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	Gases
1.	Particles are very Closely packed	Particles are loosely packed	Particles are very loosely packed
2.	Voids are extremely small	Voids are relatively larger	Voids are very large
3.	Inter particle forces are large	Inter 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 = w/M
  Where            n = number of moles

  w = mass of gas in grams

  M = 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 KP_­a = 101325 P_a = 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 ∝ 1/V  (n, T constant)

  V ∝ 1/P  (n, T constant)

  i.e. P = K/V (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

  

• • (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.

  Question.         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 litre

  T₁ = 273 K                    T₂ = 273 K

  As temperature is constant

  Then from Boyle’s law

  P₁V₁ = P₂V₂

  or 1 x 100 = P₂ x 5

  Therefore, P₂ = 20 atm

  Therefore, Additional pressure that should be applied = P₂ – P₁ = 20 – 1 = 19 atm

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 + at) or V – V₀ = V₀ µ t

If the temperature is measured in the Celsius scale and V₀ is the volume at 0°C, it is found that a = 1/273. The volume at temperature T is then;

V_t = V₀ (1 + t/273) = V₀ x ((273 + t)/273) V_T = V_oT/273

 

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:

V₁ / T₁ = V₂ / T₂       (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

\(\displaystyle \frac{{{{P}_{1}}{{V}_{1}}}}{{{{T}_{1}}}}=\frac{{{{P}_{2}}{{V}_{2}}}}{{{{T}_{2}}}}\)

The above relation is called the combined gas law.

Avagadro’s Law

Let us take a balloon containing a certain mass of gas. If we add to it more mass of gas, holding the temperature (T) and pressure (P) constant, the volume of gas (V) will increase. It was found experimentally that the amount of gas in moles is proportional to the volume. That is, \(\displaystyle V\,\,\alpha \,\,\,n\,\) (T and P are constants)

or V = An

where A is a constant of proportionality or V/n  constant.

Thus, for equal volumes of the two gases at fixed T and P, number of moles is also equal. This is the basis of Avagadro’s law which states that:

Equal volumes of gases at the same temperature and pressure contain equal number of moles or molecules.

It has been observed that 1 mole of gas at a temperature of 273 K and a pressure of 1 atmosphere (STP or NTP) occupies 22.4 Litres. This is called the molar volume of a gas.

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 law

V ∝ T [n, P constant]     Charle’s law

V ∝ 1/P [n, T constant] Boyle’s Law

The combined gas law can be written as \(\displaystyle V\propto \frac{{nT}}{P}\) or PV ∝ nT

Therefore PV = nRT

this is called ideal gas equation

where 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)

\(\displaystyle n=\frac{W}{M}\);  \(\displaystyle P=\frac{{nRT}}{V}\);  \(\displaystyle P=\frac{W}{V}\frac{{RT}}{M}\);  \(\displaystyle P=d\frac{{RT}}{M}\);  \(\displaystyle d=\frac{{PM}}{{RT}}\)

Question.         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.

\(\displaystyle \frac{{{{V}_{1}}}}{{{{T}_{1}}}}=\frac{{{{V}_{2}}}}{{{{T}_{2}}}}\) or V_­1 = \(\displaystyle \frac{{{{T}_{1}}}}{{{{T}_{2}}}}\times {{V}_{2}}\), V₁ = \(\displaystyle \frac{{320}}{{300}}\times 600\) = 640 ml

Increase in volume of air = 640 – 600 = 40

Question.         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 givenΡ = \(\displaystyle \frac{{PM}}{{RT}}\) Obviously the choice that has greater P/T

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

Gay- Lussac’s Law

         In 1802 Joseph Gay-Lussac as a result of his experiments established a general relation between the pressure and temperature of a gas. This is Gay-Lussac’s law and it states that:

At constant volume, the pressure of a fixed mass of gas is directly proportional to the Kelvin temperature or absolute temperature.

