q Gaseous state
· Intermolecular forces versus Thermal Energy
· The gaseous state
· The gas laws
· Kinetic Molecular Model of gas
· Real Gases
· Liquefaction of gases and Crirical Point
· The liquid state
· Intermolecular Forces
The intermolecular forces are those of attraction and repulsion between interacting particles (atoms and molecules). This term does not include the electrostatic forces that exist between the two oppositely charged ions and the forces that hold atoms of a molecule together i.e., covalent bonds.
Attractive intermolecular forces are known as van der Waals forces. Van der Waals forces vary considerably in magnitude and include dispersion or
forces , dipole-dipole forces, and dipole-induced dipole forces. A particularly strong type of dipole-dipole interaction is hydrogen bonding. London
Attractive forces between an ion and a dipole are known as ion-dipole forces and these are not van der Waals forces.
Dispersion Forces or
Atoms and nonpolar molecules are electrically symmetrical and have no dipole moment because their electronic charge cloud is symmetrically distributed. But a dipole may develop momentarily even in such atoms and molecules. This can be understood as follows. Suppose we have two atoms ‘A’ and ‘B’ in a close vicinity of each other.
Symmetrical distribution of electronic charge cloud
It may so happen that momentarily electronic charge distribution in one of the atoms, say ‘A’ , becomes unsymmetrical i.e., the charge cloud is more on one side than the other (Fig b and c)
Dispersion forces or
forces between atoms London
This results in the development of instantaneous dipole on the atom ‘A’ for very short time. This instantaneous or transient dipole distorts the electron density of the other atom ‘B’ , which is close to it and as a consequence a dipole is induced in the atom ‘B’.
The temporary dipoles of atom ‘A’ and ‘B’ attract each other. Similarly temporary dipoles are induced in molecules also. This force of attraction was first proposed by Fritz London. The force of attraction between the two temporary dipoles is known as
force or dispersion force. These forces are always attractive and interaction energy is inversely proportional to the sixth power of the distance between two interacting particles ( i.e., 1/ r6 where r is the distance between two particles). These forces are important only at short distances (» 500 pm) and their magnitude depends on the polarisability of the particle. London
Dipole - Dipole Forces
Dipole-dipole forces act between the molecules possessing permanent dipole. The ends of the dipoles possess ‘partial charges’ and these charges are shown by d. Partial charges are always less than the unit electronic charge (1.6 x 10-19 C). The polar molecules Fig (a) shows electron cloud distribution in the dipole of hydrogen chloride and Fig (b) shows dipole-dipole interaction between two HCl molecules.
Distribution of electron cloud in HCl – a polar molecule
Dipole-dipole interaction between two HCl molecules
This interaction is stronger than
forces but it is weaker than ion-ion interaction because only partial charges are involved. The attractive force decreases with increase of distance between the dipoles. The interaction energy is inversely proportional to distance between polar molecules. Dipole-dipole interaction energy between stationary polar molecules (as in solids) is proportional to 1 / r3 and that between rotating polar molecules is proportional to 1/ r6 where r is the distance between polar molecules. Besides dipole-dipole interaction polar molecules can interact by London forces also. Thus cumulative effect is that the total of the intermolecular molecular forces in polar molecules increase. London
Dipole – Induced Dipole Forces
This type of attractive forces operate between the polar molecules having permanent dipole and the molecule lacking permanent dipole. Permanent dipole of the polar molecule induces the dipole on the electrically neutral molecule by deforming its electron cloud (Fig).
Dipole-induced dipole interaction between permanent dipole and induced dipole.
Thus an induced dipole is developed in the other molecule. The interaction energy is proportional to1/ r6 where r is the distance between two molecules. Induced dipole moment depends upon the dipole moment present in the permanent dipole and the polarisability of the electrically neutral molecule. Molecules of larger size can be easily polarized. High polarisability increases the strength of attractive interactions.
Cumulative effect of dispersion forces and dipole-induced dipole interactions exists.
Thermal energy is the energy of a body arising from motion of its atoms or molecules. It is directly proportional to the temperature of the substance. It is the measure of average kinetic energy of the particles of the matter and is thus responsible for the movement of particles. This movement of particles is called thermal motion.
INTERMOLECULAR FORCES vs THERMAL INTERRACTIONS
Matter can be classified into three different physical states, namely, solids, liquid and gaseous states. The three states of matter are the result of a competition between molecular interaction energy, which keeps molecules together and thermal energy which moves them apart.
This competition determines whether a given substance under given conditions is a gas, liquid or a solid.
Thermal energy is the energy possessed by matter by virtue of its temperature and is also a measure of thermal motion or movement of molecules. Molecular interactions lead to intermolecular attractive and repulsive forces. In gases , the molecules have almost no molecular interactions and their thermal motions are manifested as random translatory movement of molecules. In solids , the molecular interactions are very strong. The molecules in solids have no translatory movement and only oscillate with respect to their equilibrium position. In liquids the arrangements is in between these two extremes.
When the two molecules are far apart, they move independent of each other. At a given pressure and temperature as the molecules come closer, the molecular interactions become important. The average speed of the molecules in a gas increases with temperature and their motion becomes more random. In solids, the net forces between the molecules are attractive until as they get very close, the forces become repulsive. The molecules in a solid with relatively low thermal motion cannot break free of mutual attractions. However, on heating due to relative increase in thermal motion, the solid melts.
