SOLUTIONS-Heat And Thermodynamics-PART TEST-I-XIth

SOLUTIONS TO PART TEST - 1 (PT-1) TOPIC : HEAT & THERMODYNAMICS 1. In an ideas gas internal energy = f nRT 2 U = 5 × 2RT + 4 × 2 3 RT = 11 RT. 2 2. Higher is the temperature greater is the most probable velocity. 3. According to law of equipartition of energy, energy equally distributes among its degree of freedom, Let translational and rotational degree of freedom be f1 and f2.  KT  3 and K + K = U KR 2 T R Hence the ratio of translational to rotational degrees of freedom is 3:2. Since translational degrees of freedom is 3, the rotational degrees of freedom must be 2.  Internal energy (U) = 1 × (f + f2) × 1 RT 2 1 5  8.3 100 U = 2 4. PV = n RT = U = 2075 J  12  4  8.31 300 nRT   P = V =  4 2  20 103 = 6.25 × 105 Pa  5. Total KE = U = 2 nRT In case of H2 degree of freedom is greatest and number of moles n is highest So this is the case of maximum kinetic energy. 6. P = P T = T  V1  V2 2     = 1 2 1 2 n1 n2 n1 n2  M1(2 – ) = M2  or  = 2M1 M1  M2 = 16 15 7. When speed of 5 molecules which are selected randomly are recorded, then the average is most likely to be equal to the most probable speed.  The average of these values is most likely equal to . TF  32 8. 180 TK  273 = 100   32 = 180   273 = 100  5 – 32 × 5 = 9 – 273 × 9 =  = 574.25 9. For minimum value of m, the final temperature of the mixture must be 0°C.  20 × 1 × 10 + 20 × 80 = m 540 + m.1. 100 2  m = 1700 640 = 85 32 gm 10. Applying energy conservation : mgh = mLf Lf  h = = 80cal / gm = 80  4.2  1000 J/kg  336  103  N  s2 = = 33.6 km. g 10m/ s2 10 m/s2  10  kg 11. Work done by atmosphere = P atm V = P V (i) atm 2 Initially gas in container is in thermodynamic equilibrium with its surroundings.  Pressure inside cylinder = Patm & PV = nRT  PatmV = nRT or V = nRT Patm Putting in (1), W = 12. As p  V2 T  V3 nRT 2 i.e. if temperature increases, volume also increases hence work done will be positive. 13. For process at constant pressure 5 2 Q = nCp T = 2 nR T and W= PV = nRT = 5 Q 14. Net power given to N2 gas = 100 – 30 = 70 cal/s The nitrogen gas expands isobarically.  Q = n Cp dT dt or 70 = 5 x 7 R dT  2 dt dT dt = 2k/sec 15. As PV = nRT  P V  For n = constant :   T  = constant  for all changes. Hence (C) 16. By symmetry AB = BC & AD = DC  No current in BO and OD  TB = TO = TD dQ 17. dt = – KA (1 – 2) [Newton’s law of cooling]  Rate of loss of heat is same from dQ = ms d  d =  KA ( – ) dt dt dt ms s Since, ‘m’ is larger for sphere , rate of cooling will be less for sphere. 18. The colour of an object indicates the rate at which energy is emitted and hence indicates the temperature. 19. Heat radiated (at temp same temp)  A  Q  4R2 and Q'  (4R2 + 2 × R2) Q'  6R2  Q 4R2 = 1.5 Here R2 is extra surface area of plane surface of one of the hemisphere. 20. assuming rate of heat transfer to be constant. s (m + w) T = Q where s = specific heat of water = 1 g/cc s (m + w)T = Q w = water equivalent of calorimeter Q1  Q2 (50  w)s = (100  w) s 50  w = 100  w Q1 t1 10 50  w but Q2 = t2 = 18 = 100  w 50 = 4w w = 12.5 g 21. Thermal resistance R = 𝑙 KA for same temperature difference, thermal current  1 R  i1 = R2 = i2 R1 𝑙 2 K 2 A 2 / 𝑙1 K1A1 𝑙 2 = 𝑙1 A1 K1 A 2 K 2 1 12 1 = 2 x 22 x 1 = 1/8 22. Since elasticity of balloon is negligible, pressure inside ballon –~ pressure outside baloon = Patm.  W = P V Vin = 10 litre. Vin Vfinal  VinTfinal   Tfinal   58  =  V =   litre.  