The law may be expressed mathematically as

\(\displaystyle P\,\,\alpha \,\,\,T\,\,\,\,\,\,\text{or}\,\,\,P=kT\,\,\,\,\text{or}\,\,\,\,P/T=\operatorname{constant}\,\,\,\,\text{or}\,\frac{{{{P}_{1}}}}{{{{T}_{1}}}}=\frac{{{{P}_{2}}}}{{{{T}_{2}}}}\)

Dalton’s Law

John Dalton visualized that in a mixture of gases, each component of gas exerted a pressure as if it were alone in the container. The individual pressure of each gas in the mixture is defined as its partial pressure. It states that the total pressure of a mixture of non-reacting gases is equal to the sum of the partial pressures of the individual gases present.

Mathematically the law can be expressed as \(\displaystyle P={{P}_{1}}+{{P}_{2}}+{{P}_{3}}…\) where (V and T are constant) where P₁ P₂ and P₃ are partial pressures of the three gases 1, 2 and 3; and so on :

Dalton’s law of partial pressures follows by application of the ideal gas equation   separately to each gas of the mixture. Thus we can write the partial pressures P₁, P₂ and P₃ of the three gases as :

\(\displaystyle {{P}_{1}}={{n}_{1}}(RT/V),\,\,\,{{P}_{2}}={{n}_{2}}\,(RT/V),\,\,\,\,{{P}_{3}}={{n}_{3}}\left( {RT/V} \right)\)

Where n₁, n₂ and are moles of gases 1, 2 and 3. The total pressure, P₁ of the mixture is \(\displaystyle {{P}_{t}}=\left( {{{n}_{1}}+{{n}_{2}}+{{n}_{3}}} \right)RT/V,\)

\(\displaystyle {{P}_{t}}={{n}_{t}}\,RT/V.\)

In other words the total pressure of the mixture is determined by the total number of moles present whether of just one gas or mixture of gases.

P₁ = n₁P_t/ (n₁ + n_2);  n₂P_t/(n₁ + n₂)

\(\displaystyle {{P}_{1}}={{x}_{1}}.\,{{P}_{t}}\,\,\,\text{and}\,\,\,{{P}_{2}}={{x}_{2}}.{{P}_{t}}\)

Partial pressure of gas its mole fraction xTotal pressure

% of gas in mixture \(\displaystyle =\frac{{Partial\,pressure\,of\,gas}}{{Total\,pressure}}\times 100\)

\(\displaystyle {{P}_{{Dry\,Gas}}}={{P}_{{Moist\,Gas}}}-{{P}_{{Water\,Vapour}}}={{P}_{{Observed}}}-\text{Aqueous}\,\text{Tension}\)

Graham’s Law

When two gases are placed in contact, they mix spontaneously. This is due to the movement of molecules of gases by random motion of the molecules and is called Diffusion.

Thomas Graham observed that molecules with smaller masses diffused faster than heavy molecules. In 1829 Graham formulated what is now known as Graham’s Law of Diffusion.

It states that: under the same conditions of temperature and pressure, the rates of diffusion of different gases are inversely proportional to the square root of their molecules masses or densities. Mathematically the law can be expressed as

\(\displaystyle \frac{{{{r}_{1}}}}{{{{r}_{2}}}}=\sqrt{{\frac{{{{M}_{2}}}}{{{{M}_{1}}}}=}}=\sqrt{{\frac{{{{d}_{2}}}}{{{{d}_{1}}}}}}\)

Where r₁ and r₂ are the rates of diffusion of gases 1 and 2, while M₁ and M₂ are their molecular masses.

When a gas escapes through a pin-hole into a region of low pressure of vacuum, the process is called Effusion. The rate of effusion of gas also depends on the molecular mass of the gas. Mathematically,

\(\displaystyle \frac{{{{E}_{1}}}}{{{{E}_{2}}}}=\sqrt{{\frac{{{{M}_{2}}}}{{{{M}_{1}}}}}}\)

Question        A flask of 2 dm³ capacities 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 that

\(\displaystyle {{V}_{1}}=2d{{m}^{3}}\,\,\,{{P}_{t}}=101.325\,\,\,kPa,\,\,\)

P₂= 0.1 Pa       T=300k

We have the following results.