In the liquid state a fine balance exists between attractive intermolecular forces and thermal energy, so that the molecules can break away from one another and get attracted while approaching the other molecules. In solids and liquids intermolecular forces become repulsive when molecules are pushed together by external
pressure. Therefore the compressibility of liquids and solids is very small. The average intermolecular distance is not much different in a liquid or solid, hence, small changes in volume take place on their interconversion. Continuous breaking and making of bonds between molecules in liquid gives it a non-rigid structure. This explains why liquids do not have any definite shape although they have a definite attractive interaction energy between molecules. In gases, the average distances between the molecules are large and hence the molecular interactions are weak. For gases like hydrogen and oxygen these may be neglected at high temperature and low pressure. The forces of repulsion come into play only during direct collision between the molecules. There are large empty spaces between molecules and there is nothing to stop them getting closer when pressure is applied. Hence the gases are highly compressible.
THE GASEOUS STATE
Out of the three states of matter , the most simplest one is gaseous state. Gases show maximum regularity in their behaviour irrespective of their nature. The lowermost layer of atmosphere called troposphere, which is held to the surface of the earth by gravitational force. The thin layer of atmosphere is vital to our life. This shields us from harmful radiations and contains substances like dioxygen, dinitrogen, carbondioxide, water vapour etc.
Only eleven elements exist as gases under normal conditions (Fig).
Eleven elements that exist as gases
Characteristics of gaseous state
The gaseous state is characterized by the following properties.
- Gases are highly compressible
- Gases exert pressure equally in all directions
- Gases have much lower density than solids and liquids.
- The volume and shape of gases are not fixed. These assume volume and shape of the container.
- Gases mix evenly and completely in all proportions without any mechanical aid.
THE GAS LAWS
The forces of attraction between the gas molecules are negligible. Certain generalisations were developed from the quantitative studies of the behaviour of gases. These generalisations are called Gas Laws . These laws are relationships between measurable properties of gases like pressure, volume, temperature and mass. The relationships between these variables describe state of the gas.
1. Boyle's Law ( Pressure – volume relationship)
This law describes the pressure volume relationship of gases at constant temperature. It was given by Robert Boyle. The law states that :
The volume of a given mass of a gas is inversely proportional to its pressure at constant temperature.
Mathematically, the law may be expressed as :
V a ( 1/ P ) ( Temperature and mass constant)
V = K ( 1/P)
(where K is a constant of proportionality).
or PV = K = constant.
Thus another statement of Boyle’s law may be given as follows:
For a certain amount of gas, the product of pressure and volume is constant at constant temperature.
Let V1 be the volume of a given mass of the gas having pressure P1 at temperature T. 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 = P2 V2 ( Temperature and mass constant)
Experimental verification of Boyle’s Law
The experimental verification of the law can be carried out by measuring the volumes of a given mass of gas at different pressures, keeping temperature constant. In each case the product PV is found to be a constant.
Graphical representation of Boyle’s Law
When pressure of the gas (P) is plotted against volume (V) , keeping (P) along X-axis and V along Y-axis at different temperatures, we get a curve as shown in Fig . The curve clearly shows that when pressure is increased, volume decreases and vice versa. At a particular state corresponding to pressure P1 , volume is V1 and when pressure is increased to P2, the corresponding volume V2 is smaller than V1. The value of K for each curve is different because for a given mass of gas , it varies only with temperature. Each curve corresponds to a different constant temperature and is known as an isotherm (constant temperature plot). Higher curves corresponds to higher temperature. It should be noted that volume of the gas doubles if pressure is halved.
Graph of pressure P vs volume V of a gas at different temperature.
Alternately , if graph is plotted between P and 1/V , a straight line passing through origin is obtained. However, at high pressures, gases deviate from Boyles law and under such conditions a straight line is not obtained in the graph.
Graph pf pressure of a gas P, vs 1/V
However, if the graph is plotted between the product PV along Y-axis and pressure P along X-axis, we get a straight line parallel to the pressure axis (Fig 5).
This indicates that PV remains constant even if we change the pressure.
Practical importance of Boyle's Law
The Boyles’ law expresses in a quantitative manner the important experimental fact that the gases are compressible. When a given mass of gas is compressed , the same number of molecules occupy a smaller space. This means that the gas becomes denser For example, air at sea-level is dense because it is compressed by a mass of air above it. However, the density and pressure decreases with increase in altitude.
A relationship between density and pressure of a gas can be obtained from Boyle’s law. By definition , density ‘d’ is related to the mass ‘m’ and volume ‘V’ by the relation d = m / V . If we put the value of V in this equation from Boyle’s law equation, we obtain the relationship.
This shows that at constant temperature , pressure is directly proportional to the density of a fixed mass of the gas.
1. What will be the minimum pressure required to compress 500 dm3 of air at 1 bar to 200 dm3 at 300 C ?
2. A vessel of 120 mL capacity contains a certain amount of gas at 350 C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 350 C. What would be its pressure ?
3. At a constant temperature, a gas occupies a volume of 200 mL at a pressure of 0.720 bar. It is subjected to an external pressure of 0.900 bar. What is the resulting volume of the gas ?
4. What is the volume of a sample of oxygen at a pressure of 3.5 atm. if its volume at 1 atm is 3.15 L ?
5. A gas occupies a volume of 2.5 L at 9 x 105 N m-2. Calculate the additional pressure required to decrease the volume of gas to 1.5 L.
6. A certain quantity of gas occupies a volume of 22.5 L at 760 mm and 0oC. What will be the volume, if pressure and temperature are 1,520 mm and 2730 C ?
