{"id":9453,"date":"2019-12-02T15:19:20","date_gmt":"2019-12-02T09:49:20","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=9453"},"modified":"2021-07-02T15:44:19","modified_gmt":"2021-07-02T10:14:19","slug":"1st-puc-physics-question-bank-chapter-13","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-physics-question-bank-chapter-13\/","title":{"rendered":"1st PUC Physics Question Bank Chapter 13 Kinetic Theory"},"content":{"rendered":"

You can Download Chapter 13 Kinetic Theory Questions and Answers, Notes, 1st PUC Physics 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 Physics Question Bank Chapter 13 Kinetic Theory<\/h2>\n

1st PUC Physics Kinetic Theory Textbook Questions and Answers<\/h3>\n

Question 1.
\nEstimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3 A.
\nAnswer:
\nConsider 1 mole of oxygen gas at STP
\n1 mol = 6.023 \u00d7 1023<\/sup> molecules
\n= 22400 cm3<\/sup> of gas
\nDiameter of oxygen molecule = 3 A\u00b0
\n\u2234 Radius of oxygen molecule
\n= \\(\\frac{3}{2}\\) =1.5 A\u00b0 (1 A\u00b0 = 10-8<\/sup>cm)
\nVolume of N oxygen molecules
\n\\(=\\frac{4}{3} \\pi r^{3} \\times N\\)
\n\\(=\\left[\\frac{4}{3} \\times 3.142 \\times\\left(1.5 \\times 10^{-8}\\right)^{3} \\times 6.023 \\times 10^{23}\\right] \\mathrm{cm}^{3}\\)
\n= 8.514 cm3<\/sup>
\n\u2234 fraction of molecular volume to actual volume of oxygen
\n\\(=\\frac{8.514}{22,400}\\)
\n= 3.8 \u00d7 10-4<\/sup>.<\/p>\n

Question 2.
\nMolar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP: 1 atmospheric pressure, 0 \u00b0C). Show that it is 22.4 liters.
\nAnswer:
\nP = 1 atm = 1. 013 \u00d7 105<\/sup> Nm-2<\/sup>
\nn= 1 mol
\nR = 8. 314 J mol-1<\/sup> k-1<\/sup>
\nT = 0\u00b0 C = 273. 15 k
\nThe ideal gas state equation is given by
\nPV = n RT .
\n\u21d2 \u2234 \\(V=\\frac{n R T}{P}=\\frac{1 \\times 8.314 \\times 273.15}{1.013 \\times 10^{5}}\\)
\nV = 0.0224 m3<\/sup> = 0.0224 \u00d7 106<\/sup> cm3<\/sup>
\n= 22400 cm3<\/sup>
\nV = 22. 4 litres
\n\u2234 The volume occupied by 1 mol of gas at 1 atm pressure and 0\u00b0C is 22.4 liters.<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nFigure 13.8 shows plot of PV\/T versus P for 1.00 \u00d7 10-3<\/sup> kg of oxygen gas at two different temperatures.
\n1. What does the dotted plot signify?
\n2. Which is true: T1<\/sub> > T2<\/sub> or T1<\/sub> < T2<\/sub>?
\n3. What is the value of PV\/T where the curves meet on the y-axis?
\n4. If we obtained similar plots for 1.00 \u00d7 10-3<\/sup> kg of hydrogen, would we get the same value of PV\/T at the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value of PV\/T (for low-pressure high-temperature region of the plot)? (Molecular mass of H2<\/sub> = 2.02 u, of 02<\/sub> = 32.0 u, = 8.31 J mol-1<\/sup> K-1<\/sup>.)
\n\"1st
\nAnswer:
\n1. The dotted plot signifies the behavior of equal amount of an ideal gas. Since the ideal gas satisfies the equation PV = nRT,
\nnR = \\(\\frac{P V}{T}\\) is a constant forgiven amount (n) of an ideal gas.
\nThus \\(\\frac{P V}{T}\\) is independent of pressure.<\/p>\n

2. The \\(\\frac{P V}{T}\\) curve at temperature T1<\/sub> is closer to the dotted line (ideal gas) than the \\(\\frac{P V}{T}\\) curve at T2<\/sub>. Since a real gas at higher temperature behaves more like an ideal gas than a real gas at lower temperature, T1<\/sub> >T2<\/sub>.<\/p>\n

3. Since all three curves meet on the y-axis the value of \\(\\frac{P V}{T}\\) = nR, where n is the no of moles of ideal gas.
\n\u2234 n = no. of moles of oxygen gas
\n\\(=\\frac{\\text { Given mass of oxygen }}{\\text { gram molecular mass of oxygen }}\\)
\n\\(=\\frac{1}{32}=0.31\\)
\n\u2234 \\(\\frac{P V}{T}\\) = nR =0.031 \u00d7 8.314
\n= 0.26 J K-1<\/sup><\/p>\n

4. Since the molecular mass of hydrogen is less than that of oxygen, the number of moles is 1 g of hydrogen is more than that of oxygen. There fore, the value of \\(\\frac{P V}{T}\\) at the point where the curves meet the Y-axis will not be same for oxygen and hydrogen gases.
\nn = 0.031 mol of H2<\/sub> will yield same \\(\\frac{P V}{T}\\)
\nas that of 1 g of O2<\/sub> gas.
\n\u2234 Mass of H2<\/sub> gas
\n= 0. 311 \u00d7 2.02 = 0. 0626g<\/p>\n

