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States of Matter :, Gases and Liquids, Matter : Matter exists in three states : solid,, liquid and gas., X The main differences in the three states, can be explained on the basis of their, energy content and forces of attraction, between molecules., X Energy of molecules is maximum in the, gaseous state and forces of attraction, between molecules are maximum in the, solid state., X Gases are non-metallic elements, highly, compressible, can diffuse easily and form, homogeneous mixtures, whereas vapour, is the gaseous state of any substance, which is usually a liquid or solid at room, temperature., Intermolecular forces : These are the, forces of attraction and repulsion that exist, between molecules of a compound. These, cause the compound to exist in a certain state, of matter – solid, liquid or gas and affect the, melting and boiling points of compounds, as well as the solubilities of one substance, in another. Attractive intermolecular, forces are also called van der Waals’ forces., These are weak forces. Different types of, intermolecular forces are :, X London or dispersion forces : This is, the weakest intermolecular force. It is a, temporary attractive force that results, when the electrons in two adjacent atoms, occupy positions that make the atoms, to form temporary dipoles. This force, is sometimes called an dipole-induced, dipole attraction., London or dispersion forces increase with, – increase in number of electrons in, molecules, , –, –, X, , X, , X, , increase in molecular size, increase in molecular weight., , Dipole-dipole forces : This is the second, strongest van der Waals’ force. These, forces exist between polar molecules, where the positive end of one molecule, attracts the negative end of another, molecule., , Dipole-dipole forces are characteristically, weaker than ion-dipole forces., These forces increase with, – increase in molecular size of molecule, – increase in molecular weight of, molecule, – increase in polarity of molecule., Dipole-induced dipole forces :, These forces operate between the polar, molecules having permanent dipole and, the molecules having no permanent, dipole., Hydrogen bonding : It is the electrostatic, force of attraction between hydrogen, atom and highly electronegative atom, (N, O, and F atoms) of another molecule., It is represented by dotted line. It is of two, types :, – Intermolecular hydrogen bonding :, It is the force of attraction that, exists between H-atom and highly, electronegative atom between several, molecules of the same substance. For, example, hydrofluoric acid.

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+, , –, , +, , –, , +, , –, , +, , –, , +, , –, , H—F H—F H—F H—F H—F, , –, , , , Intermolecular hydrogen bonding, , Intramolecular hydrogen bonding :, It is the force of attraction that, exists between H-atom and highly, electronegative atom within the same, molecule., For example, o-nitrophenol, OH, , O, N, , –, , O, , Boyle’s law : This law states that, ‘volume, of a given mass of a gas is inversely, proportional to its pressure, keeping the, temperature constant’, i.e.,, 1, V∝, (at constant temperature), P, or PV = constant, P 1V1 = P 2V2 (mass and, (Initial state) (Final state) temperature constant), X, , Graphical representations of Boyle’s, law :, , Intramolecular H–bonding, , Thermal energy : The energy which arises, by motions of atoms or molecules of a body is, known as thermal energy. It is the measure, of average kinetic energy of the particles of, matter and it is directly proportional to the, temperature of the substance., , PV, , P, 1, V, , Intermolecular forces vs thermal, energy :, , T 1 > T2 > T 3, , V, , K = °C + 273.15 ≈ °C + 273 or °C = K – 273, SI unit of temperature is Kelvin (K)., , T1, T2, T3, , V, , P, , Gaseous State :, X There are few parameters which are, important to understand the gaseous, state viz. mass, volume, pressure and, temperature., – Mass : Mass of a gas is expressed in, gram (g) or kilogram (kg)., – Volume : Volume of a gas is generally, expressed in litre (L) or millilitre, (mL). SI unit of volume is m3., – Pressure : There are various units, used to express pressure of a gas, like, atmosphere, bar, N m–2, Pascal, etc., 1 atm = 76 cm Hg = 760 mm Hg or, 760 torr, 1 bar = 0.987 atm ≈ 1 atm = 105 Pa, – Temperature : There are three, different scales to measure temperature, viz. Centigrade or Celsius, Fahrenheit, and Kelvin scale., 5, 9, ° C = (° F − 32) or ° F = (° C) + 32, 9, 5, , , P, , P, , X, , Relationship between pressure and, density :, According to Boyle’s law,, 1, 1, but V ∝ (where, d is density, V∝, d, P, m, ), and d =, V, 1 1, d, or d ∝ P or, \, ∝, = constant, d P, P, d, d, Thus, 1 = 2 (temperature and mass, P1 P2, constant), , X, , Significance of Boyle’s law : With the, use of Boyle’s law, it can be concluded, that gases are compressible. Since the, gas density is directly proportional to, pressure, so more the gas is compressed,, denser it becomes., , Charles’ law : This law states that, ‘at, constant pressure the volume of a given, mass of a gas increases or decreases by, 1/273 of its volume at 0 °C for every rise or, fall of one degree in temperature’., Vt = V0 +, , V0, × t (at constant pressure), 273

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t ,  273 + t , , = V0 1 +, , Vt = V0 , , , 273 , ,  273 , where, V0 = volume at 0 °C, Vt = volume at, t °C and t = temperature., From the above expression, it can be seen, that at temperature –273 °C, the volume, of gas becomes zero. This temperature is, known as absolute zero., X Absolute zero : It is the theoretical or, hypothetical temperature at which the, gas is supposed to possess zero volume., At this temperature, the total kinetic, energy of the molecules is zero and hence,, molecular motion ceases., X Absolute scale of temperature : This, is also called thermodynamic scale of, temperature., X At constant pressure, if V1 is the volume, at T1 and as temperature changes to T2,, volume will change to V2, then, , T,  273 + t , = P0, = P0 , ,  273 , 273, where, P 0 and P t are the pressures at 0, °C and t °C respectively, t and T are the, temperature in °C and Kelvin respectively., Thus, P t ∝ T (since P 0 and 273 both are, constant) or simply P ∝ T, P, P, P, = constant or 1 = 2, i.e.,, T, T1 T2, , (at constant volume), This pressure-temperature relationship is, also known as Amonton’s law., X, , Graphical representation of Gay, Lussac’s law : A plot of P vs T for a, given mass of gas at constant volume is, a straight line., – P-T curves drawn at constant volume, are known as isochor., V1, , V1 V2, =, T1 T2, , –273, , 0, T (°C), , Significance of Charles’ law : Since, density and temperature are inversely, proportional to each other. Hence, hot air, is lighter than atmospheric air. It means, air expands on heating as its density, decreases. This fact is applied in filling, hot air balloons which ultimately rise up., , Gay Lussac’s law (pressure-temperature, relationship) : This law states that ‘at, constant volume, the pressure of a given mass, of a gas increases or decreases by 1/273 of its, pressure at 0 °C for every rise or fall of 1 °C, in temperature’., Pt = P0 +, , V4, , P, , V, , Temperature, T (K), , X, , V3, , Graphical representations of Charles’, law : A plot of V vs T (temperature in, Kelvin scale) for a given mass of a gas, at constant pressure is a straight line, passing through origin., , Volume (V), , X, , V2, , P0, t , , × t = P0  1 +, , , 273, 273 , , , , V1 < V2 < V3 < V4, T, , Avogadro’s law : This law states that equal, volume of all gases under similar conditions, of temperature and pressure contain equal, number of molecules. One mole of a gas, contains 6.023 × 1023 number of molecules, (Avogadro’s number). It means one mole of, each gas has same volume under the given, conditions of temperature and pressure., It means samples of different gases which, contain the same number of molecules, occupy same volume., V1 V2, =, V ∝ n (T, P constant) =, n1 n2, Ideal gas equation :, X By combining Boyle’s law and Charles’, law, an equation can be derived that gives, the simultaneous effect of the changes of, pressure and temperature on the volume, of the gas. This is known as combined gas, equation., V ∝ 1/P (constant T) (Boyle’s law) ...(i), V ∝ T (constant P) (Charles’ law) ...(ii), V ∝ n (constant P and T) (Avogadro’s law), ...(iii)