(i)   The volume of O₂ left behind will be the same, i.e. 2 dm³

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

\(\displaystyle n={{P}_{2}}\,\,\frac{{{{V}_{1}}}}{{kT}}=\frac{{{{{10}}^{{-4}}}kpa\times 2d{{m}^{3}}}}{{\left( {8.314\,kpad{{m}^{3}}{{k}^{{-1}}}mo{{l}^{{-1}}}} \right)\left( {300K} \right)}}=8.019\times {{10}^{{-8}}}mol\)

2g of N₂= 1/14 mol Total amount of gases in flask \(=1/14\,mol\,\,+\,\,8.019\times {{10}^{{-8}}}\,mol\)

\(=1/14\,\,mol.\)

Thus, the pressure of the flask is given by

\(\displaystyle P=\frac{{nRT}}{V}=\frac{{\left( {\frac{1}{{14\,mol}}} \right)\times \left( {8.314kpad{{m}^{3}}{{k}^{{-1}}}mo{{l}^{{-1}}}\times 300K} \right)}}{{\left( {2d{{m}^{3}}} \right)}}=89.08\,\,kPa.\)

Kinetic Theory of Gases     

Assumptions of kinetic theory of gases

(i)        A gas consists of extremely small discrete particles called molecules dispersed throughout the container. The actual volume of the molecules is negligible compared to the total volume of the gas.

(ii)       Gas molecules are in constant random motion with high velocities.

(iii)      The gas molecules move freely, independent of each other.

(iv)      There is no loss of kinetic energy of a molecule during a collision. All collisions are perfectly elastic.

(v)       The pressure of a gas is caused by hits, recorded by molecules on the walls of the container.

(vi)      The average kinetic energy of molecules depend (directly proportional to) absolute temperature (Kelvin temperature). A gas which confirms to the assumptions of the kinetic theory of gases is called an ideal gas. It can be mathematically proved that for a gas PV=1/3mmu²

Where P = pressure of the gas

V= volume of the gas

m= mass of one molecule

n =no. of molecules of the gas

u²= mean square velocity of the molecules

\(\displaystyle =V_{1}^{2}+V_{2}^{2}+V_{3}^{2}…….+V{{n}^{2}}/n\) \(\displaystyle {{V}_{1}},{{V}_{2}},{{V}_{3}}…..\) are the individual velocities of the molecules. This is called the kinetic gas equation. If N is the number of molecules in a given mass of gas,

\(PV=1/3\,\,m\,\,\,N{{u}^{2}}\)

\(=2/3\,\,N\times \,\,1/2\,\,m{{u}^{2}}\)

\(=2/3\,\,N\times e\)

Where e is the average kinetic energy of a single molecule.

\(PV=2/3\,\,Ne=2/3\,\,E\)

or \(PV=2/3\,\,\,E\)

Where E is the total kinetic energy of all the N molecules.

Different Kinds of Velocities

In our study of kinetic theory we come across different kinds of molecular velocities.

1. Average velocity (V_av)
2. Root Mean Square velocity (V_rms)
3. Most Probable velocity(V_mp)

Average Velocity

Let there be n molecules of a gas having individual velocities \(\displaystyle {{V}_{1}},{{V}_{2}},{{V}_{3}}….{{V}_{{n.}}}\). The ordinary average velocity is the arithmetic mean of the various velocities of the molecules. \(\displaystyle {{V}_{{av}}}=\frac{{{{V}_{1}}+{{V}_{2}}+{{V}_{3}}+……..{{V}_{n}}}}{n}\)

It has been established by Maxwell that \(\displaystyle {{V}_{{av}}}=\sqrt{{\frac{{8RT}}{{\pi M}}}}\) where M= Molecular mass.

Root Mean square Velocity

If \(\displaystyle {{V}_{1}},{{V}_{2}},{{V}_{3}}…..{{V}_{n}}\) are the velocities of n molecules in a gas V₂ the mean of the squares of all the velocities is

\(\displaystyle {{V}^{2}}=\frac{{V_{1}^{2}+V_{2}^{2}+V_{3}^{2}+…..V_{n}^{2}}}{n}\)

\(\displaystyle {{V}_{{rms}}}=\sqrt{{\frac{{V_{1}^{2}+V_{2}^{2}+V_{3}^{2}+….V_{n}^{2}}}{n}}}\)

V is the thus the Root mean square velocity. It is denoted by V_rms.

