Solutions

Solutions - II. Ideal Solution and colligative properties Solution- II

In the last chapter, we studied about the basic terminology used in solutions and the different laws, which are very useful in understanding the nature of solutions. In this chapter, we will be studying mainly about the colligative properties of solutions and their applications such as determining the molecular weight of polymers.

Colligative properties

Non-volatile solute is that solute which undergoes neither dissociation nor association.

Volatile solute is that solute which undergoes dissociation or association.

Colligative properties of solutions are those properties, which depend only upon the number of solute particles or concentration of solute particles in the solution and not on their nature. Colligative means “depending on the collection”. Therefore, colligative properties depend on the numbers of solute particles rather than their nature. Such properties are:

(a) Relative lowering in vapour pressure,

(b) Elevation of boiling point,

(c) Depression of freezing point and

(d) Osmotic pressure.

These properties are used to determine the molecular weight of polymers. we will first discuss those solutes, which are non-volatile.

(a) Relative lowering in vapour pressure:-

The vapor pressure of a liquid is the partial pressure exerted by vapor when both are in equilibrium with each other.

When non-volatile solute particles are added into solution, they lower the vapor pressure of solvent.

According to Raoult’s law “When we increase the mole fraction of nonvolatile solute particles in a solution, the vapor pressure over the solution will be reduced. The reduction in vapor pressure is due to the total concentration of solute particles.

According to Raoult’s Law, the vapour pressure of a solution containing a non-volatile solute is given by

Where, PA = vapor pressure of solution,

PoA = vapor pressure of pure solution

XA = mole fraction of solution

Notation ‘A’ is used for solution and ‘B’ is used for solute.

Since for a binary mixture,

Thus, according to Raoult’s Law, the relative lowering of vapour pressure of a solvent is equal to the mole fraction of the solute.

Calculation of molecular mass of solute particles, from lowering of vapour pressure

For a very dilute solution,

Where,

Po A – PA / Po A = Relative lowering of vapor pressure of solvent.

WB = Weight of the solute

WA = Weight of the solvent.

MA = Molecular mass of solvent

MB = Molecular mass of solute

Measurement of relative lowering of vapour pressure (Ostwald- Walker method):

The given apparatus consists two set of bulbs. The first set is filled with solution and second set is filled with pure solvent. These sets of tube are connected with tubes filled with dehydrating agents like CaCl2. Temperature of tubes of solution and pure solvent is maintained constant.

A current of pure dry air is passed through solution. The air gets saturated with the vapours. The air takes up an amount of vapors from the solution and then pure solvent. These two sets of bulb are weighed again and dehydrating agent tubes.

Total loss in mass of both sets of tubes = gain in mass of guard tubesOr,

(b) Elevation of boiling point:

The boiling point of a liquid is the temperature at which the vapour pressure of the liquid becomes equal to the atmospheric pressure. When a non-volatile solute is dissolved in a solvent, its vapour pressure decreases, as there is no contribution from the non-volatile solute. Therefore, the boiling point of the solution is always higher than the boiling point of the pure solvent.

This figure shows the vapor pressure curves for the pure solvent, solution consisting of a nonvolatile solute. PQ is the vapor pressure curve for pure solvent, and RS is vapor pressure curve for the solution. T0 represents the boiling point of pure solvent and T1 represents the boiling point of solution.

When non-volatile solutes are added to solvent, we have to heat the solution to a slightly higher temperature so that the vapor of the solution becomes equal to atmospheric pressure and solution begins to boil. The differences

in the boiling points are represented as , which is the elevation in boiling point.

For a dilute solution, the elevation in boiling point is found to be proportional to the molality of the solutions, i.e.

Where,

is the elevation in boiling point,

‘m’ is the molatility

Kb is the Molal elevation constant

Kb is the Molal elevation constant (boiling elevation constant) which is equal to the elevation in boiling point when one mole of the solute is dissolved in 1000g of the solvent. The magnitude of Kb depends only on the solvent.

Kb (Ebulloiscopic constant) may be calculated from the following equation.

Where, T is the boiling point of the solvent

Lv is the latent heat of vaporization per gram of the solvent.

Calculation of molecular mass of solute from elevation in boiling point:

We know that,

If WB gram of solute is dissolved in WA grams of solvent, then

From (1) and (2)

Therefore, molecular mass of solute will be

(c) Depression of freezing point:x

Freezing point is the temperature at which the solid and the liquid state of the substance have the same vapour pressure. Since the presence of a non-volatile solute lowers the vapour pressure of the solvent, the freezing point of the solution is always less than that of the pure solvent.

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The curve shows the vapour pressure of pure solvent and frozen solvent. Line AB, BC, and A'B' are the vapor pressure curves of solid solvent (ice), liquid water and solution containing non-volatile solute respectively. If TO be the freezing point of pure solvent T1 is the freezing point of the solution, then depression in freezing point depends upon the molal concentration of the solute.

The depression in freezing point is proportional to the molality of the solution.

Where Kf is molal depression constant (freezing point depression constant). It is the depression in freezing point when 1 mol of a solute is dissolved in 1000 g of the solvent

Kf may be calculated from the following equation.

Where, T is the freezing point of the solvent

Lf is the latent heat of fusion per gram of the solvent.

Calculation of molecular mass of solute from depression in freezing point:

We know that,

If WB gram of solute is dissolved in WA grams of solvent, then

From (1) and (2)

Therefore, molecular mass of solute will be

Applications:-

There are many applications of the depression in freezing point. It is used to lower the freezing temperature of water and clear roads during winters. Depression in freezing point is also useful in antifreeze. Ethylene glycol is used as antifreeze.

(d) Osmotic pressure:-

The spontaneous flow of solvent molecules from a dilute solution to a concentrated solution when a perfect semi-permeable membrane separates the two is called osmosis.

Osmotic pressure (p) is the pressure, which must be applied to the solution side (more concentrated solution) to just prevent the passage of pure solvent into it through a semi permeable membrane.

An ideal semi-permeable membrane will allow the solvent molecules to pass but prevent the solute molecules from passing. The different concentrations on the two sides of the membrane will cause an initial difference in chemical potential.

From the above figure, it is clear that solvent molecules will pass from the pure solvent side to the solution side. The excess height in the column of liquid will be osmotic pressure.

Mathematically,

Where, is the osmotic pressure of the solution.

nB is the number of moles of solute,

V is the volume of the solution in liters,

R is the gas constant, and

T is the temperature on the Kelvin scale.

Difference between diffusion and osmosis:

Isotonic solutions are those solutions, which have the same osmotic pressure at the same temperature. In addition, they have same molar concentration.