7. A balloon filled with hydrogen at room temperature. It will burst if pressure exceeds 0.2 bar. If at 1 bar pressure, the gas occupies 2.27 L volume, up to what volume can the balloon be expanded ?
2. Charle's Law
( Temperature – volume Relationship)
This law describes the relationship between volume and temperature of gases at constant pressure. It was put forwarded by Jacques Charles and was further developed by Gay-Lussac. The law can be stated as :
The volume of a given mass of a gas increases or decreases by (1/273) of its volume at 0oC for each degree rise or fall of temperature, provided pressure is kept constant.
If V0 is the volume of the gas at 0oC, then
Volume of the gas V1 at 10C = V0 + (V0/273)
= V0[ 1 + (1 / 273)]
Volume of gas at 2o C, V2 = V0[ 1 + (2 / 273)]
Volume of gas at t0 C, Vt = V0[ 1 + (t / 273)] ..........(1)
Absolute Scale of Temperature
By carrying out the similar calculations, it can be shown that the volume of the gas below 0°C will be less than V0 . For example, the volume of the gas at - t°C will be :
V( - t) = V0 [ 1 - (t / 273)]
Thus decrease of temperature results in the decrease in the volume of the gas and ultimately , the volume should become zero at -273 0C. It means that any further lowering of temperature is impossible because it would correspond to the negative volume which is meaningless. Hence an important conclusion can be drawn from the above discussion that the lowest possible temperature is -273° C. The lowest possible temperature at which all the gases are supposed to occupy zero volume is called absolute zero. A scale of temperature based on the choice of zero is called Absolute scale of Temperature. Since the scale was suggested by Lord Kelvin, it is also known after his name as Kelvin Scale of Temperature.
Careful measurements have revealed that absolute zero of temperature is -273.15° C. Temperature on the Kelvin scale are indicated by writing the letter K. By convention, the degree sign (° ) is not used while expressing temperatures on Kelvin scale. For example,
-273.15°C = 0 K
The relationship between Kelvin scale and Celsius scale is :
T = t + 273.15
where T is the temperature on the Kelvin scale while ‘t’ is the temperature on Celsius scale. While solving numerical problems, the temperature on Celsius scale is converted into Kelvin by addition of 273 instead of 273.15 for the sake of simplicity.
Our conclusion that gases occupy zero volume at 0 K cannot be realized in actual practice because all the gases condense to liquids and solids before this temperature is reached. However, Kelvin scale is quite significant for scientific work and can be justified by thermodynamic arguments. For this reason it is also called Thermodynamic Scale of Temperature.
Alternate Statement For Charles' Law
The relationship between the volume of a given mass of gas at t0 C ( Vt ) and 00C ( V0 ) is given as :
Vt = V0 [ 1 + (t / 273)]
= V0 [(273 + t ) / 273 ]
Vt = [(V0 Tt / T0 ]
where T is the corresponding temperature on Kelvin scale.
Since ( V / T ) = constant at constant P.
or V a T ( at constant P ).
The Charle’s law may be stated in an alternate way as :
The volume of the given mass of gas at constant pressure is directly proportional to the temperature on Kelvin scale.
V a T ( constant P )
Let V1 be the volume of a certain mass of a gas at temperature T1 and pressure P. If the temperature is changed to T2 keeping pressure constant, the volume changes to V2. The relationship between four variables V1, T1, V2, and T2 is :
V1 / T1 = V2 / T2
The experimental verification of the law can be done by measuring the volumes of the given mass of a gas at different temperatures keeping the pressure constant. In each case the ratio V/T comes out to be constant.
Charles found that for all , at any given pressure, graph of volume vs temperature (in Celsius) is a straight line and extending to zero volume, each line intercepts the temperature axis at - 273.15°C. Slopes of lines obtained at different pressure are different but at zero volume all the lines meet the temperature axis at - 273.15°C (Fig). Each line of the volume vs temperature graph is called isobar.
Volume vs Temperature(°C) graph
Practical Importance of Charle’s Law
One of the interesting applications of Charle’s law is the use of hot air balloons in sports and for meteorological observations . According to Charle’s law, gases expand on heating. Since the mass of the gas sample is constant, the larger the volume at higher temperature will have the lower mass per unit volume and therefore lower density. Thus hot air is less dense than cool air. This causes hot air balloon to rise by displacing cool air (more dense) of the atmosphere. On this principle, hydrogen balloons (which rise higher because of low density of hydrogen) were developed as a means of transportation across
Atlantic. However, such air ships are not being used because hydrogen is inflammable and the transportation will be risky. But hydrogen balloons and hot air balloons are still used for weather observation.
8. On a ship sailing in pacific ocean where temperature is 23.4°C, a balloon is filled with 2 L air. What will be the volume of the balloon when ship reaches
Indian Ocean, where temperature is 26.1°C ?
9. What is the increase in volume, when the temperature of 600 mL of air increases from 270C to 47° C under constant pressure ?
10. A sample of gas is found to occupy a volume of 900 cm3 at 27°C. Calculate the temperature at which it will occupy a volume of 300 cm3.
11. It is desired to increase the volume of 80 cm3 of a gas by 20% without changing the pressure. To what temperature the gas be heated if its initial temperature is 25° C.
3. Gay – Lussac’s Law
(pressure – temperature Relationship)
The mathematical relationship between pressure and temperature was given by Joseph Gay Lussac and is known as Gay Lussac’s law. It states that at constant volume, pressure of a fixed amount of a gas varies directly with temperature.
P a T ( n, V constant)
This relationship can be derived from Boyle’s and Charles’ Law. Pressure vs temperature(Kelvin) graph at constant molar volume is shown in Fig. Each line of this graph is called isochore.