W = P V   1  105 × 10–2   = 200 J Tin Tfinal final  Tin  atm in  Tin   290  23. U in a – b process = 0 (isothermal process) in b – c process Q = 0 (adiabatic process)  U = – W = – 4 J U in cyclic process = 0  U in c – a process = 4J 24. Let initial temperature and volume be T0 and V0. Since the process is adiabatic, the final temperature and volume is TV –1 = T V  –1 ( = 5 for monoatomic gas) 0 0 3 2  V0  3  T = T0  V / 8  = 4T0  0   percentage increase in temperature of gas is 4 × 100 – 100 = 300% 25. Let initial pressure, volume, temperature be P0, V0, T0 indicated by state A in P-V diagram. The gas is then isochorically taken to state B (2P0, V0, 2T0) and then taken from state B to state C (2P0, 2V0, 4T0) isobarically. Total heat absorbed by 1 mole of gas Q = C (2T – T ) + C (4T – 2T ) = 5 R T + 2 0 7 R × 2T = 2 0 19 RT 2 0 Total change in temperature from state A to C is T = 3T  Molar heat capacity = Q T = 19 RT0 2 3T0 = 19 R. 6  1/  V  1/ 2 1 T2   P2  2rms   T2  V2rms  P2  2 26. T  P  and V  T     V P 1  1  1rms  1  1rms  1  1.41 = 221.4 = 21/7 27. Vr.m.s. = 28. (B) = 3.5 . PV = nRT P(aT – b) = nRT  P = nRT aT – b nR =  b  a – T    a – b  now as T increases then    increases hence P decreases  29. At equilibrium : dQ1 = dt dQ2 dt KA(100  )  t = 4nk An(  0) 2n2t   = 100/3. 30. Moles in ‘A’ initially & finally will be same PA (L / 3)  P Ax .... (1) RT RT Moles of ‘B’ remain same 2PA (L / 3)  PA (L  2x) .... (2) RT Divide (1) by (2) R T 1  x  x = L/4 2 L  2x SOLUTIONS TO PART TEST - 2 (PT-2) TOPIC : HEAT & THERMODYNAMICS 1. External Work = work done on both gases = – (work done by the gases)  4 / 3 V0 2 / 3 V0  8 9 = – P0 V0𝑙n  V0  P0 V0𝑙n  0  = – P0V0𝑙n 9 = P0V0𝑙n 8 2. B = – p  dv   v    dv = – pv  vT = pv  T = p 3. K1 0°C 70  0 B 70°C 100  70 B B K2 100°C R2 K1 R1 = R2  R1 = 3/7  K 2 = 3/7 0°C K2 T K1 100°C T  0 R2 100  T = R1  100  T = R1 T R2 K 2 = K1 = 7/3  300 – 3 T = 7T  T = 30°C 4. n = 7v – v2 – 10 for most probable velocity dn  0 dv  n is maximum at this velocity 7–2v = 0  v = 5. As dQ = msdT 7 = 3.5 m/sec 2 dQ  ms dT dt dt From question : S  T or S = K1T. (K1 being proportionality constant) d Also, dt dT = constant = K2 (say)  ms dt dT = K2  m(K1T) dt = K2  K1  T2  m   K 2  2 = t  T  Hence, the graph will be parabolic 6. AC = 3V0 AB = 3V0 cos 37º in triangle ABN AN = 3V cos2 37º P0 P = V0 V P0 PB = V0 (V0 + 3V0 cos2 37º) P0 V  3V  16  PB =  0 V0  0  25  1 48  =    25  = P0(73/25) 7. Wein's displacement law is :  .T = b 1 i.e. T  m Here,  becomes half.  Temperature doubles. Also E = T4 E  T 4  1   1  E2  T2   T 4  E =  2   T1  .E1 = (2)4 . 16 = 256 J m–2 s–1 8. dQ  KAT T 10 = = J /sec. dt 2𝑙 2𝑙 120 KA dQ New rate dt T = 𝑙 2KA 20 = t J /sec. ; 20 So time taken is t = 40 × 120 sec. = 60 sec. 9. PV = m RT M0  RT P = M0 P  R T  M0 Slope of the curve  Temperature Hence cd and ab are isothermal processes.   1 V i.e. bc and da are constant volume process (A) and (B) are true. Temp. in cd process is greater than ab. Net work done by the gas in the cycle is negative, as is clear by the PV-diagram. 10. Since , e = a = 0.2 (Since, a = (1 – r – t) = 0.2 for the body B) E = (100) (0.2) = 20 W/m2 Power emitted = E.A = 20 × 10 = 200 Watt 11. Heat obviously flows from higher temperature to lower temperature in steady state.  A is true. Temperature gradient  1 in steady state.  B is false. cross sec tion area Thermal current through each cross section area is same.  C is true. Temperature decreases along the length of the rod from higher temperature end to lower temperature end.  D is false. 