Tiêu đề 32. Kinetic Theory of Gases Medium Ans.pdf ✅

1. (a) Temperature 2. (b) Using , P1 = P2 + gh Here, P 1.013 10 atm,h 40m 5 2 =  = 3 3 10 kgm −  = (density of water), 2 g 9.8ms− = P 1.013 10 10 9.8 40 5 3  1 =  +   = 493300Pa Now, 2 2 2 1 1 1 T P V T P V = Here, T1 = (12+ 273) = 285K,T2 = (35+ 273) = 308K 6 3 V1 1 10 m − =  V2 is the volume of the air bubble when it reaches the surface 308 285 1.013 10 (493300 1 10 ) T P P V T V 5 6 1 2 1 1 2 2       = = − 6 3 6 3 5.26 10 m 5.3 10 m − − =  =  3. (d) At high temperature and low pressure the real gas behaves as an ideal gas. 4. (d) Here, diameter, d = 3Å 10 cm 1.5 10 cm 2 3 10 m 2 3 2 d r −10 −8 −8  = =  =  =  Molecular volume of oxygen gas, r N 3 4 V 3 =   N → Avogadros number Actual volume occupied by 1 mol of oxygen gas at STP = 22,400 2 cm 8 3 23 3.14 (1.5 10 ) 6.023 10 3 4 V =      − 3 = 8.51cm Therefore, ratio of the molecular volume in the actual volume of oxygen. 4 4 1 3.8 10 4 10 22,400 8.51 V V − − = =  =  5. (a) Using, Dalton’s law of partial pressures. P = P1 + P2 + P3 V nRT V nRT V nRT = + + V 3nRT = .... (i) Here, ,T 27 C 27 273 30K 2 1 n = =  = + = V 5 10 cc. −3 =  3 2 5 10 3 8.31 300 P −      = 5 2 P 7.48 10 Nm− =  6. (a) As partial pressure of a gas in a mixture is the pressure it would exert for the same volume and temperature, if it alone occupied the vessel, therefore, for common V and T. P1V = 1RTand P2V = 2RT Here, 1 and 2 represent for neon gas and oxygen gas respectively. Now, 2 1 2 1 P P   =         = =    2 3 P P Given 2 3 2 1 2 1 If N1 and N2 are number of molecules of two gases, then , 2 3 N N 2 1 2 1 =    = where A 1 1 N N  = and A 2 2 N N  = 7. (d) According to Graham’s law of diffusion, 1 2 2 1 r r   = or 1 2 2 1 M M r r = Here, 1 r = diffusing rate of hydrogen 28.7cm s , 3 −1 = r2 = diffusion rate of unknown gas 3 1 7.2cm s − = M1 = molecular mass of hydrogen = 24 2 M 7.2 28.7 2  = or 2 32 7.2 28.7 M 2 2   =      = 32 is molecular mass of oxygen. 8. (c) Using, PV = nRT Since volume is constant 2 1 1 2 1 2 1 2 P P orT T T T P P  = =  or       = + 100.4 100 T (T 1) 1 1 T1 = (T1 +1) 0.99  T1 = 250K 9. (a) As PV = nRT RT 32 5 PV 32 5 molecularmass m n       = =  = 10. (d) Here, V 1500m ,T1 27 C 300K 3 1 = =  = P1 = 4atm,T2 = −3C = 270K P2 = 2atm According to ideal gas equations 2 2 2 1 1 1 T P V T P V = 3 2 2 1 1 1 2 2700m 300 2 4 1500 270 P T T P V V =     =  = 11. (b) Molecular mass of water = 18  Number of molecular in 18g or 0.018kg of water 23 = 610 Mass of molecule of water
26 23 3 10 6 10 0.018 − =   =  Volume of a water molecule 29 3 26 10 m 100 3 10 Density ass − − =   =  = 29 3 3 10 m − =  12. (b) According to ideal gas law 1 1 2 2 2 1 2 2 2 1 1 1 P V P V orT T T P V T P V = = Here, P1 = P,V1 = V,T1 = T ,T ? 2 V ,V 2 P P2 = 2 = 2 = PV 2 V 2 P T T2              = 4 T T2 = 13. (a) Let V be the original volume of the gas. For a isothermal process PV = constant PiTi = PfVf 2 V 2P P V P P V V i i f i f i =        =        = For an adiabatic process  PV = Constant According to questions    =       P V 2 V (2P ) i f 1 0.76 2(2) 2(2) 2(0.38) P P 1,4 i f = = = = − − 14. (d) According to ideal gas law 1 1 2 2 2 1 2 2 2 1 1 1 P V P V or T T T P V T P V = = Here, P1 = P,V1 = V,T1 = T P 2P,V 2V,T ? 