Pressure vs temperature(K) graph (isochors) of a gas
AVOGADRO’S LAW (volume – amount relationship)
This law relates the volume of a gas to the number of molecules at constant temperature and pressure. It was given by Amedeo Avogadro. It states that equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules.
For example, 22.4 litres of all gases at 273 K and 1 atm pressure contain 6.02 x 1023 molecules. It follows, therefore that volume of the gas is directly proportional to the number of molecules.
V a N ( Temperature and pressure constant ).
The number of molecules (N) of any gas is directly proportional to the number of moles(n). Thus ,
V a n ( Temperature and pressure constant ).
Since volume of a gas directly proportional to the number of moles ; one mole of each gas at standard temperature and pressure(STP) will have same volume. Standard temperature and pressure means 273.15 K(0°C) temperature and 1 bar(i.e., exactly 105 pascal) pressure. These values approximate freezing temperature of water and atmospheric pressure at sea level. At STP molar volume of an ideal gas or a combination of ideal gases is 22.71098 L mol-1.
The previous standard for STP is often used. In this definition STP denotes the same temperature of 0°C (273.15 K) , but a slightly higher pressure of 1 atm( 101.325 kPa). One mole of any gas or a comination of gases occupies 22.413996 L of volume at STP.
Standard ambient temperature (SATP) , conditions are also used in some scientific works. SATP conditions means 298.15 K and 1 bar ( i.e., exactly 105 Pa). At SATP (1 bar and 298.15 K), the molar volume of an ideal gas is 24.789 L mol-1.
12. Calculate the number of nitrogen molecules present in 2.8 g of nitrogen gas.
IDEAL GAS EQUATION
The three gas laws can be combined together in a single equation which is known as ideal gas equation.
where R is proportionality constant. On rearranging the equation (2) we obtain
R is called gas constant. It is same for all gases. Therefore it is also called Universal Gas Constant. Equation (3) is called ideal gas equation.
Ideal gas equation is a relation between four variables and it describes the states of any gas, therefore it is also called equation of states.
If temperature, volume and pressure of a fixed amount of gas vary from T1, V1 and P1 to T2, V2 and P2, then we can write,
If out of the six variables, values of five variables are known, the value of the unknown variable can be calculated from equation(1). This equation is also known as combined gas law.
NATURE OF GAS CONSTANT , R
In order to understand the significance of R, let us examine the nature of quantities in the ideal gas equation.
The value of R will therefore be expressed in units of energy per Kelvin per mole.
Numerical value of R
The numerical value of R depends upon the units in which pressure , volume and temperature are measured.
The volume of one mole of an ideal gas under STP conditions (273.15 K and 1 bar pressure) is 274.71098 L mol-1. The value of R for one mole of an ideal gas can be calculated under these conditions as follows :
R = (105 Pa ) (22.71 x 10-3 m3)
( 1 mol) x ( 273.15 K)
= 8.314 Pa m3 K-1 mol-1
= 8.314 x 10-2 bar L K-1 mol-1
= 8.314 J K-1 mol-1
At STP conditions used earlier (0°C and 1 atm pressure), the value of R is 8.20578 x 10-2 L atm K-1 mol-1
R = (1 atm ) x ( 22.4 143996 L )
(1 mol ) x (273.15 K )
= 8.20578 x 10-2 L atm K-1 mol-1
An ideal gas is one which strictly obeys Boyle’s and Charle’s laws at all temperature and pressure and it strictly follows the General Gas Equation , P V = n R T .
In ideal gases:-
i) The actual volume of the molecules should be negligibly
small as compared to the volume of the container.
ii) There should not be any intermolecular attraction
between molecules of the gas.
Calculation of Molecular mass from Gas Equation
According to the general gas equation,
13. At 0° C , the density of a gaseous oxide at 2 bar is same as that of nitrogen at 5 bar. What is the molecular mass of the oxide ?
14. Pressure of 1 g of an ideal gas A at 300 K is found to be 2 bar when 2 g of another ideal gas B is introduced in the same flask at the same temperature , the pressure becomes 3 bar. Find the relationship between their molecular masses.
15. The density of a gas at sea level at 0°C is 1.29 kg m-3, what is its molecular mass ? (Assume that the pressure is equal to 1 bar).
16. Density of a gas is found to be 5.46 g/dm3 at 300K at 2 bar pressure. What will be its density at STP ?
17. 340.5 mL of phosphorus vapour weighs 0.0625 g at 5460 C and 0.1 bar pressure. What is the molecular mass of phosphorus ?
18. A student forgot to add the reaction mixture to the round bottomed flask at 300 K but put it on flame. After a lapse of time, he realized his mistake , using a pyrometer he found the temperature of the flask was 750 K. What fraction of air would have been expelled out ?
19. Calculate the temperature of 4.0 moles of a gas occupying 5 dm3 at 3.32 bar.
20. A sample of nitrogen gas occupies a volume of 320 cm3 at STP. Calculate its volume at 660C and 0.825 atm pressure.
21. An iron tank contains helium at a pressure of 2.5 atm at 250C. The tank can withstand a maximum pressure of 10 atm. The building in which tank has been placed catches fire . Predict whether the tank will blow up or melt ( melting point of iron 15350C.
22. How many moles of oxygen are present in 400 cm3 sample of gas at a pressure of 760 mm of Hg at a temperature of 27 0C.