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Combining eqns. (ii) and (iii),, V∝, , X, , nT, nT, or V = R, P, P, , PV, = nR or PV = nRT, T, For one mole (n = 1), PV = RT (Ideal gas, equation or equation of state for gases)., , ∴, , Units of, pressure, , Units Values of Units of gas conof vol- gas constant (R), ume stant (R), , atm, , L, , 0.0821, , L atm K–1 mol–1, , atm, , cm3, , 82.1, , atm cm3 K–1 mol–1, , dynes cm–2, , cm3, , 8.314 × 107 ergs K–1 mol–1, , dynes cm–2, , cm3, , 1.987, , cal K–1 mol–1, , 3, , 8.314, , J K–1 mol–1, , –2, , Pa or N m, , m, , PV, P1V1, = nR ; 2 2 = nR, T2, T1, where ‘R’ is constant of proportionality and, is called universal gas constant. Its value, varies with the units in which pressure and, volume are expressed., Now, if V 1 is the volume of a gas at, temperature T1 and pressure P1, V2 is the, volume of same amount of gas at temperature, PV, PV, T 2 and pressure P 2 , then 1 1 = 2 2 =, T1, T2, constant, X, , ∴ Pgas = PTotal × xgas, i.e., partial pressure = total pressure ×, mole fraction., Behaviour of real gases (deviation from, ideal gas behaviour) :, X Real gases do not obey ideal gas equation, under all conditions. They nearly obey, ideal gas equation at higher temperatures, and very low pressures. However, they, show deviations from ideality at low, temperatures and high pressures., X The isotherms obtained by plotting, pressure, P against volume, V for real, gases do not coincide with that of ideal, gas, as shown :, , Relationship between molar mass, and density : Let m be the mass of gas, in grams and M be the molar mass of gas., m, We know, n =, M, Now, PV = n RT =, , dT, = constant, P, d1T1 d2T2, \, =, P1, P2, Partial pressure : In a mixture of different, gases which do not react chemically, each, gas behaves independently of the other, gases and exerts its own pressure called, its partial pressure., , X, , ideal behaviour; when Vreal = Videal, negative deviation; when Vreal < Videal, , V, Vreal = Volume of the real gas at given pressure., Videal = Volume of the gas calculated by ideal gas, equation at given pressure., , P=, , \, , positive deviation; when Vreal > Videal, , P, , m, RT, M, , m RT dRT, dRT, =, or M =, V M, M, P, dT M, , since M and R are constant for a, =, P, R, gas., , Dalton’s law of partial pressures :, The total pressure exerted by a mixture, of gases is the sum of partial pressures of, each component of the mixture., PTotal = p1 + p2 + ... + pn (at constant T,, V)...(i), where PTotal is the total pressure exerted, by the mixture of gases and p1, p2, ....., pn are the partial pressures of n gases, present in the mixture., , X, , The deviation from ideal gas behaviour, can also be expressed by compressibility, factor, Z., – Compressibility factor (Z) : The, ratio of volume of real gas, V real to, the ideal volume of that gas, V ideal, calculated by ideal gas equation is, known as compressibility factor., Z=, , –, –, , PVreal, V, = real, nRT, Videal, , nRT , ,  ∵ Videal = P , , For ideal perfect gases, the, compressibility factor, Z = 1., But for real gases, Z ≠ 1.

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S.No., If Z > 1, 1. Vreal > Videal, 2., 3., , 4., , X, , X, , X, , X, , X, , Repulsive forces, > attractive forces, Gas cannot be, compressed, easily., , 223 K, , If Z < 1, Vreal < Videal, , 323 K, Z, , Attractive forces >, repulsive forces, Gas can be, compressed easily., , For permanent For gases, gases like He, CH4, CO2., H 2., , 1, , 0, , For gases like He, H 2 the Z value, increases with increase in pressure, (positive deviation). It is because, the, repulsive forces become more significant, and the attractive forces become less, dominant. Hence, these gases are difficult, to impress., For gases like CH4, CO2, NH3, etc., the, Z value decreases initially (negative, deviation) but increases at higher, pressure. It is because at low pressures,, the attractive forces are more dominant, over the repulsive forces, whereas at, higher pressures the repulsive forces, become significant as the molecules, approach closer to each other., But for all the gases, the Z value, approaches one at very low pressures,, indicating the ideal behaviour., Graph of Z vs P for N2 gas at different, temperatures is shown as :, , P, , In the given graph, the curves are, approaching the horizontal line with, increase in the temperature i.e., the, gases approach ideal behaviour at higher, temperatures., , like, , The isotherms for one mole of different, gases, plotted against the Z value and, pressure, P at 0 °C are shown as :, , 373 K, , van der Waals’ equation : The deviations, from ideal gas behaviour can be ascertained, to the following faulty assumptions by, kinetic theory of gases :, X The real volume of the gas molecules is, negligible when compared to the volume, of the gas., X There are no forces of attraction or, repulsion between the gas molecules., Hence, van der Waals suggested the, following corrections :, – Volume correction, Videal = V – nb, – Pressure correction :,  n2 , Pideal = Preal + a ,  V 2 , X, , Now, the ideal gas equation can be, modified by introducing this volume and, , an2 , pressure correction as :  P + 2  (V – nb), , V , = nRT, , X, , (van der Waals equation of state), , For 1 mole of gas,  P + a  (Vm – b) = RT, , Vm2 , where, Vm = molar volume of a real gas,, a = van der Waals constant of attraction., a ∝ attraction between gas molecules, ∝ compressibility ∝ ease of liquefaction., b = excluded volume of the gas, X Units of ‘a’ and ‘b’ :, Unit of a = atm L 2 mol –2 , Unit of b, = L mol–1

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OBJECTIVE TYPE QUESTIONS, , Multiple Choice Questions (MCQs), , 2. Which of the following is true about gaseous, state?, (a) Thermal energy = molecular attraction, (b) Thermal energy >> molecular attraction, (c) Thermal energy << molecular attraction, (d) Molecular forces >> those in liquids, 3. Intermolecular forces in solid hydrogen are, (a) covalent forces, (b) van der Waals forces or London dispersion, forces, (c) hydrogen bonds, (d) all of these., 4. For the type of interactions; (I) covalent bond,, (II) van der Waals forces, (III) hydrogen bonding,, (IV) dipole-dipole interaction, which represents, the correct order of increasing stability?, (a) (I) < (III) < (II) < (IV), (b) (II) < (III) < (IV) < (I), (c) (II) < (IV) < (III) < (I), (d) (IV) < (II) < (III) < (I), 5. The types of attractive forces between a polar, molecule and a non-polar molecule are, (a) dipole-dipole forces, (b) hydrogen bonds, (c) dipole-induced dipole forces, (d) dispersion forces., 6. Dipole-induced dipole interactions are, present in which of the following pairs?, (a) HCl and He atoms (b) SiF4 and He atoms, (c) H2O and alcohol, (d) Cl2 and CCl4, , 7. If P, V and T represent pressure, volume and, temperature of the gas, the correct representation, of Boyle’s law is, (a) V ∝ 1/P (P is constant), (b) PV = RT, (c) V ∝ 1/P (at constant T), (d) PV = nRT, 8. At 25°C and 380 mm pressure, 400 mL of, dry oxygen was collected. If the temperature is, constant, what volume will the oxygen occupy at, 760 mm pressure?, (a) 200 mL, (b) 400 mL, (c) 569 mL, (d) 621 mL, 9. Use of hot air balloons in sports and, meteorological observations is an application of, (a) Boyle’s law, (b) Newton’s law, (c) Kelvin’s law, (d) Charles’ law., 10. Equal volumes of gases at the same, temperature and pressure contain equal number, of particles. This statement is direct consequence, of, (a) perfect gas law, (b) partial law of volumes, (c) Charles’ law, (d) ideal gas equation., 11. A plot of volume (V) versus temperature (T), for a gas at constant pressure is a straight line, passing through the origin. The plots at different, values of pressure are shown in the figure. Which, of the following order of pressure is correct for, this gas?, p1, p2, , Volume (mL), , 1. Which one of the following statements is, wrong for gases?, (a) Gases do not have a definite shape and, volume., (b) Volume of the gas is equal to volume of, container confining the gas., (c) Confined gas exerts uniform pressure on the, walls of its container in all directions., (d) Gases are not compressible., , p3, p4, Temperature (K)

