{"id":9002,"date":"2019-11-28T11:00:55","date_gmt":"2019-11-28T05:30:55","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=9002"},"modified":"2021-07-02T15:44:32","modified_gmt":"2021-07-02T10:14:32","slug":"1st-puc-chemistry-question-bank-chapter-5","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-chemistry-question-bank-chapter-5\/","title":{"rendered":"1st PUC Chemistry Question Bank Chapter 5 States of Matter"},"content":{"rendered":"

You can Download Chapter 5 States of Matter Questions and Answers, Notes, 1st PUC Chemistry Question Bank with Answers<\/a> Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 1st PUC Chemistry Question Bank Chapter 5 States of Matter<\/h2>\n

1st PUC Chemistry States of Matter One Mark Questions and Answers<\/h3>\n

Question 1.
\nWrite the ideal gas equation for one mole of a gas.
\nAnswer:
\nPV = RT<\/p>\n

Question 2.
\nWrite the Vander Waals1 equation for one mole of a real gas.
\nAnswer:
\n\"1st<\/p>\n

Question 3.
\nWrite Vander Waals equation constants ‘a’ and ‘b’.
\nAnswer:
\nUnits for ‘a’: L2<\/sup> atm mol-2<\/sup>
\nUnits for ‘b’: Litre mole-1<\/sup><\/p>\n

Question 4.
\nGie the value of universal gas constant in SI units.
\nAnswer:
\n8.3124 J K-1<\/sup> mol-1<\/sup><\/p>\n

Question 5.
\nGive the combined gas equation (or general equation for gases).
\nAnswer:
\nPV = nRT for n moles.<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nWrite kinetic equation for gases.
\nAnswer:
\nPV = \\(\\frac { 1 }{ 3 }\\) mnc2<\/sup><\/p>\n

Question 7.
\nWrite an equation for the pressure exerted by a gas in a container.
\nAnswer:
\n\\(\\mathrm{PV}=\\frac{1}{3} \\frac{\\mathrm{mnc}^{2}}{\\mathrm{V}}\\)<\/p>\n

Question 8.
\nWrite an equation for the root mean square velocity of a gas.
\nAnswer:
\n\\(V_{\\mathrm{rms}}=\\sqrt{\\frac{3 \\mathrm{RT}}{\\mathrm{M}}}\\)<\/p>\n

Question 9.
\nLiquid ammonia bottle is cooled before opening the seal. Why?
\nAnswer:
\nNH3 is liquified at high temperature and pressure. It is cooled so as to reduce temperature so that it does not burst.<\/p>\n

Question 10.
\nThe size of weather balloon becomes larger and larger as it ascends up into higher altitudes. Why?
\nAnswer:
\nAt higher altitudes, atmospheric condition is less, therefore, air inside balloon exerts less pressure released and it becomes larger and larger.<\/p>\n

Question 11.
\nWhich property of liquid is responsible for spherical shape of liquid drops?
\nAnswer:
\nSurface tension and viscosity.<\/p>\n

\"KSEEB<\/p>\n

Question 12.
\nWhat is the effect of temperature on viscosity and why?
\nAnswer:
\nViscosity decrease in the temperature because intermolecular force of attraction decreases.<\/p>\n

Question 13.
\nWhat is the effect of pressure on (i) viscosity, (ii) surface tension, (iii) density of liquid? –
\nAnswer:
\nIncrease in pressure increases viscosity, surface tension and density because inter molecular force of attraction increases.<\/p>\n

Question 14.
\nWhat is the boiling point of water at (i) higher altitudes, (ii) in pressure cooker?
\nAnswer:
\n(i) >100 \u00b0C (ii) < 100 \u00b0C.<\/p>\n

Question 15.
\nName four measurable properties of gases.
\nAnswer:<\/p>\n

    \n
  1. Volume<\/li>\n
  2. Pressure<\/li>\n
  3. Temperature<\/li>\n
  4. Mass<\/li>\n<\/ol>\n

    Question 16.
    \nName the SI unit of pressure and give its definition.
    \nAnswer:
    \nPascal is SI unit of pressure. It is defined as pressure exerted when 1 newton force is acting per square metre area.<\/p>\n

    \"KSEEB<\/p>\n

    Question 17.
    \nDefine an ideal gas.
    \nAnswer:
    \nIdeal gas is gas which follows all the gas laws at all temperature and pressure.<\/p>\n

    Question 18.
    \nGive various forms of ideal gas equation.
    \nAnswer:
    \n\"1st<\/p>\n

    Question 19.
    \nWhat are real gases?
    \nAnswer:
    \nReal gases are those which do not follow all the gas laws at all temperature and
    \npressure.<\/p>\n

    Question 20.
    \nUnder what conditions of T and P, most of gases deviate from ideal gas behavior?
    \nAnswer:
    \nAt low temperature and high pressure most of gases deviate from ideal gas behavior.<\/p>\n

    Question 21.
    \nState Dalton\u2019s law of partial pressure.
    \nAnswer:
    \nIt states that whenever two or more gases, which do not react chemically, are
    \nenclosed in vessel, the total pressure is equal to sum of partial pressure of each gas.<\/p>\n

    Question 22.
    \nWhat is meant by compressibility factor?
    \nAnswer:
    \nCompressibility factor is ratio of PV to nRT.
    \nZ = \\(\\frac{\\mathrm{PV}}{n \\mathrm{RT}}b\\)<\/p>\n

    Question 23.
    \nDefine boiling point of a liquid.
    \nAnswer:
    \nBoiling point is temperature at which vapour pressure of liquid becomes equal to atmospheric pressure.<\/p>\n

    Question 24.
    \nHow is density of gas related to its molar mass?
    \nAnswer:
    \nDensity of gas is directly proportional to molar mass.<\/p>\n

    \"KSEEB<\/p>\n

    Question 25.
    \nHow is molar mass of gas related to rate of diffusion?
    \nAnswer:
    \nThe rate of diffusion is inversely proportional to molecular mass
    \n\\(\\frac{r_{1}}{r_{2}}=\\sqrt{\\frac{M_{2}}{M_{1}}}\\)<\/p>\n

    Question 26.
    \nWhat is meant by elastic collision?
    \nAnswer:
    \nElastic collision is a collision in which there is no net loss energy rather there is transfer of energy.<\/p>\n

    Question 27.
    \nDefine aqueous tension.
    \nAnswer:
    \nAqueous tension is pressure of water vapour at particular temperature.<\/p>\n

    Question 28.
    \nState Graham\u2019s law of diffusion.
    \nAnswer:
    \nIt states that the rate of diffusion of gases is inversely proportional to square root of molecular masses at constant temperature and pressure.<\/p>\n

    Question 29.
    \nDefine “effusion”.
    \nAnswer:
    \nEffusion is the process of diffusion taking place from a tiny hole.<\/p>\n

    Question 30.
    \nArrange the following gases in the increasing orderof their van der waals1 constant ‘a’ values : CO2<\/sub>, H2<\/sub>> C6<\/sub>H6<\/sub>
    \nAnswer:
    \nH2<\/sub> < CO2<\/sub> < C6<\/sub>H6<\/sub><\/p>\n

    Question 31.
    \nWhat is SI unit of (i) viscosity, (ii) surface tension?
    \nAnswer:
    \n(i) Pascal second, (ii) Nm1<\/sup><\/p>\n

    Question 32.
    \nDefine critical temperature of the gas. Give its expression.
    \nAnswer:
    \nCritical temperature is a temperature above which a gas cannot be liquefied.
    \n\\(\\mathrm{T}_{\\mathrm{c}}=\\frac{8 \\mathrm{a}}{27 b \\mathrm{R}}\\) where \u2018a\u2019 and \u2018b\u2019 are Van der waal\u2019s constants, R is gas constant.<\/p>\n

    \"KSEEB<\/p>\n

    Question 33.
    \nWhat do you mean by Boyle temperature? Give its expression and its relation with inversion temperature.
    \nAnswer:
    \nBoyle temperature is a temperature at which most of real gases show ideal gas\u00a0behavior over wide range of pressure. \\(\\mathrm{T}_{b}=\\frac{\\mathrm{a}}{b \\mathrm{R}} \\mathrm{T}_{\\mathrm{i}}=2 \\mathrm{T}_{b}\\)<\/p>\n

    Question 34.
    \nDefine inversion temperature of a gas.
    \nAnswer:
    \nInversion temperature is a temperature below which if gas is allowed to expand, it causes cooling effect.<\/p>\n

    Question 35.
    \nDefine (i) critical pressure, (ii) critical volume and give their expression.
    \nAnswer:
    \n(i) Critical pressure, it is pressure required to liquefy the gas at critical temperature.
    \n\\(\\mathrm{P}_{\\mathrm{c}}=\\frac{\\mathrm{a}}{27 \\mathrm{b}^{2}}\\)
    \n(ii) Critical volume, it is volume occupied by 1 mole of gas at critical temperature and pressure. Vc<\/sub> = 3b<\/p>\n

