UNIT 5 STATES OF MATTER Concept- Intermolecular forces and gas laws KNOWLEDGE

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1 UNIT 5 STATES OF MATTER Concept- Intermolecular forces and gas laws KNOWLEDGE One Mark Questions Q1. What will be the molar volume of nitrogen and argon at K and 1atm pressure?( L-2) Ans. At K and 1atm pressure every gas has molar volume = 22.4L. Q 2.What do you mean by thermal energy? ( L-1) Ans. Thermal energy is the energy of a body arising from motion of its atoms or molecules. Q 3.Name two intermolecular forces that exist between HF molecules in liquid state.( L-2 ) Ans. Dipole-dipole interaction and hydrogen bonding. Q 4.Define Boyle s law. How is it represented mathematically? (L-1) Ans.The volume of a given mass of a gas is inversely proportional to its pressure at constant temperature t. Two Marks Questions Q 1.How is the strength of hydrogen bond determined? ( L-2) Ans. Strength of H bonding determined by the coulombic interaction between the electronegative atom of one molecule and H atom of other molecule. Q 2.Why liquids diffuse slowly as compared to gases? ( L-2) Ans.In liquid the intermolecular forces (vander waal s forces) are greater than in gas. Three Mark Questions Q 1.Define (a) Dipole-Dipole Forces (b) Avogadro law (c) Intermolecular forces ( L-1) Ans.(a) Dipole-dipole forces : The force of attraction which act between two molecules having permanent dipoles. (b) Avogadro Law : Equal volumes of all gases under the similar condition of temperature and (c) Intermolecular Forces: The forces of attractions and repulsion between interacting particles. Q 2.Describe Charle s Law. In terms of Charle s Law explain why C is the lowest temperature? Ans. The volume of a fixed mass of a gas is directly proportional to its absolute temperature at constant Pressure. V At C, volume of the gas approaches to zero i.e the gas ceases to exist. Q 3.Define dispersion forces. Write characteristic features of dispersion forces.( L-2) Ans. Dispersion forces: The forces of attraction between two temporary dipoles. Characteristics of dispersion forces: (i) These forces are always attractive. (ii) These forces are effective only at short distance. (iii) Their magnitude depends on the polarisability of the particles

2 UNDERSTANDING One Mark Question Q1. Write various means to easily liquify the gases. (L-1) Ans. By lowering the temperature and increasing the pressure. Q2. According to you, what are the factors that determine the state of matter ( L-2) Ans. The factors that determine the different states of the matter are pressure, temperature. Q 3.Arrange solid, liquid and gas in order of energy. Give reason.(l-1) Ans. Solid < liquid < gas. This is because solid absorb energy to change into a liquid which further absorb energy to change into a gas. Two Marks Questions Q 1 Give one example for each of the following types of inter-particle forces (L-2) (a) London dispersion forces (b)dipole-dipole forces (c) Hydrogen bond (d) Ion-dipole forces Ans. (a) noble gases. (b) between HCl molecules. (c) Between water molecules or HF molecules. (d)between NO - 3 and water molecules. Q 2.Why liquids do not have definite shape although they have definite volume? (L-3) Ans.Because inter molecular forces are strong enough to hold the molecules but not so strong as to fix them into definite position. Q 3.Which type of intermolecular forces are effected in each of the following process:( L-2) (i)sublimation of iodine. (ii)evaporation of water. Ans. (i) Sublimation of iodine : Dispersion forces. (ii)evaporation of water : Hydrogen bond. Three Mark Questions Q 1.Which type of intermolecular forces exists among the following molecules?( L-2) (a) H 2 S molecules (b) H 2 O (c) Cl 2 and CCl 4 (d) SiH 4 (e) Helium (f) He atoms and HCl molecules Ans. (a) Dipole-Dipole interactions. (b)h-bond. (c)dispersion forces. (d)dispersion forces. (e)dispersion forces. (f)induced dipole-dipole interactions.

Q 2 Elaborate the statement that physical state of existence at given conditions is a balance between the thermal energy and intermolecular forces. (L-3) Ans.

