Tiêu đề 32. Kinetic Theory of Gases Hard and Answer.pdf ✅

1. (d) As (8.99 10 ) (3180) 3.03 10 N/m 3.0atm 3 1 v 3 1 P 2 2 2 5 2 =  rms =   =  = − 2. (c) Due to increase in temperature root mean square velocity of gas molecules increases. So they strike the wall more often with higher velocity. Hence the pressure exerted by a gas on the walls of the container increases. 3. (d) Kinetic energy E = J 5 1.5  10 , volume V = 20 litre = 3 3 20 10 m −  Pressure V E 3 2 = 6 2 3 5 5 10 / 20 10 1.5 10 3 2 =  N m           = − . 4. (d) Mean square velocity of molecule m 3kT = For gas A, x component of mean square velocity of molecule 2 = w  Mean square velocity m kT w 3 3 2 = = .....(i) For B gas mean square velocity m kT v 2 2 3 = = .....(ii) From (i) and (ii) 1 3 2 2 2 = v w so 3 2 2 2 = v w . 5. (a) 3 3 V 10 m − = , 22 N = 3.0 10 , m kg 26 5.3 10 − =  , v m s rms = 400 / 2 3 26 22 2 rms (400) 10 5.3 10 3.0 10 3 1 v V mN 3 1 P − −    = =  . 4 2 = 8.4810 N/m 6. (c) 2 3 1 rms v V M P =  V MT P  At constant volume and temperature, if the mass of the gas is doubled then pressure will become twice. 7. (c) From ideal gas equation PV = RT we get 6 2 3 1 1 1 1 1 2 1 2 1 2 =                 =                 = V V P P V V P P T T  T2 = 6T1 = 6  300 = 1800 K = 1527 C. 8. (c) From PV = RT we get 5 9 0.5 1 300 270 2 1 1 2 1 2  =            =                 = P P T T V V  3 2 900 5 9 V = 500  = m 9. (c) From PV = RT we get 4 1 2 / 2 1 1 1 1 2 1 1 2 1 2 =                 =                 = V V T T V V T T P P  4 1 2 P P = 10. (d) Quantity of gas in these bulbs is constant i.e. Initial No. of moles in both bulb = final number of moles ' 2 ' 1 +  2 = 1 +  ( ) 1.5 (273) 1.5 (273) (273) R T PV R PV R PV R PV + = +  T 1.5 273 1.5 273 2 = +  T = 819 K = 546 C . 11. (d) Number of moles in first vessel 1 1 1 RT P V  = and number of moles in second vessel 2 2 2 RT P V  = If both vessels are joined together then quantity of gas remains same i.e  = 1 +  2 2 2 1 1 (2 ) RT P V RT P V RT P V = + 2 2 1 1 2 2T P T P T P = + 12. (b) 3 3 V1 2.4 10 m − =  , 2 5 1 0 10 m N P = P = and T1 = 300 K (given) If area of cross-section of piston is A and it moves through distance x then increment in volume of the gas = Ax and if force constant of a spring is k then force F = kx and pressure = A kx A F = 3 3 3 2 1 2.4 10 8 10 0.1 3.2 10 − − − V = V + Ax =  +   =  and 5 3 5 2 0 2 10 8 10 8000 0.1 10 =    = + = + − A kx P P From ideal gas equation 2 2 2 1 1 1 T P V T P V =  2 5 3 5 3 2 10 3.2 10 300 10 2.4 10 T − −    =    T2 = 800 K 13. (a) P1 T1 V Initially P2 T2 V P T V P T V Finally
Ideal gas equation, in terms of density = = 2 2 2 1 1 1 T P T P   constant  1 2 2 1 2 1 T T P P =     76 75 280 300 76 70 Top Bottom Bottom Top Bottom Top =  =  = T T P P   14. (c) At low pressure and high temperature real gas obey PV = RT i.e. they behave as ideal gas because at high temperature we can assume that there is no force of attraction or repulsion works among the molecules and the volume occupied by the molecules is negligible in comparison to the volume occupied by the gas. 15. (d) Because there is zero attraction between the molecules of ideal gas. 16. (a) Root means square velocity m s M RT vrms 1930 / 3 = = (given)  kg gm RT M 2 10 2 1930 1930 3 8.31 300 (1930 ) 3 3 2 =  =    = = − i.e. the gas is hydrogen. 17. (a) As M RT vrms 3 =  4 2 / / = = = A B A B B A M M T T C C       As = 4 given B A B A M M T T 18. (c) Root mean square velocity does not depends upon the quantity of gas. For a given gas and at constant temperature it always remains same. 19. (b) M RT vrms 3 =  2 2 2 2 2 2 O O H H O H T M M T v v =   900 32 2 1930 300 2 =  vO  vO 836 m/s 4 1930 3 2 =  = . 20. (a) For oxygen 2 2 2 3 O O O M RT v = and For hydrogen 2 2 2 3 H H H M T v = R According to problem 2 2 3 O O M RT = 2 2 3 H H M T = R  2 2 2 2 H H O O M T M T =  32 2 47 273 H2 T = +  TH 2 20 K 32 320 2 =  = . 21. (c) If a lorry is moving with constant velocity then the rms v of gas molecule inside the container will not change and we know that 2 rms T  v . So temperature remains same. 22. (d) 5 2 3 4 5 6 5 2 2 2 2 2 2 5 2 4 2 3 2 2 2 1 + + + + = + + + + = v v v v v vrms 20 4.24 5 100 = = = 23. (b) 2 3 1 rms v V mN P =  2 rms P  m v so 2 / 2 2 2 1 1 1 1 2 1 2 1 2 1 2 =         =         =  v v m m v v m m P P  P2 = 2P1 = 2P0 24. (c) E  T  1 2 1 2 T T E E =  (20 273) 2 2 1 1 + = T E E  T2 = 293  2 = 586 K = 313 C . 25. (a) Kinetic energy per degree of freedom kT 2 1 = As diatomic gas possess two degree of freedom for rotational motion therefore rotational K.E. kT  = kT      = 2 1 2 In the problem both gases (oxygen and nitrogen) are diatomic and have same temperature (300 K) therefore ratio of average rotational kinetic energy will be equal to one.