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(a) p1 > p2 > p3 > p4, (c) p1 < p2 < p3 < p4, , (b) p1 = p2 = p3 = p4, (d) p1 < p2 = p3 < p4, , 12. When the product of pressure and volume is, plotted against pressure for a given amount of, gas, the line obtained is, (a) parallel to x-axis, (b) parallel to y-axis, (c) linear with positive slope, (d) linear with negative slope., 13. Containers A and B have same gases., Pressure, volume and temperature of A are all, twice as that of B, then the ratio of number of, molecules of A and B are, (a) 1 : 2, (b) 2 : 1, (c) 1 : 4, (d) 4 : 1, 14. A gas cylinder can withstand a pressure of, 15 atm. The pressure of cylinder is measured 12, atm at 27°C. Upto which temperature limit the, cylinder will not burst?, (a) 375°C, (b) 102°C, (c) 33.75°C, (d) 240°C, 15. Select the correct statement. In the gas, equation, PV = nRT, (a) n is the number of molecules of a gas, (b) n moles of the gas have a volume V, (c) V denotes volume of one mole of the gas, (d) P is the pressure of the gas when only one, mole of gas is present., 16. Molar volume of CO2 is maximum at, (a) NTP, (b) 0°C and 2.0 atm, (c) 127°C and 1 atm, (d) 273°C and 2.0 atm., 17. The volume occupied by 1.8 g of water vapour, at 374°C and 1 bar pressure will be, [Use R = 0.083 bar L K–1mol–1], (a) 96.66 L, (b) 55.87 L, (c) 3.10 L, (d) 5.37 L, 18. Dimension of universal gas constant (R) is, (a) [VPT –1n–1], (b) [VP –1Tn–1], (c) [VPTn–1], (d) [VPT –1n], 19. The mole fraction of dioxygen in a, neon‑dioxygen mixture is 0.18. If the total, pressure of the mixture is 25 bar, the partial, pressure of neon in the mixture would be, (a) 25.18 bar, (b) 25.82 bar, (c) 4.5 bar, (d) 20.5 bar, 20. 25 g of each of the following gases are taken, at 27°C and 600 mm pressure. Which of these, will have the least volume?, , (a) HBr, (c) HF, , (b) HCl, (d) HI, , 21. At high pressure, the compressibility factor Z, is equal to, Pb, (a) unity, (b) 1 −, RT, Pb, (c) 1 +, (d) zero., RT, 22. Gas deviates from ideal gas nature because, molecules, (a) are colourless, (b) attract each other, (c) contain covalent bond, (d) show Brownian movement., 23. van der Waals equation of state is obeyed by, real gases. For n moles of a real gas, the expression, will be,  P na   V , (a)  + 2  , = RT,  n V   n − b, a , , (b)  P + 2  (V – b) = nRT, , V , na , , (c)  P + 2  (nV – b) = nRT, , V , , n2 a , (d)  P +,  (V – nb) = nRT, V2 , , 24. The van der Waals equation reduces itself to, the ideal gas equation at, (a) high pressure and low temperature, (b) low pressure and low temperature, (c) low pressure and high temperature, (d) high pressure and high temperature., 25. Maximum deviation from ideal gas is expected, from, (a) CH4(g), (b) NH3(g), (c) H2(g), (d) N2(g), 26. A gas such as carbon monoxide would be most, likely to obey the ideal gas law at, (a) low temperatures and high pressures, (b) high temperatures and high pressures, (c) low temperatures and low pressures, (d) high temperatures and low pressures., 27. The correction factor ‘a’ to the ideal gas equation, corresponds to, (a) density of the gas molecules, (b) volume of the gas molecules, (c) electric field present between the gas molecules, (d) forces of attraction between the gas molecules.

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28. Which of the following plots is not according, to Boyle’s law?, (a), , (c), , (b), , (d), , 29. Which of the following expressions does not, represent Charles’ law?,  273.15 + t°C , (a) Vt = V0 , ,  273.15 , , (b) Vt = a + bt, V0, , , (c) Vt = , t, 273, ., 15, K, , , (d) Vt = V0t, 30. To raise the volume of a gas by four times,, the following methods may be adopted. Which of, the methods is wrong?, (a) T is doubled and P is also doubled., (b) Keeping P constant, T is raised four times., (c) Temperature is doubled and pressure is, halved., (d) Keeping temperature constant, pressure is, reduced to 1/4 of its initial value., , Case Based MCQs, Case I : Read the passage given below and, answer the following questions from 31 to 35., , (a) 1/r4, (c) 1/r5, , Intermolecular forces are the forces of attraction, and repulsion that exist between molecules, of a compound. These cause the compound to, exist in a certain state of matter – solid, liquid, or gas and affect the melting and boiling points, of compounds as well as the solubilities of one, substance in another. Attractive intermolecular, forces are also called van der Waals’ forces. These, are weak forces., , 34. Attractive intermolecular forces known as, van der Waals forces do not include which of the, following types of interactions?, (a) London forces, (b) Dipole-dipole forces, (c) Ion-dipole forces, (d) Dipole-induced dipole forces, , 31. Dipole-dipole forces act between the, molecules possessing permanent dipole. Ends, of dipoles possess ‘partial charges’. The partial, charge is, (a) more than unit electronic charge, (b) equal to unit electronic charge, (c) less than unit electronic charge, (d) double the unit electronic charge., 32., is, (a), (b), (c), (d), , The nature of inter-particle forces in benzene, dipole-dipole interaction, dispersion force, ion-dipole interaction, H-bonding., , 33. The interaction energy between two, temporary dipoles is proportional to (where r is, the distance between the two particles), , (b) 1/r2, (d) 1/r6, , 35. In which of the following molecules, the, van der Waals forces are likely to be the most, important in determining the m.pt. and b.pt?, (a) CO, (b) H2S, (d) HCl, (c) Br2, Case II : Read the passage given below and, answer the following questions from 36 to 40., If a hydrogen atom is bonded to a highly, electronegative element such as fluorine, oxygen,, nitrogen, then the shared pair of electrons lies, more towards the electronegative element. This, leads to a polarity in the bond in such a way, that a slight positive charge gets developed on, H-atom, viz,, Hd+ : Od–, Hd+ : Fd–, Hd+ : Nd–, Such a bond between the hydrogen atom of one, molecule and the more electronegative atom of, the same or another molecule is called hydrogen, bond.