12. no. of collisions per unit area  1 time between two collision  area Vrms   n• /A  Vr m s distance b/w walls  A • distance b/w walls × A = volume So A • if both T and V are halved, increases A T1/ 2 , V n•  P A T1/ 2 13. An isothermal process is an example to show the possibility "for both the pressure and volume of a monoatomic ideal gas to change simultaneously without causing the internal energy of the gas to change". Hence Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. 14. For monoatomic gas the degrees of freedom are only translational and hence do not depend on temperature. Therefore molar heat capacity of an ideal monoatomic gas at constant volume is a constant at all temperatures. 15. Heat energy stored in 1 g of steam at 100°C is more in comparison to 1 g of water at 100°C. Hence burns caused by steam are more severe. 16. v rms = , if the absolute temperature T doubles, v rms must increase by (m and k are constants). 17. In 40 min. temperture of water has come down by 40°C. mS  T Therefore rate P = t 0.60  4200  40 = 40  60 = 42.0 W 18. Sample of ice has been receiving heat at constant rate P from water. Its temperature has increased by 30°C in time 60 min. Therefore mi si Ti P = 60 min.  mi = Alternate : (60  60s)  (42 W) (2100J / kg).(30C) = 2.4 kg mwSwTw + m'wL = miSi 30° (m'w = mass of water which is converted to ice) 19. Thermal equilibrium reaches after 60 min. Ice conversion takes place for 20 min. During this time water at 0°C continues to give heat at rate P. m'w × Lf = P × (20 × 60s)  m'w = 42  20  60 3.3  105 kg = 0.15 kg 20. W = Area under the curve = 3 P V 2 1 1 V2  2 V1 and P1V1 = nRT1 3 . P V P2  2P1 W Therefore nRT1 = 2 1 1 P V 1 1 21. Q = dU + W dU = nCv dT For final state P2V2 = 2P1 2V1 = 4P1V1 = nR(4T1) Hence final temp. is 4T1 dU = n . 3 R . 3T = 2 1 9 nRT 2 1 Q = 3 . nRT + 2 1 9 nRT 2 1 = 6nRT1 Q nRT1 = 6 22. nC T = Q  nCT = 6n RT1 dT = 4T1 – T1 = 3T1 n . C . 3T1 = 6nRT1 C = 2 R 23. In (A), V is on vertical axis. As V is icreasing, W is positive. As negative work in part-II is greater than positive work in part-I, net work during the process is negative. Using PV = nRT and as Vremains same for initial and final points of the process, it is obvious that final temp. is greater than initial temperature as pressure has increased. Therefore dU is positive. Hence option (S) is connected with (A). Similar arguments can be applied to other graphs. 24. (A) Initially more heat will enter through section A but the metal will absorb some heat and less heat will leave from C. (B) At steady state heat accumulatron = ) dQ  dt is same for all sections dQ 1  dQ  (C) At steady state = kA or =   dt kA  dt  is inversely proportional to area of cross-section. Hence is maximum at B and minimum at C (D) At steady state heat accumulation = 0 dT So dt = 0 for any section. SOLUTIONS TO PART TEST - 3 (PT-3) TOPIC : HEAT & THERMODYNAMICS 1. Heat released by steam inconversion to water at 100°C is Q1 = mL = 4 × 540 = 2160 cal. Heat required to raise temperature of water from 46°C t 100°C is Q = mS  = 20 × 1 × 54 = 1080 J Q1 > Q2 Q1 and Q2 = 2 Hence all steam is not converted to water only half steam shall be converted to water  Final mass of water = 20 + 2 = 22 gm 2. Q =  U + W WAB = 0 AB = AB =  nR  T  2 5 ( PV) 2  ( PV) 2 QAB = 2.5 P0 V0 Process BC BC BC BC QBC = 0 + 2P0 V0 𝑙n 2 = 1.4 P0 V0 Qnet = QAB + QBC = 3.9 P0 V0 3. Y = Stress Strain = T / A 𝑙 / 𝑙 T = Y. 