Question 4.
\nAn oxygen cylinder of volume 30 liters has an initial gauge pressure of 15 atm and a temperature of 27 \u00b0C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 \u00b0C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31 J mol-1<\/sup> K-1<\/sup>, molecular mass of o2<\/sub> = 32 u).
\nAnswer:
\nInitial Volume, V1<\/sub> = 30 litres
\n= 30 \u00d7 10-3<\/sup> m3<\/sup>
\nInitial Pressure P1<\/sub> = 15 atm
\n= 15 \u00d7 101.3 KPa
\nInitial Tamperature, T1<\/sub> = 27\u00b0C \u2243 300K
\nLet \u2018n\u2019 be the initial number of moles of oxygen.
\nFrom ideal gas law,
\nP1<\/sub> V1<\/sub> = n1<\/sub> RT1<\/sub>
\n\u21d2 \u2234 n1<\/sub> = \\(\\frac{P_{1} V_{1}}{R T_{1}}\\)
\n\\(=\\frac{\\left(15.195 \\times 10^{5}\\right) \\times 30 \\times 10^{-3}}{8.314 \\times 300}\\)
\n= 18.276
\nFinal volume, V2<\/sub> = 30 litres = 30 \u00d7 10-3<\/sup>
\nFinal pressure, P2<\/sub> = 11 atm = 11.143 \u00d7 105<\/sup>
\nFinal temperature, T2<\/sub> = 17\u00b0C = 290 K
\n\u2234 Final no. of moles, n2<\/sub> is given by.
\nn2<\/sub> = \\(\\frac{P_{2} V_{2}}{R T_{2}}\\)
\n\\(=\\frac{\\left(11.143 \\times 10^{5}\\right) \\times 30 \\times 10^{-3}}{8.314 \\times 290}\\)
\n= 13.865
\n\u2234 Moles of oxygen taken out,
\n\u2206n = n1<\/sub> – n2<\/sub>
\n= 18.276-13.865
\n= 4.411
\n\u2234 Mass of oxygen taken out
\n= \u2206n \u00d7 M
\nwhere M is the molecular mass of oxygen.
\n\u2234 mass of oxygen taken out = 4.411 \u00d7 32
\n= 141.152 g
\n\u2234 141.152 g of oxygen was taken out of cylinder.<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\nAn air bubble of volume 1.0 cm3 rises from the bottom of a lake 40 m deep at a temperature of 12 \u00b0C. To what volume does it grow when it reaches the surface, which is at a temperature of 35 \u00b0 C?
\nAnswer:
\nInitial volume of air bubble,
\nV1<\/sub> = 1.0 cm3<\/sup>
\n= 10-6<\/sup>m3<\/sup>
\nInitial temperature, T1<\/sub> = 12\u00b0C = 285K
\nInitial Pressure, P1<\/sub>= Ps<\/sub> + \u03c1gh
\nwhere P1<\/sub> = depth of air bubble = 40m
\nPs<\/sub> = Pressure at lake surface
\n= 1atm
\n\u03c1 = density of water = 102<\/sup>kg\/cm3<\/sup>
\ng = acceleration due to gravity
\n= 9.8 ms-2<\/sup>
\n\u2234 P1<\/sub> = 1.013 \u00d7 105<\/sup>+103<\/sup> \u00d7 9.8 \u00d7 40
\n=\u00a0 493300 Pa
\n=\u00a0 4.933 \u00d7 105<\/sup> Pa.
\nLet final volume of air bubble = V2<\/sub>
\nFinal Pressure = Ps<\/sub> = 1.013 \u00d7 105<\/sup> Pa
\nFinal Temperature, T2<\/sub> = 35\u00b0C = 308 K
\nFrom ideal gas law,
\nPV = n RT
\n\u2234 \\(\\frac{P V}{T}\\) = n R
\n\u2234 for given n, \\(\\frac{P V}{T}\\) = constant
\n\u21d2 \\(\\frac{P_{1} V_{1}}{T_{1}}=\\frac{P_{2} V_{2}}{T_{2}}\\)
\n\u2234 V2<\/sub> = \\(\\frac{P_{1} V_{1} T_{2}}{T_{1} P_{2}}\\)
\n\\(=\\frac{4.933 \\times 10^{5} \\times 10^{-6} \\times 308}{285 \\times 1.013 \\times 10^{5}}\\)
\n= 5.263 \u00d7 10-6<\/sup> m3<\/sup>
\n= 5.263 cm3<\/sup>.
\n\u2234 The volume of air bubble grows to 5. 263 cm3<\/sup> when it reaches the lake surface.<\/p>\n

Question 6.
\nEstimate the total number of air molecules (Inclusive of oxygen, nitrogen, water vapour, and other constituents) in a room of capacity 25.0 m3<\/sup>at a temperature of 27 \u00b0C and 1 atm pressure.
\nAnswer:
\nFrom ideal gas law
\nPV = nRT
\nHere P = atm = 1. 013 \u00d7 105<\/sup> Pa.
\nV = Volume = 25 m3<\/sup>
\nT = Temperature = 27\u00b0C = 300k
\nn = no of moles of gas
\nR = Gas constat = 8. 314 J mol-1<\/sup> k-1<\/sup>
\n\u2234 n = \\(\\frac{P V}{R T}=\\frac{1.013 \\times 10^{5} \\times 25}{8.314 \\times 300}\\)
\n= 0.01015 \u00d7 105<\/sup> =1015 mol
\nSince 1 mol \u2245 6. 023 \u00d7 1023<\/sup> molecules
\nN = 1015 mol \u2245 6.023 \u00d7 1023<\/sup> \u00d7 1015
\nN = 6.113 \u00d7 1026<\/sup> molecules.
\n\u2234 The total no of molecules in the given room is 6.113 \u00d7 1026<\/sup>.<\/p>\n