    Question 36.
    \nWhat type of graph will you get when PV is plotted against P at constant temperature.
    \nAnswer:
    \nA straight line parallel to pressure axis.<\/p>\n

    1st PUC Chemistry States of Matter Two Marks Questions and Answers<\/h3>\n

    Question 1.
    \nState Boyle\u2019s law of gases. Give its mathematical expression.
    \nAnswer:
    \nBoyle\u2019s law states volume of fixed mass of gas in inversely proportional to pressure at constant temperature. (P1<\/sub>V1<\/sub> = P(2<\/sub>V2<\/sub>)T<\/sub><\/p>\n

    Question 2.
    \nState Charle\u2019s law. Give the mathematical expression.
    \nAnswer:
    \nCharle\u2019s law states that volume of fixed mass of gas is directly proportional to absolute temperature at constant pressure.
    \n\\(\\left(\\frac{\\mathrm{V}_{1}}{\\mathrm{T}_{1}}=\\frac{\\mathrm{V}_{2}}{\\mathrm{T}_{2}}\\right)_{\\mathrm{P}}\\)<\/p>\n

    \"KSEEB<\/p>\n

    Question 3.
    \nGive any two differences between ideal and real gas.
    \nAnswer:
    \n\"1st<\/p>\n

    Question 4.
    \nMention the causes for the deviation of real gas from ideal behaviour.
    \nAnswer:<\/p>\n

      \n
    • There is no intermolecular force of attraction within gaseous molecules (at high pressure and low temperature there is appreciable intermolecular force of attraction).<\/li>\n
    • The actual volume of the molecule is negligible as compared to total volume of the gas. (At high pressure the gaseous molecule occupy minimum volume, which is not negligible).<\/li>\n<\/ul>\n

      Note: Because of this deviation Van der waal modified the ideal gas pressure and ideal gas volume respectively.<\/p>\n

      Question 5.
      \nWhat will be the density of CO2<\/sub> in kg\/m3<\/sup> at 323 K and 101.3 kPa pressure?
      \nAnswer:
      \nT = 323 K, P = 101.3 kPa, M = 44 \u00d7 10-3<\/sup>kg mol-1<\/sup>, d = ?,
      \nR= 8.314 \u00d7 10-3<\/sup> kPa m3<\/sup>K-1<\/sup>mol-1<\/sup>
      \n\"1st<\/p>\n

      Question 6.
      \nHow many molecules of an ideal gas are there in 1 \u00d7 10-3<\/sup> dm3<\/sup> at STP?
      \nAnswer:
      \n22.4 dm3<\/sup> of an ideal gas has 6.023 \u00d7 1023<\/sup> molecules at STP
      \n1 \u00d7 l0-3<\/sup>dm3<\/sup> of an ideal gas contains = \\(\\frac{6.023 \\times 10^{23}}{22.4} \\times 1 \\times 10^{-3}=2.69 \\times 10^{19}\\)molecules.<\/p>\n

      Question 7.
      \nCalculate the number of moles of hydrogen present in 500 cm3<\/sup> of a gas under a pressure of 101.3 kPa at a temperature of 300 K. (R = 8.314 J K-1<\/sup> mol-1<\/sup>).
      \nAnswer:
      \nn = ?
      \n\"1st<\/p>\n

      Question 8.
      \n50 ml of oxygen were collected at 100<\/sup> C under 750 mm pressure. Calculate volume at STP.
      \nAnswer:
      \nV1<\/sub> = 50ml V2<\/sub>=?
      \nP1<\/sub>= 750mm P2<\/sub> = 760mm
      \nT1<\/sub> =10 + 273 = 283K T2<\/sub> =273.15K
      \n\"1st<\/p>\n

      \"KSEEB<\/p>\n

      Question 9.
      \nWhy ethyl alcohol has lower boiling point than water?
      \nAnswer:
      \nEthyl alcohol has higher vapour pressure than water due to less intermolecular forces of attraction.<\/p>\n

      Question 10.
      \nA human adult breathes in approximately 0.50 L of air at 1 atm with each breath. If an air tank holds 10 L of air at 200 atm, how many breaths the tank will supply?
      \nAnswer:
      \nP1<\/sub>V1<\/sub>=P2<\/sub>V2<\/sub>
      \n200 \u00d7 10 = 1 \u00d7 V2<\/sub>
      \nV2<\/sub> = 2000 L
      \nNumber of Breaths = \\(\\frac{2000 \\mathrm{L}}{0.5 \\mathrm{L}}\\) = 4000<\/p>\n

      Question 11.
      \nWhat will happen to volume of fixed amount of gas at a certain T and P if :
      \n(a) T is kept constant but pressure is decreased to \\(\\frac { 1 }{ 4 }\\) th<\/sup> of original value?
      \n(b) Pressure is halved and temperature in Kelvin is doubled?
      \nAnswer:
      \n(a) The volume will become 4 times
      \n(b) The volume will remain same<\/p>\n

      Question 12.
      \n10 g of O2<\/sub> were introduced into an evacuated vessel of 5 L capacity at 27 \u00b0C,
      \nCalculate pressure of gas in bar in the container [R = 0.083 bar L K-1<\/sup> mol -1<\/sup>] [At. mass of O = 16u]
      \nAnswer:
      \nPV = raRT bar
      \nP \u00d7 5 L = \\(\\frac { 10 }{ 32 }\\) \u00d7 0.083 \u00d7 300 K
      \nP = \\(\\frac{24.9 \\times 10}{160}=\\frac{249}{160}\\)\u00a0 = 1.556 bar<\/p>\n

      Question 13.
      \nCalculate the density of Rn gas at 298 K and 755 mm Hg pressure. [Atomic mass of Rn = 222u]
      \nAnswer:
      \n\"1st<\/p>\n

      Question 14.
      \n2NO (g) + O2<\/sub> (g) \u2192 2NO2<\/sub> (g). A 1.98 L reaction vessel is filled with a mixture of NO and O2<\/sub> to a pressure of 4.33 bar at 310 K. The reaction is allowed to continue until the total pressure decreases to 4.00 bar. What is the final pressure of NO2<\/sub> ?
      \nAnswer:
      \nReaction 2NO + O2<\/sub> \u2192 2NO2<\/sub>
      \n\"1st
      \nQuestion 15.
      \nCalculate the total pressure in a 10 L cylinder which contains 0.4 g of helium, 1.6 g of oxygen and 1.4 g of nitrogen at 27 \u00b0C. Also calculate the partial pressure of helium gas in the cylinder. Assume ideal behaviour for gases.
      \nAns:
      \nTotal number of moles = \\(\\frac{0.4}{4}+\\frac{1.6}{32}+\\frac{1.4}{28}\\) = 0.1 + 0.05 + 0.05 =0.2 moles.
      \npV = nRT
      \np \u00d7 10 L = 0.2 \u00d7 0.082 \u00d7 300 \u21d2 p = \\(\\frac{24.6 \\times 0.2}{10}\\)
      \np = 0.492 atm
      \nPHe<\/sub> = \\(\\frac{0.1}{0.2}\\) \u00d7 0.492 = 0.246 atm<\/p>\n

      \"KSEEB<\/p>\n

      Question 16.
      \nCalculate the number of moles of hydrogen (H2<\/sub>) present in a 500 cm3<\/sup> sample of hydrogen gas at a pressure of 760 mm Hg and 27 \u00b0C.
      \nAnswer:
      \nPV = nRT, P = 1 atm, V = 500 cm3<\/sup>, n = ?, R = 82.1 atm cm3<\/sup>K-1<\/sup>mol-1<\/sup>, T = 300 K.
      \n\"1st<\/p>\n

      Question 17.
      \n34.05 ml of phosphorus vapour weighs 0.0625 g at 0.1 bar pressure. What is the molar mass of phosphorus?
      \nAnswer:
      \npV = nRT
      \n\"1st<\/p>\n

      Question 18.
      \nIn terms of Charle’s law explain why -273 \u00b0C is the lowest possible temperature.
      \nAnswer:
      \nCharles Law plotted graphically with volume against temperature in \u00b0C. These plots when extended intersect the temperature axis at the same point -273 \u00b0C. Charles concluded that all gases at this temperature could have zero volume and below this temperature volume would be negative. It shows -273 \u00b0C is lowest temperature attainable.<\/p>\n