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36. Which of the following compounds can form, hydrogen bond?, (a) CH4, (b) H2O, (c) NaCl, (d) CHCl3, 37. The boiling point is not affected due to, hydrogen bonding in, , 1, P, (at constant n and T), , From Boyle’s law we get, V ∝, , From Charles’ law we get, V ∝ T, (at constant n and P), From Avogadro’s law we get, V ∝ n (at constant T and P), , (a) water, , (b) ammonia, , Combining the above three equations we get, , (c) methyl alcohol, , (d) hydrogen chloride., , V∝, , 38. Unusual high b.p. of water is result of, (a) intermolecular hydrogen bonding, (b) intramolecular hydrogen bonding, (c) both intra and intermolecular hydrogen, bonding, (d) high specific heat., 39. Boiling point of hydrogen fluoride is highest, amongst HF, HCl, HBr and HI. Which type of, intermolecular forces are present in hydrogen, fluoride?, (a) H—F has highest van der Waals forces and, dipole moment., (b) H—F has highest London forces., (c) H—F has highest dipole moment hence has, dipole-dipole, London forces and hydrogen, bonding., (d) H—F has strong intermolecular interactions, like dipole-induced dipole., 40. Which of the following statements is not true?, (a) Intermolecular hydrogen bonds are formed, between two different molecules of compounds., (b) Intramolecular hydrogen bonds are formed, between two different molecules of the same, compound., (c) Intramolecular hydrogen bonds are formed, within the same molecule., (d) Hydrogen bonds have strong influence on the, physical properties of a compound., Case III : Read the passage given below and, answer the following questions from 41 to 45., An ideal gas is a gas to which the laws of Boyle and, Charles are strictly applicable under all conditions, of temperatures and pressures., , nT, nT, or, V = R, [where R = ideal gas constant], P, P, , or PV = nRT, Ideal gas equation is a relation between four, variables and it describes the state of any gas., For this reason, it is also called equation of state., 41. At 25°C and 730 mm pressure, 380 mL of dry, oxygen was collected. If the temperature is constant,, what volume will the oxygen occupy at 760 mm, pressure?, (a) 365 mL, (b) 449 mL, (c) 569 mL, (d) 621 mL, 42. 7.0 g of a gas at 300 K and 1 atm occupies a, volume of 4.1 litre. What is the molecular mass of, the gas?, (a) 42, , (b) 38.24, , (c) 14.5, , (d) 46.5, , 43. If P is the pressure and r is the density of a, gas, then P and r are related as, (a) P ∝ r, , (b) P ∝ r2, , (c) P ∝ 1/r, , (d) P ∝ 1/r2, , 44. I, II, III are three isotherms respectively at, T1, T2 and T3. Temperature will be in order, (a) T1 = T2 = T3 , (b) T1 < T2 < T3, (c) T1 > T2 > T3, (d) T1 > T2 = T3, 45. If volume of 2 moles of an ideal gas at 540 K is, 44.8 litre, then its pressure will be, (a) 1 atm, , (b) 3 atm, , (c) 2 atm, , (d) 4 atm

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Case IV : Read the passage given below and answer, the following questions from 46 to 48., Real gases do not obey ideal gas equation under, all conditions. They nearly obey ideal gas equation, at higher temperatures and very low pressures., However, they show deviations from ideality at, low temperatures and high pressures., The isotherms obtained by plotting pressure, P, against volume, V for real gases do not coincide, with that of ideal gas, as shown :, P, , positive deviation; when Vreal > Videal, ideal behaviour; when Vreal = Videal, negative deviation; when Vreal < Videal, , V, Vreal = Volume of the real gas at given pressure., Videal = Volume of the gas calculated by ideal gas, equation at given pressure., , The deviation from ideal gas behaviour can also, be expressed by compressibility factor, Z., , 46. The gas equation PV = nZRT becomes ideal, gas equation when, (a) Z = 0, (b) Z = 0.5, (c) Z = 1, (d) Z = 2, 47. The units of van der Waals’ constants a and b, respectively are, (a) L atm2 mol–1 and mol L–1, (b) L atm mol2 and mol L, (c) L2 atm mol–2 and mol–1 L, (d) L–2 atm–1 mol–1 and L mol–2, 48. The correction factor ‘b’ to the ideal gas, equation corresponds to, (a) density of the gas molecules, (b) excluded volume or covolume, (c) electric field present between the gas molecules, (d) forces of attraction, molecules., , between, , the, , gas, , Assertion & Reasoning Based MCQs, For question numbers 49-55, a statement of assertion followed by a statement of reason is given. Choose, the correct answer out of the following choices., (a) Assertion and reason both are correct statements and reason is correct explanation for assertion., (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion., (c) Assertion is correct statement but reason is wrong statement., (d) Assertion is wrong statement but reason is correct statement., 49. Assertion : Three states of matter are the, result of balance between intermolecular forces, and thermal energy of the molecules., Reason : Intermolecular forces tend to keep, the molecules together but thermal energy of, molecules tends to keep them apart., 50. Assertion : At constant temperature, PV vs, V plot for real gases is not a straight line., Reason : At high pressure all gases have Z > 1, but at intermediate pressure most gases have Z < 1., 51. Assertion : The plot of volume (V) versus, pressure (P) at constant temperature is a, hyperbola in the first quadrant., Reason : V ∝ 1/P at constant temperature., , 52. Assertion : Compressibility factor (Z) for, for non ideal gases is always greater than 1., Reason : The gases which lave Z > 1 are, difficult to compress., 53. Assertion : van der Waals equation is, applicable only to non-ideal gases., Reason : Ideal gases obey the equation, PV = nRT., 54. Assertion : Vapour pressure of NH 3 is, higher than C2H5OH., Reason : H-bonding is observed in both the, molecules., 55. Assertion : The graph between P v/s 1/V is, a straight line., Reason : At constant temperature, P ∝ 1/V.

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SUBJECTIVE TYPE QUESTIONS, , Very Short Answer Type Questions (VSA), 1. Calculate the volume occupied by 4.0 mole of, an ideal gas under NTP condition., , 4., , 2. How is the partial pressure of a gas in a, mixture related to the total pressure of the, gaseous mixture?, , ideal behaviour?, , 3. The variation of pressure with volume of the, gas at different temperatures can be graphically, represented as shown in figure., , of water in all the three states., , Under which of the following two conditions, , applied together, a gas deviates most from the, 5., , Physical properties of ice, water and steam are, , very different. What is the chemical composition, 6., , Name two intermolecular forces that exist, , between HF molecules in liquid state., 7., , Define the partial pressure of gas., , 8., , Define compressibility factor., , 9., , Using the equation of state PV = nRT; show, , that at a given temperature, density of a gas is, On the basis of this graph given, how will the, volume of a gas change if its pressure is increased, at constant temperature?, , proportional to gas pressure P., 10. Why is boiling point of hydrogen fluoride, higher than that of hydrogen chloride?, , Short Answer Type Questions (SA-I), 11. 40 mL of O2 was collected at 100 °C and 1 bar, pressure. Calculate its volume (in mL) at 273 K, and 1.013 bar., 12. 0.068 dm3 of a sample of nitrogen is collected, over water at 20 °C and 0.92 bar. What will be the, volume of dry nitrogen at STP (in mL) (Aqueous, tension of water at 20 °C = 0.023 bar)?, 13. Calculate the volume (in m3) occupied by 2, moles of an ideal gas at 25 × 105 Nm–2 pressure, and 300 K temperature., , (R = 0.083 bar L K–1 mol–1), 17. Which type of intermolecular forces exist, among the following molecules?, (i) He atoms and HCl molecules, (ii) HF molecules, (iii) N2 molecules, (iv) HCl molecules, 18. Explain the physical significance of van der, Waals’ parameters., , 14. At 0°C, the density of a certain oxide of a gas, at 2 bar is same as that of dinitrogen at 5 bar., What is the molecular mass of the oxide?, , 19. Calculate the total pressure (in bar) in a, mixture of 8 g of dioxygen and 4 g of dihydrogen, confined in a vessel of 1 dm3 at 27°C., , 15. How much time (in years) would it take to, distribute one Avogadro number of wheat grains,, if 1010 grains are distributed each second?, , R = 0.083 bar dm3 K–1 mol–1, , 16. Calculate the volume (in litres) occupied by, 8.8 g of CO2 at 31.1°C and 1 bar pressure., , 20. 1 mole of sulphur dioxide occupies a volume, of 350 mL at 27 °C and 5 × 106 Pa pressure., Calculate the compressibility factor of the gas. Is, it less or more compressible than an ideal gas?