It can be seen that \(PV=1/3\,\,mn{{u}^{2}}\)

\({{u}^{2}}=3\,\,PV/mn\)

For one mole of gas \(PV=RT\)

\({{u}^{2}}=3RT/M\)

\(\displaystyle U=\sqrt{{\frac{{3RT}}{M}}}=\sqrt{{\frac{{3PV}}{M}}}=\sqrt{{\frac{{3P}}{d}}}\)

Most probable Velocity

Most probable velocity is possessed by the largest number of molecules in a gas. According to the calculations made by Maxwell, the most probable velocity, Vmp, is given by the expression

\(\displaystyle {{V}_{{mp}}}=\sqrt{{\frac{{2RT}}{M}}}\)

Relation between average velocity, RMS velocity and most probable velocity

We know that the average velocity. V_av is given by the expression

\(\displaystyle {{V}_{{av}}}=\sqrt{{\frac{{8RT}}{{\pi M}}}}\) and \(\displaystyle {{V}_{{rms}}}=\sqrt{{\frac{{3RT}}{M}}}\)

\(\displaystyle \frac{{{{V}_{{av}}}}}{{{{V}_{{rms}}}}}=\sqrt{{\frac{{8RT}}{{\pi M}}}}\times \sqrt{{\frac{M}{{3RT}}}}=\sqrt{{\frac{8}{{3\pi }}}}=0.9213\)

The expression for the most probable velocity, V_mp is

\(\displaystyle {{V}_{{mp}}}=\sqrt{{\frac{{2RT}}{M}}}\,and\,{{V}_{{rms}}}=\sqrt{{\frac{{3RT}}{M}}}\)

\(\displaystyle \frac{{{{V}_{{mp}}}}}{{{{V}_{{rms}}}}}\sqrt{{\frac{{2RT}}{M}}}\times \sqrt{{\frac{M}{{3RT}}}}=\sqrt{{\frac{2}{3}}}=0.8165\)

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 Z = PV/n RT

• 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

Therefore, Ideal volume v_i = (V – v);       Now v α n or v = nb

[n = no. of moles of real gas; b = constant of proportionality called Van der Waal’s constant]

Therefore, V_i = V – nb; b = 4 x volume of a single molecule.

Pressure Correction

Let the correction term be P

Ideal pressure P_i = (P + p);    Now, \(\displaystyle P\propto {{\left( {\frac{n}{V}} \right)}^{2}}\)  = \(\displaystyle \frac{{a{{n}^{2}}}}{{{{V}^{2}}}}\)

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

Hence ideal pressure, P_i = \(\displaystyle \left( {P+\frac{{a{{n}^{2}}}}{{{{V}^{2}}}}} \right)\)

Here,    n = Number of moles of real gas;            V = 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

\(\displaystyle \left( {P+\frac{{a{{n}^{2}}}}{{{{V}^{2}}}}} \right)(v-nb)=nRT\)

Question.         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 litre;   n = 1 mole

T = 300L ; Z = \(\displaystyle \frac{{PV}}{{nRT}}\)

Therefore Z = \(\displaystyle \frac{{50\times 0.350}}{{1\times 0.082\times 300}}=0.711\)

Thus SO₂ is more compressible than expected from ideal behaviour.

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:

\(\displaystyle \left( {P+\frac{a}{{{{V}^{2}}}}} \right)V=RT\); PV + \(\displaystyle \frac{a}{V}\) =RT

PV = RT -\(\displaystyle \frac{a}{V}\) or PV < RT

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

At fairly high pressures a/V2 may be neglected in comparison with P. The Vander Waals equation becomes P (V – b) = RT PV – Pb = RT PV = RT + Pb or PV > RT This accounts for the rising parts of the PV vs P isotherm at high pressures.