23. A discharge tube containing nitrogen gas at 25 oC is evacuated till the pressure is 2 x 10-2 mm .If the volume of discharge tube is 2 litres, calculate the number of molecules still present in the tube.
24. Calculate the pressure of 1 x 1022 molecules of sulphur dioxide when enclosed in a vessel of 2.5 L capacity at a temperature of 27 0 C.
25. A gas having molecular mass 84.5 g mol-1 enclosed in a flask at 270C has a pressure of 1.5 atmosphere. Calculate the density of the gas under these conditions.
26. How many grams of potassium chlorate must be decomposed to produce 2.40 L of oxygen at 740 Torr and 250 C
27. Calculate the volume at STP of a gas whose volume is 200 cm3 at 270 C and 70 cm pressure.
28. What temperature would be necessary to double the volume of a gas at STP if the pressure is decreased by 25% ?
29. A weather balloon has a volume of 175 L, when filled with hydrogen at a pressure of 1atm. Calculate the volume of the balloon when it rises to a height of 2,000 m, when the atmospheric pressure is 0.80 atm. Assume the temperature is constant.
30. At 298 K and 760 mm Hg a gas occupies 600 mL volume. What will be its pressure at a height where temperature is 283 K and volume of the gas is 640 mL.
This law describes the relation between the pressure of the mixture of non-reacting gases enclosed in a vessel to their individual pressures. The total pressure exerted by a mixture of two or more non-reacting gases in a definite volume is equal to the sum of partial pressures of the constituent gases.
The partial pressure of a gas in a mixture is defined as the pressure which the gas would exert, if it is allowed to occupy the whole volume of the mixture at the same temperature.
Consider three vessels each of 1 L capacity. Let the first vessel contain gas ‘A’ having a pressure of 300 mm Hg and the second contains a gas ‘B’ having a vapour pressure 400 mm Hg.
’s Law of partial pressure Dalton
Now, if we put both the gases A and B in the third vessel at the same temperature, the total pressure in the third vessel would be ;
P = PA + PB
= 300 + 400 = 700 mm Hg
Consider a mixture of gases containing n1 , n2 and n3 moles of three gases 1, 2, and 3 respectively. Let the corresponding partial pressures be P1, P2 and P3. According to the law, the total pressure is given by :
Ptotal = P1 + P2 + P3
If the gases present in the mixture behave ideally, then, it is possible to write separately for each gas,
P1 V = n1 R T............. (1)
P2 V = n2 R T..............(2)
P3 V = n3 R T..............(3)
Ptotal = P1 + P2 + P3
= (n1 RT / V ) + ( n2 RT / V ) + ( n3 RT / V )
= ( n1 + n2 + n3 ) ( RT / V )
Ptotal = ( ntotalR T ) / V .........(4)
where ntotal is the amount of gases in the mixture.
Dividing equation (1) by (4) we get:
P1 / Ptotal = n1 / ntotal
P1 = ( n1 / ntotal )Ptotal
= X1 Ptotal
P2 = X2 Ptotal
where X1, X2, etc represent the mole fractions of component gases. The mole fraction of a constituent in a mixture is defined as the ratio of the number of moles of that constituent to the total number of moles of all the constituents in the mixture.
's Law Dalton
This law is useful in knowing the pressure of the gas collected by the down ward displacement of water as shown in Fig8. The gas being collected over water also contains water vapours.
The collection of oxygen gas over water
The observed pressure of the moist gas is equal to the sum of the pressure of dry gas and the pressure of the water vapours. The observed pressure of the moist gas is equal to the sum of the pressure of the dry gas and the pressure of water vapours. The pressure of the water vapours is constant at a particular temperature and is known as aqueous tension at that temperature.
Pobserved = Pgas + aqueous tension.
Pgas = Pobserved - aqueous tension.
31. What will be the pressure exerted by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide contained in a 9 dm3 flask at 300 K ?
32. What will be the pressure of the gas mixture when 0.5 L of H2 at 0.8 bar and 2.0 L of oxygen at 0.7 bar are introduced in a 1 L vessel at 300 K ?
33. A 2.5 L flask contains 0.25 mol each of sulphur dioxide and nitrogen at 300 K. Calculate the partial pressure exerted by each gas and also the total pressure.
34. A 5 L flask contains 19.5 g of SO3 and 1 g of He. The temperature of the flask is 200C. Calculate the partial pressure exerted by SO3 and He and the total pressure of the gaseous mixture.
35. A 200 cm3 capacity flask contained oxygen at 200 mm pressure. Another flask of 300 cm3 capacity contained nitrogen gas at 100 mm Hg pressure. The two flasks were then connected so that both the gases are filled in the combined volume. What is the partial pressure of each gas in the final mixture and the total pressure ?
36. Calculate the total pressure in a mixture of 4 g of oxygen and 2 g of hydrogen contained in 1 L at 0° C.
37. From 200 mg of CO2 1021 molecules are removed. How many moles of CO2 are left ?
38. An open vessel at 270 C is heated until 3/5th of the air in it has been expelled. Assuming that the volume of the container remains constant, find the temperature at which the vessel has been heated ?
39. Calculate the total pressure in a mixture of 8 g of oxygen and 4 g hydrogen confined in a vessel of 1 dm3 at 300 K.
40. Pay load is defined as the difference between mass of displaced air and the mass of the balloon. Calculate the pay load when balloon of radius 10 m , mass 100 kg is filled with helium at 1.66 bar at 300 K (density of air = 1.2 kg m-3 and R = 0.083 bar dm3 K-1 mol-1).