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Short Answer Type Questions (SA-II), 21. Describe London forces or dispersion forces, with example., , 23. On the basis of intemolecular forces and, thermal energy explain why substances exist in, three different states of matter?, 24. Two moles of ammonia gas are enclosed in, a vessel of 5 litre capacity at 27ºC. Calculate the, pressure exerted by the gas, assuming that, (i) the gas behaves like an ideal gas (using ideal, gas equation), (ii) the gas behaves like a real gas (using van, der Waals equation), Given that for ammonia, a = 4.17 atm litre2 mol–2, and b = 0.037 litre mol–1., 25. 2.9 g of a gas at 95 °C occupied the same volume, as 0.184 g of dihydrogen at 17 °C at the same, pressure. What is the molar mass of the gas?, 26. On the basis of their interaction energy and, thermal energy explain why, (i) a solid has high rigidity?, (ii) In gas, molecules are sufficiently apart from, one another?, (iii) liquid has no definite shape?, 27. Explain the following :, (i) Boyle’s law, (ii) Avogadro’s law, 28. (a) Which gas law is shown by the following, graph?, , PV, P, , (b) At 25°C and 760 mm Hg pressure, a gas, occupies 600 mL volume. What will be its pressure, (in mmHg) at a height where temperature is, 10°C and volume of the gas is 640 mL?, 29. Pressure versus volume graph for a real gas, and an ideal gas are shown in the figure., , Pressure, , 22. The drain cleaner, Drainex contains small, bits of aluminium which react with caustic, soda to produce dihydrogen. What volume of, dihydrogen (in mL) at 20°C and one bar will be, released when 0.15 g of aluminium reacts?, , Real gas, Ideal gas, , 0, , Volume, , Answer the following questions on the basis of, this graph., (i) Interpret the behaviour of real gas with, respect to ideal gas at low pressure., (ii) Interpret the behaviour of real gas with, respect to ideal gas at high pressure., (iii) Mark the pressure and volume by drawing a, line at the point where real gas behaves as, an ideal gas., 30. A perfectly elastic spherical balloon of 0.2, m diameter was filled with hydrogen at sea, level. What will be its diameter (in m) when it, has risen to an altitude where the pressure is, 0.65 atm? (Assume no change in temperature, and atmospheric pressure at sea level)., 31. (i) Out of CO2 and He, which gas have, higher value of van der Waals’ constant ‘b’?, (ii) At 27°C density of a gaseous substance at 3, bar is same as that of hydrogen at 9 bar. What is, molar mass of the substance?, 32. What will be the pressure (in bar) of a gas, mixture when 0.5 L of H2 at 0.8 bar and at 2.0 L of, oxygen at 0.7 bar are introduced in a 1 L vessel at, 27 °C?, 33. (a) 22 g of dry ice is placed in an evacuated, bottle of 1 litre capacity and tightly stoppered., What would be the pressure inside the bottle in, atm, when it is heated to 37°C?, (b) 3.12 g of sulphur is vapourised at 427°C and, 760 mm pressure, when the vapours occupy a, volume of 700 mL. Find the molecular formula of, sulphur. (atomic mass of sulphur = 32)., 34. An open beaker at 27°C is heated to 477°C., What percentage of air would have been expelled, out?, 35. Explain the difficulties faced by the, mountaineers with respect to the air present, around them. How is this difficulty solved?

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Long Answer Type Questions (LA), 36. Payload is defined as the difference between, the mass of displaced air and the mass of the, balloon. Calculate the payload (in kilograms), when a balloon of radius 10 m, mass 100 kg is, filled with helium at 1.66 bar at 27°C. (Density of, air =1.2 kg m–3 and R = 0.083 bar dm3 K–1 mol–1), 37. A liquefied petroleum gas (LPG) cylinder, weighs 14.8 kg when empty. When full it, weighs 29.0 kg and shows a pressure of, 2.5 atm. In the course of use at 27 °C, the mass, of the full cylinder is reduced to 23.2 kg. Find, out the volume of the gas in cubic metres used, up at the normal usage conditions and the final, pressure inside the cylinder. Assume LPG to be, n-butane with normal boiling point of 0 °C., 38. At temperature, T and pressure P, two, ideal gases A and B are mixed. Show that, the density d of the mixture is given by, , OBJECTIVE TYPE QUESTIONS, 1., , (d) : Gases are highly compressible., , 2. (b) : For gaseous state, thermal energy >> molecular, attraction., 3. (b) : Solid hydrogen, H2 is non-polar compound and, possesses London dispersion forces. Infact these are the only, attractive forces which a non-polar compound can have., 4., , (c), , 5. (c) : It is the type of force between the polar molecule, and a non-polar molecule. Dipole of polar molecule induces, dipole on the electrically neutral molecule., , 6. (a) : HCl is polar (m ≠ 0) and He is non-polar (m = 0)., Therefore, HCl and He atoms will posses dipole-induced dipole, interaction., 7., , (c) : Boyle’s law relates pressure and volume of a gas at, 1, constant temperature i.e., V ∝ (at constant T)., P, 8. (a) : Applying P1V1 = P2V2, P1 = 380 mm, V1 = 400 mL, P2 = 760 mm, V2 = ?, , P, where XA and XB, RT, are the mole fractions and MA and MB are, the molecular weights of the gases A and B, respectively., d = ( X A M A + X B MB ) ×, , 39. The volume of a gas is to be increased by, 20% without changing the pressure. To what, temperature (in °C) the gas must be heated if, the initial temperature of the gas is 27 °C?, 40. Answer the following :, (a) How are the van der Waals’ constants ‘a’ and, ‘b’ related to the molecular size?, (b) Using van der Waals’ equation, calculate, the constant ‘a’ when two moles of a gas, confined in a 4 L flask exerts a pressure of, 11.0 atmospheres at a temperature of 300 K., The value of ‘b’ is 0.05 L mol–1., , V2 =, , PV, 380 × 400, 11, =, = 200 mL, P2, 760, , 9. (d) : According to Charles’ law V ∝ T i.e., air expands on, heating, its density decreases. Hence hot air is lighter., 10. (d) : Ideal gas equation PV = nRT, If P, V and T are same n will also be same., 1, hence, volume decreases, 11. (c) : Since pV ∝ T and p ∝, V, with pressure. The values of p1, p2, p3, p4 show decrease in, volume. Hence, the order of pressure is p1 < p2 < p3 < p4., 12. (a) : PV = constant (at a given temperature), , 13. (b) : PAVA = nARTA and PBVB = nBRTB, PAVA 2PB × 2VB, n A RTA, 2TB, =, =, =2, PB × VB, nB PBVB, RTB, TB, , Thus number of molecules are also in the ratio 2 : 1.