Question 7.
\nEstimate the average thermal energy of a helium atom at
\n1. room temperature (27 \u00b0C),
\n2. the temperature on the surface of the Sun (6000 K),
\n3. the temperature of 10 million kelvin (the typical core temperature in the case of a star).
\nAnswer:
\nAverage thermal energy of an atom
\nE= \\(\\frac{3}{2}\\) KT
\nwhere, K = Boltzmann constant
\n= 1.38 \u00d7 10-23<\/sup>m2<\/sup>kgs-2<\/sup>K-1
\n<\/sup> T = absolute temperature
\n1. room temperature (27\u00b0C)
\nT = 27\u00b0C = 300k
\n\u2234 E= \\(\\frac{3}{2}\\) \u00d7 1.38 \u00d7 10-23<\/sup> \u00d7 300
\n= 1.242 \u00d7 10-21<\/sup> J<\/p>\n

2. Surface of the sun (6000 K)
\nT = 6000K
\n\u2234 E = \\(\\frac{3}{2}\\) \u00d7 (1.38 \u00d7 10-23<\/sup>) \u00d7 6000
\nE = 1.242 \u00d7 10-19<\/sup> J<\/p>\n

3. core of a star (10 million K)
\nT = 10 \u00d7 106<\/sup> = 107<\/sup>K
\n\u2234 E = \\(\\frac{3}{2}\\) \u00d7 (1.38 \u00d7 10-23<\/sup>) \u00d7 107<\/sup>
\nE =2.07 \u00d7 10-16<\/sup>J.<\/p>\n

\"KSEEB<\/p>\n

Question 8.
\nThree vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equal number of respective molecules? Is the root mean square speed of molecules the same In the three cases? If not, in which case is Vrms<\/sub> the largest?
\nAnswer:
\nSince all vessels are of same capacity and are at same pressure and temperature,
\nP = P1<\/sub> = P2 <\/sub>= P3<\/sub> ……..(1)
\nV = V1<\/sub> = V2<\/sub> = V3<\/sub> ……..(2)
\nT = T1 <\/sub>= T2<\/sub> = T3<\/sub> ……..(3)
\nwhere Pi<\/sub> Vi<\/sub> and Ti<\/sub> are pressure, volume and temperature of ith<\/sup> vessel respectively.
\nFrom ideal gas law
\nPV = nRT
\n\u2234 Pi<\/sub> Vi<\/sub> = ni<\/sub> RTi<\/sub> i = 1,2, 3
\nFrom equation (1), (2) and (3).
\nPV = ni<\/sub> RT
\nor ni<\/sub> = \\(\\frac{P V}{R T}\\) = constant for all the three vessels.
\n\u2234 All three vessels have same no of moles of respective gases. From Avagadro\u2019s law, equal mols of gases contain equal no of molecules. The root mean square (RMS) speed of a molecule is given by
\n\\(V_{r m s}=\\sqrt{\\frac{3 k_{B}}{m}}\\)
\nwhere\u00a0 kB = Boltzmann constant
\nT = temperature
\nm = molecular mass of the gas.
\nAs Vrms<\/sub> \u03b1 \\(\\frac{1}{\\sqrt{\\mathrm{m}}}\\) the rms speed of gas molecules in the three vessels are different. Since neon has the least molecular mass of the three gases, it has the highest rms speed.<\/p>\n

Question 9.
\nAt what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at – 20 \u00b0 C? (atomic mass of Ar<\/sub>= 39.9 u, of He<\/sub> = 4.0 u).
\nAnswer:
\nTemperature of helium gas,
\nT1<\/sub>= – 20\u00b0C = 253 k
\ngram molecular mass of Helium, m1<\/sub> =4.0
\ngram molecular mass of Argon, m2<\/sub> = 39.9 gm2<\/sub>
\nLet temperature of Argon = T2<\/sub>
\nThe rms velocity of a gas atom is given by
\n\\(\\mathrm{V}_{\\mathrm{rms}}=\\sqrt{\\frac{3 \\mathrm{RT}}{\\mathrm{M}}}\\)
\nwhere T = Temperature of the gas
\nR = Universal Gas constant
\nM = Molar mass of the gas.
\nGiven Vrms(Helium)<\/sub> = Vrms(Argon)<\/sub>
\n\u2234 \\(\\sqrt{\\frac{3 \\mathrm{RT}}{\\mathrm{M}}}\\)(Helium)<\/sub> = \\(\\sqrt{\\frac{3 \\mathrm{RT}}{\\mathrm{M}}}\\)(Argon)<\/sub>
\n\\(\\frac{T_{1}}{M_{1}}=\\frac{T_{2}}{M_{2}}\\)
\n\u2234 T2<\/sub> = \\(\\frac{T_{1} M_{2}}{M_{1}}\\)
\n\\(=\\frac{253 \\times 39.9}{4}\\)
\n= 2523.675 K
\n\u2234 At 2523.675 K, the rms speed of an atom of argon will be equal to that of helium at – 20\u00b0C.<\/p>\n