2 = 2 = 2 = 4T PV T(2P)(2V) T2 = = 15. (b) rms speed 3 V V V V 2 3 2 2 2 1 rms + + = 3 (0.5) (1) (2) 2 2 2 + + 1 1.3228kms − = Average speed 3 v v v v 1 2 3 av + + = 1 1.17kms 3 0.5 1 2 − = + + = 1.1306 .14 1.17 1.3228 V V av rms = = =  16. (c) The helium molecule is monoatomic and hence its internal energy per molecule is k T 2 3 B (where B k is the Boltzmann, constant). The internal energy per mole is therefore is RT. 2 3 One gram of helium is one fourth mole and hence its internal energy is R 100 300J, 2 3 4 1   = taking the value of R to be approximately 8J mol K . −1 −1 17. (d) The kinetic energy of 1g molecule of a gas at temperature T. 8.31 273 3.4 10 J 2 3 RT 2 3 3 = =   =  18. (a) Using, rms 2 rms 2 v V mN 3 1 v 3 1 P =  = rms 2  P  mv As m is halved vrms is doubled P0 becomes twice = 2P0 19. (a) Pressure, E 3 2 v V M 3 1 P rms 2 = = Here, rms 2 M v 2 1 E = is total internal energy of the gas. 20. (b) If the container is suddenly stopped loss in kinetic energy of gas 2 v0 (mn) 2 1 = , where n is number of moles of gas Let T is the fall in temperature of gas. Then, 2 mv0 2 1 R T 2 3 n  =       3R mv T 2 0  = 21. (d) Since there is no loss of energy in the process, therefore, sum of kinetic energy of gases A and B = Kinetic energy of mixture. Temperature of the mixture, 1 2 1T1 2T2 T  +   +  = 2 3 2(27 273) 3(37 273) + + + + = 5 1530 5 600 930 = + = =360K = 360− 273= 33C 22. (d) Average kinetic energy per molecular of a gas
= KT = 2 3 a constant at a given temperature 23. (b) M 3RT Vrms = Now, rms velocity of H2 molecule = rms velocity of O2 molecules 32 3R (47 273) 2 3R T  + =  20K 32 2 320 T =  = 24. (b) According of Kinetic theory, average K.E. of a gas molecule k T 2 3 Mv 2 1 B rms 2 = 17 23 B rms 5 10 3 1.38 10 273 m 3k T V − −      = = 2 1 1.5 10 ms − − =  25. (a) if an ideal gas is compressed adiabatically, its temperature rises, because heat produced cannot be lost to the surrounding. Each molecules has more KE than before because of collisions of molecules with moving parts of the wall (ie., piston compressing the gas). 26. (a) Here V 100ms ,T1 27 C (27 273)L 300K 1 rms1 = =  = + = − P1 =1atmVrms2 = ?,T2 =127C= (127+ 273)K = 400K P2 = 2atm From 2 3 400 300 2 T T P P V V ; T P V T P V 2 1 1 2 2 1 2 2 2 1 1 1 = = =  Again rms1 2 1 1 v V M 3 1 P = and 2 rms 2 2 v V M 3 1 P = 1 2 2 1 rms 2 rms 2 P P V V V V 1 2   = 3 2 (100) 2 V V P P V v 2 1 2 1 2 rms 2 rms 2 2 = 1   =   1 rms ms 3 200 V 2 − = 27. (d) Molecules of an ideal gas move randomly with different speeds. 28. (d) M 3RT Vrms = % increase in 100 M 3RT M 3RT M 3/ RT V 1 2 1 rms  − = 100 300 400 300 100 T T T 1 2 1  −  = − = 100 15.5% 17.32 20 17.32  = − = 29. (a) The wok done is zero in this process since volume is kept fixed. 30. (a) For a diatomic gas, Molar heat capacity at constant pressure, R 4 7 Cp = Molar heat capacity at constant volume R 2 5 Cr = 7 / 5 5 / 2R 7 / 2R C C V P  = = 31. (a) We treat water like a solid. For each atom average energy is 3k T. B Water molecular has three atoms, two hydrogen and one oxygen. The total energy of one mole of water is U = 33kBT NA = 9RT         = A B N R k  Heat capacity per mole of water is 9R T U T Q C =   =   = 32. (c) For a monoatomic gas like helium 3 5  He = For a diatomic gas like oxygen 5 7 O2  = (3 2) 3 O 2 e mix 2 +   +    =  1.5 15 5 113 5 3 10 5 21 5 2 5 2 5 7 3 =  = + =   = 33. (c) Since one mole of any ideal gas at NTP occupies a volume of 22.4 litre. Therefore, cylinder of fixed capacity 44.8 litre must contain 3 moles of helium at NTP. For helium, R 2 3 CV = (monoatomic)  Heat needed to raise the temperature, Q = number of moles × molar specific heat × raise in temperature.