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14. (b) : Cylinder will burst at that temperature when it, attains the pressure of 15 atm, P1 = 12 atm; T1 = 27°C = 27 + 273 = 300 K; P2 = 15 atm; T2 = ?, P1 P2, =, T1 T2, ⇒ T2 =, , 15 × 300, = 375 K = (375 − 273)°C = 102° C, 12, , 15. (b) : From ideal gas equation, PV = nRT, it can be, concluded that n moles of a gas occupy the volume V at, pressure P and temperature T., 1, and V ∝ T, P, \ CO2 has maximum volume at minimum pressure and, maximum temperature., 16. (c) : We know V ∝, , 22. (b) : Unlike postulates of kinetic molecular theory,, intermolecular forces in a gas are not negligible., 23. (d) : van der Waal’s equation is, , an 2 , P + 2  (V – nb) = nRT, V , , 24. (c), 25. (b) : It is a polar molecule, thus more attractive forces, between its molecule., 26. (d) : Real gases show ideal gas behaviour at high, temperatures and low pressures., 27. (d), 28. (d) : P ∝, , 17. (d) : m = 1.8 g, m 1.8, = 0.1 mol, n= =, M 18, , through origin., , T = 374°C = 647 K, P = 1 bar, R = 0.083 bar L K–1 mol–1, nRT 0.1 × 0.083 × 647, V=, =, = 5.37 L, 1, P, 18. (a) : From the gas equation, PV = nRT, P ×V, R=, = [VPT −1n −1], n ×T, 19. (d) : Given Ptotal = 25 bar, xO2 = 0.18, \ xNe = 1 – 0.18 = 0.82, pNe = xNe × Ptotal = 0.82 × 25 = 20.5 bar, , 30. (a) :, , 20. (d) : Ideal gas equation is PV = nRT, If pressure and temperature are same for all the gases then, V ∝ n (from above equation), wt., n = no. of moles of the gas =, Mol. wt., (here weight of all gases are equal), 1, ∴ V∝, Mol. wt., From the options, HI has more molecular weight than remaining, gases, so it has least volume., a , , 21. (c) : P + 2  (V − b ) = RT, , V , At high pressure b cannot be neglected in comparison to V., Further though V becomes small, a/V2 is large but as P is very, high, a/V 2 can be neglected in comparison to P. Hence, P(V – b) = RT or PV = RT + Pb, PV, Pb, Pb, or, = 1+, i .e. Z = 1 +, RT, RT, RT, , 1, , the graph will be a straight line passing, V, , V, 29. (d) : Vt = 0 × t, t0, PV, = constant, only (a) is wrong., T, , 31. (c) : Partial charge is a small charge developed by, displacement of electrons. It is less than unit electronic charge, and is represented as d+ or d–., 32. (b) : Benzene is non-polar compound and exhibits, London or dispersion forces., 33. (d) , , 34. (c), , 35. (c) : In the molecules the van der Waals force are likely to, determine the m.pt. and b.pt. Greater the mass of the molecule, greater will be its van der Waals force and higher will be its, m.pt. and b.pt. Br2 has highest m.pt., 36. (b) : Oxygen has high electronegativity and small size,, thus forms H-bond., 37. (d) : HCl does not undergo H-bonding and its boiling, point is not affected by H-bonding., 38. (a) : Due to intermolecular hydrogen bond in H2O, its, molecules are associated with each other which is responsible, for unusual high b.p. of water., 39. (c) : H—F has dipole-dipole interaction, London forces, and hydrogen bonding due to highest electronegativity of F., Hence, boiling point of H—F is highest., 40. (b) : Intramolecular hydrogen bonds are formed within, the same molecule., 41. (a) : Applying P1V1 = P2V2, P1 = 730 mm, V1 = 380 mL, P2 = 760 mm, V2 = ?, PV 730 × 380, V2 = 1 1 =, = 365 mL, P2, 760

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m, RT, M, P = 1 atm; V = 4.1 L; m = 7.0 g ;, R = 0.0821 L atm K–1 mol–1; T = 300 K, RT, 0.0821 × 300, × 7.0 = 42.05 ≈ 42 g mol−1, M=, ×m =, PV, 1 × 4.1, ρRT, 43. (a) : P =, i .e., P ∝ ρ, M, 44. (c) : Draw a line at constant pressure parallel to volumeaxis. Take volume corresponding to each temperature., From volume axis, V1 > V2 > V3, Hence T1 > T2 > T3., 42. (a) : Applying, PV =, , 45. (c) : Number of moles, n = 2;, Temperature, T = 540 K, Volume, V = 44.8 L; P = ?, [R, gas constant = 0.0821 L atm K–1 mol–1], According to ideal gas equation PV = nRT, P=, , nRT 2 × 0.0821 × 540, =, = 2 atm, V, 44.8, , 46. (c) : The ideal gas equation is PV = nRT, When Z (the compressibility factor) is one, the given equation, PV = nZRT becomes ideal gas equation., 47. (c) : a and b are expressed in terms of the units of P and V., Pressure correction = P ′ =, a=, , P ′V, n2, , 2, , =, , n 2a, , V2, pressure × (volume)2, (mole)2, , Unit of a = atm × (L)2mol–2, Unit of b is the same as for the volume. i.e., L mol–1, 48. (b) : The correction factor ‘b’ represents excluded volume, or covolume., 49. (a), 50. (b) : At constant temperature, plot of PV vs V for real, gases is not linear because real gases have intermolecular, forces of attraction., 1, 51. (a) : According to Charles’s law V ∝ ., P, \, , P, V, , 52. (d) : Compressibility factor (Z ) can be greater, equal or, lesser than 1., , 53. (b) : The van der Waals equation is applicable to real, gases only, while PV = nRT is applicable to ideal gases., 54. (b) : H - bonding in liquid NH3 is weaker than C2H5OH., Thus, escaping tendency is higher and hence, the vapour, pressure of NH3 is also higher., 55. (a), , SUBJECTIVE TYPE QUESTIONS, 1., or, , PV = nRT, nRT 4 × 0.082 × 273, = 89.54 L, V =, =, 1, P, , 2., , Partial pressure of a gas, , = Mole fraction of that gas × Total pressure, 3. The volume of a gas decreases if the pressure on the gas, is increased keeping the temperature constant., 4. At high pressure and low temperature, the gases deviate, from ideal behaviour due to significant intermolecular forces, and more than negligible size of the molecules., 5. The chemical composition of water remains same in all, the physical states i.e., solid, liquid and gas., 6. Dipole-dipole interactions and hydrogen bonding exist, between HF molecules in liquid state., 7. In a mixture of gases, the pressure exerted by the, individual gas is called its partial pressure., 8. The extent to which a real gas deviates from ideal, behaviour can be conveniently studied in terms of quantity ‘Z’, PV, called the compressibility factor, which is defined as Z =, nRT, For an ideal gas, as PV = nRT, Z = 1, 9., , PV = nRT, , wRT, w, , ,  Since n = M , MV, w, , or P = dRT , Since d = V , M, or, P ∝ d, Hence, density (d) of a gas ∝ P, because R, T and M are, constants., or, , P=, , 10. Boiling point of HF is higher than HCl due to extensive, hydrogen bonding between H — F molecules., H—F, H—F, H—F, H—F, 11. Given, P1 = 1 bar, V1 = 40 mL = 0.04 L = 0.04 dm3,, T1 = 373 K, P2 = 1.013 bar, T2 = 273 K, V2 = ?, PV, 1 1 = P2V2, T1, T2