Hypertonic and Hypotonic solutions:

When two solutions have different osmotic pressures, the solution of lesser osmotic pressure is called hypotonic solutions and the solution of higher osmotic pressure is called hypertonic solutions.

Reverse Osmosis:

The movement of solvent particles from higher concentration to lower concentration through semi-permeable membrane on applying pressure is known as reverse osmosis. If the external pressure greater than osmotic pressure is applied on more concentrated solution side, the solvent molecules start passing through semi-permeable membrane from this solution to the solvent or less concentrated solution. This is known as reverse osmosis. It is used to purify the seawater.

Abnormal colligative properties:

When solute particles undergo association or dissociation, the number of solute particles change and therefore

colligative properties will change because colligative properties depend upon number of solute particles.

Van’t Hoff factor, ‘i’ is used to express the extent of association or dissociation of solutes in solution. It is the ratio of the normal and observed molar masses of the solute, i.e.,

In case of association, value of observed molar mass will be less than the normal, the factor ‘i’ has a value less than one. But in case of dissociation, the Van’t Hoff factor is more than one because the observed molar mass has higher value.

In case of solutes, which do not undergo any association or dissociation in a solvent, the vant Hoff factor, ‘i’, will be equal to one because the observed and normal molar masses will be same.

Inclusion of van’t Hoff factor, ‘i’, modifies the equations for colligative properties as follows:

1. Degree of dissociation: It is defined as the fraction of total substance that undergo dissociation into ions.

2. Degree of association: It is defined as the fraction of total number of molecules, which combine to form associated molecules.

Solved Examples:]

Example 1: Calculate the lowering in vapour pressure when 100 g of a hydrocarbon (molecular weight =200 g) is added to 360 g of water if the vapor pressure of pure water is 24 mm Hg at 25 oC .

Solution:-

According to Raoult’s law;

Relative lowering in vapor pressure is equal to mole fraction of solute.

Given that,

nB = 100/200 = 0.5 moles

nA = 360 /18 = 20 moles

PoA = 24 mm of Hg

Substituting these values in the formula;

Example 2: The vapour pressure of pure benzene at room temperature is 500 mm Hg. A non-volatile solid (2 g) is added to 50 g of benzene. The vapor pressure of the solution is 450 mm Hg. What is the molecular mass of solute?

Solution:-

According to Raoult’s law;

Relative lowering in vapor pressure is equal to mole fraction of solute.

Given that,

nA = 50/78 = 0.64 moles

nB =2/MB

PoA = 500 mm of Hg

PA = 450 mm of Hg

Substituting these values in the formula;

Example 3:- What mass of non-volatile solute (carbon) needs to be dissolved in 90 g of water in order to decrease the vapour pressure of water by 10 %. What will be mole fraction of solute and molality of solution?

Answer:

According to Raoult’s law;

Relative lowering in vapor pressure is equal to mole fraction of solute.

Given that,

nB = m/12 moles

nA = 90 /18 = 5 moles

PoA = 100 mm of Hg

PA = 90 mm of Hg

Substituting these values in the formula;

Molality = 6.67 * 1000/ 12 * 90= 12.18 m

Example 4: - A very small amount of non-volatile solute is dissolved in 56.8 cm3 of benzene (density = 0.889g/cm3). At room temperature, vapour pressure of this solution is 98.88 mm Hg while that of benzene is 100 mm Hg. Find the molality of this solution. If the freezing temperature of this solution is 0.73 degree lower than that of benzene, what is the value of molal freezing-point depression constant of benzene? (IIT-JEE – 1997)

Answer: -

We know that, for dilute solution;

Molality = moles of solute / weight of solvent (kg)

Molality = (0.00725 * 1000) / (56.8 * 0.889)

Molality = 0.1435 m

Further, We know that,

Example 5:- Calculate the osmotic pressure of a 0.5 M solution of cane sugar at 0oC.

Solution:

Example 6:- A solution of 2 g of ‘A’ in 100 cc of water is found to be isotonic with a 5 % (weight by volume) solution of hydrocarbon (molecular mass = 160 g). Calculate molecular weight of ‘A’.

Answer: - Isotonic solutions have the same molar concentration.

Molar concentration of hydrocarbon = (5 /160) * (1000/100)

= 0.3125 M

Molar concentration of ‘A’ = (2 /M) * (1000/100) = 20 /M

Since, ‘A’ is isotonic with hydrocarbon solution.

20/M = 0.3125

M = 64

Example 7: A mixture, which contains 0.5 g of camphor and 0. 1 g of hydrocarbon freezes at 350 k. The solute contains 96 % of C and 4 % H by weight. What is the molecular formula of compound? ( Kf = 30, Tf = 380 K)Answer: By depression in freezing point:

If the molecular weight of solute is M , then

Molality = moles of solute / weight of solvent (kg)

Molality = (0.1 / M )* 1000 / 0.5 = 200 /M

Since, m =1

Therefore, M = 200

Now, it is given that weight of C = 96 % and H = 4 %

Moles of C = 96 /12 = 8

Moles of H = 4 /1 = 4

Molar ratio of C and H = 2: 1

Empirical formula would be = CH2

Weight of empirical formula = 12 + 2 = 14

Therefore molecular formula = C14H28

The change in the molecular mass due to some molecular change of solute ( dissociation).

Example 8: If 50 g of a solute is dissolved in 500 g of benzene, what will be the boiling point of the solution at 1 atm pressure. The molal boiling constant elevation is 4oC m-1 and the boiling point of pure benzene is 76.33oC. (Molecular weight of solute 120 g)

Answer:

By boiling point elevation:

Moles of solute is 50/120 = 5/12 moles

Molality = moles of solute / weight of solvent (kg)

Molality = (5/12) / 0.5

= 10/12 -------------------------------(2)

From (1) and (2)

Thus, boiling point of solution is 76.33 + 3.33 = 79.66oC.

Example 9: A 0.1 M aqueous solution of NaCl freezes at 0.22oC. Calculate ‘i’ and osmotic pressure at 0oC. Kf of the solution is 1.50. Consider that solution is pure.

Solution:-

Therefore, i = 0.22/ 0.15 = 1.47

Now , observed osmotic pressure ( considering effect of i)

Example 10: When 20 g of naphthoic acid (molecular weight =172) is dissolved in 50 g of benzene (Kf = 1.72 K kg mol-1), a freezing point depression of 2K is observed. The van’t Hoff factor is(IIT-JEE – 2007)Solution:

By depression in freezing point:

Example 11: The Henry’s law constant for the solubility of N2 gas in water at 298 K is 0.1 * 105 atm. The mole fraction of N2 in air is 0.8. The number of moles of N2 from air dissolved in 10 moles of water of 298 K and 5 atm pressure is

(IIT-JEE-2009)

Solution; We know that,

From Henry law;

P = KH X

Given that,

P = 5 atm

KH = 105 atm

5 atm = 105 atm (X /10 + X)

Since, X is very small. So, X+ 10 ~ 10

Therefore, X = 5 * 10-4

This is the moles of air in water.