Question 10.
\nEstimate the mean free path and collision frequency of a nitrogen molecule in a atm and temperature of 17 \u00b0C. Take the radius of a nitrogen molecule to be roughly 1.0 A. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N2<\/sub> = 28.0 u).
\nAnswer:
\nThe mean free path is given by
\n\\(\\ell=\\frac{\\mathrm{k} \\mathrm{T}}{\\sqrt{2} \\pi \\mathrm{d}^{2} \\cdot \\mathrm{P}}\\)
\nwhere k = Bottzmann constant
\n= 1.38 \u00d7 10-23<\/sup>kg2<\/sup>m2<\/sup>s-2<\/sup>k-1<\/sup>
\nd = diameter of the atom \/ molecule
\n= 2 \u00d7 1 A\u00b0 = 2 \u00d7 10-10<\/sup> m
\nP = Pressure of the gas = 2 atm
\n= 2.026 \u00d7 105<\/sup> Pa
\n\\(\\therefore \\ell=\\frac{1.38 \\times 10^{-23} \\times 290}{\\sqrt{2} \\times \\pi \\times\\left(2 \\times 10^{-10}\\right)^{2} \\times\\left(2.026 \\times 10^{5}\\right)}\\)
\n= 11.12 \u00d7 10-8<\/sup> m
\n= 1.112 \u00d7 10-7<\/sup> m
\n\u2234 mean free path = 1.112 \u00d7 10-7<\/sup> m
\nRMS speed of nitrogen molecules is given by
\n\"1st
\nVrms<\/sub> = 508.26 ms-1<\/sup>
\n\u2234 collision frequency, f = \\(\\frac{V_{rms}}{\\ell}\\)
\n\\(=\\frac{508.26}{1.112 \\times 10^{-7}}\\) = 4.57 \u00d7 109<\/sup>s-1<\/sup>
\n\u2234 collision time Tc<\/sub> = \\(\\frac{\\mathrm{d}}{\\mathrm{V}_{\\mathrm{rms}}}\\)
\n\\(=\\frac{2 \\times 10^{-10}}{508.26}\\)
\n= 3.935 \u00d7 10-13<\/sup> s.
\nTime gap between successive collisions,
\nTg<\/sub> = \\(\\frac{1}{f}\\)
\n\\(=\\frac{1}{4.57 \\times 10^{9}} \\mathrm{s}\\)
\n= 2.188 \u00d7 10-10<\/sup>s
\n\\(\\therefore \\frac{T_{9}}{T_{c}}=\\frac{2.188 \\times 10^{-10}}{3.935 \\times 10^{-13}}=555 \\simeq 500\\)
\n\u2234 the time taken between successive collisions is about 500 times the time spent on collision.<\/p>\n

Question 11.
\nA metre long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread, which traps a 15 cm column of air. What happens if the tube Is held vertically with the open end at the bottom?
\nAnswer:
\nTotal length of bore = 1 m = 100 cm.
\nlength of air column = 15cm
\nlength of mercury thread = 76cm.
\n\u2234 length of open space = 100 – (15 + 76)
\n= 9 cm
\nInitial pressure P1<\/sub> = 76cm of mercury
\nInitial volume\u00a0 \u00a0V1<\/sub>= 15. A cm3<\/sup>
\nwhere, A = sectional area of the bore in cm2<\/sup>
\nLet h cm3<\/sup> of mercury flow out of tube when held vertically
\n\u2234 New length of air column = 15 + 9 + h = 24 + h cm
\nNew length of mercury = 76 – h cm
\n\u2234 Final pressure, P2<\/sub> = 76 – (76-h)
\n= h cm of mercury
\nFind volume, V2<\/sub> = (24 + h) \u00d7 A cm3<\/sup>
\nSince temperature is constant and no air escapes the bore,
\nP1<\/sub> V1<\/sub> = P2<\/sub> V2<\/sub>
\n\u2234 76 \u00d7 15A = h (24+h) A
\n\u2234 76 \u00d7 15 = h (24+h)
\n\u21d2 h = \\(\\frac{-24 \\pm \\sqrt{576+4560}}{2}\\)
\n\u2234 h = \\(\\frac{-24 \\pm 71.67}{2}\\)
\nh = 23. 83, -47.83
\nSince height can only be positive, the amount of mercury that flows out of bore is 23. 83cm. There fore, mercury left in the bore = 52.2 cm and length of air column = 47.8 cm.<\/p>\n

\"KSEEB<\/p>\n

Question 12.
\nFrom a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm3<\/sup> s-1<\/sup>. The diffusion of another gas under the same conditions ie measured to have an average rate of 7.2 cm3<\/sup> s-1<\/sup>. Identify the gas. [Hint: Use Graham\u2019s law of diffusion: R1<\/sub>\/R2<\/sub> = ( M2<\/sub>\/M1<\/sub>)1\/2<\/sup> , where R1<\/sub>, R2<\/sub> are diffusion rates of gases 1 and 2, and M1 <\/sub>and M2<\/sub> their respective molecular masses. The law is a simple consequence of kinetic theory.]
\nAnswer:
\nAccording to graham\u2019s law of diffusion,
\n\\(\\frac{\\mathrm{R}_{1}}{\\mathrm{R}_{2}}=\\left(\\frac{\\mathrm{M}_{1}}{\\mathrm{M}_{2}}\\right)^{1 \/ 2}\\)
\nR1<\/sub> = rate of diffusion of hydrogen
\n= 28.7cm3<\/sup> S-1<\/sup>
\nR2<\/sub> = Rate diffusion of unknown gas
\n= 7.2cm3<\/sup> s-1<\/sup>
\nM1<\/sub> = Molecular mass of hydrogen = 2.02
\nM2<\/sub>= Molecular mass of unknown gas.
\n\u2234 M2<\/sub> = \\(M_{1}\\left(\\frac{R_{1}}{R_{2}}\\right)^{2}\\)
\n= \\(2.02\\left(\\frac{28.7}{7.2}\\right)\\)
\n= 32.09
\nThe molecular mass of oxygen gas is 32. Hence, The given unknown gas is oxygen.<\/p>\n