      Question 19.
      \n(i) Why is Boyle\u2019s law is obeyed by N2<\/sub>,O2<\/sub> or CO2<\/sub> only at low pressure and high temperature?
      \n(ii) Compare the rate of diffusion of HCl and NH3<\/sub>. (Atomic masses of H = 1u, Cl = 35.5 u, N = 14u)
      \nAnswer:
      \n(i) It is because at low pressure and high temperature these gases follow ideal gas behavior due to negligible force of attraction and volume occupied by gas molecules can be ignored.
      \n(ii)
      \n\"1st<\/p>\n

      Question 20.
      \nArrange solid, liquid and gas in order of energy giving reasons.
      \nAnswer:
      \nSolid<liquid<\/p>\n

      Question 21.
      \nThe Van der Waal\u2019s constants for two gases are as follows:-
      \n\"1st
      \nWhich of them is more easily liquefiable and which has greater molecular size?
      \nAnswer:
      \nGreater the value of \u2018a\u2019, more easily the gas is liquefiable. Similarly, greater the value of V, greater is the molecular size. Hence, gas Y will be more easily liquefiable and will have greater molecular size.<\/p>\n

      \"KSEEB<\/p>\n

      Question 22.
      \nCompare the rates of diffusion of 235<\/sup>UF6<\/sub> and 238<\/sup>UF6<\/sub>
      \nAnswer:
      \nMolecular mass of 235<\/sup>UF6<\/sub> = 235 + 6 \u00d7 19 = 349
      \nMolecular mass of 238<\/sup>UF6<\/sub> = 238 + 6 \u00d7 19 = 352
      \n\"1st<\/p>\n

      Question 23.
      \nAt a certain altitude, the density of air is 1\/10 th of the density of the earth\u2019s atmosphere and temperature is -10 \u00b0C. What is the pressure at that altitude? Assume that air behaves like an ideal gas, has uniform composition and is at STP at the earth\u2019s surface.
      \nAnswer:
      \n\"1st<\/p>\n

      Question 24.
      \nWhat is the ratio of average kinetic energy of oxygen molecules to that of ozone molecules at 27 \u00b0C?
      \nAnswer:
      \nAverage kinetic energy of any gas depends only on temperature and not upon the nature of the gas. Hence, both the gases will have same average kinetic energy at 27 \u00b0C, ie., the ratio will be 1: 1.<\/p>\n

      Question 25.
      \nWhat is the difference between total kinetic energy and translational kinetic energy? For what type of molecules, the two are equal?
      \nAnswer:
      \nTotal kinetic energy is the sum of translational, vibrational and rotational kinetic energies. The total kinetic energy is equal to the translational kinetic energy for monoatomic gases (He, Ne, etc.,) as they do not possess vibrational and rotational motion but have only translation motion.<\/p>\n

      Question 26.
      \nOut of N2<\/sub> and NH3<\/sub>, which one will have greater value of \u2018a\u2019 and which one will have greater value of ‘b\u2019 ?
      \nAnswer:<\/p>\n

        \n
      1. As NH3<\/sub> is more easily liquefiable (due to hydrogen bonding), intermolecular forces of attraction are stronger then in N2<\/sub>. Hence, NH3<\/sub> will have greater value of\u2018a\u2019.<\/li>\n
      2. As NH3<\/sub> molecule is larger in size than N2<\/sub> , hence NH3<\/sub> will have greater value for \u2018b’.
        \n(For NH3<\/sub>, a = 4.17 L2<\/sup> atm mol-2<\/sup> , b = 0.0371 L mol-1<\/sup>
        \nFor N2<\/sub> , a = 1.39 L2<\/sup> atm mol-2<\/sup>, b = 0.0319 L mol-1<\/sup>)<\/li>\n<\/ol>\n

        \"KSEEB<\/p>\n

        Question 27.
        \nWhat would have happened to the gas if the molecular collisions were not elastic?
        \nAnswer:
        \nOn every collision, there would have been loss of energy. As a result, the molecules would have slowed down and ultimately settle down\u2019 in the vessel. Moreover, the pressure would have gradually reduced to zero.<\/p>\n

        Question 28.
        \nCO2<\/sub> is heavier than O2<\/sub> and N2<\/sub> gases present in the air but it does not form the lower layer of the atmosphere. Why?
        \nAnswer:
        \nGases possess the property of diffusion which is independent of the force of gravitation. Due to diffusion, the gases mix into each other and remain almost uniformly distributed in the atmosphere.<\/p>\n

        Question 29.
        \nN2<\/sub>O and CO2<\/sub> have the same rate of diffusion under same conditions of temperature and pressure. Why?
        \nAnswer:
        \nBoth have same molar mass (= 44 g mol-1<\/sup>). According to Graham\u2019s law of diffusion, rates of diffusion of different gases are inversely proportional to the square root of their molar masses under same conditions of temperature and pressure.<\/p>\n

        Question 30.
        \nWhy liquids have a definite volume but no definite shape?
        \nAnswer:
        \nThis is because the intermolecular forces of attraction are strong enough to hold the molecules together but not so strong as to fix them into definite positions (as in solids). Instead, they possess fluidity and hence no definite shape.<\/p>\n

        Question 31.
        \nAt a particular temperature, why vapour pressure of acetone is less than that of ether?
        \nAnswer:
        \nThis is because the intermolecular forces of attraction in acetone are stronger than those present in ether.<\/p>\n

        Question 32.
        \nA liquid is transferred from a smaller vessel to a bigger vessel at the same temperature. What will be the effect on the vapour pressure?
        \nAnswer:
        \nNo effect, as it depends only on the nature of the liquid and temperature.<\/p>\n

        \"KSEEB<\/p>\n

        Question 33.
        \nWhy vegetables are cooked with difficulty at a hill station?
        \nAnswer:
        \nThe atmospheric pressure is less and so the boiling point is lowered.<\/p>\n

        Question 34.
        \nWhat is the approximate relationship between heat of vaporization and boiling point of a liquid?
        \nAnswer:
        \n\u2206Hvap <\/sub>\/ Tb<\/sub>\u00a0=21 cal K-1<\/sup> mol-1<\/sup> (Trouton\u2019s rule)<\/p>\n

        Question 35.
        \nExplain Pressure coefficient of a gas.
        \nAnswer:
        \nPressure coefficient (\u03b1p<\/sub>). At constant volume, the increase in the pressure of a gas per degree rise of temperature divided by its pressure at 0 \u00b0C is called pressure coefficient of the gas. Mathematically, \\(\\alpha_{\\mathrm{P}}=\\frac{\\mathrm{P}_{\\mathrm{t}}-\\mathrm{P}_{0}}{\\mathrm{P}_{0} \\times \\mathrm{t}}\\). For all gases, value of aP is \u03b1p<\/sub> is \\(\\frac { 1 }{ 273 }\\)<\/p>\n

        \"KSEEB<\/p>\n

        Question 36.
        \nExplain Atmolysis.
        \nAnswer:
        \nAtmolysis is the process of separation of two gases on the basis of their different rates of diffusion due to difference in their densities is called atmolysis. It has been applied with success for the separation of isotopes and other gaseous mixtures.<\/p>\n

        Question 37.
        \nWrite any two difference between evaporation and boiling.
        \nAnswer:
        \n\"1st<\/p>\n

        Question 38.
        \nDefine surface tension.
        \nAnswer:
        \nSurface tension is defined as force per unit length acting perpendicular to the tangential line on the surface. It is due to imbalanced attractive forces acting downwards and tends to reduce the surface area of a liquid to minimum.<\/p>\n

        Question 39.
        \nDefine viscosity of a liquid.
        \nAnswer:
        \nThe internal resistance to the flow of the liquids which one layer offers to another layer trying to pass over is called its viscosity. It depends upon the nature of the liquid and temperature.<\/p>\n

        \"KSEEB<\/p>\n

        Question 40.
        \nWrite an expression for kinetic gas equation.
        \nAnswer:
        \nPV = \\(\\frac { 1 }{ 3 }\\)mnc2<\/sup><\/p>\n

        Question 41.
        \nBoth N2<\/sub>O and CO2<\/sub> have similar rates of diffusion under similar conditions of temperature and pressure. Explain.
        \nAnswer:
        \nGram molar mass of both these gases is the same (44 g). According to Graham\u2019s law of diffusion, the rates of diffusion of the gases are inversely proportional to the square root of their molar masses under similar conditions of temperature and pressure. Therefore, both these gases diffuse at the same rate.<\/p>\n

        Question 42.
        \nIf the density of a gas at the sea level at 0 \u00b0C is 1.29 kg mm-3<\/sup>. What is its molar mass? (Pressure = 1 bar)
        \nAnswer:
        \n\"1st<\/p>\n

        Question 43.
        \nWhat will be the minimum pressure required to compress 500 dm3<\/sup> of air at 1 bar to 200 dm3<\/sup> at 30 \u00b0C.
        \nAnswer:
        \nP1<\/sub>V1<\/sub> = P2<\/sub>V2<\/sub>
        \n1 \u00d7 500 = P2<\/sub> \u00d7 200 P2<\/sub> = 2.5 bar<\/p>\n