Since, N2 has mole fraction in air = 0.8

Therefore, moles of N2 = 5 * 10-4 * 0.8 = 4 * 10-4

States of matter - Introduction

CONTENTS1. Introduction2. Chemistry&Gases3. Kinetic Theory of Gases4. Real & Ideal Gases1. INTRODUCTION1.1 The three states of matterThe three states of matter are solid, liquid and gas. Almost everything we see about us appears to fit neatly into one of these categories. There is one key idea that you should understand if you are to appreciate the differences between the three states of matter. It is this: the state in which a substances exists is the result of the competition between intermolecular forces, which keep molecules together, and heat energy, which moves them apart (Figure 1.1).

Heat energy, which we can also call thermal energy, is a measure of the amount of random movement of molecules. We should not say that heat causes the random movement of molecules; rather thermal energy is the random movement of molecules. The more thermal energy a substance has, the greater is the tendency for its molecules to be jumbled up, i.e. to be more disordered. The most disorderly arrangement that molecules can achieve is in a gas. At the other extreme the most orderly arrangement is in a solid. Liquids are somewhere in between (Figure 1.2).Intermolecular forces tend to hold molecules together. There are intermolecular forces between all molecules; but between some they are very weak and between others they are quite strong. When

the forces are weak, the molecules are not likely to cling together to make a liquid or solid unless they have very little thermal energy. The noble gases are excellent examples. For instance, helium will not liquefy until the temperature is almost as low as – 269° C, or 4 K. On the other hand the intermolecular forces between water molecules are very strong.Intermolecular forces tend to bring order to the movements of molecules; heat energy points in the direction of randomness or chaos.Whether a substance exists as a solid, liquid or gas depends on where the balance between these two opposing influences lies.

Gas is a state of matter without a definite shape or volume, and the volume of which changes with the change in temperature and pressure. The molecules of a gas are in random motion .Gases have weaker intermolecular forces (called van der Waals forces) but the average speed of the molecules is quite high. Gases tend to occupy the whole volume of the container they are kept in as they don’t have a boundary like liquids and solids do.The average distance between the molecules in gas are large that’s why they have weaker intermolecular forces and the average molecular speed is high. Since, the average distance between the molecules it makes them highly compressible or therefore volume decreases with the increase in pressure. And with the increase in the temperature the average speed of the molecules increases and their kinetic energy goes up.There is definite relationship between the mass, pressure, volume, density and temperature of the gaseous state. An equation relating pressure, volume, temperature and amount of gas is known as equation of state. We will come to this later.Therefore, in measuring volume or amount of the gas the temperature and volume should also be specified.There are two standard temperature and pressure conditions normally used to measure the data of gases. They are:

ConditionTemperatur

ePressur

eMolar mass

STP/NTP (Standard temperature and Pressure/Normal temperature and Pressure)

273.15 K 1 atm 22.414 L

SATP Standard Ambient Temperature and Pressure

298.15 K 1 atm 22.788 L

Pressure of a gas is pressure exerted by the molecules on the walls of the container. The molecules are numerous in number and the pressure exerted by each molecule is also different but as a whole the force applied is uniform.Pressure is defined as force applied per unit area. Its SI unit is Pascal, which in 1 Newton per m2. Mathematically, Pressure = (force / area)

p = F/A or dF/dAWhere, p = pressure, F = normal force,A is the area.

Other units of pressure are:

Atmospheric pressure is also measured in terms of height of mercury column. 1 atm is also expressed as 76cm of mercury column. Though, dimensionally cm has the dimensions of length and not pressure. We can find the corresponding pressure from height of mercury column by the formula:

P = hdg

Where,h = height of the mercury column supported by the barometer.d = density of mercury,g = acceleration due to gravity.Depending upon the units of h, d and g used we get pressure in one of its units. Pressure is measured using the manometer which has the desired fluid (gases in most cases) on one end and its open on the other end , and pressure of the fluid is measured by change in the length of the mercury (or any other liquid used).

(Image taken from www.efunda.com)

Solved ExampleExample1Convert the given values in atm.a) 38 cm Hg b) 400 Torr. Solution a) 76 mm Hg is = 1 atm

Therefore, 38 cm of Hg = (1/ 76) X 38 atm = 0.5 atmb) 760 Torr = 1 atmTherefore, 400 Torr = (1 / 760) X 400 atm= 0.52 atm.

Solved ExampleExample2:If the density of mercury becomes 13 X 103 Kg/m3 find the drop in pressure of 76 cm of mercury (Assuming other factors remain unchanged).Solution:Pressure = hdg. ΔP = Pi - Pf = hdig - hdfg = hg(di - df)= 76 X 10-2 X 9.8 X ( 13.6 – 13 ) X 103 Pa. = 4468.8 Pa

1.2 How do we know that gases are disorderly?One piece of evidence comes indirectly from the experiments first performed by Robert Brown in 1827. He observed the movement of pollen grains on the surface of water, which he found to be completely unpredictable. The random movements of the grains, known as Brownian motion, were finally given a mathematical explanation by Albert Einstein in 1905. He showed that the grains went on a random walk (Figure 1.3). A random walk is the sort of walk that a very drunk person would go on if put out in an open space. If we assume that the drunk found it impossible to make a conscious choice, he (or she) would be as likely to walk in one direction as any other. The reason why the grains behave in this way is that they are being bombarded by molecules in the liquid, which are themselves moving in a perfectly random way.

Around 1908 Jean Perrin made observations of Brownian motion in gases. He showed that small particles, much larger than individual molecules but still very small (less than 10-6m in diameter), also went on random walks. This could only be explained along the same lines as Brownian motion in liquids. The particles were being struck by the randomly moving gas molecules.1.3 Differences in properties of solids, liquids and gasesThe molecules in a gas are, on average, much further apart than they are in a liquid or solid. Also, the molecules in a gas travel very much faster than those in a liquid. The molecules in a solid vibrate about the same average position rather than traveling from place to place. The difference in spacing, and in speed, are the main reason for the different properties of the three states of matter. For example, gases are not very good conductors of heat. For heat to be conducted the movement energy of the molecules must be passed on from one to another. This requires the molecules to collide, which happens less easily in a gas than in a liquid.