𝑙 A 𝑙 = Y.A  T In both the rods tension will be same so T1 = T2 Hence Y A  T = Y A  T A1 A 2 Y22 = Y11 4. V2 rms = < V2 > =  V2dN V 2  V 2  V 2  ........... N dN = dN here dV = N(V) 2 rms  1 N(V) V2 dV 0 1 V0  3N  3 =  0 .V2  V2 dV = V 2  V = .V . N0  V 3  5 0 rms 0 0  0  2𝑙 d𝑙 5. As ; d𝑙 =  𝑙 dT  𝑙 𝑙 T = aT dT 0 T2  𝑙n4 1/ 2 𝑙n2 = a 2  T =  a  6. For a block body, wavelength for maximum intensity :   1 T & P  T4  P  1 4  P = 16 P. ∴ P T = 32PT  P’ = 16 P. 7. PV = nRT PdV = nRdT  = 1 dV and dV  nR  = 1 V dT dT P T 1 For given temperature T ,  = T0 8. The rate of heat loss by a thin hollow sphere of thickness ‘x’ , mean radius ‘r’ and made of density ‘’ is given by dT mS dt = –  A(T4 – T 4 ) ( 4r2x) S dT =    4r2 (T4 – T 4 ) dt  dT dt  (T 4  T 4 ) = – 0 Sx is independent of radius Hence rate of cooling is same for both spheres. P2 9. Equation of process   = constant = C (1) Equation of State P  R T (2)  M From 1 and 2 PT = constant  C is false, D is true.  As -changes to 2 Hence T changes to  P changes to 2T .  B is true. from equation (1)  A is false. A 10. A × 100 =  𝑙  2 A  × 100  % increase in Area = 2 × 0.2 = 0.4   V V × 100 = 3 × 0.2 = 0.6 % Since l = l  T 𝑙 × 100 =  T × 100 = 0.2   = 0.25 × 10–4 / ºC 𝑙 11. Area under the curve is equal to number of molecules of the gas sample. Hence N = 1 . a . V 2 0   aV0 = 2N V0  a  V V = 1 v N(V)dV = 1 C . .V dV = 2 V  avg 2 = avg N 0 N 0  v0  3 0 V0 3  v0  a  2 V 1 2 r ms = 1 V 2 N(V)dV N 0 = 1 V 2  N 0  V0 .V  dV = V0  2 rms V0 = Area under the curve from 0.5 V0 to V0 is 3 of total area. 4 12 V r.m.s. = Since PV = nRT therefore P and V both can change simultaneously keeping temperature constant. 13. Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1 14. From Wein's law  T = constant i.e., peak emission wavelength   1 . Hence as T increase  decreases. m m T m 15. to 17. TP(1 – /) = K. (given) dT Lapse Rate is dy , therefore differentiating w.r.t. y 1    1    dT  1      1 dP dT T  1    dP . P  T. dy  .P  .  0  dy    dy P  .   dy Using PV = M RT M  P =  RT M T  M P R and dP dy = – g dT M    1  1  Mg  . dy R  ( g)   1   R     dT using given value dy = – 9.8 K/km. 18. (A) PV = nRT P = (nRT) 1 = (constant) 1 V V T = constant i.e. isothermal process  V increases, W is positive and Q = U + U = W > 0 (B) Q = 0 pdV = W = positive (C) PV = nRT As volume increases, T also increases i.e., U > 0 PdV = W > 0 So Q > 0 (D) For cyclic process U = 0 W < 0 (anticlockwise) Q < 0. 𝑙 1 dx 19. dR =  k (r  x tan )2 O 1 𝑙 Solving R = k  r1 r2  G = 1 = R k  r1 r2 𝑙 20. Due to the heating pressure inside is not changed. Let inside pressure be . Then for equilibrium of the system : P(A1 – A2) = P0(A1 – A2) + (m1 + m2)g  PV = (p0A + mg)𝑙 𝑙 is displacement of the piston. P . V = nRT. 1g T = PV nR = (p0A  mg)𝑙 nR PA1 T PA2 (105Pa x 103 m2  5 x 10)(50 102 ) = 1 x R m2g P0A2 T = 75 K. R 21. The maximum temperature of the gas will be during process BC. Process BC can be represented by straight line, y = m x + C So P = m V + C Putting point B & C gives 3 P = 2 m V + C  (1) P = 6 m V + C  (2) P So subtracting 2 P =  4 m V So m =  2 V P From (2) P =  2 V . 6 V + C  C = 4 P  P  Hence we get equation as, y =  2 V  x + 4 P  (4)   where y is pressure and x is volume of gas. Putting y from above, P now we have x y = n R T  2 V x2 + 4 P x = n R T  (3) d T For maximum temperature d x Hence x = 4 V = 0   2 V x + 4 P = 0 8 PV Putting in (3) we get n R Tmax = 2 P (4 V) = 8 P V So, Tmax = n R 22. Before heating let the pressure of gas be P1 from the equilibrium piston, PA = kx1 PA  nRT  A 1 8.3  100 102  x = K =   =   K 0.83  200 = 0.05 m Since during heating process, The spring is compressed further by 0.1 m  x = 0.15 m work done by gas = 1 .200((0.15)2 – (0.05)2) = 2 1 .200 (.02) = 2J 2