        Question 44.
        \nCalculate the molar mass of an unknown gas which diffuses 1.117 times faster than oxygen gas through the same aperture under the same conditions of temperature and pressure.
        \nAnswer:
        \n\"1st<\/p>\n

        1st PUC Chemistry States of Matter Three Marks Questions and Answers<\/h3>\n

        Question 1.
        \nHow vapour pressure of liquid is related to (i) temperature, (ii) nature of liquid, (iii) boiling point, (iv) atmospheric pressure?
        \nAnswer:
        \n(i) Vapour pressure is directly proportional to temperature.
        \n(ii) If intermolecular force of attraction is less in the liquid, its vapour pressure will be high.
        \n(iii) Higher the vapour pressure, lower will be boiling point.
        \n(iv) If atmospheric pressure is low, boiling point will be less and higher will be the vapour pressure.<\/p>\n

        Question 2.
        \nWhich type of intermolecular forces exist among the following molecules?
        \n(i) H2<\/sub>S molecules (ii) H2<\/sub>O molecules (iii) Cl2<\/sub> and CCl4<\/sub> molecules (iv) SiH4<\/sub> molecules (v) Helium atoms (vi) He atoms and HCl molecules.
        \nAnswer:
        \n(i) Dipole-dipole interactions (because H2<\/sub>S is polar)
        \n(ii) Hydrogen bonding
        \n(iii) London dispersion force (because both are non-polar)
        \n(iv) London dispersion forces (because SiH4<\/sub> is non-polar)
        \n(v) London dispersion forces (because He atoms have symmetrical electron clouds)
        \n(vi) Dipole-induced dipole forces (because HCl is polar while He atom has symmetrical r electron cloud)<\/p>\n

        \"KSEEB<\/p>\n

        Question 3.
        \nExplain Relative humidity and percentage of humidity.
        \nAnswer:
        \nRelative humidity is a method of expressing the extent of moisture (water vapour) present in the air. It is defined as the ratio of the partial pressure of water vapour in the air to the vapour pressure of water at that particular temperature.
        \n\"1st
        \nThus, if the percent relative humidity of air is 60%, it means that partical pressure of water vapour in air is 0.6 times more than the vapour pressure of water at that temperature.<\/p>\n

        Question 4.
        \nGive a reason for the following, (i) The size of weather balloon becomes larger and larger as it ascends into higher altitudes, (ii) Tyres of automobiles are inflated to lesser pressure in summer than in winter.
        \nAnswer:
        \n(i) As we go to higher altitudes, the atmospheric pressure decreases. Then the pressure outside the balloon decreases. To regasm equilibrium with the external pressure, the gas inside expands to decrease its pressure. Hence, the size of the balloon increases.<\/p>\n

        (ii) In summer, due to higher temperature, the average kinetic energy of the air molecules inside the tyre increases, ie., molecules start moving faster. Hence, the pressure on the walls of the tube increases. Its pressure inside is not kept low at the time of inflation, at higher temperature, the pressure may become so high that the tyre may burst.<\/p>\n

        Question 5.
        \nThe molecular speeds of gaseous molecules are analogous to those of rifle bullets. Then why is the odour of a gas not detected so fast?
        \nAnswer:
        \nThough the molecules of a gas travel at high speeds but they do not travel in straight lines in one direction like bullets. As they travel, they collide with the molecules of the gases present in the air. As a result, they are deflected. Thus, they follow a zig-zag path, i.e., the net distance travelled in a particular direction is quite small in a given time. That is why the odour is not detected so fast.<\/p>\n

        Question 6.
        \nThe average kinetic energy of a gas molecule at 0 \u00b0C is 5.621 \u00d7 10-27<\/sup> J. Calculate Boltzmann constant. Also calculate the number of molecules present in one mole of the gas.
        \nAnswer:
        \n\"1st<\/p>\n

        1st PUC Chemistry States of Matter Four \/ Five Marks Questions and Answers<\/h3>\n

        Question 1.
        \nWrite the postulates of Kinetic theory of gases.
        \nAnswer:<\/p>\n

          \n
        1. Gases are made up of large number of the minute particles.<\/li>\n
        2. Pressure is exerted by a gas<\/li>\n
        3. There is no loss of kinetic energy.<\/li>\n
        4. Molecules of gas attract on one another.<\/li>\n
        5. Kinetic energy of the molecule in directly proportional to absolute temperature.<\/li>\n
        6. Actual volume of the gaseous molecule very small.<\/li>\n
        7. Gaseous molecules are at always in motion.<\/li>\n
        8. There is more influence of gravity on movement of gaseous molecule.<\/li>\n<\/ol>\n

          Question 2.
          \nDerive the expression for the pressure exerted by gas (Derive the Kinetic gas
          \nequation PV = \\(\\frac { 1 }{ 3 }\\) mn2<\/sup>)
          \nAnswer:<\/p>\n

          Step 1 :
          \n\"1st
          \nLet the mass of each molecule of gas be \u2018m\u2019 kg and all are moving with root mean square velocity \u2018c’.<\/p>\n

          Step 2 :
          \nAll gaseous molecules move in all direction randomly. Hence at a given
          \ninterval of time \\(\\frac { n }{ 3 }\\) molecules moves either along X or Y or Z axis.<\/p>\n

          Step 3 :
          \nConsider n number of molecules of gas in a cube of length T within a 3 dimensional (X, Y and Z axis) network.
          \nThen the area of the cube = l2<\/sup> ……………..(1)
          \nThe volume of the cube = V = l3<\/sup> ……………..(2)<\/p>\n

          Step 4 :
          \nCalculation of change in momentum w.r.t. one axis.
          \nConsider the side \u2018A\u2019 of the cube. Let one (n = 1) molecule of a gas moving along the axis towards the side \u2018A\u2019 with a velocity \u2018c\u2019 and strikes the wall \u2018A\u2019.
          \nInitial momentum before striking = me ……………..(3)
          \nDistance travelled = 1 units ……………..(4)
          \nAfter striking the’wall \u2018A\u2019 molecule rebounds back with same speed in opposite direction.
          \n\u2234 Final momentum after striking = -me ……………..(5)
          \nDistance travelled = 1 units ……………..(6)
          \n\u2234 Total change in momentum = (3) – (5) = mc – (- mc) = 2mc
          \nTotal distance travelled = (4) + (6) = 1 + 1 = 2l
          \nNumber of collisions per sec = C\/21 ……………..(7)
          \n\u2234 Total change in momentum per sec per molecule on face
          \n\u2018A\u2019 = collision per second \u00d7 change in momentum \\(\\frac{\\mathrm{C}}{2 \\mathrm{l}} \\times 2 \\mathrm{m} \\mathrm{C}=\\frac{\\mathrm{m} \\mathrm{C}^{2}}{1}\\)<\/p>\n

          Step 5 :
          \nTotal change in momentum per sec for one molecule strikes on the wall A = \\(\\frac{m c^{2}}{1}\\)
          \n\u2234The change in momentum per sec for \\(\\frac { n }{ 2 }\\) molecules strikes on wall A = \\(\\frac{n}{3} \\times \\frac{m c^{2}}{1}\\)
          \n= \\( \\frac{\\mathrm{mnc}^{2}}{31}\\) …………(9)<\/p>\n

          Step 6 :
          \nThe change in momentum per sec = force. And the force in gaseous molecule is only due to pressure,
          \n\"1st<\/p>\n

          Question 3.
          \nDerivation of Vander waal\u2019s equation
          \nAnswer:
          \nPressure correction (getting corrected ideal pressure)
          \n\"1st
          \nLet us consider a container containing \u2018n\u2019 number of gaseous molecule. Now consider two gaseous molecule A and B inside it.<\/p>\n

          At \u2018A\u2019 : The gaseous molecule \u2018A\u2019 present at the interior of the container and is attracted (disturbed) by all neighboring gaseous directions. Hence the net force exerted by this molecule is zero.<\/p>\n

          At \u2018B\u2019 : The gaseous molecule \u2018B\u2019 is about to strike the walls of the container, it is influenced by other neighboring molecule as follows.
          \n(a) It is attracted by large number of molecules inward and net effect is an inward pull. This inward force (pressure) is directly proportional to the number of molecules per unit volume (at the bulk of the container) ie., P \u221d \\(\\frac{1}{V}\\) …………(1)
          \n(b) Within the container there are some molecules which are already striking the wall of the container. Thus there exist pulling of B inward by the molecules striking on the wall. This pressure (force) is directly proportional to the number of molecules striking on unit volume of the walls of the container. P \u221d \\(\\frac{1}{V}\\) …………(2)<\/p>\n