Question 13.
\nA gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so called law of atmospheres
\nn2<\/sub> = n1<\/sub> exp [ -mg (h2<\/sub> – h1<\/sub>) \/KB<\/sub>T] where n2<\/sub>, n1<\/sub> refer to number density at heights h2<\/sub> and h1<\/sub> respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column:
\nn2<\/sub> = n1<\/sub> exp [-mg NA<\/sub> \\(\\left(\\rho-p^{\\prime}\\right)\\) (h2<\/sub>-h1<\/sub>\/ \\(\\boldsymbol{\\rho}\\) RT)]
\nwhere \u03c1 is the density of the suspended particle, and \\(\\rho^{\\prime}\\) that of surrounding medium. [NA<\/sub> is Avogadro\u2019s number, and R the universal gas constant.] [Hint: Use Archimedes principle to find the apparent weight of the suspended particle.]
\nAnswer:
\nFrom the law atmospheres , we have n2<\/sub> = n1<\/sub> exp (-mg (h2<\/sub> -h1<\/sub>)\/ KB<\/sub>T ]\u00a0 …..(1)
\nwhere n2<\/sub> = number density at height h2<\/sub>
\nn1<\/sub> = number density at height h1<\/sub>
\nm = mass of the particle suspended in the column
\nLet
\n\u03c1 = density of suspended particle
\n\\(\\rho^{\\prime}\\) = density of the medium
\nm2<\/sub> = mass of medium displaced
\nFrom Archimedes\u2019 principle,
\nEffective weight of the suspended particle = Actual weight of the particle – weight of the medium displaced.
\ni.e., We<\/sub> = mg – \\(m^{\\prime}\\)g
\nLet V be the volume displaced.
\n\u2234 \\(\\mathrm{m}^{1}=\\mathrm{v}_{\\mathrm{f}}^{\\prime}\\) and \\(\\mathrm{V}=\\left(\\frac{\\mathrm{m}}{\\mathrm{f}}\\right)\\)
\n\u2234 \\(\\mathrm{m}^{\\mathrm{l}}=\\left(\\frac{\\mathrm{m}}{\\rho}\\right) \\rho^{\\mathrm{l}}\\)
\n\u2234 effective weight, We<\/sub> = mg – m \\(\\left(\\frac{\\rho^{\\prime}}{\\rho}\\right) \\mathrm{g}\\)
\nWe<\/sub> = mg [1- \\(\\rho^{\\prime}\\)\/\u03c1]\u00a0 \u00a0……(2)
\nWe know that
\nR = NA<\/sub>.kB<\/sub> (by definition)
\nwhere R = Universal Gas constant
\nNA<\/sub> = Avogadro number
\nkB<\/sub> = Boltzmann constant
\n\u2234 kB<\/sub> = R\/NA<\/sub> …………(3)
\nSubstituting the effective weight in (2) inplace of mg and using (3) in equation (1),
\n\\(n_{2}=n_{1} \\exp \\left[m g\\left(1-\\frac{\\rho^{1}}{\\rho}\\right)\\left(h_{2}-h_{1}\\right) \/\\left(R \/ N_{A}\\right) T\\right]\\)
\n\\(=n_{1} \\exp \\quad\\left[\\mathrm{mg} \\mathrm{N}_{\\mathrm{A}}\\left(\\rho-\\rho^{\\prime}\\right)^{\\left(\\mathrm{h}_{2}-\\mathrm{h}_{1}\\right)} \/ \\rho \\mathrm{RT}\\right]\\)<\/p>\n

Question 14.
\nGiven below are densities of some solids and liquids. Give rough estimates of the size of their atoms :<\/p>\n\n\n\n\n\n\n\n\n
Substance<\/td>\nAtomic mass(u)<\/td>\nDensity (103 <\/sup>kg nr3<\/sup>)<\/td>\n<\/tr>\n
Carbon (diamond)<\/td>\n12.01<\/td>\n2.22<\/td>\n<\/tr>\n
Gold<\/td>\n197.00<\/td>\n19.32<\/td>\n<\/tr>\n
Nitrogen (liquid)<\/td>\n14.01<\/td>\n1.00<\/td>\n<\/tr>\n
Lithium<\/td>\n6.94<\/td>\n0.53<\/td>\n<\/tr>\n
Fluorine (liquid)<\/td>\n19.00<\/td>\n1.!4<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

[Hint: Assume the atoms to be \u2018tightly packed\u2019 in a solid or liquid phase, and use the known value of Avogadro\u2019s number. You should, however, not take the actual numbers you obtain for various atomic sizes too literally. Because of the crudeness of the tight packing approximation, the results only indicate that atomic sizes are in the range of a few A].
\nAnswer:
\nAssuming all molecules to be spherical, and solids and liquids are tightly packed,
\nVolume, V= \\(\\frac{4}{3} \\pi r^{3}\\)
\nwhere r = radius of a molecules
\nAlso, V = \\(\\frac{\\text { mass }}{\\text { density }}\\)
\nFor 1 mole of a substance,
\n\\(\\mathrm{V}=\\frac{4}{3} \\pi \\mathrm{r}^{3} \\cdot \\mathrm{N}_{\\mathrm{A}}=\\frac{\\mathrm{M}}{\\mathrm{J}}\\)
\nwhere NA<\/sub> = Avogadro number = 6. 023 \u00d7 1023<\/sup>
\nM = gram atomic mass
\n\u03c1 = density of the substute
\n\u2234 \\(r=\\left(\\frac{3 M}{4 \\pi \\int N_{A}}\\right)^{1 \/ 3}\\)<\/p>\n