In a solid the molecules are held in position by the overall effects of the attractions and repulsions of their neighbours. Even so, the molecules do have some movement. They vibrate to-and-fro, although on average they keep the same position. As the temperature gets higher they vibrate more violently, and they can move from place to place. Especially, metals have many free electrons that can carry their movement energy with them even though the atoms themselves are stuck in one place.Owing to the large amount of empty space in a gas, it is fairly easy to squeeze the molecules into a smaller volume. That is, gases are easily compressed. Liquids and solid have their molecules already very close together, so they are very difficult to compress. We should think about this difference in compressibility more carefully. It seems obvious that there is a limit to how close molecules can get to one another, but why is there a limit? The answer lies in the structure of molecules.1.4 The potential energy curve for two neighbouring moleculesWhen two molecules are far apart they move completely independently; neither will feel the presence of the other. However, if they come closer together then the intermolecular forces get to work. They will attract one another. The amount of the attraction depends on several factors, which are described in Unit 20. The attractions tend to bring the molecules together; but now think about them coming very close together. The outside of a molecule is really a layer of negatively charged electrons: the electron cloud. When molecules approach closely, the electron clouds repel one another. It is the great strength of the repulsion that puts a limit on how close the molecules can get.Now you need to remember that attraction means a lowering of energy, repulsion means an increase in energy. This convention allows us to explain the energy diagram of Figure 1.4. You can see that there is a minimum in the curve. This is when the attractive and repulsive forces just balance. The molecules are then at their equilibrium distance apart. The normal equilibrium distance between molecules is of the order of 200 to 800 pm. The shape of the curve gives us an idea of why it is that gases can be difficult, and sometimes impossible, to liquefy if the temperature is too high. At high temperatures the average speed of the molecules in a gas is much higher than at a low temperature. If two molecules hit one another at great speed, their electron clouds become squashed together; this is rather like two springs being pushed together. This brings them high up the repulsion part of the curve. Then they fly apart and go off to make further collisions. At lower temperatures the average speed of the molecules in a gas is much higher than at a low temperature. If two molecules hit one another at great speed, their electron clouds become squashed together; this is rather like two springs being pushed together. This brings them high up the repulsion part of the curve. Then they fly apart and go off to make further collisions. At lower temperatures, when their speeds are much lower, the force of the collisions can be very much less. Now the interaction of their electron clouds may take them only part of the way up the repulsion part of the curve. If they do not go too high, they will not spring apart. Rather, they will stick together and oscillate around their equilibrium position.

1.5 Some remarkable substancesIn this section we shall briefly look at some substances that are hard to classify as a solid, or gas.(a) Liquid crystalsIt seems a contradiction to call a crystal ‘liquid’. We expect crystals to be solids, and certainly not liquids. Essentially, liquid crystals are liquids that can have sufficient long range order in them to make them behave like a solid. However, they will only behave like a solid over a certain range of temperatures. Usually a liquid crystal is made from molecules that are long, thin and not very symmetrical. You will find some examples in Figure 1.5.The intermolecular forces must be strong enough to hold the molecules together, but not so strong as to restrict their movement too much. The unsymmetrical nature of the molecules leads to an unsymmetrical packing of the molecules. When the packing is unsymmetrial we say that the arrangement is anisotropic. (Isotropic means the same in every direction; anisotropic is the opposite – not the same in every direction.)The very useful property of liquid crystals is that the arrangement of the molecules can be upset by very slight changes in their surroundings. Especially, in the liquid molecules rearrange themselves when the crystal is subjected to a small electric field. The change from anisotropic to a more isotropic arrangement changes the way the crystal absorbs light.(b) GlassIt may surprise you to know that glass is best considered as a liquid. It happens to be an extremely viscous liquid. Indeed, it is so viscous that many people find it very difficult to think of glass as anything but a solid. The basic building block of ordinary glass is a tetrahedron built from a silicon atom with four oxygen atoms around it (Figure 1.6). The tetrahedron join to give a three dimensional interlocking structure that gives glass. Its has no long range order in its structure. Given a long enough time, glass will flow like a liquid. For example, stained glass put into cathedrals during the fifteenth century is thicker at the bottom than at the top.

(c) ColloidsColloids can take on the appearance of solids, liquids, or gases, although they are invariably mixture of some kind. Smoke from a fire is a colloidal system. It is made of tiny particles of solid that float in air. Colloidal particles are generally between 5 and 1000nm in diameter. This means that they are much larger than atoms or small molecules such as water, but smaller than particles that we can see clearly with the naked eye. The air in which they float is called the continuous phase; the particles themselves make up the disperse phase. (We can talk about the disperse phase existing in the continuous phase.)There are eight types of colloidal system. The disperse phase and the continuous phase can be a solid, a liquid, or a gas. Table 1.1 will give you an idea of the wide variety of colloidal systems that are to be found, whether in nature or made artificially. The only combination of disperse and continuous phase that does not occur is gas in gas.Eight Different Types of Colloidal Dispersions are:1. Foam2. Solid foam3. Liquid Aerosol4. Emulsions5. Gels6. Solid Aerosol7. Sol (Colloidal suspension)8. Solid sol (Solid suspension)

There are two approaches to making colloids; either you can take large particles and make them smaller, or you can take small ones and make them larger. The first method is dispersal, the second is aggregation. An example of dispersion is making salad dressing out of oil and vinegar. Vigorous shaking of the two together breaks up the two layers of liquid into tiny droplets, many of which can be

of a colloidal size. You can use an aggregation technique in some simple chemical reactions. For example brushed or stirred too vigorously, the gel can be disrupted and a sol re-formed. The viscosity of a sol is much less than that of a gel so the paint will begin to drip. The decrease in viscosity of many colloidal systems with the increased rate at which they are agitated is also responsible for the deadly nature of quicksand. The more that a creature trapped in the sand struggles, the more the structure of the colloidal quicksand is changed, and the faster the creature will sink.Whether a sol is negatively or positively charged depends on its nature, and the way it is make. If the charges are neutralized than there is little to stop the particles coming together and coagulating. The coagulation of some sols cannot be reversed. These are the lyophobic (solvent hating) sols (Figure 1.8).We can make use of the fact that many sols contain charged particles in an electrophoresis experiment (Figure 1.9). Here, a little of the sol is placed between two electrodes. Depending on their charge, the colloidal particles travel to the positively charged or the negatively charged electrode.

Some colloids are lyophilic (solvent loving) sols (Figure 1.10). Normally the coagulation of a lyophilic sol can be reversed. Gelatin in water is a lyophilic sol that you may have used in cooking, and almost certainly you will have eaten. The change from the (almost) solid gel (jelly) into liquid sol can be achieved by dissolving in hot water. The process is reversed by cooling.Biologically important colloids, often made out of proteins, are also lyophilic. The colloidal particles have a tendency to adsorb water molecules onto their surface owing to their ability to hydrogen bond with water molecules. In addition they can trap ions such as H+ or OH- that may be in the water.Both lyophilic and lyophobic colloids can be coagulated by adding inorganic salts, but for different reasons. For example, proteins can be coagulated by adding large quantities of ammonium sulphate to an aqueous solution of the protein. Ammonium sulphate is very soluble in water, and when large numbers of ammonium and sulphate ions are released into water, the water molecules are pulled way from the surface of the protein to make hydration spheres around the ions. When the protecting layer of water molecules around then protein is removed, the protein molecules begin to stick together and coagulate. A positively or negatively charged hydrophobic sol can be precipitated by adding ions of the opposite charge.