41. Calculate the volume occupied by 8.8 g of CO2 at 31.1°C and 1 bar pressure. (R = 0.083 bar K-1 mol-1)
42. 2.9 g of a gas at 95°C occupied the same volume as 0.184 g of hydrogen at 17°C , at the same pressure. What is the molecular mass of the gas ?
KINETIC MOLECULAR THEORY OF GASES
The gas laws are based on experiments quite independent of any theory of the nature of the gas. But a simple theory that a gas consists of a very large number of molecules in rapid motion in vacant space can be easily used to derive and justify these laws mathematically. This theory is known as Kinetic theory of Gases. The various postulates of the theory are :
i) A gas consists of a large number of identical molecules of mass m. The dimensions of these molecules are very small compared to the space between them. Hence molecules are treated as point masses.
ii) There are practically no attractive forces between the molecules. The molecules therefore , move independently.
iii) The molecules are in a state of ceaseless and random motion, colliding with each other and with the walls of the container. The direction of their motions changes only on collision. These collisions are known as elastic collisions in which the energy and momenta of the molecules are conserved. In non-elastic collisions these quantities are not conserved.
iv) The pressure of a gas is the result of collisions of molecules with the walls of the container.
v) The average kinetic energy of the colliding molecules is directly proportional to its temperature.
The above postulates are applicable only to ideal gases and are approximately valid for real gases.
Explanation of gas Laws on the basis of Kinetic Theory.
The kinetic theory satisfactorily explains the gas laws described as below :-
1. Bolyle's Law
The kinetic molecular theory assumes that the pressure exerted by a gas is due to the bombardment of its molecules on the walls of the container. Thus, the pressure depends upon the number of molecular impacts or collision per unit area of the walls per second. If the temperature is kept constant, the average speed ( ½ m u2 a T ) remains the same. Now if the volume is increased there will be less number of molecules colliding each unit area of the vessel in a unit time. Consequently, the pressure will decrease. On the other hand , if the volume is decreased, there will be more molecules in a given volume and hence pressure will increase. Thus, it can be concluded that at a given temperature, the pressure exerted by a gas is inversely proportional to the volume.
2. Charle's Law
When temperature of the gas is increased, the kinetic energy of the molecules ( ½ m u2 a T ) increases. As a result , the velocity of the molecules increases. Consequently, they collide with the walls of the container more frequently and more vigorously . Therefore, the pressure of the gas is expected to increase. However, if pressure is kept constant, the force due the molecular collision on the walls of the container must be kept the same. This is possible only if the volume of the gas increases, ie., gas expands. On the other hand if the temperature is decreased, the average velocity of the molecules decreases and therefore the pressure decreases. To keep the pressure constant, volume of the gas must decrease so that the force due to collisions per unit area remains the same. Thus, it can be concluded that at given pressure, volume of the gas increases with rise in temperature and decreases with decrease in temperature. Hence volume is directly proportional to the temperature at constant pressure. This is Charle’s law.
's Law of Partial Pressures Dalton
In absence of attractive forces, the particles of the gas behave independent of one another. The same is true even if there are more than one type of molecules as happens in the mixture of non-reacting gases. Being independent, the number of molecules colliding the unit area of the wall per second at a given temperature, for a fixed amount of the gas is the same. This implies that the partial pressure of the gas will be unaffected by the presence of the molecules of other gases. However, the total pressure exerted is due to the impact of the molecules of all gases. Accordingly, the total pressure would be the sum of the partial pressures of the gases. This is
’s law. Dalton
4. Graham’s law
Average kinetic energy of any gas is the same at the same temperature. Therefore the two gases the one with lower molecular mass will have higher speed, hence it will effuse/diffuse faster.
KINETIC GAS EQUATION
Based on the postulates of kinetic theory of gases, it is possible to derive an equation ( known as Kinetic gas Equation ) which connects different characteristics of an ideal gas. The equation is :
PV = (1/3) m
where P = the pressure of the gas.
V = the volume of the gas.
m = the mass of a molecule of the gas.
N = the number of molecules of the gas. and
U2 = the mean square velocity of gaseous molecules
defined as :
Average Kinetic Energy of Molecules
The average kinetic energy is defined as :
K.E = (1/2) m U2 .........(1)
According to kinetic gas equation:
PV = (1/3) m
m U2 = ( 3 P V ) / N .......(2)
Substituting the values of m U2 in equation ( 1 ) we get :
K.E = ( 3 P V ) / 2 N
For one mole of a gas P V = RT
and N = NA ( Avogadro’s number ).
Hence we have ;
K.E = ( 3 R T ) / 2 NA
= ( 3 / 2 ) k T
where k = R / NA and is known as Boltzmann constant.
Its value is :
k = R / NA = ( 8.314 J /K/mol) / 6.023 x 1023 mol-1
= 1.38 x 10-23 J / K
The total kinetic energy for one mole of an ideal gas :
Etotal = NA ( K E ) = ( 3 / 2 ) R T
The average kinetic energy as well as molar kinetic energy depends only on the temperature of the gas.
Calculation of Molecular velocities
From the kinetic gas equation:
PV = ( 1 / 3 ) m
For one mole of the gas, P V = R T
and also P V = ( 1 / 3 ) m NA U2
( where NA is the Avogadro’s number)
P V = ( 1 / 3 ) M U2
(where M is the mass of one mole of the gas = m NA )
( 1 / 3 ) MU2 = P V = R T
U2 = ( 3 P V ) / M
= ( 3 R T ) / M
= ( 3 P ) / d
( since M / V = density )
Any one of the formula can be used to calculate the value of U , the RMS velocity in m/s ( SI unit ), P must be expressed in Nm-2 and d in kg m-3.