          From the above two points A and B it is clear that the actual pressure exerted on the wall due to gaseous molecular is less than the ideal pressure.
          \nActual Pressure (P) = Ideal Pressure (Pi) – Pressure Correction (Pc)
          \nP = Pi – Pc
          \nPi = P + Pc …………….(3)
          \n\"1st<\/p>\n

          Volume correction : At high pressure the molecules are closer and covering\/ occupying a considerable minimum area or volume. (Because the space available for the molecule to move is less than the actual volume of the container. Hence Vander waal accounted for this by subtracting volume correction \u2018b\u2019 from the observed volume V.
          \nIdeal Volume (Vi) = V – b ………………(6)
          \nWe know that if Pi and Vi are pressure and volume of ideal gas then it satisfies PiVi = RT
          \nBut from (5) and (6) the ideal pressure and ideal Volumes are substituted.
          \n\"1st
          \nEquation (7) is called Vander waal equation for one mole of gas. This equation for \u2018n\u2019 moles of gas is
          \n\"1st<\/p>\n

          \"KSEEB<\/p>\n

          Question 4.
          \nInterpret PV – P curves based on Vander waals equation.
          \nAnswer:
          \nConsider Vander waals eqution for one mole of a gas \\(\\left(p+\\frac{q}{V^{2}}\\right)(V-b)=R T\\) ……………(1)<\/p>\n

          Case 1: At low pressure:-
          \nWhen the pressure is low b = 0
          \n\u2234 Equation becomes
          \n\"1st
          \nPV > RT
          \nExplains gradual increase in PV < RT. Hence the curve first descends at lower pressure.<\/p>\n

          Case 2 ; At high pressure : At high pressure, \\(\\frac{\\mathbf{a}}{\\mathbf{V}^{2}}\\) = 0 \u2234 Equation becomes P(V – b) = RT or PV – Pb = RT PV > RT
          \nExplains gradual increase in PV with simultaneous increase is pressure.<\/p>\n

          Case 3 : Change in temperature :
          \nAt low temperature the value of P and V are very small, hence deviate from ideal behavior.
          \nAt high temperature P and V are very large such that \\(\\frac{\\mathbf{a}}{\\mathbf{V}^{2}}\\) and b are negligible hence
          \nVander waals Eqn reduces to PV = RT.
          \n\u2234 At high temperature almost all gases behaves like ideal gas.<\/p>\n

          Question 5.
          \nTwo flasks \u2018A\u2019 and <B\u2019 have equal volumes. Flask \u2018A\u2019 contains H2<\/sub> and is maintained at 300 E while ‘B\u2019 contains equal mass of CH4<\/sub> gas and is maintained at 600 E.
          \n(i) Which flask contains greater number of molecules? How many times more?
          \n(ii) In which flask pressure is greater? How many times more?
          \n(iii) In which flask molecules are moving faster?
          \n(iv) In which flask the number of collisions with walls are greater?
          \nAnswer:
          \n(i) Number of moles in flask A = 8 \u00d7 number of moles in flask B.
          \nLet the equal mass be x. Ratio of number of moles => \\(\\frac{x}{2}: \\frac{x}{16}\\) = 8:1<\/p>\n

          (ii) Pressure \u03b1 number of moles Pressure \u03b1 temperature
          \nIf temperature had been equal, pressure in \u2018A\u2019 would have been 8 times but in \u2018A\u2019
          \ntemperature is \\(\\frac{1}{2}\\) of that in B
          \n\u2234Pressure will be 4 times more
          \n\"1st
          \nPressure in flask A = 4 x Pressure in flask B.<\/p>\n

          (iii) Flask B, because velocity is directly proportional to square root of temperature.
          \n(iv) Flask A, because number of molecules of flask A is more.<\/p>\n

          \"KSEEB<\/p>\n

          Question 6.
          \nCalculate the pressure exerted by 1.00 mol of CO2<\/sub> (g) at 298 K that occupies 65.4 ml using Van der waal\u2019s equation.
          \n\u2018a\u2019 for CO2<\/sub> is 3.592 L2<\/sup> bar\/mol2<\/sup>, \u20186\u2019 = 0.0427 L mol-1<\/sup>. Compare it with the pressure predicted by ideal gas equation for same conditions of T and P?
          \nAnswer:
          \n\"1st<\/p>\n

          Question 7.
          \nA gaseous mixture contains 2.2 bar He, 1.1 bar H2<\/sub> and 4.2 bar N2<\/sub> . What is mole fraction of N2<\/sub> ?
          \nAnswer:
          \n\"1st<\/p>\n

          Question 8.
          \nAccount for the following:
          \n(a) The size of weather balloon becomes larger and larger as it ascends up into higher altitudes.
          \n(b) Copper is malleable and ductile whiles sulphur is not.
          \nAnswer:
          \n(a) At higher altitudes, atmospheric pressure decreases, therefore, air inside the weather balloon exerts pressure and its size becomes larger and larger as it ascends higher and higher.
          \n(b) Copper is metal and has strong metallic bond, that is why it is malleable and ductile. Sulphur is non-metal and has weak intermolecular force of attraction therefore it is brittle.<\/p>\n

          Question 9.
          \n(a) Explain Boyle\u2019s law with the help of kinetic theory of gases.
          \n(b) An open beaker at 27 \u00b0C is heated to 477 \u00b0C. What fraction of air would have been expelled out?
          \nAnswer:
          \n(a) The kinetic theory of gases assumes that pressure of gas is due to collision of gas molecules with the walls of the container. The more will be frequency of collision, more will be pressure. The reduction in volume (v) of gas increases no. of molecules per unit volume to which pressure is directly proportional. Therefore, the volume of the gas is reduced if pressure is increased or we can say pressure is inversely propotionsl to volume P \u221d \\(\\frac{1}{V}\\)<\/p>\n

          (b)
          \n\"1st<\/p>\n

          Question 10.
          \n(a) What do you mean by \u2018Surface Tension\u2019 of a liquid?
          \n(b) Equal volumes of two gases A and B diffuse through a porous pot in 20 and 10 seconds respectively. If the molar mass of A be 80, find the molar mass of B.
          \nAnswer:
          \n(a) Surface Tension of a liquid is the force by which surface molecules are attracted towards the bulk.
          \n\"1st<\/p>\n

          \"KSEEB<\/p>\n

          Question 11.
          \n(a) Why air is dense at the sea level? Explain.
          \n(b) Calculate the total pressure in a mixture of 4g of O2<\/sub> and 2g of H2<\/sub>
          \nconfined to a total volume of 1L at 0 \u00b0C. (R = 0.821 L atm mol-1<\/sup>)
          \nAnswer:
          \n(a) Heavier gas will come down and lighter air goes up. Air at sea level is denser due to compression by mass of air above it.
          \n\"1st<\/p>\n

          Question 12.
          \nShown below is a volume vs temperature graph at different pressures P1<\/sub>>P2<\/sub>>P3<\/sub> an<P4<\/sub>.
          \n(a) Which gas law does this depict?
          \n(b) What does each line of this graph called?
          \n(c) What is the value of temperature at the point where the lines intercept at
          \ntemperature axis and what is the specific name given to this hypothetical temperature?
          \n(d) Giving reason arrange the pressures, p1<\/sub>,p2<\/sub>,p3<\/sub> and p4<\/sub> in increasing order.
          \nAnswer:
          \n(a) Charle\u2019s law.
          \n(b) Isobar.
          \n(c) – 273.15 \u00b0C, absolute zero is a lowest hypothetical temperature at which gases are supposed to occupy zero volume.
          \n(d) p1 <\/sub>< p2 <\/sub>< p3 <\/sub>< p4<\/sub>
          \nReason : For a particular value of temperature : If volume is more it means pressure is less hence p1<\/sub> is lesser than p2<\/sub> and so on.<\/p>\n

          Question 13.
          \nDensity of a mixture of gases CO and CO2<\/sub> is 1.5 g\/L at 303 K temperature and 73 cm of Hg. What is the mole percentage of two gases in the mixture?
          \nAnswer:
          \n\"1st
          \nM = 38.85 (Molecular mass of the mixture)
          \nLet total moles = 100
          \nMoles of CO = a and Moles of CO2<\/sub> = 100 – a
          \nMass of CO = 28a
          \nMass of CO2<\/sub> =(100 – a)44
          \nTotal Mass = 28 a + (100 – a)44
          \nMolecular mass of the mixture\u00a0 = \\(=\\frac{28+(100-a) 44}{100}\\) = 38.85(evaluated)
          \na =32.19
          \nMolecular percentage of CO = 32.19
          \nMole percentage of CO2<\/sub> = (100 \u03b1) = 100 – 32.19 = 67.81<\/p>\n