15 45R 2 3 = 2  = = 458.31J = 373.95J 34. (a) Let 1 and 2 represent for Argon atom and Helium atom. rms speed of Argon 1 1 rms M 3RT v 1 = rms speed of Helium , 2 2 rms M 3RT V 2 = According to questions, 1 2 Vrms = Vrms 2 2 1 1 M 3RT M 3RT  = 2 2 1 1 M T M T = or 39.9 4 253 M M T T 1 2 2 1 =  =  2.52 10 K 3 =  35. (c) Here, ,c 3J g K ,M 4 2 1 n 1 1 = v = = − − 1 1 CV Mcv 4 3 12mol K − −  = =  = At constant volume P  T. 2 1 1 2 1 2 2,T 2T T T P P  = = = Rise in temperature T = T2 −T1 = 2T1 −T1 = T1 = 273K Heat required, Q = nCVT 12 273 1638J 2 1 =   = 36. (a) In a process = x PV constant, molar heat capacity is given by 1 x R 1 R C − +  − = As the process is V P = constant, i.e., = −1 PV constant, therefore, x = −1. For an ideal monoatomic gas, 3 5  = 2R 2 R R 2 3 1 ( 1) R 1 3 5 R C = + = − − + −  = Q nC( T) 1(2R)(2T T ) 2RT .  =  = 0 − 0 = 0 37. (d) n 2  =1+ 38. (a) For a gas, we know 1 C R V =  − or 0.67 =  −1 or,  =1.67 Hence the gas is monoatomic. 39. (b) 120 480 T T V V 1 2 rms rms 1 2 = = V 2(V ) rms = rms1 40. (c) 2 3 41. (d) Here, P1 =1atm,T1 = 27C = 27 + 273= 300K 2 3 P 8P ,T ?, 2 = 1 2 =  = As change are adiabatic, − − − −  = 2 1 1 2 1 P1 T P T 1 2 1 1 2 P P T T − −         =         (1.5 1)/1.5 1/3 1 1 2 2 1 300(8) 300(8) P P T T = =         = −  − T2 = 600K = (600− 273)C = 327C 42. (d) As , V nRT P = it (i.e., P) remains unaffected as n,R. T and V. 43. (d) The pressure on EFGH would be half that on ABCD. 44. (b) Boyle’s law is applicable to an isothermal process where temperature remains constant . 45. (c) Pressure = force/are = Mg / area of the piston = constant.If the temperature is increased , only the volume increases as the piston moves up without frictions. Recall V  T at constant pressure. 46. (a) According to Charle’s law V  T or T V = constant = P 1 As in graph, slope at P2 is more than slope at P , 1 P1  P2 47. (d) According to gas equations PV = Nrt or 1 2 2 1 1 2 1 1 1 2 2 2 T T , V V P P or T P V T P V = = Here, T2 = 3000K,T1 = 300K Since H2 splits into hydrogen atoms, therefore volume becomes half i.e., 2 V1 2 1 V =