1. Carbon (diamond):
\nM = 12.01g = 12.01 \u00d7 10-3<\/sup>Kg
\nf = 2.2 \u00d7 103<\/sup> kgm-3<\/sup>
\n\u2234 \\(r_{c}=\\left(\\frac{3 \\times 12.01 \\times 10^{-3}}{4 \\pi \\times 2.2 \\times 10^{3} \\times 6.023 \\times 10^{23}}\\right)\\)
\n\u2243 1.29 \u00d7 10-10<\/sup>m
\nrc<\/sub> \u2243 1.29 A\u00b0<\/p>\n

2. Gold :
\nM = 197 g = 197 \u00d7 10-3<\/sup> kg
\nf= 19.32 \u00d7 103<\/sup> kg m-3<\/sup>
\n\\(\\therefore r_{9}=\\left(\\frac{3 \\times 197 \\times 10^{-3}}{4 \\pi \\times 19.32 \\times 10^{3} \\times 6.027 \\times 10^{23}}\\right)^{1 \/ 3}\\)
\n= 1.59 \u00d7 10-10<\/sup>m
\nrg<\/sub> = 1.59 A\u00b0<\/p>\n

3. Nitrogen (liquid):
\nM = 14.01g = 14.01 \u00d7 10-3<\/sup>kg
\nf = 103<\/sup>kgm-3<\/sup>
\n\\(r_{n}=\\left(\\frac{3 \\times 6.94 \\times 10^{-3}}{4 \\pi \\times 0.53 \\times 10^{3} \\times 6.023 \\times 10^{3}}\\right)^{1 \/ 3}\\)
\n= 1.73 A\u00b0<\/p>\n

4. Fluorine (liquid):
\nM = 19 \u00d7 10-3<\/sup>kg
\nf = 1.14 \u00d7 103<\/sup>kgm-3<\/sup>
\n\\(r_{n}=\\left(\\frac{3 \\times 19 \\times 10^{-3}}{4 \\pi \\times 1.14 \\times 10^{3} \\times 6.023 \\times 10^{23}}\\right)^{1 \/ 3}\\)
\n= 1.876 A\u00b0<\/p>\n\n\n\n\n\n\n\n\n
Substance<\/td>\nRadius<\/td>\n<\/tr>\n
Carbon (diamond)<\/td>\n(A\u00b0)
\n1.29<\/td>\n<\/tr>\n
Gold<\/td>\n1.59<\/td>\n<\/tr>\n
Nitrogen (liquid)<\/td>\n1.77<\/td>\n<\/tr>\n
Lithium<\/td>\n1.73<\/td>\n<\/tr>\n
Fluorine (liquid)<\/td>\n1.88<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

1st PUC Physics Kinetic Theory One Mark Questions and Answers<\/h3>\n

Question 1.
\nDefine mean free path.
\nAnswer:
\nThe average distance covered by a moving molecule between successive collisions is called the mean free path.<\/p>\n

Question 2.
\nState Avagadro\u2019s hypothesis.
\nAnswer:
\nAvagadro\u2019s hypothesis states that under fixed conditions of temperature and pressure, equal volumes of all gases contain equal number of molecules.<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nState Boyle\u2019s law.
\nAnswer:
\nBoyle\u2019s law states that under constant temperature, the pressure of given mass of a gas is inversely proportional to its volume.<\/p>\n

Question 4.
\nState Charle’s low.
\nAnswer:
\nCharle\u2019s law states that under constant pressure, the volume of a given mass of any gas is directly proportional to its temperature.<\/p>\n

Question 5.
\nState law of equipartition of energy.
\nAnswer:
\nThe law of equipartition of energy states that in thermal equilibrium, the total energy of a particle is equally distributed in all possible modes of energy, with each mode having an average energy equal to
\n\\(\\frac{1}{2}\\) kB<\/sub> T.<\/p>\n

Question 6.
\nDefine the Avogadro number.
\nAnswer:
\nAvogadro number, NA<\/sub>, is defined as the number of molecules contained in 22.4 litres of any gas at STP, Its value is equal to 6.02 \u00d7 1023<\/sup>.<\/p>\n

Question 7.
\nCalculate the number of molecules contained in 448 ml of an ideal gas at STP.
\nAnswer:
\n22.4 L contains 6.02 \u00d7 1023<\/sup> molecules therefore 448 ml contains
\nN = \\(\\frac{448}{22400} \\times 6.02 \\times 10^{23}\\)
\n\u2234 N = 12.04 \u00d7 1021<\/sup>
\n= 1.2 \u00d7 1022<\/sup>
\n\u2234 448 ml of ideal gas at STP contains 1.2 \u00d7 1022<\/sup>molecules.<\/p>\n

\"KSEEB<\/p>\n

Question 8.
\nIf the temperature (absolute scale) of an ideal gas is doubled keeping pressure and volume constant, find the increase in its rms speed.
\nAnswer:
\nFor an ideal gas,
\n\\(\\mathrm{V}_{\\mathrm{ms}} \\alpha \\sqrt{\\mathrm{T}}\\)
\nThere fore, if the temperature is doubled, the rms speed increases by a factor of \\(\\sqrt{2}\\).<\/p>\n

1st PUC Physics Kinetic Theory Two Or Three Marks Questions and Answers<\/h3>\n

Question 1.
\nWhat is an ideal gas? Mention the ideal gas equation.
\nAnswer:
\nAn ideal gas is a gas that obeys all the gas laws at all temperatures and is given by
\nPV = \u00b5RT
\nwhere P = pressure of the gas
\nV = Volume of the gas
\n\u00b5 = number of mols of the gas
\nR = Universal gas constant
\nT = absolute temperature .<\/p>\n