          Question 14.
          \nThe van der waal’s equation of state for 1 mole of gas is \\(\\left(\\mathbf{P}+\\frac{a}{\\mathbf{V}^{2}}\\right)(\\mathbf{V}-\\mathbf{b})=\\mathbf{R} \\mathbf{T}\\)
          \nArrange the following gases in order of increasing value of \u2018a\u2019. The gases are CO2<\/sub>, H2<\/sub>, He and N2<\/sub>. How does the value of V related to the ease of liquifaction?
          \nAnswer:
          \nH2<\/sub> < He < N2<\/sub> < CO2<\/sub> is the order of increasing value of \u2018a\u2019. Measure of \u2018a\u2019 is proportional to the amount of force of intermolecular attraction. CO2<\/sub> molecule being larger in size, has higher magnitude of van der Waal’s force of attraction and hence highest value of \u2018a\u2019.
          \nAlso higher is the intermolecular force of attraction, easier it is to liquefy a gas ie., higher the value of \u2018a\u2019, higher will be the case of liquefaction.<\/p>\n

          \"KSEEB<\/p>\n

          Question 15.
          \nIn an attempt to record critical temperature of some substances, the value of water has been missed out.
          \n\"1st
          \nWill the value of x be more or less than (a) CO2<\/sub> ? (b) NH3<\/sub> ? Reason out your answers. Also arrange CO2<\/sub>, H2<\/sub>O and NH3<\/sub> in the order of ease of liquefaction.
          \nAnswer:
          \nValue of x will be more than for both CO2<\/sub> and NH3<\/sub>. This is because in CO2<\/sub> only weak Van der waals forces exist but in H2<\/sub>O intermolecular H-bonding exist. Also magnitude of H-bonding in H2<\/sub>O is more than in NH3<\/sub>.<\/p>\n

          H2<\/sub>O has more intermolecular force of attraction therefore it should be easily liquefiable. It is already liquid at room temperature and atmospheric pressure.
          \nEase of liquefaction CO2<\/sub> < NH3<\/sub> < H2<\/sub>O.<\/p>\n

          Question 16.
          \nAn open vessel at the temperature 27 \u00b0C is heated until \\(\\frac { 3 }{ 5 }\\) parts of the air in it has been expelled. Assuming that the volume of the vessel remains constant, find the temperature to which the vessel has been heated.
          \nAnswer:
          \nFor the open vessel, pressure and volume remains contant. If rej moles me present at T1<\/sub> and n2<\/sub> moles are present at T2<\/sub> then
          \npV = n1<\/sub>RT1<\/sub> and pV = n2<\/sub>RT2<\/sub> => n1<\/sub>RT1<\/sub> = n2<\/sub>RT2<\/sub>
          \nn1<\/sub>T1<\/sub> = nT2<\/sub> => \\(\\frac{n_{1}}{\\mathrm{T}_{1}}=\\frac{n_{2}}{\\mathrm{T}_{2}}\\)<\/p>\n

          Let number of moles of air originally present = n
          \nAfter heating, number of moles of air expelled = \\(\\frac { 1 }{ 2 }\\) n
          \n\"1st<\/p>\n

          Question 17.
          \nIn a hospital, an oxygen cylinder of capacity 10 L, oxygen is filled at 200 atmospheric pressure. If a patient breathes in 0.50 ml of oxygen at 1.0 atm with each breath, for how many breaths the cylinder will be sufficient? Assume temperature to be constant.
          \nAnswer:
          \n10 L at 200 atm pressure = V2<\/sub>L at 1 atm pressure
          \nP1<\/sub>V1<\/sub> = p2<\/sub>V2<\/sub> => 200 \u00d7 10 = 1 \u00d7 V2<\/sub>
          \nV2<\/sub> = 2000 L
          \n\"1st<\/p>\n

          Question 18.
          \nTwo flasks of equal volume are connected by a narrow tube of negligible volume and are filled with N2 gas when both the flasks are immersed in boiling water the gas pressure inside the system is 0.5 atm. Calculate the pressure of the system when one of the flasks is immersed in ice water while the other flask in boiling water. ‘
          \nAnswer:
          \nTemperature of the gas when flasks are in boiling water = 100 \u00b0C = 100 + 273 = 373 K
          \nPressure = 0.5 atm
          \nAverage temperature of the gas when one flask is in ice and other in boiling water. =50\u00b0C = 50 + 273 = 323 K
          \n\"1st<\/p>\n

          Question 19.
          \n(i) Which of the liquids in each of the following pairs has higher vapour pressure : (a) Alcohol, glycerine ; (b) Petrol kerosene; (c) Mercury, water
          \n(ii) Which one in each of the following pairs JU m ire viscous (a) coconut oil, castor oil ; (b) glycerine, kerosene ; (c) soft drinks, aerated water (soda water)?
          \n(iii) Separate portions of chloroform and water at the same temperature are poured on your hands. The chloroform will feel colder. Account for this in terms of attractive forces.
          \nAnswer:
          \n(i) (a) Alcohol (b) Petrol (c) Water
          \n(ii) (a) Castor oil (b) Glycerine (c) Soft drink
          \n(iii) (a) Chloroform molecules have weak van der waal\u2019s forces between them, while water molecules have hydrogen bonding between them. Since van der waal\u2019s forces are weaker chloroform evaporates easily and causes cooling.<\/p>\n

          Question 20.
          \nDeduce the relation PV = nRT, where R is a constant called universal gas constant.
          \nAnswer:
          \nAccording to Boyle\u2019s law, V \u221d \\(\\frac { 1 }{ P }\\) (at constant T and n)
          \nAccording to Charle\u2019s law, V \u221d \\(\\frac { 1 }{ T }\\) (at constant P and n)
          \nAccording to Avogadro\u2019s law, V \u221d (at constant T and P)
          \nCombining the three laws, V \u221d \\(\\frac { nT }{ P }\\) or PV \u221d nRT
          \nOr PV = nRT where R is a constant called universal gas constant.<\/p>\n

          Question 21.
          \n(i) What is the effect of temperature on (a) density, (b) surface tension, (c) viscosity and (d) vapour pressure of a liquid?
          \n(ii) What is the effect of pressure on (a) volume, (b) boiling point and
          \n(c) viscosity of a liquid?
          \nAnswer:
          \n(i) (a) Density decreases (b) Surface tension decreases (c) Viscosity decreases (d) Vapour pressure increases with rise in temperature.
          \n(ii) (a) Volume decreases (b) B. pt increases (c) Viscosity increases with increase in pressure.<\/p>\n

          Question 22.
          \nAn LPG cylinder when full contains 14.2 kg gas and exerts pressure of 2.5 atm. If half of its gas is consumed what will be the pressure of the gas in the cylinder?
          \nAnswer:
          \nIn a LPG cylinder, the gas is held as a liquid. There is an equilibrium between the LPG (liquid) and its vapour. The equilibrium vapour pressure does not depend upon the quantity of liquid present there as long as liquid ……….. vapour equilibrium is maintained. Therefore, even when half the gas has been used, the pressure excited by \u2019 the vapour of the gas remains equal to 2.5 atm.<\/p>\n

          \"KSEEB<\/p>\n

          Question 23.
          \nWhy at extremely low pressures, the real gases obey the ideal gas equation?
          \nAnswer:
          \nAt low pressures, volume V is very large and hence the correction term b (a constant of small value) can be neglected in comparison to very large value of V. Thus the van der waal\u2019s equation for 1 mole of a real gas.
          \n\"1st<\/p>\n

          Question 24.
          \nExplain the following observations.
          \n(a) Aerated water bottles are kept under water during summer.
          \n(b) Liquid ammonia bottle is cooled before opening the seal.
          \n(c) The tyre of an automobile is inflated at lesser pressure in summer than in winter.
          \n\"1st
          \nAnswer:
          \n(a) Aerated water contains C02 gas dissolved in aqueous solution under pressure and bottles are well stoppered. As in summer the temperature increases and we know that the solubility of the gases decreases with increase of temperature and as a result more of gas is expected to be generated in the bottle thereby pressure exerted by the gas in the bottle will increase. If the bottles are not kept under water, the gas generated may be large in quantity and hence pressure exerted by the gas may be very high and the bottle may explode. So, to decrease the temperature and hence to avoid explosion of the bottles.<\/p>\n

          (b) Liquid Ammonia bottle contains the gas under very high pressure. If the bottle is opened as such, then the sudden decrease in pressure will lead to a large increase in the volume of the gas. As a result, the gas will come out the bottle with force. This will lead to the breakage of the bottle. Cooling under tap water will result in the decrease of volume. It reduces the chances of accident.<\/p>\n

