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@ pundamental Physics (XD, , 13/58 Pradecfe, , , , , , , , by oxygen gas at STP, T;, Q. 1. Estimate the fraction of molecular volume to the actual volume occupied by ‘ake, the diameter of an oxygen molecule to be 3 A., , d 3 3 Sol.. Here, diameter, d= 3 A, regeng hap 8 om., , Molecular volume, V = Srv , where N is Avogadro’s number, , te asx10%)3 x (6023x1073) = 8-52 ce., , aaka, = 22400 ce, , Actual volume occupied by 1 mole of oxygen at STP, V’, V . 82 39x 10404 x 104, v’ 22400, 1 mole of any (ideal) gas at standard temperature and, , Q. 2. Molar volume is the volume occupied by aa pressure (STP : 1 atmospheric pressure, 0°C). Show that it is 22°4 litres., , RT, Sol. For one mole of an ideal gas, PV=RT_ -- V= a, Put R=8-31 J mole-! K+, T= 273 K, P= 1 atmosphere = 1-013 x 105 Nm, Pes 0.0224 m3 = 0.0224 x 106 cc = 22400 22-4 litr, rorssa0® = OO ar Oe =e e, , Q. 3. Fig. 13(N).1 shows plot of PV/T versus P for 1-00 x 10- kg of oxygen gas at two different temperatures., (a) What does the dotted plot signify ?, (b) Which is true : 7; < Tz or T, <1?, (c) What is the value of PV/T where the curves meet, on the Y-axis ?, (d) If we obtained similar plots for 1-00 x 103 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, PVIT (for low pressure high temperature region of, the plot) ? (Molecular mass of H = 2.02 u, of, O = 32.0 u, R = 8.31 J mol"! K“), , FIGURE 13(N).1, , , , , , PV, Sol. (a) The dotted plot shows that —- (= i seep. | ;, p ws that T (=R) is a constant quantity, independent of pressure P. This signifies, the ideal gas behaviour., , (b) The curve at temperature T; is closer to the dotted plot than the curv mperature behaviour f, a real gas approaches the behaviour of a perfect gas when fasiponene incre; — T,> “1, ased, a

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KINETIC THEORY, , , , , , 13/59, (c) Where the two curves meet, the value of PV/T on Y-axis is equal to }1 R, As mass of oxygen gas = 1.00 x 10-3kg=1g. - o =pR= (35) x 8-31 JK“! = 0-26 JK+, , @ If we obtained similar plots for 1-00 x 10-3 kg of hydrogen, we will not get the same value of PV/T at, , the point, where the curves meet on the Y-axis. This is because molecular mass of hydrogen is different, from that of oxygen., , PV, For the same value of “Fr + Mass of hydrogen required is obtained from, , PV, , m, — = UR = ——x831 = 0-26, r ne 2:02, 2:02 x0:, m = ee gram = 6-32 x 10-7 gram, , Q. 4. An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm. and a temperature of, 27°C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm. and its, , temperature drops to 17°C. Estimate the mass of oxygen taken out of the cylinder. (R = 8-1 J mole! K-1,, molecular mass of O, = 32 u)., , Sol. Initially in the oxygen cylinder, V, = 30 litres = 30 x 10-3 m? ;, P,=15 atm. = 15x 1-01 x 10°Pa; 1, =27 + 273 = 300K., If the cylinder contains n, mole of oxygen gas, then P; Vj = 1, RT,, 2AM 3 (15x 101x105) x (30x1073) ees, RT, 83x 300 oa, For oxygen, molecular weight, M = 32 g, Initial mass of oxygen in the cylinder, m, =n, M = 18-253 x 32 = 584-1 g., Finally in the oxygen cylinder, let n, moles of oxygen be left ;, Here, — V = 30x 10° m3 ; Py = 11 x 1-01 x 10° Pa ; Tp = 17 + 273 = 290K, _ PyVy _ (11x101x10°) x (30x 10-3), ~ RT, 8-3x 290, .. Final mass of oxygen gas in the cylinder, m, = 13-847 x 32 = 453-1 g, .. Mass of the oxygen gas withdrawn = m, — my = 584-1 - 453-1 = 131-0 g, , Q.5. An air bubble of volume 1-0 cm? rises from the bottom of a lake 40 m deep at a temperature of 12°C., , To what volume does it grow when it reaches the surface, which is at a temperature of 35°C ? Given, 1 atm. = 1-01 x 10° Pa., , Sol. V, = 1-0 cm? = 1-0 x 10 m? ; Ty = 12°C = 12 + 273 = 285 K;, P, = 1atm. +h, p g= 1-01 x 10° + 40x 10° x 9-8 = 493000 Pa,, When the air bubble reaches at the surface of lake, then, Vp = 2; Ty = 35°C = 35 + 273 = 308 K ; P) = 1 atm. = 1.01 x 105 Pa, , or, , Now, ny = 13-847, , BY, PV;, Now, tiebec22, _ Ff, BY,T., aT1t2, or y=, Th, ‘ 5) x 308, ee )x = 5-275 x 10% m3, , 285x1-01x10°, , cy ae ESS GPR [EXOKOLZS EXERCISES