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PV T, 1× 0.04 × 273, = 0.029 dm3 or 29 mL, or V2 = 1 1 2 =, T1P2, 373 × 1.013, 12. Given P1 = 0.92 – 0.023 = 0.897 bar,, V1 = 0.068 dm3 , T1 = 293 K, P2 = 1 bar (Pressure of dry gas),, V2 = ?, T2 = 273 K, PV, 1 1 = P2V2 or V = PV, 1 1T2, 2, T1, T2, P2T1, 0.897 × 0.068 × 273, = 0.057dm3 or 57 mL, 1 × 293, 13. According to ideal gas equation, PV = nRT, =, , or, , V=, , nRT 2 × 8.314 × 300, =, ≈ 2 × 10–3 m3, P, 25 × 105, , d, RT, 14. M oxide = oxide ...(1), Poxide, ...(2), d N2 RT, M N2 =, , PN2, Dividing equation (1) by equation (2) gives, M oxide d oxide × PN2, =, d N2 × Poxide, M N2, But d oxide = d N , 2, 5 bar, × 28 g mol−1, 2 bar, Moxide = 70 g mol–1, ∴ M oxide =, , 15. Time taken to distribute 1010 grains = 1 sec., Time taken to distribute 6.023 × 1023 grains, =, , 1 × 6.022 × 1023, 1010, , = 6.022 × 1013 sec, , 6.022 × 1013, =, = 1.90956 × 106 years, 60 × 60 × 24 × 365, 16. According to ideal gas equation, PV = nRT, n=, , 8.8, moles, P = 1bar,, 44, , T = 273 + 31.1 = 304.1 K,, V=?, R = 0.083 bar L K–1 mol–1, nRT, Now, V =, P, ,  8.8, moles  × (0.083 bar L K −1 mol−1) × (304.1K), , , 44, =, 1bar, = 5.05 L, , 17., (ii), (iii), (iv), , (i) Dipole-induced dipole forces, Hydrogen bonding, Dispersion forces, Dipole-dipole forces, , 18. ‘a’ measures the intermolecular forces of attraction. The, greater the value of 'a', the more will be intermolecular forces, of attraction. ‘b’ measures volume occupied by molecules of a, gas., 19. Partial pressure of oxygen gas,, 8, nRT, P=, , n = mol, V=1 dm3, T = 300 K, 32, V, P(O2 ) =, , 8 × 0.083 × 300, = 6.225 bar, 32 × 1, , Partial pressure of hydrogen gas,, nRT, 4, P=, , n = = 2 mol, V, 2, 2 × 0.083 × 300, P(H2 ) =, = 49.8 bar, 1, Total pressure = P( O2 ) + P(H2 ) = 6.225 + 49.8, , , = 56.025 bar, , PV, nRT, n = 1 mol, P = 5 × 106 Pa, V = 350 mL = 0.350 × 10–3 m3, R = 8.314 N m K–1 mol–1, T = 27 + 273 = 300 K, , 20. Compressibility factor, Z =, , 5 × 106 × 0.350 × 10 −3, = 0.702, 1.0 × 8.314 × 300, Thus, SO2 is more compressible than an ideal gas (which has, Z = 1)., ∴, , Z=, , 21. London or dispersion forces : This is the weakest, intermolecular force. It is a temporary attractive force that, results when the electrons in two adjacent atoms occupy, positions that make the atoms to form temporary dipoles. This, force is sometimes called an dipole-induced dipole attraction., Because of the constant motion of the electrons, an atom, or molecule can develop a temporary (instantaneous) dipole, when its electrons are distributed unsymmetrically about the, nucleus., , Atom A, , Atom B, , symmetrical distribution of electronic charge cloud, (a)

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Predominance of thermal energy and the intermolecular forces, in the three state of matter is as follows :, Gas → Liquid → Solid, Atom ‘A’ with instantaneous, dipole, more electron density, on the right hand side, , Predominance of intermolecular interactions, , Atom ‘B’ with induced dipole,, , Gas ←   Liquid ←   Solid, Predominance of thermal energy, , (b), , Atom ‘A’ more electron, density on the left hand side, , Atom ‘B’ with induced dipole, , (c), Dispersion forces or London forces between atoms., , 22. The reaction between aluminium and caustic soda is, 2Al + 2NaOH + 2H2O → 2NaAlO2 + 3H2, 2 × 27, = 54 g, , , , 3 × 22.4 L, at STP, , \ 54 g of Al produces H2 at S.T.P. = 3 × 22.4 L, 0.15 g of Al will produce H2 at S.T.P., = 3 × 22.4 × 0.15 = 0.186 L, 54, At STP , P1 = 1 atm, V1 = 0.186 L, T1 = 273 K, , Given conditions, P2 = 1 bar = 0.987 atm, V2 = ?, T2 = 273 + 20 = 293 K, , 1 1 = P2V2, Applying ideal gas equation, PV, T1, T2, 1 × 0.186 0.987 × V2, =, 273, 293, 293 1 × 0.186, V2 =, ×, = 0.2030 L = 203 mL, 0.987, 273, , 23. The intermolecular forces tend to keep the molecules, together but thermal energy tend to keep them apart. Thus,, these two compete and the competition between these two, (i.e., intermolecular forces and thermal energy) results in three, states of matter., (i) In a solid, the intermolecular forces predominate over the, thermal energy and hence, the particles are held together in, rigid, highly-oriented and close-packed structure., (ii) In liquids, the intermolecular forces are no longer strong, enough, however, these are still sufficient so that particles, remain in each other’s environment, hence, liquids have, sufficient mobility., (iii) In gases, the thermal energy dominates the effect of, intermolecular forces, thus, the gas molecules acquire the, unrestricted and independent mobility in the vapour state., , 24. Given, n = 2 moles, V = 5 litres, T = 27ºC = (27 + 273), K = 300 K, a = 4.17 atm litre2 mol–2, b = 0.037 litre mol–1, Also, we know that R = 0.0821 litre atm K–1 mol–1, (i) If the gas behaves like an ideal gas, we have PV = nRT, nRT 2 × 0.0821 × 300, = 9.85 atm, =, \ P=, 5, V, (ii) If the gas behaves like a real gas, we apply van der Waals’, equation i.e.,, , an 2 , nRT, an 2, − 2, P + 2  (V − nb ) = nRT or P =, V − nb V, V, =, , 2 × 0.0821 × 300 4.17 × (2)2, −, = 9.33 atm., 5 − 2 × 0.037, (5)2, , 25. Case I : Let molar mass of gas be M g mol–1, Weight of gas = 2.9 g, Weight, 2.9, No. of moles =, =, Molar mass M, T = 273 + 95 = 368 K, 2.9, PV =, × R × 368 ...(i), M, Case II : Mass of dihydrogen = 0.184 g, 0.184, No. of moles of H2 =, 2, T = 273 + 17 = 290 K, 0.184, PV =, × R × 290 ...(ii), 2, From equations (i) and (ii),, 2.9, 0.184, × 368 =, × 290, 2, M, 2.9 × 368 × 2, M=, = 40 g mol–1, 0.184 × 290, 26. (i) A solid has high rigidity because thermal motion is, too weak to overcome the strong intermolecular forces of, attraction., (ii) In a gas, thermal energy is so high that the molecules, cannot come close together. Hence, there are large empty, spaces between them.