Physical Chemistry: Gases - States of matterSTUDY OF DIFFERENT LAWS OF GASES

In the previous chapter, we studied about the states of matter. We had studied the number of ways in which matter can exist on this planet. We had also discussed how they differ in their chemical and physical properties.

Now in this chapter we will talk about different laws of gases. What will be the impact on the gas when any of the parameters (pressure, temperature, volume etc) are changed? That is what we will elaborate upon.

1. Gay-Lussac’s law of combining volumes

Gay-Lussac (1808) announced the law that is now known as Gay-Lussac’s law of combining volumes. It states that:

“When gases combine, they do so in volumes those are in a simple whole number ratio to each other and to that of the product (if it is a gas). It is assumed that the gas volumes are measured at the same temperature and pressure.”

For example:

1 liter of nitrogen gas reacts with 3 liters of hydrogen gas to produce 2 liters of ammonia gas

N2 (g) + 3H2 (g) -----> 2NH3 (g)

Since all the reactants and products are gases, the mole ratio of N2 (g):H2 (g):NH3 (g) of 1:3:2 is also the ratio of the volumes of gases

Therefore, 100mL of nitrogen gas would react with 100 x 3 = 300mL of hydrogen gas to produce 100 x 2 = 200mL ammonia gas.

Gay-Lussac was also correct about the results of his experiments leading to the time when it became possible to predict not just, what would be made in a chemical reaction, but how much would be made. However, even as great a chemist as Dalton was not entirely convinced that Gay-Lussac was correct.

In 1810, Dalton wrote:

The truth is, I believe, that gases do not unite in equal or exact measures in any one instance; when they appear to do so, it is owing to the inaccuracy of our experiments.

However, Dalton pointed out that:

In fact, his [Gay-Lussac’s] notion of measures is analogous to mine of atoms; and if it could be proved that all elastic fluids [gases] have the same number of atoms in the same volume, or numbers that are as 1, 2, 3, etc. the two hypotheses would be the same, except that mine is universal, and his applies only to elastic fluids.

It is clear that Dalton saw a link between Gay-Lussac’s work on volumes of gases and his own idea that all chemicals were built from individual atoms. Dalton was also claiming that these ideas were the more powerful of the two.

2. Avogadro’s law:-

Amadeo Avogadro gave the idea that equal volumes of gases might contain equal numbers of atoms in 1811. He made more exact the connection between gas volumes and atoms. The title of Avogadro’s paper was ‘Essay on a manner of determining the relative masses of the elementary molecules of bodies, and the proportions in which they enter into these compounds’.

He said: It must then be admitted that very simple relations also exist between the volumes of gaseous substances and the numbers of simple or compound molecules, which form them. The first hypothesis to present itself in this connection, and apparently even the only admissible one, is the supposition that the number of integral molecules in any gases is always the same for equal volumes, or always proportional to the volumes.

The thing to notice about this passage is Avogadro’s use of the term molecule. In his writing, he tended to use this work in a variety of different senses, but he certainly used it at times in the same way that we do now. The leap in imagination that Avogadro had made was to realize that a single atom was not necessarily the basic building block of gases:

We suppose that the constituent molecules of any simple gas whatever are not formed of a solitary elementary molecule, but are made up of a certain number of these molecules united by attraction to form a single one.

In this passage, he uses the term ‘elementary molecule’ to mean a single atom. His ‘compound molecule’ is two or more atoms joined.

Let us see how Avogadro and Gay-Lussac’s ideas work out in our modern language of chemistry. The results of Gay-Lussac’s investigations had shown that hydrogen and chlorine gas reacted in the ratio of one volume of hydrogen to one volume of chlorine to give two volumes of hydrogen chloride:

Hydrogen + oxygen ----> water

2 L 1 L 2L

Source: - www.chem.ufl.edu

If, as was originally thought, the smallest part of each of these gases was one atom, and if, as Dalton wondered, gases ‘have the same number of atoms in the same volume’, then we should have

Hydrogen + oxygen ---> water

1 atom ½ atom 1 atom

However, one of the key assumptions of the atomic theory was (and is) that fractions of atoms cannot exist. Avogadro saw the way round the problem. In effect, his suggestion was that the smallest part of each of the gases was a molecule; and the molecules were made out of atoms joined together. This allows us to write the reaction in a way that might be familiar to you:

Hydrogen + chlorine -----> hydrogen chloride

1 volume 1 volume 2 volume

H2 (g) + Cl2 (g) -----> 2 HCl (g)

Now we can say that one molecule of hydrogen will be made from half molecules, i.e. atoms, of hydrogen and chlorine. If this seems obvious, please do appreciate that it was not always so.

In its modern version, Avogadro’s theory says that:

Under the same conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules.

The Avogadro constant (NA) is our measure of the number of particles in each mole of a gas. Its value is approximately 6.023 * 1023 mol-1. Notice the units of NA. It is not a pure number, although you may find books call it the ‘Avogadro number’. You will find the definition of the Avogadro constant in Unit 37.

3. Dalton’s law of partial pressures

It says that:

“The total pressure exerted by a mixture of gases is the sum of the pressures that the individual gases would exert if they were to occupy the same volume alone.”

In symbols, for a mixture of gases A and B, the total pressure P total is given by

P total = PA + PB Dalton’s law of partial pressures

This is law is not obeyed by all mixtures of gases, but it is often a very good approximation. In the last unit, we saw how this law could be explained by the kinetic theory of gases.

Source:- cwx.prenhall.com

In this figure, it is clear that, partial pressures exerted by hydrogen and helium gas are 2.9 atm and 7.2 atm respectively. When same amount of both used in mixture, then total pressure of mixture is 10.1 atm i.e. sum of partial pressure of individual gas.

Example: 7.96 grams of NH3 is collected by the downward displacement of water at a total pressure of 2.68 atm, at a total volume of 4.55L at a temperature of 270C. Find the vapor pressure of water vapor.

Solution: Using the ideal gas equation, P = nRT / V,

Or, n = 7.96/17 = 0.47 moles.

P = .47 X 0.0821 X (273+27) / 4.55 = 2.53 atm.