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13/60 Pradeep 's Fundamental Physics (X1) (nay, , Q. 6. Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other, constituents) in a room of capacity 25-0 m? at a temperature of 27°C and 1 atm pressure. ( Boltzmann, is, , constant = 1-38 x 10-23 JK~, Sol. Here, V = 25-0 m3 ; T= 27 + 273 = 300 K ; k = 1-38 x 10°23 JK" ,, Now, PV = nRT = n(NK)T = (nN) kT = N’ kT where, nN = N’ = total no. of air molecules in the given gay, , nie Pe (1:01 105) x 25, AT (138% 1073) x 300, Q. 7. Estimate the average thermal energy of a helium atom at (i) room temperature (27°C) (ii) the, temperature on the surface of the Sun (6000K), (iii) the temperature of 10 million kelvin (the typicaj, core temperature in the case of a star)., Sol. (i) Here, T= 27°C = 27 + 273 = 300 K, , = 6-10 x 107, , 3, Average thermal energy = 3it = ; x 1:38 x 10-23 x 300 = 62 x 10°21 J, , w, , (ii) At T= 6000 K, Average thermal energy = Ser == x 1:38 x 10-3 x 6000 = 1-24 x 10-1 J, , N, , (ii) At T= 10 million K = 107 K, , Average thermal energy = Sur = ; x 1:38 x 10-73 x 107 = 21 x 10-16 J, , Q. 8. Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel, contains neon (monoatomic), the second contains chlorine (diatomic), and the third contains uranium, hexafluoride (polyatomic). Do the vessels contain equal number of respective molecules ? Is the root, square speed of molecules the same in the three cases ? If not, in which case is Urms the largest ?, , Sol. All the three vessels (at the same temperature and pressure) have same volume. So, in accordance with the, Avogadro’s law, the three vessels will contain equal number of respective molecules, being equal to, Avogadro’s number N = 6-023 x 103., , _ Ber, , As, Uns =4{] > Le, atagiventemp. U,,. © =,, mi te mT, , Therefore, rms speed of molecules will not be the same in the three cases., As neon has the smallest mass, therefore, rms speed will be the largest in case of neon., , Q. 9. At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the, rm.s. speed of a helium gas atom at - 20°C ? (Atomic mass of Ar = 39-9 u, of He = 4-0 u)., , Sol. Let C and C ’ be the r.m.s. velocity of argon and a helium gas atoms at temperature TK and T’ K respectively., Here, M=39:9;M’=40;T=?;T’ =-20+ 273 =253 K, , 3RT 3RT 3RT’ 3Rx 253, c=,/— =,/—— =,j—— =, Now, Vm ~V309 0M ORG, 3RT 3Rx 253 39-9 x 253, : =C’ ee | T =———— = 2523-7K, Since, C=C’, therefore 30.9 4 or 4, , Q. 10. Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing, , 4 nitrogen at 2 atm. and temperature 17°C, 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 nitrogen = 28-0 u)., , Sol. Here,A=?, f=?; p=2atm=2x1013x10°Nm*; T= 17°C =(17 + 273) K=290K, O=2x1=2A=2x10!%m; — k= 1.38 x 10-3 J molecule! K-!, M = 28 x 10° kg, aa 138 x 10-73 x 290 ___138x29x1077, , V2m0? p 1-414x314(2x10-)2 x2.026x10° 1414x314x4x 2026, , , , CASAAS EXERCISES, , , , , , =1-41x 107m, , , , NCERT’,

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KINETIC THEORY, , Qu., , Q. 13., , . When the tube is held horizontally, the mercury thread of, , 13/61, , ? rms = re pe pera 508-24 m/s, M x10 ee, , isi v 508-24 ,, Collision frequency = no. of collisions per second = Oar: = iixio7? = 458 x 10, at one end) contains a 76 cm long mercury, , A meter long narrow bore h nd closed, w bore held horizontally (a! if the tube is held vertically with the open, , thread which traps a 15 cm column of air. What happens, , end at the bottom ?, length 76 cm traps a length of air = 15 cm. A, , length of 9 cm of the