          (c) The pressure of the air is directly proportional to the temperature. During summer
          \ndue to high temperature the pressure in the tyre will be high as compared to that in water. The tube may burst under high pressure in summer. Therefore, it is advisable to inflate the tyres to lesser pressure in summer than in winter.<\/p>\n

          Question 25.
          \n(i) State the law depicting the volume temperature relationship, (ii) Name the temperature at which the volume of the gas becomes equal to zero.
          \nAnswer:
          \n(i) The law is known as Charle\u2019s law. Charle\u2019s Law : Pressure remaining constant, the volume of a given mass of a gas increases or decreases by \\(\\frac { 1 }{ 273 }\\) of its volume at 0 \u00b0C for every one degree centigrade rise or fall in temperature.
          \n\"1st
          \nWhere Vt<\/sub> = Volume of the gas at t \u00b0C and V0<\/sub> is its volume at 0 \u00b0C.
          \n(ii) The temperature at which the volume of a gas becomes equal to zero is called absolute zero.
          \n\"1st
          \nAbsolute zero (0 K) = -273 \u00b0C
          \nie., the volume of gas = 0; because, before attaining this temperature, molecular motion of the gas is negligible, ie., they freeze. Gas no longer remains as gas.<\/p>\n

          Question 26.
          \nExplain the following statements :
          \n(i) Sodium chloride pieces are harder than sodium metal.
          \n(ii) Copper is ductile and malleable but brass is not.
          \n(iii) Latent heat of fusion of solid carbon dioxide is much less than that of silicon dioxide.
          \n(iv) Water has its maximum density near 277 K.
          \n(v) Ice floats on surface of water near the melting point.
          \nAnswer:
          \n(i) It has stronger electrostatic forces than metallic bond in sodium.
          \n(ii) Copper is a pure element while brass is an alloy. Pure metal is more malleable.
          \n(iii) SiO2<\/sub> has covalent bonds while CO2<\/sub> has only Van der waal\u2019s forces (molecular solid). Hence SiO2<\/sub> has higher heat of fusion.
          \n(iv) Below 277 K and above 277 K, water expands hence its volume increases, density decreases and is maximum at 277 K
          \n(v) At 0 \u00b0C or 273 K, its density is less than that of water hence floats.<\/p>\n

          Question 27.
          \nWhat is the concept of Maxwell-Boltzmann distribution of molecular speeds, at a given temperature in a gas sample?
          \nAnswer:
          \nConcept s A gas consists of tiny particles (atom or molecules) separated from one another by large spaces. These particles are constantly moving in all directions. . During their motion, they collide with one another and also with the walls of the container. As a result of collisions, the speed and the direction of the molecules go on changing. Thus, all the molecules present in given sample of a gas do not possess the same speed. The speeds of individual molecules are different and are distributed over a wide range. The speeds of different molecules go on changing. However, the distribution of speed among different molecules remains the same at a particular temperature although the individual speeds of the molecules may change.<\/p>\n

          The distribution of molecules between different possible speeds was given by Maxwell and Boltzmann. He plotted the fraction of molecules, ie.,\\(\\frac{\\Delta \\mathrm{N}}{\\mathrm{N}}\\) (along y-axis) having different speeds against the speeds of the molecules (along x-axis). The curve so obtained is shown in Fig. below and is known as Maxwell\u2019s distribution curve.<\/p>\n

          The important features of Maxwell\u2019s distribution curve can be summarized as follows:<\/p>\n

            \n
          • The fraction of molecules with very low or very high speeds is very small.<\/li>\n
          • The fraction of molecules possessing higher and higher speeds goes on increasing till it reaches the peak and then it starts decreasing.<\/li>\n
          • The maximum fraction of molecules possesses a speed, corresponding to the peak in the curve. The speed corresponding to the peak in the curve is referred to as most probable speed.<\/li>\n<\/ul>\n

            \"KSEEB<\/p>\n

            1st PUC Chemistry States of Matter Numerical Problems and Answers<\/h3>\n

            Question 1.
            \nCalculate the RMS velocity of oxygen at STP.
            \nAnswer:
            \nAt STP P = 1.013 \u00d7 105<\/sup> Pa
            \nV = 22.4 \u00d7 l0-3<\/sup>m3<\/sup>
            \nMolecular weight of oxygen = 32 \u00d7 10-3<\/sup> kg
            \n\"1st
            \n= 461.23 metres per second, c = 461.23 ms-1 <\/sup><\/p>\n

            Question 2.
            \nCalculate the RMS velocity of a molecule of nitrogen at STP.
            \nAnswer:
            \nAt STP P = 1.013 \u00d7 105<\/sup> Pa
            \nv = 22.4 \u00d7 10-3<\/sup>m3<\/sup>
            \nMolecular weight of nitrogen = 28 \u00d7 10-3<\/sup> kg
            \n\"1st
            \nc = 493.07ms-1<\/sup><\/p>\n

            Question 3.
            \nCalculate the RMS velocity of methane molecules at 298 K. R = 8.3 14 J K-1<\/sup>mol-1<\/sup>. Molecular weight of methane =16 \u00d7 10-3<\/sup> kg.
            \nAnswer:
            \n\"1st<\/p>\n

            Question 4.
            \nThe density of oxygen gas at STP is 1.429 gL-1<\/sup>. Find the RMS velocity.
            \nAnswer:
            \nAt STP, P = 1.013 \u00d7 105<\/sup>Pa; d = 1.429gL-1<\/sup> in mters it is
            \n\"1st<\/p>\n

            Question 5.
            \nAt what temperature will the rms velocity of hydrogen be the same as that of nitrogen at 350 K?
            \nAnswer:
            \n\"1st<\/p>\n

            Question 6.
            \nAt what temperature will the RMS velocity of oxygen be twice at 200 K?
            \nAnswer:
            \nLet c1<\/sub> be the rms velocity at 200 K (T1<\/sub>)
            \nAt temperature T2<\/sub>, c2<\/sub> = 2c1<\/sub>
            \n\"1st<\/p>\n

            Question 7.
            \nCalculate the total kinetic energy of 320 g of sulphur dioxide at 273 K.
            \nAnswer:
            \n(KE) n mole <\/sub> = \\(\\frac { 3 }{ 2 }\\) \u00d7 nRT
            \nT = 273K n = \\(\\frac { 320g }{ 64g }\\) = 5
            \n\\(\\frac { 3 }{ 2 }\\) \u00d7 5 \u00d7 8.314 \u00d7 273 = 17020J<\/p>\n

            Question 8.
            \nCalculate the total kinetic energy of 2 moles of chlorine at 300 K.
            \nAnswer:
            \n(KE)nmoies = \\(\\frac { 3 }{ 2 }\\) \u00d7 nRT;
            \nn = 2; T = 300 K
            \n= \\(\\frac { 3 }{ 2 }\\) \u00d7 2 \u00d7 8.314 \u00d7 300 =7482.0 J<\/p>\n

            Question 9.
            \nCalculate the kinetic energy of one mole of helium having RMS velocity K2<\/sup>ms-1<\/sup>.
            \nAnswer:
            \nHelium molecule is monoatomic, M = 4g = 4 \u00d7 10-3<\/sup> kg
            \n(KE)mole<\/sub> = \\(\\frac { 1 }{ 2 }\\) Mc2<\/sup> = \\(\\frac { 3 }{ 2 }\\) x 4 x 10-3<\/sup> x (102<\/sup>)2<\/sup> = \\(\\frac { 1 }{ 2 }\\) \u00d7 4 \u00d7 10 = 20J<\/p>\n

            Question 10.
            \nCalculate the kinetic energy of one mole of nitrogen at 400 K.
            \nAnswer:
            \n(KE)mol<\/sub> =\\(\\frac { 3 }{ 4 }\\) RT=\\(\\frac { 3 }{ 2 }\\) \u00d7 8.314 x JK-1<\/sup>mol-1<\/sup> \u00d7 400K = 498.9 J mol-1<\/sup>.
            \nNote: From problems 16 and 19, it can be observed that the kinetic energy of one mole of any two gases, at the same temperature will be the same.<\/p>\n

            \"KSEEB<\/p>\n

            Question 11.
            \nCalculate the kinetic energy of 300 molecules of CO at 30 \u00b0C.
            \nAnswer:
            \n\"1st<\/p>\n

            Question 12.
            \nCalculate the total kinetic energy of the molecules of nitrogen per cm3<\/sup> at STP and per m3<\/sup> at STP.
            \nAnswer:
            \n1 mole of a gas occupies 22400 cc at STP
            \nKE of 22,400 cc of a gas at STP = KE of 1 mole of a gas at STP
            \n= \\(\\frac { 3 }{ 2 }\\) x RT = \\(\\frac { 3 }{ 2 }\\) x 8.314 \u00d7 273 =3404.58 J
            \nKE of 1 cc of a gas = \\(\\frac { 3404.58 }{ 22400 }\\) = 0.152 J
            \nKE of a m3<\/sup> of a gas = 0.152 \u00d7 106<\/sup> = 152 \u00d7 103<\/sup> J.<\/p>\n