• At very low pressures: V becomes so large that both b and a/V² 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 a/V² is negligible even at ordinary temperatures. Thus PV > 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,

\(\displaystyle \left( {P+\frac{a}{{V_{m}^{2}}}} \right)\left( {{{V}_{m}}} \right)=RT\) or Z = \(\displaystyle \frac{{{{V}_{m}}}}{{RT}}=1-\frac{a}{{{{V}_{m}}RT}}\)

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 V_m, Z has more value at higher temperature.

At high pressures, the gas equation can be written as

P (V_m – b) = RT

Z = \(\displaystyle \frac{{P{{V}_{m}}}}{{RT}}\) = 1 + \(\displaystyle \frac{{Pb}}{{RT}}\)

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.

Question.         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

\(\displaystyle \frac{{{{P}_{2}}{{V}_{2}}}}{{{{T}_{2}}}}\times \frac{{{{T}_{1}}}}{{{{P}_{1}}{{V}_{1}}}}=\frac{{{{Z}_{2}}}}{{{{Z}_{1}}}}\) or V2 = \(\displaystyle \frac{{{{Z}_{2}}}}{{{{Z}_{1}}}}\times \frac{{{{T}_{2}}}}{{{{T}_{1}}}}\times \frac{{{{P}_{1}}{{V}_{1}}}}{{{{P}_{2}}}}\) = \(\displaystyle \frac{{1.375}}{{1.072}}\times \frac{{273}}{{473}}\times \frac{{300\times 1}}{{600}}\) = 370.1 ml

Boyle’s Temperature (T_b)

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 [T_b(O₂)] = 46 K] whereas the gases which are difficult to liquefy have a low Boyle’s temperature [T_b(He) = 26K].

Boyle’s temperature T_b = \(\displaystyle \frac{a}{{Rb}}=\frac{1}{2}{{T}_{i}}\)

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

Critical Constants

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

\(\displaystyle {{T}_{c}}=\frac{{8a}}{{27Rb}}\)

• Critical Pressure (P_c): It is the minimum pressure required to cause liquefaction at T_c

\(\displaystyle {{P}_{c}}=\frac{a}{{27{{b}^{2}}}}\)

• Critical Volume: It is the volume occupied by one mol of a gas at T_c and P_c

V_c = 3b

Molar 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 x 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 C_p and C_v. C_p is greater than C_v and C_p-R = 2 cal mol^-1 K^-1.

From the ratio of C_p and C_v, we get the idea of atomicity of gas.

For monatomic gas C_p = 5 cal and C_v =3 cal

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 oil

O_2→ alkaline pyrogallol

NO→FeSO₄ solution

CO₂,SO₂ →alkali solution (NaOH, KOH, Ca(OH)₂, HOCH₂CH₂NH₂, etc.)

NH₃ → acid solution or CuSO₄ solution

Cl₂ →  water

Equation for combustion of hydrocarbons.

C_xH_y + \(\displaystyle \left( {x+\frac{y}{4}} \right)\) O₂→xCO₂ + \(\displaystyle \frac{y}{2}\) H₂O

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 (T_b) of the liquid is nearly two – third of its critical temperature (T_c) when both are expressed on the absolute scale.

\(\displaystyle {{T}_{b}}=\frac{2}{3}\,{{T}_{c}}\)

• 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.

\(\displaystyle \frac{{\Delta {{H}_{\nu }}}}{{{{T}_{{b.p.}}}}}=88\,J{{K}^{{-1}}}\,\,mo{{l}^{{-1}}}\)

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. CCl_{4 ,}CHCl₃ 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,

\(\displaystyle surface\,\,tension=\frac{{Work\,\,done}}{{Change\,\,in\,\,area}}\)

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, Nm^-1 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 Nm^-2s  or Pas (pascal second).

\(\displaystyle 1\,\,Pas=1\,\,N{{m}^{{-2}}}\,s=({{10}^{5}}\,\,dyn)\,{{({{10}^{{+2}}}\,\,cm)}^{{-2}}}\,s=10\,\,dyn\,\,c{{m}^{{-2}}}\,s=10\,\,poise\)

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.

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

\(\displaystyle N{{H}_{3}},\,S{{O}_{2}},\,\,HCl,\,\,HF\) 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.