tube will be left at the open end, Fig. 13(N).2(a). The pressure of air enclosed in tube, , will be atmospheric pressure, Let area of cross-section of the tube be 1 sq. cm., , & P, = 76 cm and V, = 15 cm}. FIGURE 13(N).2,, , When the tube is held vertically, 15 cm air gets another, , 9 cm of air (filled in the right handside in the horizontal Mercury (24+ hyem, Position) and let A cm of mercury flow out to balance, , the atmospheric pressure, Fig. 13(N).2(b). Then the, , heights of air column and mercury column are (24 + h) wZ-, cm and (76 — A) cm respectively. k#— 11, , The pressure of air = 76 — (76 — h) = h cm of mercury. Bre S78 et er (8- Bice, x V2 = (24 + A) cm? and P, = hm, , If we assume that temperature remains constant, then oe, , , , , , P,V,=P,V, or 76x 15=hx (24+h), , i 2, r24t yOAY + 4x10 ong cm of -47-8em, , 2, , Since h cannot be negative (because more mercury cannot flow into the tube), therefore h = 23-8 cm. Thus,, in the vertical position of the tube, 23-8 cm of mercury flows out., , From a certain apparatus, the diffusion rate of hydrogen has an average value of 28-7 cm? s+. The, diffusion of another gas under the same conditions is measured to have an average rate of 7-2 em? s!,, , Identify the gas., , or h?+24h-1140=0 or h=, , . According to Graham’s law of diffusion,, , 5 _ [M2, % “yu 1, where, r, = diffusion rate of hydrogen = 28-7 cm? s"! , r = diffusion rate of unknown gas = 7-2 cm? s“!, , M, = molecular mass of hydrogen = 2 u, , M,=?, 287 [M, 28.7 P, , ee | = = 31-78 =, 72 2 or 2 (FF x2 =31-78 = 32, , This is the molecular mass of oxygen gas., , A 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’ : Mz = ny exp [- mg (hy - hy )/kp T), , where, m1, m2 refer to number density at heights 4, and h, respectively. U:, , equation for sedimentation equilibrium of a suspension in Hquld ee ee, ny =n, exp [- mg Ny (Pp -p’) (hy - hy )(p RT)], , where, p is the density of the suspended particle, and p’ th:, number, and R is the universal gas constant]. P’ that of surrounding medium, [N, is Avogadro’s |

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LACRUIED, , i ll, , Serer ets, , 13/62 Pradeep's Fundamental Physics (XI) Weta, , m, Sol. According to the law of atmospheres, "2 =" €XP- = (hy ~h, } +i), , where n,n, refer to number density of particles at heights hy and h, respectively., If we consider the sedimentation equilibrium of suspended particles in a liquid, then in place of mg, we, will have to take effective weight of the suspended particles., Let V = average volume of a suspended particle, p = density of suspended particle, p’ = density of liquid,, m = mass of one suspended particle, m’ = mass of equal volume of liquid displaced., According to Archimede’s principle, effective weight of one suspended particle, = actual weight — weight of liquid displaced = mg - m’g, , , m)., 9”, = mg -—Vp'g = mg -| — p’g =mg | 1-—, (F}remme('-5), , Also, Boltzmann constant, kp = =, A, where, R is gas constant and N, is Avogadro’s number., , Putting, mg (: <I in place of mg and value of ky in (i), we get, , RT, Q. 14. Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms :, , mg Ny p’ ee Z n, =m, exp.| -—— ry (h, — hy) which is the required relation., , , , , , , , , , , , “Atomic Wass ()_[ Density (0° m3), 12-01 2-22, 1970 19-32, Nitrogen (Liquid) 14-01 1-00, Lithium 6-94 0-53, Fluorine (liquid) 19-00 1-14, , , , , , , , , , , , 4, Sol. If r is radius of the atom, then volume of each atom = 5 % r?, , , , 1/3,, 3M, <hieie ai elk aneinn ia Gee tacks ob wheteace «2 a AN se ee |, 3 p 4npN, , For carbon, M = 12-01 x 1073 kg, p = 2:22 x 10° kg m=>, 3x12-01x103, , 7) = 1.29 x 10710 m = 129A, 4x = x(222x10?)x (602310), , , , r=, , Similarly, for gold r= 1-59 A, For liquid nitrogen, r= 1-77 A, for lithium, r=1-73A and _ for liquid fluorine, r = 1-88 A