3 Q 2 Elaborate the statement that physical state of existence at given conditions is a balance between the thermal energy and intermolecular forces. (L-3) Ans. Intermolecular forces tend to keep the molecules together but thermal energy of the molecules tends to keep them apart. The three states of matter are the result of the balance between intermolecular forces and thermal energy of the molecules. Q 3 What are isochores? From the three isochores I, II, III given below, for a certain amount of ideal Gas,What is the correct arrangement of volumes V 1, V 2 and V 3. ( L-3) Ans. Isochore is the plot beween P and T for a definite amount of a gas at a constant volume. From different isochors at different volumes draw a line parallel to Temperature axis represent a constant P and cutting the three isochors at T 1,T 2 and T 3 respectively.fr om the graph we find T 1,>T 2 >T 3. 1 > V 2 > V 3. APPLICATION One Mark Questions Q 1. What would happen if dipole develops momentarily in atoms and non polar molecules which are electrically symmetrical? (L-2) Ans. London forces or dispersion forces are developed. Q 2. On the basis of Boyle s Law explain why mountaineers carry oxygen cylinder with them. (L-2) Ans. At high altitude as the atmospheric pressure is low, the air is less dense. As a result less oxygen is available for breathing. Q 3. Predict what will happen if intermolecular forces between the molecules are very weak (L-2) Ans. When intermolecular forces are very weak, molecules do not cling together t o make liquid and solid. Two Marks Questions Q 1. In a hospital an oxygen cylinder holds 10 L of oxygen at 200atm pressure. If a patient breathes in 0.50ml of oxygen at 1atm with each breath, for how many breaths the cylinder will be sufficient. (Assume that all the data is at 30 0 C.) (L-3)

4 Ans. Apply Boyle s Law P 1 V 1 = P 2 V x 10 = 1 x V 2 V 2 =2000 L No. of breaths = Total volume/ vol. consumed per breath =2000/0.5x10-3 L =4x10 6 Q 2. What will be the volume of hydrogen when 3 litres of it are cooled from 15 0 C to C at constant pressure?(l-2) Ans. T 1 = = 288K, V 1 = 3L T 2 = = 200 K, V 2 =? V 1 /T 1 = V 2 / T 2, V 2 = ( 3 x 200) / 288 = 2.08 L. Q 3. Explain the absolute zero (in terms of volume) with the help of isobar. (L-3) Ans: At absolute zero,volume approach to zero and below this temperature, the volume will be negative, which is meaningless. Hence absolute zero is the lowest possible temperature. Three Marks Questions Q1 Use the information and data given below to answer the questions (L-3) Stronger intermolecular forces result in higher boiling point Strength of London forces increases with the number of electrons in the molecule Boiling point of HF,HCl, HBr and HI are 293K,189K,206K and 238K respectively (a) Which type of intermolecular forces are present in the molecules HCl, HBr and HI? (b) Looking at the trend of boiling points of HCl, HBr and HI, explain, out of dipole interaction and London interaction, which one predominant here? (c) Why is the boiling point of HF highest while that of HCl lowest? Ans :( a) In the molecules HCl, HBr and HI, dipole-dipole interaction and London forces are present. (b) Dipole moment of HCl > HBr > HI but boiling point of HCl < HBr <HI.It means London forces are predominant over dipole-dipole interaction which depends upon surface area.hcl has least surface area, Whereas HI has maximum surface area. (c) HF has highest dipole moment due to highest difference in electronegativity, therefore hydrogen bonding is also present due to which force of attraction will increase, hence it has higher boiling point. Q 2. A gas is enclosed in a room.the temperature, pressure, density, and number of moles respectively is T 0 C,P atm, d g cm -3 n moles. (a) What will be the pressure, temperature, density and number of moles in each compartment, if the room is partitioned into four equal compartments? (b)what will be the values of pressure, temperature, density and number of moles if an equal volume of the gas at pressure P and temperature T is let in the same room? (L-2) Ans: (a)pressure =Patm, Temp. =T o C, density =d g/cm 3 & No. of moles=n/4 (b) Pressure =2Patm, Temp. = T o C,

5 density =2d g/cm 3 & No. of moles=2n. Q 3. An open vessel contains 200mg of air at 17 0 C.What weight percentage of air would be expelled if the vessel is heated to C? (L-3) Ans: Suppose vol. of 200mg of air at 17 0 C =V ml Apply Charle s Law V 1 / T 1 =V 2 /T 2 V 1 / 290 = V 2 /390 V 2 =1.34 V 1, volume of air expelled = 1.34V 1 V 1 =o.34v 1 ml % of air expelled: 0.34V 1 x 100 /V 1 = 34% Q 4. In a J tube partially filled with mercury the volume of an air column 4.2ml and the mercury level in two limbs is same. Some mercury is now added to the tube so that the volume of the air is enclosed in shorter limb is now 2.8ml.What is the difference in the levels of mercury in this situation (atm pressure=1bar.) (L-3) Ans: P 1 = 1bar= mmHg P2 = ( P 1 V 1 =4.2mL,V 2 =2.8mL P 1 V 1 = P 2 V 2 or P 2 = P 1 V 1 / V 2 = x4.2 /2.8 = mmHg h = = mmhg or cm. Q 5 An iron tank contains helium at a pressure of 2.5atm at 25 0 C.The tank can withstand a maximum pressure of 10atm.the building in which tank has been placed catches fire. Predict whether, the tank will blow up first or melt (m.p of Fe= C.) (L-3) Ans :. P 1 = 2.5 atm P 2 =? T 1 = 25 o C =298 K T 2 = C =1808 K According to Gay Laussac law, P 1 / T 1 = P 2 / T 2 P 2 = 2.5 x 1080/ 298 =15.16 atm. Since pressure of gas in the tank is much more than 10 atm at the m.p. of Fe.Thus tank will blow up before reaching m.p. Concept Ideal gas Equation, kinetic molecular theory of gases, ideal gas and real gases, liquefaction of gases, liquid state KNOWLEDGE One mark questions Q1.What is Boyle s point?(l1) Ans. The temperature at which the real gas obeys ideal gas law for an appreciable range of pressure is called Boyle s point. Q2.Why liquid drops are spherical in shape? (L2)