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(iii) In a liquid, there is a reasonable balance between the, attractive intermolecular forces and thermal energy. Hence,, molecules in a liquid exist together, i.e., it is a condensed state, of matter but there is no rigidity. That is why they have no, definite shape., 27. (i) Boyle’s law : According to Boyle’s law, at constant, temperature, the volume of a fixed amount of a gas is inversely, proportional to its pressure, i.e., if volume increases, the, pressure would decrease. This is because as volume increases,, the number of molecules striking the walls of a container in a, given time decreases leading to decrease in pressure., 1, Constant, P ∝ ;P =, V, V, ⇒ PV = Constant ⇒ P1V1 = P2V2, (ii) Avogadro’s law : This law states that equal volume of all, gases under similar conditions of temperature and pressure, contain equal number of molecules., , Pressure at the altitude (P2) = 0.65 atm (Given), As temperature remains constant, applying Boyle’s law,, P1V1, , =, , P2V2, , (At sea level)    (At altitude), 4, 4, 1 atm × π(0.1 m)3 = 0.65 atm × πr23, 3, 3, 3, (0.1 m), or r23 =, = 1.54 × 10 −3 m3, 0.65, \, , r2 = (1.54 × 10–3)1/3 m = 0.1154 m, , \, , Diameter of the balloon at altitude, , , , = 2 × 0.1154 m = 0.2308 m, , 31. (i) Since, CO2 molecules have larger size than that of He, molecules, hence, CO2 has larger value of van der Waals’ constant, ‘b’., d RT, (ii) M1 = 1 ...(i), P1, [where,, , V V, V ∝ n (T, P constant) = 1 = 2, n1 n2, , M1 = Molecular weight of the substance, , 28. (a) Boyle’s law, , P1 = Pressure of the substance], d H RT, ...(ii), M H2 = 2, PH2, , (b) Applying gas equation (combined gas law),, PV, 1 1 = P2V2 ⇒ 760 × 600 = P2 × 640, T1, T2, 298, 283, or, , P2 = 676.6 mm of Hg, , 29. (i) At low pressure real gas starts behaving like an ideal gas., , d1 = Density of the substance, , Dividing equation (i) by equation (ii), PH, M1 d1 × PH2, M, =, or 1 = 2 [Given, d1 = d H2], M H2 d H2 × P1, M H2 P1, , (ii) At high pressure gases deviate from ideal behaviour., (iii), , or, , M1 =, , M H2 × PH2 2 × 9, =, =6, P1, 3, , 32. Partial pressure of H2 :, V1 = 0.5 L, V2 = 1 L,, P1 = 0.8 bar, P2 = ?, By Boyle’s law, P1V1 = P2V2, , Suppose the radius of the balloon at altitude = r2, , 0.8 × 0.5, = 0.4 bar, 1, Partial pressure of O2 :, V1 = 2.0 L, V2 = 1 L, P1 = 0.7 bar, P2 = ?, By Boyle’s law, P1V1 = P2V2, 0.7 × 2.0, P2 =, = 1.4 bar, 1, Pressure of the gas mixture,, Pmix = pH2 + pO2 = 0.4 + 1.4 = 1.8 bar, , 4, Then, volume of the balloon at altitude (V2 ) = πr23, 3, , 33. (a) W = 22 g CO2, V = 1 L, M = 44, T = 37 + 273, = 310 K, P = ?, , At point A real gas behaves as an ideal gas., 30. If r1 is the radius of the balloon at sea level, then, volume, 4 3 4, 3, of the balloon at sea level = πr1 = π(0.1 m), 3, 3, 4, 3, i.e., volume of the gas at sea level (V1) = π(0.1 m), 3, Pressure at the sea level (P1) = 1 atm, , P2 =

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Dry ice is solid CO2, which when heated in an evacuated bottle, it is converted into gaseous CO2., From ideal gas equation,, WRT, 22 × 0.082 × 310, ⇒P=, = 12.71 atm., PV =, M, 44 × 1, Now, pressure inside the bottle is 12.71 atm., (b) For sulphur,, W = 3.12 g, T = 427 + 273 = 700 K, P = 760 mm = 1 atm., V = 700 ml = 0.7 L., Now molecular mass of sulphur,, WRT 3.12 × 0.082 × 700, = 255.84, M=, =, PV, 1 × 0.7, As atomic mass = 32, so no. of atoms in one molecule of, 255.84, sulphur =, =8, 32, Hence, molecular formula of sulphur is S8., 34. Suppose the number of moles of gas present at 27°C in, flask of volume V at pressure P is n1, then assuming ideal gas, behaviour,, PV = n1R × 300...(i), Suppose n2 = number of moles at 477°C, then, PV = n2R × 750...(ii), From equation (i) and equation (ii), we get, 300, × n 1 = 0.4 n1, 750, So, (1 – 0.4) × 100% i.e. 60% is expelled out., n2 =, , 35. At altitude, the atmospheric pressure is low. Hence, air is, less dense. As a result, less oxygen is available for breathing., The person feels uneasiness, headache, etc. This is called altitude, sickness. This difficulty is solved by carrying oxygen cylinders, with them., 36. Volume of balloon =, , 4 3, πr, 3, , Radius of balloon, r = 10 m, 4, V = × 3.14 × (10)3 = 4186.7 m 3, 3, Mass of displaced air = 4186.7 m3 × 1.2 kg m–3 , , = 5024.04 kg, PV, Moles of gas present =, RT, 1.66 × 4186.7 × 103, =, = 279.11 × 103 moles, 0.083 × 300, , Mass of helium present = 279.11 × 103 × 4, , , = 1116.44 × 103 g, , , , = 1116.44 kg, , Mass of filled balloon, , , = 100 + 1116.44 , = 1216.44 kg, , Payload = Mass of displaced air – Mass of balloon, , , = 5024.04 – 1216.44 = 3807.6 kg, , 37. Weight of LPG originally present, = 29.0 – 14.8 = 14.2 kg, Pressure = 2.5 atm, Weight of LPG present after use = 23.2 – 14.8, , = 8.4 kg, Since volume of the cylinder is constant, applying, P n w / M w1, =, PV = nRT ⇒ 1 = 1 = 1, P2 n2 w 2 / M w 2, ⇒, ..., , 2.5 14.2, 2.5 × 8.4, =, or P2 =, = 1.48 atm, P2 8.4, 14.2, Weight of used gas = 14.2 –8.4 = 5.8 kg, , Moles of gas =, , 5.8 × 103, = 100 mol, 58, , Normal conditions : P = 1 atm, T = 273 + 27 = 300 K, Volume of 100 mol of LPG at 1 atm and 300 K, nRT 100 × 0.082 × 300, V=, =, = 2460 L = 2.460 m3, P, 1, 38. At temperature T and pressure P, two ideal gases, A and, B are mixed., nA and nB are the number of moles of A and B respectively., n is total moles of A and B present in the mixture., n = nA + nB., Let the gas mixture has a volume V., VA and VB are volume of A and B respectively., From ideal gas equation, PV = nRT., n RT, For gas A, PVA = nART. \ VA = A, P, n RT, For gas B, PVB = nBRT. \ VB = B, P, MA and MB are molecular weights of the gases and XA and XB, are the mole fractions of the gases A and B respectively., mass, Density of a gas, d =, volume, d=, , M n A M A + nB M B, P  n A M A + nB M B , =, =, , , V n ART + nB RT RT  n A + nB , P, P

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=, , (b) The van der Waals’ equation is, , P  n A M A + nB M B  P  n A, nB, , , =,  M A + MB , RT , n, n,  RT  n, , , , n 2a , P + 2  (V − nb ) = nRT, v, Given, n = 2 mol, V = 4 L,, , P, { X AM A + X B M B }, RT, 39. Suppose volume of gas at 27°C = V cm3, 20, × V = 0.2 V, Increase in volume desired = 20% of V =, 100, \ Final volume = V + 0.2 V = 1.2 V, Now, V1 = V cm3, T1 = 300 K,, V2 = 1.2 V, T2 = ?, At constant P,, V, 1.2 V, V1 V2, =, ⇒, =, 300, T2, T1 T2, d=, , P = 11.0 atm, T = 300 K, Substituting the values in the above equation,, 2 , ,  2 mol , –1, 11.0 atm +  4 L  ⋅ a  (4 L – 2 mol × 0.05 L mol ), , , = 2 mol × 0.082 L atm K–1 mol–1 × 300 K, a, , 2 −2  (4 L – 0.1 L) = 49.2 L atm, 11.0 atm + 4 mol L , a, , 2 −2  (3.9 L) = 49.2 L atm, 11.0 atm + 4 mol L , , ⇒ T2 = 360 K = 360 – 273 = 87°C, 40. (a) The van der Waals’ constant ‘a’ is measure of, intermolecular attractions. Therefore, the value of ‘a’ reflects, the tendency of the gas to liquefy. The gas having larger value, of ‘a’, will liquefy more easily. The van der Waals’ constant, ‘b’ is a measure of the close-packed molecular volume. Thus, the molecule of a gas having greater value of ‘b’ has bigger, size., , 42.9 L atm +, , , , 3 .9 a, mol2 L–1 = 49.2 L atm, 4, , 3 .9, a mol2 L–1 = (49.2 L atm – 42.9 L atm), 4, = 6.3 L atm, a=, , 6.3 L atm × 4, 2 −1, , 3.9 mol L, , = 6.46 atm L2 mol–2