26. (a) The rms speed depends upon the molecular mass M vrms 1  but kinetic energy does not depends on it 0 E  M In the problem m1  m2  m3  1 2 3 ( ) ( ) ( ) rms rms rms v  v  v but ( ) ( ) ( ) K1 = K2 = K3 27. (d) 3 8.31 273 3.4 10 2 3 2 3 E = RT =   =  Joule 28. (c) Average translational kinetic energy does not depends upon the molar mass of the gas. Different gases will possess same average translational kinetic energy at same temperature. 29. (d) E  T but vrms  T i.e. if temperature becomes twice then energy will becomes two time i.e. 2  6.21  10–21 = 12.42  10–21 J But rms speed will become 2 times i.e. 484  2 = 684 m/s. 30. (b) Kinetic energy of N molecule of gas E NkT 2 3 = Initially 1 1 1 2 3 E = N kT and finally 2 2 2 2 3 E = N kT But according to problem E1 = E2 and N2 = 2N1  1 1 1 2 (2 ) 2 3 2 3 N kT = N kT  2 1 2 T T = Since the kinetic energy constant 1 1 2 2 2 3 2 3 N kT = N kT  N1T1 = N2T2  NT = constant From ideal gas equation of N molecule PV = NkT  P1V1 = P2V2  P1 = P2 As 1 2 [ V = V and NT = constant] 31. (b) Average speed of gas molecule m kT v av  8 = . It depends on temperature and molecular mass. So the average speed of oxygen will be same in vessel A and vessel C and that is equal to V1 . 32. (a) As V  T  1 2 1 2 T T V V =  2 1 293 313 V  V      = Fraction of gas comes out 0.07 293 293 20 313 1 1 1 1 2 1 = =  −      = − = V V V V V V . 33. (c) In the given graph line have a positive slop with X-axis and negative intercept on Y-axis. So we can write the equation of line y = mx – c...... (i) According to Charle’s law 0 0 273 t V V Vt = + , by rewriting this equation we get 273 273 0 −         = Vt V t ......(ii) By comparing (i) and (ii) we can say that time is represented on Y-axis and volume in X-axis. 34. (b) When temperature of gas increases it expands. As the cross-sectional area of right piston is more, therefore greater force will work on it (because F = PA). So piston will move towards right. 35. (c) From ideal gas equation PV = RT .....(i) or PV = RT .....(ii) Dividing equation (ii) by (i) we get T T V V  =   = =    V T T V 1 (given)  T 1  = . So the graph between  and T will be rectangular hyperbola. 36. (d) From Gay Lussac’s law 273 373 0 273 100 273 T T P P 1 2 1 2  =      + + = =  P 760 1038 mm 273 373 2   =      = . 37. (a) P1 = P , T1 = T , P2 = P + (0.4% of P) 100 250 0.4 P = P + P = P + T2 = T + 1 From Gay Lussac's law 2 1 2 1 T T P P =
 1 250 + = + T T P P P [As V = constant for closed vessel] By solving we get T = 250 K. 38. (a) From ideal gas equation PV = RT  T V R P  = Comparing this equation with y = mx Slope of line V R m  tan  = = i.e. tan  1 V  It means line of smaller slope represent greater volume of gas. For the given problem figure Point 1 and 2 are on the same line so they will represent same volume i.e. V1 = V2 Similarly point 3 and 4 are on the same line so they will represent same volume i.e. V3 = V4 But V1  V3 (= V4 ) or V2  V3 (= V4 ) as slope of line 1-2 is less than 3-4. 39. (d) As argon is a monoatomic gas therefore its molecule will possess only translatory kinetic energy i.e. the share of translational and rotational energies will be 100% and 0% respectively. 40. (d) Mean kinetic energy for  mole gas RT f 2 = .  E RT 2 7 =  NkT M m 2 7       = NkT      = 2 7 44 1 NkT 88 7 = [As f = 7 and M = 44 for CO 2 ] 41. (c) Given velocity of sound sec 330 m v s = , Density of gas 3 1.3 m kg  = , Atomic pressure 2 5 1.01 10 m N P =  Substituting these value in   P vsound = we get  = 1.41 Now from f 2  = 1 + we get 5. 1.4 1 2 1 2 = − = − =  f 42. (b) As d P kT 2 2 1   =   1 P  i.e. by increasing  two times pressure will become half. 43. (b) Mean free path  = 0.8  10–7 m number of molecules per unit volume 25 n = 2.7  10 per m 3 Substituting these value in 2 2 1 nd  = we get d m 19 10 1.04 10 3.2 10 − − =  =  = 3.2 Å 44. (a) For polyatomic gas ratio of specific heat  < 1.33 Because we know that as the atomicity of gas increases its value of  decreases. 45. (c) By comparing with relation − 1 =  R Cv we get  − 1 = 0.67 or  = 1.67 i.e. the gas is monoatomic. 46. (a) At constant pressure (Q)p =  CpT = 1  Cp (30 − 20) = 40  mole kelvin calorie Cp = 4  Cv = Cp − R mole kelvin calorie  = 4 − 2 = 2 Now (Q) = C T = 1  2  (30 − 20) v  v = 20 calorie 47. (c) Specific heat at constant volume 2 3 1 R R Cv = − =  (given)  3 2  − 1 =  3 5  = . 48. (b) P T  V = constant