6 Ans. Because of surface tension, the molecules tend to minimize the surface area and sphere has minimum surface area. Q3.State with expression Dalton s law of partial pressure?(l1) Ans. The total pressure exerted by the mixture of non reactive gases is equal to the sum of the partial pressures of individual gases. P total = p 1 + p 2 + p 3 + Two marks questions:- Q1. What would have the effect on the gas pressure if collisions between the gas molecules are not elastic?(l2) Ans. The gas pressure would gradually become zero as molecules will gradually slow down and ultimately settle down due to constant loss of energy. Q2.What are the two faulty assumptions of kinetic theory of gases?(l3) Ans. i) There is no force of attraction and repulsion between the molecules of a gas. ii) Volume of the molecules of a gas is negligibly small in comparison to the empty space between them. Q3. Give the units of Vander Waal s constants a and b.(l2) Ans. Unit of a is bar L 2 mol -2 and that of b is L mol -1. Three mark questions Q1. Define i) surface tension ii) viscosity. Give their units.(l1) Ans.i) The force acting parallel to the surface and perpendicular to a line of unit length anywhere in the surface is called surface tension. Its unit is Nm -1. ii) The force of friction which one part of the liquid offers to another part of the liquid is called viscosity. Its unit is poise. Q2.Give three differences between evaporation and boiling.(l1) Ans.i) Evaporation occurs at all temperatures while boiling occurs at boiling point. ii) Evaporation is surface phenomenon while boiling occurs throughout the bulk of liquid. iii) Evaporation is slow while boiling is rapid. Q3. Give Vander Waal s equation of state for real gases? What is the significance of constants a and b in this equation?(l3) Ans. ( P + an 2 /V 2 ) (V nb) = nrt a gives the idea of magnitude of attractive forces between the molecules of the gas while b is a measure of effective size of gas molecules. UNDERSTANDING One Mark questions Q1. Explain how can gases be liquefied?(l1) Ans. Gases can be liquefied by decrease of temperature and increase of pressure. Q2.What is the effect of temperature on surface tension and viscosity of a liquid?(l1) Ans. Both decrease by increasing temperature.

7 Q3..Name the property of liquids which i)causes internal resistance to flow ii)causes capillary action. Give their units.(l2) Ans. i) viscosity. Its unit is poise. ii) surface tension. Its unit is Nm -1. Two marks questions Q1. In the plot of Z(compressibility factor) vs P, Z attains a value of unity at a certain pressure. What does this signify?(l3) Ans. It implies that at this value of pressure attractive and repulsive forces balance each other. Below this pressure attraction dominates and Z < 1. Above this pressure repulsion dominates and Z > 1. Q2. Account for the following properties of gases on the basis of kinetic molecular theory of gasesa) high compressibility b) gases occupy whole of the volume available to them.(l2) Ans. a) High compressibility is due to large empty space between the gas molecules. b) Due to absence of attractive forces between the molecules, they can easily separate from one another. Q3. Explain the pressure and volume corrections in ideal gas equation?(l2) Ans. At low temperature and high pressure intermolecular attractions cannot be neglected, the observed pressure P is smaller than the ideal pressure P i. Therefore, P i = P + p = P + an 2 /V 2 As volume of gas molecules is not negligible as compared to total volume of the gas the ideal volume V i is smaller than the observed volume. Therefore, V i = V + v = V - nb Three Marks questions Q1.i) Can a gas with a = 0 be liquefied?(l2) Ans. A gas with a = 0 means absence of intermolecular forces. Hence such a gas cannot be Liquified. ii) Out of NH 3 and N 2 which will have (L3) a) larger value of a b)larger value of b.explain. Ans. a) NH 3 will have larger value of a because of H bonding. b) NH 3 will have larger value of b due to larger molecular size. Q2. What is meant by compressibility factor of gases? How does its value deviate from that of an ideal gas in case of real gases and what does it indicate?(l2) Ans. It is the ratio of the product PV to nrt. Z = PV/nRT. For ideal gas its value is unity. Z < 1 indicates negative deviation i.e. gas is more compressible due predominance of attractive forces. Z > 1 indicates positive deviation i.e. gas is less compressible due predominance of repulsive forces.