            Question 13.
            \nCalculate the RMS speed of a molecule of hydrogen chloride at STP (The molar volume of any gas at STP = 22.4dm3<\/sup>
            \nAnswer:
            \n\\(\\mathrm{c}=\\sqrt{\\frac{3 \\mathrm{PV}}{\\mathrm{M}}}\\)
            \nwhere p = 1.01325 \u00d7 1015<\/sup>Nm-2<\/sup>
            \n1.01325 \u00d7 1015<\/sup>kgm-1<\/sup>s-2<\/sup>
            \nV = 22.4dm<sup<3mol-1<\/sup> = 0.0224m3<\/sup>mol-1<\/sup>
            \nM = 0.03645kgmol-1<\/sup>
            \n\"1st
            \nThis can be worked out using the relation \\(\\mathrm{c}=\\sqrt{\\frac{3 \\mathrm{RT}}{\\mathrm{M}}}\\)
            \nR = 8.314 J K-1<\/sup> mol-1<\/sup>
            \nM = 0.03645 kg mol-1<\/sup>
            \n\"1st<\/p>\n

            Question 14.
            \nCalculate the Root Mean Square velocity of a nitrogen molecule at 300 K and at a pressure of 2 \u00d7 10sNm–<\/sup>2
            \nAnswer:
            \n\"1st<\/p>\n

            Question 15.
            \nCalc\u00f9late the Root Mean Square velocity of a nitrogen molecule at 300 K and at a pressure of 2 \u00d7 105<\/sup> Nm-2<\/sup>
            \nAnswer:
            \n1 mole of nitrogen at STP occupies a volume of 0.0224 m3<\/sup> and weighs 0.028 kg.
            \n\"1st<\/p>\n

            \"KSEEB<\/p>\n

            Question 16.
            \nThrough the two ends of a glass tube of length 200 cm, hydrogen chloride gas and ammonia are allowed to enter. At what distance does ammonium chloride first appear?
            \nAnswer:
            \n\"1st
            \nThus NH3<\/sub> travel 1.465 times faster than HCl. In other words, NH3<\/sub> will travel 1.465 cm in the same time in which HCl travels 1 cm.
            \nLength of the tube = 200 cm
            \n\"1st
            \nThus, NH4<\/sub>Cl will first appear at a distance of 118.9 cm from NH3<\/sub> end or 81.1 cm from HCl end.<\/p>\n

            Question 17.
            \nCalculate the pressure exerted by 110 g of carbon dioxide in a vessel of 2 L capacity at 37 \u00b0C. Given that the van der waals constants are a = 3.59 L2<\/sup>atmmol-2<\/sup> and b = 0.0427 L mol-1<\/sup>. Compare the value with the calculated value if the gas were considered as ideal.
            \nAnswer:
            \n\"1st
            \n\"1st<\/p>\n

            Question 18.
            \nOne mole of SOa gas occupies a volume of 350 ml at 27 \u00b0C and 50 atm pressure. Calculate the compressibility factor of the gas. Comment on the type of deviation shown by the gas from ideal behavior.
            \nAnswer:
            \nCompressibility factor, Z = \\(\\frac{\\mathrm{PV}}{\\mathrm{nRT}}\\)
            \nSubstituting, n = 1 mol, P = 50 atm, V = 350 \u00d7 10-3<\/sup> L = 0.35 L
            \nR = 0.0821 L atm K-1<\/sup>mol-1<\/sup>, T = 27 + 273 = 300 K, we get
            \n\"1st
            \nFor ideal gas, Z = 1. As for th given gas Z < 1, it shows negative deviation ie., it is more compressible than expected from ideal behavior.<\/p>\n

            Question 19.
            \nThe Van der waals constant ‘b\u2019 for oxygen is 0.0318 L mol-1<\/sup>. Calculate the diameter of the oxygen molecule.
            \nAnswer:
            \n\"1st<\/p>\n

            Question 20.
            \nThe boiling point of n – hexane is 68.9 \u00b0C. Calculate its approximate critical temperature.
            \nAnswer:
            \nAccording to Goldberg\u2019s rule, for a liquid Tb<\/sub> = \\(\\frac { 2 }{ 3 }\\)Tc<\/sub>
            \nHence Tc<\/sub> = \\(\\frac { 2 }{ 3 }\\)Tb<\/sub> = \\(\\frac { 2 }{ 3 }\\) (38.9 + 273)K = 512.85 – 273\u00b0C<\/p>\n

            \"KSEEB<\/p>\n

            Question 21.
            \nAssuming CO2<\/sub> to be Van der waal’s gas, calculate its Boyle temperature. Given a = 3.59 L2<\/sup>atm mol-2<\/sup> and b = 0.0427 Lmol-1<\/sup>.
            \nAnswer:
            \nBoyle temperature in terms of Van der waal’s constants is TB<\/sub> = \\(\\frac{\\mathrm{a}}{\\mathrm{Rb}}\\)
            \n\"1st<\/p>\n

            Question 22.
            \nA cylinder of 20.0 L capacity contains 160 g of oxygen gas at 25 \u00b0C. What mass of oxygen must be released to reduce the pressure of the cylinder at 1.2 atm?
            \nAnswer:
            \nNumber of moles of oxygen gas present initially in the cylinder is
            \n= \\(\\frac{160 \\mathrm{g}}{32 \\mathrm{g} \\mathrm{mol}^{-1}}\\) = 5 moles
            \n(Molar mass of 02 = 32 g mol-1<\/sup>)
            \nTo calculate the number of moles now present, we have P = 1.2 atm, T = 298 K, V = 20.0 L
            \nApplying the relation, FV = nRT, we have,
            \n\"1st
            \n\u2234 Number of moles of 02 required to be released = 5 – 0.98 = 4.02 mol
            \nOr Mass of to be released = 4.02 \u00d7 32 g = 128.64 g.<\/p>\n

            Question 23.
            \n50 litre of dry N2<\/sub> is passed through 36 g H2<\/sub>O at 27 \u00b0C. After the passage of the gas, the mass of water was reduced to 34.80 g. Calculate the aqueous tension of water at 27 \u00b0C.
            \nAnswer:
            \nWater carried away by N2<\/sub> gas = 36 – 34.80 = 1.20 g
            \nAs this water is carried away by 50 L of N2<\/sub> gas, this means that volume occupied by water vapours = 50 L. Thus, now we have n = \\(\\frac { 1.20 }{ 18 }\\) = 0.0667mol, V = 50 L, T = 300
            \nK, P = ?
            \nApplying the gas equation, PV = nRT
            \n\"1st
            \n= 0.03286 \u00d7 760 mm = 24.97 mm.<\/p>\n

            Question 24.
            \nA 5 L vessel contains 1.4 g of nitrogen. When heated to 1800 K, 30% of molecules are dissociated into atoms. Calculate the pressure of the gas at 1800 K.
            \nAnswer:
            \n\"1st
            \n\u2234 Total number of moles = 0.035 + 0.030 = 0.065
            \ni.e., n = 0.065 mol, V = 5L, T – 1800 K, P = ?
            \nApplying PV = nRT, we get
            \n\"1st<\/p>\n

            Question 25.
            \nThe volume expansion of a gas under constant pressure is 0.037. Calculate its volume at – 100 \u00b0C if its volume at 100 \u00b0C is 685 cm3<\/sup>.
            \nAnswer:
            \nVolume expansion of a gas means increase or decrease in volume per degree rise or fall in temperature of its volume at 0 \u00b0C.
            \nVt<\/sub> = V0 <\/sub>+ 0.0037 \u00d7 V0 <\/sub>\u00d7 t = V0<\/sub>(l + 0.0037t)
            \nAt 100 \u00b0C, V1oo\u00b0C<\/sub> = v0<\/sub> (1 + 0.0037 \u00d7 100) = 685cm3<\/sup> (Given)
            \nOr V0<\/sub>= \\(\\frac{685}{1.37}\\) = 500cm3<\/sup>
            \nAt -100\u00b0C, V-1oo\u00b0C<\/sub>=500[1 + 0.0037 \u00d7 (-100)] = 315cm3<\/sup><\/p>\n

            \"KSEEB<\/p>\n","protected":false},"excerpt":{"rendered":"

            You can Download Chapter 5 States of Matter Questions and Answers, Notes, 1st PUC Chemistry Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. Karnataka 1st PUC Chemistry Question Bank Chapter 5 States of Matter 1st PUC Chemistry States of Matter One Mark …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[81],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/9002"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=9002"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/9002\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=9002"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=9002"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=9002"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}