boyle’s law (B) Charle’s law(C) Avogadro’s law (D) Dalton’s law

10. When the product of pressure and volume is plotted against pressure for a given amount of a gas, the line obtained is:(A) Parallel to Xaxis (B) Parallel to Yaxis(C) Linear with positive slope (D) Linear with negative slope

11. The rms velocity of CO2 at a temperature T (T in Kelvin) is X cm sec–1. At what temperature (in Kelvin) the rms velocity of N2O would be 4x cm sec–1?(A) 16T (B) 2T(C) 4T (D) 32T

12. The following graph illustrates

(A) Dalton’s law (B) Charles’s Law(C) Boyle’s law (D) Gay Lussac’s Law

13. The ratio of rates diffusion of SO2, O2 and CH4 is(A) 1 : : 2 (B) 1 : 2 : 4(C) 2 : : 1 (D) 1 : 2 :

14. A bottle of cold drink contains 200 ml liquid in which CO2 is 0.1 molar. Suppose CO2behaves like an ideal gas, the volume of the dissolved CO2 at STP is(A) 0.224 litre (B) 0.448 litre(C) 22.4 litre (D) 2.24 litre

15. In the equation PV = nRT, which one cannot be the numerical value of R?(A) 8.31 107 erg K–1 mol–1- (b) 8.31 107 dyne cm K–1 mol–1

(C) 8.31 J K–1 mol–1 (D) 8.31 atm K–1 mol–1

16. A bottle of ammonia and a bottle of dry hydrogen chloride connected through a long tube are opened simultaneously at both ends, the white ammonium chloride ring first formed will be(A) at the centre of the tube (B) near the hydrogen chloride bottle(C) near the ammonia bottle (D) throughout the length of the tube

17. Root mean square velocity of a gas molecule is proportional to(A) m1/2 (B) m0

(C) m–1/2 (D) m18. A sample of a given mass of a gas at a constant temperature occupies 95 cc under a

pressure of 9.962 104 Nm–2. At the same temperature, its volume at a pressure of 10.13 104 N m–2 is:(A) 190cc (B) 93cc(C) 46.5 cc (D) 47.5 cc

19. Equal volumes of the gases which do not react together are enclosed in separate vessels. This pressures of 100 mm, and 400 mm respectively. If the two vessels are joined together, then what will be the pressure of the resulting mixture (temperature remaining constant)?

(A) 125 mm (B) 500mm(C) 1000 mm (D) 250 mm

20. There are 6.02 1022 molecules each of N2, O2, and H2 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) 7

ANSWERS TO ASSIGNMENT PROBLEMS

Subjective:

Level - O

3. 0.2, 0.3, 0.5 4. Lighter, 1:1.61

5. 0.711 6. 21.51 atm

11. – 155.54°C 12. 6.239 litre

13. 2.68 1010 14. 11.2 L

15. 54.6°C

Level – I

1. 732mm 2. 1.67g

3. 25.215 atm 4. 7:32

5. 9.081 1042 molecules 6. 3.9 104 cm sec–1

7. PA = 8PB 8. 2240.8

9. 1256.93°C 10. 0.35 cm sec–1

11. (a) 0.571 atm, 0.3 and 0.4 moles 12. 23.3(b) 44 % Mg, 56 % Al

13. 1.41 1023 cc/molecule, 99.96% 14. 7volume, C2H6N2

15. 0.30 atm, 0.402g

Level - II

1. (a) 25.65 litre 2. (a) 29.55 g(b) 1.5124 g/litre (b) 1051 mm

3. (a) H2 He N2 4. (a) 1.16 107 cm/sec(b) 1:24 (b) 1.632

5. (a) 900 cals (b) 51.75 cal

6. 1.62Å 7. 1.47 10–8 cm

8. C2H4Cl2 9. 45% CH4, 40% C2H4, 15% CO2

10. 60% N2, 40% NO

11. 82.836 atm, 98.52 atm

12. 81.12 cm

13. (a) C5H8, C5H12 0.432, 0.167 moles (b) O2 = 4ml, N2 = 64ml, CO2 = 12 ml

14. 750 K 15. 52.3%, 68.7%

Objective:

Level – I

1. C 2. B 3. C

4. B 5. A 6. D

7. D 8. D 9. B

10. B 11. C 12. D

13. A 14. C 15. B

16. A 17. A 18. C

19. D 20. A

Level – II

1. C 2. D 3. C

4. B 5. C 6. C

7. C 8. C 9. A

10. A 11. A 12. B

13. A 14. B 15. D

16. B 17. C 18. B

19. D 20. D