8 Q3. Critical temperature for CO 2 is o C. What does this mean? Explain other two critical constants.(l2) Ans. It means that CO 2 cannot be liquefied above o C which is its critical temperature(t c ). Critical pressure(p c ) it is the pressure required to liquefy the gas at critical temperature. Critical volume( V c ) it is the volume of one mole of the gas at T c and P c. APPLICATION One mark questions Q1. How is the partial pressure of a gas related to the total pressure in a mixture of gases?(l2) Ans. Partial pressure = mole fraction x total pressure. For gas A p A = x A x P Q2. When do real gases deviate from ideality?(l2) Ans. At high pressure and low temperature. Q3. Why excluded volume b is four times the actual volume of the molecules?(l3) Ans. The two molecules cannot come closer than distance 2r and the volume of sphere with radius 2r is four times the volume of one molecule. Two marks questions Q1. Critical temperature of two gases A and B are 30 o and -50 o respectively. Which of them has strong intermolecular forces and why?(l2) Ans. A will have stronger inter particle forces as it can be liquefied at a higher temperature. Q2.Explain: (L1) i)vapour pressure increases with increase in temperature. ii)giycerine is more viscous than water. Ans. i) At higher temperature inter particle attraction weakens and more number of mole cules escape to vapour. ii) As interparticle forces are stronger in glycerine. Q3. Derive relation between density and pressure of a gas.(l2) Ans. Ideal gas equation PV= nrt or n/v = P/RT or m/mv = P/RT (as nm = m/m) or d/m = P/RT or P = drt/m Three marks questions:- Q1.At what temperature 128 g of SO 2 confined in vessel of 5 dm 3 capacity will exhibit a pressure of 10.0 bar? Given a = 6.7 bar L 2 mol -2 and b= L mol -1.(L2) Ans. w = 128 g V = 5 dm 3 P = 10.0 bar M SO2 = 64 g n = w/m = 128/64 = 2 Putting values in equation : ( P + an 2 /V 2 ) (V nb) = nrt We get T = 329 K

9 Q2. Calculate the pressure exerted by 8.5 g of NH 3 contained in a 0.5 L vessel at 300 K. For ammonia a = 4.0 atm L 2 mol -2 and b = L mol -1.(L2) Ans. V =0.5 L, T = 300 K n= 8.5/17 = 0.5 mol ( P + an 2 /V 2 ) (V nb) = nrt Putting values, we have P = atm Q3. 20 mol of Chlorine gas occupies a volume of 800 ml at 300 K and 5 x 10 6 Pa pressure. Calculate the compressibility factor of the gas. Comment on the compressibility of the gas under these conditions(.l3) Ans. P = 5 x 10 6 Pa = 5 x 10 6 / 10 5 bar n = 20 T = 300 K R = L bar K -1 mol -1 V real = 800 ml V ideal = nrt/p Putting values, we have V ideal = 1004 ml Z = V real / V ideal = 800/1004 = As Z is less than 1, it means gas is more compressible under these conditions. Q4. Explain the effect of increasing the temperature of a liquid, on intermolecular forces operating between its particles. What will happen to the viscosity of a liquid if its temperature is increased? Arrange the following in increasing order of viscosity-water, hexane, glycerine.( L2) Ans. On increasing the temperature, the kinetic energy of liquid molecules increases so that it can overcome the attractive forces between the molecules and hence liquid can flow more easily i.e. viscosity decreases. Order of increasing viscosity : hexane < water < glycerine. Q5. i) Derive ideal gas equation. ii) At 25 o and 760 mm of Hg pressure a gas occupies 600 ml volume. What will be its pressure at a height where temperature is 10 o C and volume of the gas is 640 ml.(l2) Ans. i)at constt. T and n ; V 1/P ( Boyle s law ) at constt. P and n ; V T (Charle s law ) at constt. P and T; V n ( Avogadro s law) thus, V nt/p or V = nrt/p or PV = nrt ii) Given p 1 = 760 mm Hg V 1 = 600 ml T 1 = = 298 K V 2 = 640 ml and T 2 = = 283K According to combined gas law : p 1 V 1 /T 1 = p 2 V 2 /T 2 or p 2 = p 1 V 1 T 2 /T 1 V 2 putting values and calculating, we have p 2 = mm Hg

S OF MATTER TER. Unit. I. Multiple Choice Questions (Type-I)

S OF MATTER TER. Unit. I. Multiple Choice Questions (Type-I) Unit 5 STATE TES TE S OF MATTER MA TER I. Multiple Choice Questions (Type-I) 1. A person living in Shimla observed that cooking food without using pressure cooker takes more time. The reason for this observation