, Aqueous tension = (2.68 - 2.53) atm = 0.15 atm.

Example: A 900 ml of a mixture of oxygen and ozone weigh 1.4 gram at STP. Calculate the volume of oxygen present in the mixture.

Solution: Let the volume of oxygen present be V.

, Volume of ozone present = (900 - V).

At STP, 1 mole of any gas occupies 22400 mL, so we can find out the number of moles of both gases.

Moles of oxygen = V / 22400

Mass of oxygen = (V / 22400) * 32..................... (i)

Moles of ozone = (900 - V). / 22400

Mass of ozone = (900 - V). / 22400 * 48 ..................... (ii)

Now, given mass of mixture = 1.4 grams.

Using (i) and (ii) we get,

(V / 22400) X 32 + (900 - V). / 22400 X 48 = 1.4

So, V = 740 mL.

4. Boyle’s Law:-

If temperature of a gas is constant throughout the experiment, the pressure of gas is inversely proportional to the volume, provided number of moles of the gas does not change during the experiment.

PV = constant

At a constant temperature and number of moles, when pressure of a gas is increased from P1 to P2, then the volume of gas will decrease from V1 to V2.

Form Boyle’s law; P1V1= P2V2

Different graphical form of Boyle’s law:-

Graphical representation of Boyle’s law at different temperature

Relationship between density and pressure:

5. Charles’s Law

At the constant pressure, the volume, for a given number of moles of gas is directly proportional to its absolute temperature.

V/T = constant

If at a constant pressure p, volume of gas is increased from V1 to V2, then temperature will also increase from T1 to T2.

From the graph, it is clear that the volume of the gas should be zero 0K and this temperature is known as absolute zero of temperature. If we decrease the temperature beyond this, the volume of gas would be zero. That is not possible.

What does it mean???

It means, at absolute zero temperature, the kinetic energy of gaseous molecule is zero; the heat in the gas is zero. It is therefore impossible to cool the gas below absolute zero.

6. Amagat’s Law of partial volume:-

The total volume of a gas mixture at a given temperature is the sum of volumes of individual constituent particles.

V = v1 + v2 + v3 + …………………………..

Where, V is the Total volume of the system.

The partial volume of a constituent gas is defined as the mole fraction of gas multiply by total volume.

Partial volume = (moles of constituent gas / Total moles) * Total volume

7. Graham’s law of diffusion

In your laboratory, you may find a small jar called a porous pot. It is made of a clay-like material, which has millions of tiny channels through it. If you were to put water in it, you would see the outside become damp, but the water would not pour through in large amounts.

Likewise, if you were to put oxygen in the jar (and close it with a lid) the oxygen would escape, but over a period. The movement of the molecules through the channels in the pot is called diffusion.

A similar effect takes place if oxygen is trapped in a gas jar closed with a piece of card. If a pinhole were made in the card, the oxygen would gradually escape. The movement of a gas through a small hole is sometimes called effusion, but we shall continue to call it diffusion.

If a gas is allowed to pour out all at once from a container, mass (or bulk) flow takes place. This is quite different to diffusion.

In 1829, Thomas Graham published ‘A short account of experimental researches on the diffusion of gases through each other and their separation by mechanical means’. He filled tubes like those shown in Figure 36.1 with a measured amount of various gases. After leaving them for 10 hours, he determined the amount of gas left. He said:

Graham’s law states that at a constant temperature and pressure gradient the rates of effusion or diffusion are inversely proportional to the square root of their densities.

Let r1 and r2 the rates of diffusion of two gases having density d1 and d2, then

Let m1 and m2 are the number of moles of the gases and t1 and t2 are the time of flow for equal volumes of the gases. Then,

NOTE:-

1. Rate of effusion is directly proportional to square root of molar mass.

2. Rate of effusion is directly proportional to the number of moles.

3. Rate of effusion decreases with time.

Graham realized the importance of this result. In the title to his publication, he referred to the separation of gases. The idea is that if a mixture of gases is allowed to diffuse through a barrier, then the lighter gases will escape from the mixture more rapidly than the heavier ones. This fact has been used in the separation of the isotopes 238

92U and 23592U. The isotope 235

92U is the one that was the basis of the first atomic bombs. To separate this isotope from the very much more abundant 238

92U (roughly 1% compared to 99%), the uranium was converted to gaseous uranium hexafluoride. The difference in rates of diffusion of 235UF6 and 238UF6 is small, but after many diffusion cycles, the 235UF6 can be separated.

Example: An unknown gas ‘A’ has molar mass 16 and it effuses at the rate of 60mL s-1. 10 Liters of another gas took 1000 seconds to effuse. How many molecules would 28.8 grams of this gas contain?

Solution: First, we will have to find out the molar mass of the gas. Now, we know that:

r1 / r2 = (V1t2 / V2t1)

And also r1 / r2 = (M2 / M1)1/2

Combining both, we get (M2 / M1)1/2 = (V1t2 / V2t1).

Now, M1 = 16,

V1 = 60

t1 = 1,

M2 = M2,

V2= 10,000,

t2 = 100.

Putting all the values we get, (M2 / M1)1/2 = (60 * 1000/10,000) = 6

M2 = 36 X 16 = 576 g mol-1

Hence, the molar mass is 576,

i.e., 576 grams of the gas contains 1 mole of the gas = N0 molecules. (N0 = 6.023 * 1023)

28.8 grams will contain (28.8 * N0 / 576) molecules

= N0/20 molecules.

= 3.011 X 1023 molecules.

Example: A 0.05 mol sample COCl2 gas is placed in a 0.3L container at the temperature of 1270C. The final pressure in the container is 6 atm. Find the fraction of COCl2 which has dissociated. The dissociation reaction is:

COCl2 (g) --> CO (g) + Cl2 (g) Solution:

First, we find out the total number of moles present inside the container. From the ideal gas equation, we get:

n = PV/RT

= 6 X 0.3 / 0.0821 * 400

= 0.055

Now, from the dissociation reaction we get:

COCl2 (g) --> CO (g) + Cl2 (g)

0.05 - x x x

, Total number of moles available in the container: 0.05 + x = 0.055 (calculated from above)

x = 0.005.

Hence, fraction of COCl2 dissociated is 0.005/ 0.05 = 0.1 or 10%.

8. Ideal Gas Equation:-

Combining Boyle’s and Charles’s law,

PV = nRT

For n = 1 mole,

PV = RT

Or, PV/T = R = constant

The constant R is known as universal gas constant.

If R is divided by Avogadro constant, then new constant is known as Boltzmann constant,

k = (R/NA)Or,

Value of R

R = 8.314 Joules per degree per mole

R = 1.987 ~ 2 Calories per degree per mole

R = 0.0821 Liter-atm per degree per mole

Example: A balloon containing 20 L of He gas was connected to a empty cylinder of volume 80L .If the initial pressure was 1 atm , what should be the final pressure.