1 atm = 1.01325 x 105 Nm-2.
43. (a) Calculate the total and average kinetic energy of 32 g methane molecules at 27°C.
Distribution of Molecular Velocities
In a gas , we have a large number of molecules which constantly undergo collisions among themselves and with the walls of the container and there is an exchange of energy, thereby changing their speed and kinetic energy. Therefore at any instant different molecules in a gas have different speeds and hence different kinetic energies. We have molecules ranging from very small to very high speed. Eventhough the speed of individual molecule is constantly changing at a given temperature, the fraction of the molecules with a particular speed remains constant. This is known as distribution of speeds. This is also known as Maxwell – Bolzmann Distribution law. The maximum in distribution curve corresponds to the speed possessed by the highest fraction of molecules and is called the most probable speed, Ump. The Fig gives the distribution of speed of chlorine and nitrogen at a particular temperature.
Distribution of molecular speeds for chlorine and nitrogen at 300 K
The maximum in the each curve represents the most probable speed of the two gases respectively. The speed of a gas molecule at a given temperature also depends upon its mass. Lighter nitrogen molecules move faster than heavier chlorine molecules. Hence nitrogen molecules have higher value of most probable speed than chlorine molecules at any given temperature.
The temperature dependence of distribution of speed for carbon dioxide at three different temperatures is shown in Fig.
Distribution of molecular speeds for carbon dioxide at different temperatures.
It is obious from distribution curves that the fractions of molecules with higher speed increases with increase in temperature. The value of most probable speed also increases at higher temperature. Besides the most probable speed , we also define the average speed and root mean square speed.
In general there are three types of velocity of molecules are commonly used. These are described as follows:
1. Most Probable Velocity Ump
This velocity is the velocity possessed by maximum number of molecules in the gaseous sample. It can be shown that :
2. Average Velocity Uav
This is the average speeds possessed by molecules i.e.,
It can be shown that :
3. Root Mean square Velocity Urms
It is the square root of the mean of the squares of velocities.
and it is given by the expression :
The mean square speed is a direct measure of the average kinetic energy of a gas molecule. The three speeds are related by the following relationships.
Urms > Uav > Ump
Ump : Uav : Urms :: 1 : 1.128 : 1.224
43. (b) A mixture of hydrogen and oxygen contains 20% by weight of hydrogen exerts a pressure of 1 atm. Calculate partial pressure of hydrogen.
44. Calculate average, RMS and most probable velocities of CO2 at 270C.
45. Two flasks A and B each of 1 L capacity are filled with ideal gas .
The temperature of flask A is 300 K , while that of B is 600 K.
Compare the two samples with reference to :
i) Number of molecules present.
ii) Molecular velocities in the samples.
iii) Average kinetic energy per molecule.
DIFFUSION / EFFUSION OF GASES
The ability of a gas to spread and occupy the whole available volume, irrespective of other gases present in the container is called diffusion. The aroma of the food you smell near the kitchen is due to the diffusion of the molecules with aroma through the air.
Effusion is the process by which a gas under pressure escapes from one chamber of a vessel through a small opening or an orifice . Escaping of air through a punctured tyre, escaping of perfume molecules through the atomizer are examples of effusion. The process of effusion is always followed by the process of diffusion.
GRAHAMS LAW OF EFFUSION / DIFFUSION
Thomas Graham observed that lighter gases diffuse faster than the denser gases. In 1831 , he proposed the law of effusion.
The law states that under identical conditions of temperature and pressure, the rate of effusion / diffusion of a gas is inversely proportional to the square root of its density.
where r is the rate of diffusion and ‘d’ is the density of the gas.
Now if there are two gases A and B having rates of diffusion r1 and r2 and d1 and d2 their densities respectively. Then,
( at the same T and P)
Molecular mass is twice the vapour density. Therefore , the above expression may be written as :
where M1 and M2 are the molecular masses of the gases having densities d1 and d2 respectively. Thus Graham’s law may be stated as :
Under similar conditions of temperature and pressure the rates of diffusion of gases are inversely proportional to the square root of their molecular masses.
The rate of diffusion of the gas is equal to the volume of the gas which diffuses per unit time i.e.,
If V1 and V2 are the volumes of the gases diffusing in the time t1 and t2 respectively, then :
r1 = V1 / t1 and r2 = V2 / t2
Thus putting the values of r1 and r2 we arrive at the following formula:
If V2 = V1 = V , we get
This implies that the time taken for the diffusion of equal volumes of the two gases under similar conditions of temperature and pressure is directly proportional to the square root of their densities or molecular masses.
Similarly, if t1 = t2 = t
It means that volumes of the two gases which diffuse in the same time under similar conditions are inversely proportional to the square roots of their densities or molecular masses.
46. Through the two ends of a glass tube of length 200 cm hydrogen chloride gas and ammonia are allowed to enter. At what distance ammonium chloride will first appear ?
47. A teacher enters the class room from first door, while a student from the back door. There are 13 equidistant rows of benches in the class room. The teacher releases N2O gas, the laughing gas from the first bench while student releases the weeping gas C6H11OBr from the last bench. At which row will the student start laughing and weeping simultaneously ?
48. For 10 minutes each , at 27°C , from the two identical holes nitrogen and an unknown gas are leaked into a common vessel of 3 L capacity. The resulting pressure is 4.18 bar and the mixture contains 0.4 mole of nitrogen. What is the molar mass of the unknown gas ?