Question 2.
\nA sample of gas is at 300K. Find the temperature at which the r.m.s speed of its molecules is doubled without changing the pressure.
\nAnswer:
\nInitial temperature, Ti<\/sub> = 300k
\nlet final temperature, Tf<\/sub> = T
\nlet initial rms speed, Vi<\/sub> = v1<\/sub>
\n\u2234 final rms speed, Vf<\/sub> = 2V1<\/sub>
\nFor a gas of mass m and temperature T, the rms speed is given by
\n\\(V_{\\mathrm{rms}}=\\sqrt{\\frac{3 \\mathrm{k}_{\\mathrm{B}} \\mathrm{T}}{\\mathrm{m}}}\\)
\nwhere kB<\/sub> = Boltzman constant.
\n\u2234 \\(\\frac{2 V_{1}}{V_{1}}=\\sqrt{\\frac{T}{300}}\\)
\n\u2234 T = 300\u00d7 (2)2<\/sup>
\n\u2234 T = 1200 k.<\/p>\n

Question 3.
\nThe temperature of an ideal gas is increasing from 100K to 500K if the rms speed of molecules at 100k Is V, what Is the rms speed of the molecules at 500K?
\nAnswer:
\nInitial temperature, Ti<\/sub> = 100K
\nfinal temperature of the gas Tf<\/sub> = 500K
\nInitial rms speed of molecules, Vrms,i<\/sub> = V
\nlet final r.m.s speed of molecules Vrms,f<\/sub> = V’
\nfor ideal gas,
\n\\(V_{\\mathrm{rms}}=\\sqrt{\\frac{3 \\mathrm{k}_{\\mathrm{B}} \\mathrm{T}}{\\mathrm{m}}}\\)
\nwhere kB = Boltzmann constant
\nT = absolute temperature
\nm = mass of the gas
\n\"1st
\nThere fore, the r.m.s speed of molecules of the given ideal gas at 500k is \\(\\sqrt{5}\\) V.<\/p>\n

\"KSEEB<\/p>\n

Question 4.
\nAt a given temperature T, the rms speed of Hydrogen molecules is V. Find rms speed of oxygen molecules at the same temperature.
\nAnswer:
\nMolecular mass of Hydrogen, MH2<\/sub>= 2
\nMolecular mass of Oxygen, MO2<\/sub> = 32
\nrms speed of hydrogen gas VH2<\/sub> = V
\nlet rms speed of Oxygen gas, Vo2<\/sub> = v’
\nThe rms speed of gas is given by
\n\"1st
\n\u2234 rms speed of oxygen at temperature T is \\(\\frac{\\mathrm{V}}{16}\\).<\/p>\n

Question 5.
\nA box contains N molecules of an ideal gas at temperature T and pressure P. The number of molecules in the box is increased to two times the original value. Find the final temperature and pressure if the kinetic energy of the gas is kept constant.
\nAnswer:
\nInitial no. of molecules ni<\/sub> = N
\nInitial temperature, Ti<\/sub> = T
\nInitial Pressure, Pi<\/sub> = P
\nFinal no. of molecules, nf<\/sub>= 2N
\nLet final temperature, Tf<\/sub>=T’
\nlet final pressure, Pf<\/sub> = p’
\nFor ideal gases,
\nPV = \\(\\frac{2}{3}\\)E ……..(1)
\nwhere P = pressure of the gas
\nV = volume of the gas
\nE = kinetic energy of molecules of the gas
\nSince E is constant, P is also a constant for the given case
\n\u2234 P’ = p
\nFor \u2018n\u2019 mols of an ideal gas of molecular mass M,
\n\\(\\frac{1}{2} \\mathrm{nM} \\mathrm{V}_{\\mathrm{rms}}^{2}=\\frac{3}{2} \\mathrm{nRT}\\)
\n\u2234 E = \\(\\frac{3}{2}\\) NKB<\/sub>T
\n\"1st<\/p>\n

Question 6.
\nEstimate the fraction of molecular volume to the actual volume occupied by Nitrogen gas at STP. Assume the radius of N2<\/sub> = 300 pm.
\nAnswer:
\nConsider one mole of N2<\/sub> gas at STP
\n\u2234 Actual volume, vact<\/sub> = 22.4L
\nvact<\/sub> = 22.4 \u00d7 10-3<\/sup>m3<\/sup>
\nmolecular volume, Vmol<\/sub>= NA<\/sub> \u00d7 \\(\\frac{4}{3}\\) \u00d7 \u03c0 \u00d7R3<\/sup>
\n\u2234 Vmol<\/sub> = 6.81 \u00d7 10-5<\/sup>m3<\/sup>
\n\u2234 Ratio of molecular volume to actual volume,
\n\\(\\frac{V_{\\text {mol }}}{V_{\\text {act }}}=\\frac{6.81 \\times 10^{-5}}{22.4 \\times 10^{-3}}\\)
\n= 3.04 \u00d7 1 O-3<\/sup><\/p>\n

Question 7.
\nThe molecules of a given mass of an ideal gas have rms speed 100 ms-1<\/sup>at 273K and 100kPa calculate the rms speed of the molecules if the temperature is doubled and pressure heated.
\nAnswer:
\nInitial temperature of gas, Ti<\/sub> = 273K
\nFinal temperature of gas, Tf<\/sub> = 2 \u00d7 Ti<\/sub>
\n= 546K
\nInitial rms speed of gas
\nmolecules, Vi<\/sub> = 100 m\/s
\nlet final rms speed of molecules, Vf<\/sub> = V
\nFor an ideal gas,
\n\"1st
\n\u2234 V = 100 \\(\\sqrt{2}\\)
\n\u2234 V = 141.4 m\/s.<\/p>\n