Solution: Applying, Boyle’s Law, as the temperature of the system is constant:

P1V1 = P2V2

P1 = 1 atm

V1 = 20 L

V2 = (20 + 80) L = 100 L

P2 = (20 X 1 /100) = 0.2 atm.

Example: - Imagine there is a system, which maintains constant pressure irrespective of the amount of gases inside it. Now suppose at Instant A, it contains n1 moles of a gas X at temperature t1. At instant B, n2 moles of other non-reacting gas Y were injected into the system. Find the overall change in the temperature of the system.

Solution:

Since, the pressure of the system is constant therefore; we can apply Charles’s law here.

i.e., V1/T1 = V2/T2

Now, volume of any gas is directly proportional to the number of moles.

Therefore, the above equation can be written as:

ni / T1 = nf / T2

Now, ni = n1

nf = n1 + n2

The equation becomes, n1 / t1 = (n1 + n2) / t2

Alternatively, t2 = (n1 + n2) t1 / n1.

Change in temperature,

Example: - Equal weights of methane and oxygen are mixed in an empty container at 25oC. The fraction of the total pressure exerted by oxygen is (IIT-JEE-1981, 1984,1993)

(a) 1/3

(b) 1 /2

(c) 2/3

(d) 1/3 * ( 273/298)

Answer:- (a)

Let the weight of both gases are x g.

Moles of oxygen = x /32

Moles of methane = x/16

Total moles = x/32 + x/16 = 3x/32

Total pressure = p

Partial pressure exerted by the oxygen = mole fraction * total pressure

pO2 = 1/3 * p

Fraction of total pressure exerted by the oxygen gas is ( pO2/p)= 1/3

Example:- Rate of diffusion gas is (IIT-JEE 1985)

(a) Directly proportional to its density

(b) Directly proportional to its molecular weight

(c) Directly proportional to the square root of its molecular weight

(d) Inversely proportional to the square root of its molecular weight

Answer:- (d)

Example:- The rate of diffusion of methane at a given temperature is twice that of a gas X. The molecular weight of X is (IIT JEE 1990)

(a) 64

(b) 32

(c) 4

(d) 8

Answer:- (a)

We know that rate of diffusion:

Let molecular weight of gas X is M .Then,

Example:- 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 (where, p and M are pressures and molecular weights of gases A and B respectively)

Answer:- (d)

Question :- At 100oC and 1 atm if the density of the liquid water is 1g/cm3 and that of water vapor is 0.0006g /cm3 ,Then the volume ( cm3) occupied by water molecules in 1L of steam at this temperature is ( IIT-JEE 2000)

(a) 6

(b) 60

(c) 0.6

(d) 0.06

Answer: - Let us consider, 1L of water is converted into steam.

Volume of water (liquid phase) = 1L = 1000 g

Volume of 1000 g steam = 1000/ 0.006 cm3

Volume of molecules in 1000/ 0.006 cm3 steam = 1000 cm3

Volume of molecules in 1000 cm3 steam is = (1000/1000) * 0.006 * 1000 = 0.6cm3

Example:- If the helium and methane are allowed to diffuse out of the container under the similar conditions of temperature and pressure, then the ratio of rate of diffusion of helium and methane is ( IIT –JEE 2005)(a) 2

(b) 1

(c) ½

(d) 4

Answer:- ( a)

Kinetic theory of gases - States of MatterKINETIC THEORY OF GASES

It is a theoretical model to explain the behavior of an ideal gas. This theory follows some following assumptions.

1. There are no intermolecular forces between gas molecules.

2. Gas molecules have high translational energy. Therefore, they move randomly in all directions.

3. The volume of an individual gas molecule is negligible compared to the volume of the container holding the gas. This means that individual molecules do not have its volume.

4. When a gas molecule collides with each other or with a container wall, the collisions are perfectly elastic. It means there is no loss of kinetic energy.

5. The average kinetic energy of gas molecule is directly related to temperature. The greater the temperature, the greater the average motion of molecules and the greater the kinetic energy.

No real gas is ideal, although some come close to ideal behavior, e.g. helium. The power of the theory is that is allows us to account for the behavior of ideal gases and, by making a few other assumptions, the properties of real gases as well. In the rest of this unit, we shall see how this is done.

The pressure of an ideal gas

The pressure of an ideal gas:

p = 1/3Nmv2 ------------ (1)

This equation says that the pressure depends on N, the number of molecules per unit volume of the gas, on their mass, m, and on the mean square speed, v2.

Now, suppose the molecules were not all of the same mass. If we had two types of molecule, say A and B, the pressure would be due to both of them bombarding the walls, and

Total pressure = 1/3NAmAv2A + 1/3NBmBv2B

This is easier to write as

P total = pA + pB

This is one way of writing Dalton’s law of partial pressures. In words, the law says that:The total pressure of a mixture of gases is the sum of the pressures that each gas would have if it were on its own.

Strictly, the law is only valid for ideal gases. Real gases do not obey the law perfectly.

Speed of gaseous molecule

(a) RMS speed: - The rms speed of gaseous molecule is the hypothetical speed which all the molecules will move if the kinetic energy is equally distributed among them.

For n molecules having speeds v1, v2, v3 …, the rms speed will be

Now equation (1) will be

(b) Average speed:-

For n molecules having speeds v1, v2, v3 …, the average speed will be

(C) Most probable speed: - Speed posses by the most fractions of molecules.

Therefore, rms speed: Average speed: Most probable speed1 : 0.9211 : 0.8165

As the temperature of a gas sample increases from left to right, the molecular speed distribution increases.

Example: - Calculate the rms speed of oxygen gas at STP. The density of oxygen gas in these conditions is 1.2kg/m3.

Solution: - At STP; p = 1atm = 105 N/m2

Translation kinetic energy of a gas:-

The total translation kinetic energy of all the molecules of gas is

The average kinetic energy of the molecule is

We know that

For n moles of gas

The connection between energy and temperature

Another important result from kinetic theory is that the average kinetic energy of a molecule is an ideal gas is given by

½mv2 = 3/2 kT

Where k is Boltzmann’s constant, of value 1.38 X 10-23 J K-1. The value of Boltzmann’s constant multiplied by the Avogadro constant is the same as the gas constant, i.e. R = LK, which has the value 8.314 J K-1 mol-1.