49. Equal volumes of two gases A and B diffuse through a porous pot in 20 and 10 seconds respectively. If molar mass of A be 80 , find the molar mass of B.
50. A gas diffuses 1.27 times as rapidly as chlorine under the same conditions of temperature and pressure. Answer the following :-
i) Is the gas heavier or lighter than Chlorine ?
ii) What is the ratio the molecular mass of the gas to that of chlorine ?
51. A certain gas diffuses four times as long as to effuse out as hydrogen . What is the molecular mass of the gas ?
52. The volumes of a gas A and chlorine diffusing at the same time are 35 ml and 29 ml respectively. Calculate the molecular mass of A.
53. Which of the two gases , ammonia and hydrogen chloride, will diffuse faster and by what factor ?
Importance of Graham's Law
1. It forms the basis for separating the isotopes of some elements.
2. It provides a simple method for determining the
densities and molecular weights of unknown gases by
comparing their rates of diffusion or effusion with those
of known gases.
3. It is also useful in separating gases having different
Deviations from Ideal gas behaviour - Real Gases
A gas which obeys the general gas equation ( P V = n R T ) at all temperatures and pressures is called an ideal gas . However, none of the actually known gases called real gases obey this equation over the entire range of temperature and pressure. These gases show ideal behaviour at low pressures and high temperatures. As the pressure becomes high and temperature becomes low, more and more deviations are observed from general gas equation and other gas laws. Amagat , Andrews and others made extensive study of the behaviour of various gases and concluded that real gases do not obey the general gas equation under all conditions of temperature and pressure.
A convenient way to study deviations of real gases from ideal behaviour is to plot a graph between (PV/nRT) and P. The quantity (PV / n RT) is called compressibility factor and is denoted by Z. For ideal gases , Z = 1 under all conditions.
Fig 10 Plot of the compressibility factor ( Z = PV/nRT) as a function of P for gases
Thus , the simplest way to compare the behaviour of a real gas is to measure P, V, T for one mole of a gas and then plot PV/RT against P. The ideal behaviour is shown in Fig. for comparison. The behaviour of some gases such as H2, N2 ,He, and CO2 is shown in Fig. 10.
From the figure we find that the curves obtained on plotting Z against pressure for different gases have the following characteristics :
(i) All curves approach the ideal value as the pressure approaches zero. Thus at low pressures, all gases behave as ideal gases.
(ii) At moderate pressures, Z < 1 i.e., there is negative deviation. This means that the gas is more compressible than expected from ideal behaviour.
(iii) At high pressures , Z > 1 , i.e., there is a positive deviation. It means that the gas is less compressible than expected from ideal behaviour.
(iv) For H2 and He, Z is always greater than one. This means that these gases are less compressible than expected from ideal behaviour at all pressures.
(v) Extent of deviation at any temperature or pressure depends upon the nature of gas. Gases like CO2 which can be liquefied easily, show larger deviations.
It may be noted that deviations decrease with increase in temperature. At best real gases show ideal behaviour only at low pressure and high temperature. The temperature above which a gas behaves like an ideal gas and obeys the gas law is called Boyle’s temperature or Boyle point. Boyle point of a gas depends upon its nature. Above their Boyle point , real gases show positive deviations from ideality and Z values are greater than one. The forces of attraction between the molecules are very feeble. Below Boyle temperature real gases first show decrease in Z value with increasing pressure, which reaches a minimum value. On further increase in pressure, the value of Z increases continuously. Above explanation shows that at low pressure and high temperature gases show ideal behaviour. These conditions are different for different gases.
More insight is obtained in the significance of Z can be obtained as follows. Consider the following relation,
If the gas shows ideal behaviour then ,
On putting this value of ( n R T / P ) in equation (1) , we can see that compressibility factor is the ratio of actual molar volume of a gas to molar volume of it, if it were an ideal gas at that temperature and pressure.
Causes of Deviation from Ideal behaviour
The causes of deviations from ideal behaviour may be attributed to the two faulty assumptions of kinetic theory of gases. The assumptions are :
The causes of deviations from ideal behaviour may be attributed to the two faulty assumptions of kinetic theory of gases. The assumptions are :
i) The volume occupied by the gas molecules is negligibly small as compared to the total volume occupied by the gas.
ii) The forces of attraction between gas molecules are
Let us discuss the two faulty assumptions of kinetic theory .
i) The kinetic theory assumes that the volume occupied by molecules of a gas is negligible as compared to the total volume of the gas. This assumption is quite valid under ordinary conditions of temperature and pressure. For example, it has been calculated that the volume of the molecule is about 0.1% of the total volume of the gas under ordinary conditions of temperature and pressure. However as the temperature is considerably decreased or pressure is considerably increased the total volume of the gas decreases appreciably whereas the volume occupied by molecules remain the same because the molecules are incompressible. Therefore the fraction of the volume of the molecules cannot be neglected under these conditions. This proves the invalidity of this postulate at high pressure and low temperature.
ii) According to kinetic theory, the forces of attraction between the gas molecules are negligible. The assumption is quite valid at low pressure and high temperature because , the molecules lie apart from one another. On the other hand at high pressure and low temperature, the volumes of the gas is quite small. Consequently, the distance between the molecules decreases and attractive forces , though very small, exist between them. The direct proof of the presence of intermolecular forces is that the temperature of the gas falls when compressed gas is allowed to expand. This fall in temperature is due to the fact that some work has to be done to overcome the attractive forces between molecules during expansion. Thus, this postulate is also invalid at low temperature and high pressure.