Question 8.
\nState and explain Boyle\u2019s law.
\nAnswer:
\nBoyle\u2019s law states that at constant temperature, the volume of a given mass of an ideal gas is inversely proportional to its pressure.
\nLet the volume of the gas be \u2018V\u2019 and pressure \u2018P\u2019 and temperature, T. Then according to Boyle\u2019s law,
\n\\(\\mathrm{V} \\propto \\frac{1}{\\mathrm{P}}\\)
\nor V =\\(\\frac{k}{P}\\) \u21d2 V.P = k.
\nwhere k = proportionality constant i.e, according to Boyle’s law, if the temperature is kept constant, for a given change in volume the pressure changes such that the product of pressure and volume is a constant.<\/p>\n

\"KSEEB<\/p>\n

Question 9.
\nState and explain Charles law.
\nAnswer:
\nCharles\u2019 law states that at constant pressure, the volume of a given mass of an ideal gas is directly proportional to its temperature.
\nconsider a gas at pressure \u2018P\u2019, temperature \u2018T\u2019 having volume \u2018V\u2019, Then Charles\u2019 law states that
\n\\(\\mathrm{V} \\propto \\mathrm{T}\\)
\nor V = kT where k is the proportionality constant.<\/p>\n

Question 10.
\nAt what temperature the molecules of nitrogen will have the same rms velocity as the molecules of oxygen at 400K?
\nAnswer:
\nMolecular mass of Nitrogen, M1<\/sub> =28
\nMolecular mass of Oxygen, M2<\/sub> = 32
\nTemperature of the Oxygen, T2<\/sub> = 400
\nlet temperature of nitrogen, T1<\/sub>= T
\nThe rms speed of gas molecules given by,
\n\"1st
\n\u2234 T = 350 K<\/p>\n

1st PUC Physics Kinetic Theory Four\/Five Marks Questions and Answers<\/h3>\n

Question 1.
\nDerive the ratio of specific heats for a mono atomic and diatomic gases
\nAnswer:
\n1. Monatomic gas :
\nNo of degrees of freedom = 3 (Translational) of each molecule
\n\u2234 Average kinetic energy of a molecule at temperature T, E = 3 \\(\\frac{1}{2}\\) \u00d7 kB<\/sub>T
\nwhere kB<\/sub> = Boltzmonn constant
\nU = \\(\\frac{3}{2}\\) kB<\/sub>T. NA<\/sub> [Total internal energy of the gas]
\n\u2234 U = \\(\\frac{3}{2}\\) R T
\nThe molar specific heat at constant volume,
\nCv<\/sub> = \\(\\frac{\\mathrm{d} \\mathrm{U}}{\\mathrm{d} \\mathrm{T}}\\)
\n= \\(=\\frac{d}{d T}\\left(\\frac{3}{2} R T\\right)=\\frac{3}{2} R\\) ……..(1)
\nFor ant ideal gas
\nCp<\/sub> – Cv<\/sub> = R …………(2)
\nUsing equation (1) in equation (2),
\nCp<\/sub> = R + \\(\\frac{3}{2}\\) R = \\(\\frac{5}{2}\\) R
\n\u2234 ratio of specific heats,
\n\\(\\gamma=\\frac{\\mathrm{C}_{\\mathrm{p}}}{\\mathrm{W}}=\\frac{\\left(\\frac{5}{2}\\right) \\mathrm{R}}{\\left(\\frac{3}{2}\\right) \\mathrm{R}}=\\frac{5}{3}\\)<\/p>\n

2. Diatomic gas:
\nCase 1:
\nmolecules are rigid
\nNumber of degrees of freedom of each molecules = 5 (3 translational + 2 rotational)
\n\u2234 Average energy per molecule
\n= 5 \u00d7 \\(\\frac{1}{2}\\) kB<\/sub>T = \\(\\frac{5}{2}\\) kB<\/sub>T
\n\u2234 Total internal energy
\nU = \\(\\frac{5}{2}\\) RT
\nCv<\/sub> \\(\\frac{d}{d T}\\left(\\frac{5}{2} R T\\right)\\)
\n= \\(\\frac{5}{2}\\) R …….(3)
\n\u2234 Using equation (3) in equation (2),
\nCp<\/sub> – \\(\\frac{5}{2}\\) R = R
\n\u2234 Ratio of specific heats for diatomic gas,
\n\"1st
\ncase 2:
\ndiatomic molecules are not rigid,
\nAverage energy per molecule
\n= 5 \u00d7 \\(\\frac{1}{2}\\) kB<\/sub>T + kB<\/sub>T =\\(\\frac{7}{2}\\) kB<\/sub>T
\n\u2234 Internal energy of the gas, U = \\(\\frac{7}{2}\\) RT
\n\\(C_{v}=\\frac{d}{d T}\\left(\\frac{7}{2} R T\\right)=\\frac{7}{2} R\\) ……(4)
\nUsing equation (4) in equation (2),
\nCp<\/sub> = \\(\\frac{9}{2}\\) R
\n\u2234 Ratio of specific heats,
\n\u03b3 = \\(\\frac{9}{7}\\).<\/p>\n","protected":false},"excerpt":{"rendered":"

You can Download Chapter 13 Kinetic Theory Questions and Answers, Notes, 1st PUC Physics Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. Karnataka 1st PUC Physics Question Bank Chapter 13 Kinetic Theory 1st PUC Physics Kinetic Theory Textbook Questions and Answers Question …<\/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\/9453"}],"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=9453"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/9453\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=9453"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=9453"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=9453"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}