The importance of the equation for the average kinetic energy of gas molecules is its link with the temperature of the gas. The factor 3 makes its appearance in the equation owing to the three directions in which a molecule can move. If it could only move in one direction, its average kinetic energy would only be KT/2. A molecule’s ability to move in any one of a set number of ways is known as its degrees of freedom. A molecule in an ideal gas has three degrees of freedom. Non-ideal gases may have other degrees of freedom, for example due to their vibrations and rotations. The sharing out of energy, giving KT/2 to each degree of freedom, is called the principle of equi-partition of energy.

Now, consider a mixture of two gases A and B. Let m1 be the mass of a molecule of the first gas and m2 be that of the second. As the molecules collide with each other, they exchange energy. The molecules having higher kinetic energy will lose energy and the molecules having lower kinetic energy will gain this energy. In thermal equilibrium, the average kinetic energy of all molecules is equal.

Example: -If the rms speed of oxygen molecule is 500 m/s at 273K, find the rms speed of hydrogen molecules at the same temperature.

Answer:- The molecular weight of oxygen = 32 g/mole

The molecular weight of hydrogen = 2 g/mole

Given that, temperature is same

Therefore,

Collision Frequency (n):- The number of collisions occurred between molecules per cc of a gas per second is known as collision frequency.

Total number of molecules colliding per cc per second is

Collision frequency (n) = Z/2

Or,

- It is average distance travelled by a molecule between two successive collisions.

This figure shows how a gaseous molecule moves when it is enclosed in a container. It collides with so many other molecules, loses its energy, and follows a path that described in above figure.

Assuming that, the collision diameter is independent of temperature. And p = NkTTherefore,

Viscosity of gases: - viscosity is the resistance, which allows the flow of fluid. Gases also show viscous behavior. For gases,

Where, k = Boltzmann constant equal to R/N, and m are diameter and mass of molecule respectively. The unit of viscosity is poise.

Viscosity of gases increases as the temperature increases and is independent of temperature.

Molar heat capacity of ideal gases:- Specific heat is the amount of heat required to raise the temperature of a g

of substance through 1oC. The unit of specific heat is calorie.

1 calorie:- It is the amount of heat required to raise the temperature of 1 mole of gas through 1oC.

Molar heat capacity is the amount of heat required to raise the temperature of 1 mole of gas through 1oC.

Molar Heat capacity = Specific heat * Molecular weight of gas

1. Molar heat capacity at constant volume ( Cv)

2. Molar heat capacity at constant pressure ( Cp)

For Mono-atomic gases,

The spread of energies in a gas

One of the nice things about the kinetic theory is that we can use it to estimate the speeds of molecules in a gas. For example, let us use

½mv2 = 3/2kT

For helium, at 25°C (298 K). In this case m = 6.64 X 10-27 kg and a short calculation shows that

v2 = 1.86 X 106 m2 s-2

If we take the square root of v2 we have found the value of the root mean square speed of the molecules. For helium, it is 1.36 X 103 m s-1. Alternatively, this is approaching 5000 kph (kilometers per hour). The average speed of the molecules is about 8% less than the root mean square speed, so the average speed of a helium atom at 25° C is approximately 1.25 X 103 m s-1.

Of course, it is one thing to calculate the speeds of gas molecules from theory, quite another to measure them by experiment. Zartman performed one type of experiment that does just this in 1931 (although he was not the first to think of the method).

Zartman evaporated bismuth by placing a small amount of the element into a specially designed oven at over 800° C. Some of the bismuth atoms were able to escape through a small slit. The second slit ensured that the atoms were traveling in one direction. In front of the beam was a rotating drum, which also had a small slit in it. Once in every revolution the three slits were lined up, so bismuth atoms could enter the drum. If you imagine yourself to be an atom moving very quickly through the slit in the drum, you would see the far surface of the drum moving clockwise (as we have drawn it). You would hit the surface at a point not quite opposite the slit. However, if you were an atom with a much slower speed, you would see the drum move a considerable distance before you collided with the far surface.

If the drum is kept rotating at a constant speed, you should realize that where an atom (or molecule) hits the inside of the drum depends on the atom’s (of molecule’s) speed. By measuring the degree of darkening of the surface (which was made of glass), it is possible to estimate the number of atoms (or molecules) that enter the drum with a particular speed.

A graph of the number of bismuth atoms plotted against their speed is shown in following figure. This graph shows us the distribution of the speeds (or velocities) of the atoms. Notice that the distribution is not symmetrical. There is a longer ‘tail’ at lower speeds than at higher speeds. James Clerk Maxwell had worked the shape of this distribution out many years before in 1860. For this reason, we speak about a Maxwellian distribution of speeds (or velocities).

In a later unit, you will find that knowledge of the Maxwellian distribution is very helpful in explaining the rates of chemical reactions.

Kinetic theory and Avogadro’s theory

In the following unit, we shall find that Avogadro’s theory (equal volumes of gases contain equal numbers of molecules) was based on Gay-Lussac’s experimental evidence. Here we can show that this theory is correct. If we have equal volumes of two gases, A and B, at the same temperature, they must be in equilibrium with each other. Their pressures will be the same, so we can put

1/3NAmAv2A = 1/3NBmBv2A

In addition, because they are at the same temperature, the average kinetic energies of their molecules must be the same, i.e.

½mAv2A = 1/3mBmBv2

B

Putting the two equations together, we must have NA = NB. That is, we have shown that, under the same conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules.

Solved Examples:-

Example:- The ratio of root mean square velocity to average velocity of a gas molecule at a particular temperature is (IIT-JEE 1981)(a) 1.086 :1

(b) 1:1.086

(c) 2:1.086

(d) 1.086:2

Answer:- (a)

From theory, rms speed: Average speed1 : 0.9211

Or , RMS/ Average = 1/0.9211 = 1.086 :1

Question: - The rms velocity of hydrogen is times the rms velocity of nitrogen. If T is the temperature of the gas (II-JEE 2000)

(a) T(H2) = T(N2)

(b) T(H2) > T(N2)

(c) T(H2) < T(N2)

(d) T(H2) = T(N2)

Answer:- (C)

We know that rms speed of gas,

Example: - The root mean square velocity of an ideal gas at constant pressure varies with density (d) as

(IIT-JEE 2001)

Answer d:

Example: - For a mono-atomic gas kinetic energy = E. The relation with rms velocity is

(IIT-JEE 2004)

Answer: - (a)

Example: - Kinetic energy of a molecule is zero at 0oC. (IIT-JEE 1985)

(a) True

(b) False

(c) May be true or false

(d) None of these

Answer: - ( b)

At 0oc, value of T = 273K

Therefore, K.E will not be zero.

Example: - 8 g each of oxygen and hydrogen at 27oC will have the total kinetic energy in the ratio of

(IIT-JEE- 1989)

(a) 1:16

(b) 16:1

(c) 1:8

(d) 8:1

Answer:- (a)

Kinetic energy of gas is given by;