Thermodynamics and Statistical Mechanics NET-JRF PDF 2011

fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics THERMODYNAMICS AND STATISTICAL PHYSICS NET/JRF (JUNE-2011) Q1. Consider the transition of liquid wa...

fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics THERMODYNAMICS AND STATISTICAL PHYSICS NET/JRF (JUNE-2011) Q1. Consider the transition of liquid water to steam as water boils at a temperature of 1000 C under a pressure of 1 atmosphere. Which one of the following quantities does not change discontinuously at the transition? (a) The Gibbs free energy (b) The internal energy (c) The entropy (d) The specific volume Ans. : (a) Solution: In first order transition Gibbs free energy is continuous. Q2. A particle is confined to the region x  0 by a potential which increases linearly as u x   u 0 x. The mean position of the particle at temperature T is k BT k BT (b) k B T  / u 0 2 (a) (c) (d) u 0 k B T u0 u0 Ans. : (a) p2 u x 1   0 Solution: Partition function Z   e 2 mkBT dp  e kBT dx and x   xp  x dxdpx h  0 x 2  p2 u0 x   k BT   xe  te k BT t   dx  u  dt  xe dp  e 2 mk BT k BT dx k BT  x   0  x  0 0    p2 u x  0   0  k BT  u0 e e t  e dt k BT 2 mk BT dp  e k BT dx dx  u  0 0 0 Q3. A cavity contains blackbody radiation in equilibrium at temperature T. The specific heat per unit volume of the photon gas in the cavity is of the form CV  T 3 , where  is a constant. The cavity is expanded to twice its original volume and then allowed to equilibrate at the same temperature T. The new internal energy per unit volume is T 4 (a) 4T 4 (b) 2T 4 (c) T 4 (d) 4 Ans. : (d) T 4 Solution: du  C v dT   T 3dT  u  4 H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 252  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics Q4. Consider a system of N non-interacting spins, each of which has classical magnetic moment of magnitude . The Hamiltonian of this system in an external magnetic field  N    H is   i.H , where  i is the magnetic moment of the i th spin. The magnetization per i 1 spin at temperature T is  2H   H  k B T  (a) (b)  coth     k BT   k B T  H   H   H  (c)  sinh  (d)  tanh   k BT   k BT  Ans. : (b) 2   H cos     cos exp kT sin  d d Solution: For classical limit M  0 0  H cos   exp kBT sin  d d   H  k BT  M   coth      k BT  H  Q5. Consider an ideal Bose gas in three dimensions with the energy-momentum relation   p s with s  0. The range of s for which this system may undergo a Bose-Einstein condensation at a non-zero temperature is (a) 1  s  3 (b) 0  s  2 (c) 0  s  3 (d) 0  s   Ans. : (a) NET/JRF (DEC-2011) bS 3 Q6. The internal energy E of a system is given by E  , where b is a constant and other VN symbols have their usual meaning. The temperature of this system is equal to 2 bS 2 3bS 2 bS 3 S (a) (b) (c) 2 (d)   VN VN V N N Ans. : (b)  E  3bS 2 Solution: TdS  dE  PdV  dE  TdS  PdV     T  T   S V VN H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 253  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics Q7. Consider a Maxwellian distribution of the velocity of the molecules of an ideal gas. Let Vmp and Vrms denote the most probable velocity and the root mean square velocity, respectively. The magnitude of the ratio Vmp / Vrms is (a) 1 (b) 2 / 3 (c) 2/3 (d) 3 / 2 Ans. : (c) 2kT 3kT V 2 Solution: For Maxwellian distribution Vmp  , Vrms   mb  m m Vrms 3 Q8. If the number density of a free electron gas in three dimensions is increased eight times, its Fermi temperature will (a) increase by a factor of 4 (b) decrease by a factor of 4 (c) increase by a factor of 8 (d) decrease by a factor of 8 Ans. : (a) 2  3N   2 3 N Solution: Fermi energy E F    , where is number density and g is degeneracy  4Vg  2m V 2 2 n n  TF1 n1 EF  TF K  TF     TF  n   3 2 3 3   1   4 since  8. V  TF2  n2  n2 1 Q9. A system of N non-interacting spin - particles is placed in an external magnetic field H. 2 The behavior of the entropy of the system as a function of energy is given by (a) S (b) S E  BH BH  B H E B H S (c) S (d)  B H E BH  B H E Ans. : (a) H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 254  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics S  N  U  N  U  N  U  N  U  Solution:  ln   ln   , where   H. S is symmetrical Nk 2N  2  2 N  2N  about E. Q10. A gas of N non-interacting particles is in thermal equilibrium at temperature T. Each particle can be in any of the possible non-degenerate states of energy 0, 2 and 4. The average energy per particle of the gas, when   1 , is (a) 2 (b) 3 (c) 2 / 3 (d)  Ans. : (a) 0  e  o  2 e2  4 e4 Solution: E1  0, E 2  2 , E 3  4 , Z  e 0   e 2   e 4   E  e 0   e 2  e 4 1 2e 2   4e 4  2 1  2....  4 1   4 .... 2  4 6  E      2 1  e  2   e  4  1  1  2....  1  4..... 111 3 where   1. Q11. A one-dimensional chain consists of a set of N rods each of length a. When stretched by a load, each rod can align either parallel or perpendicular to the length of the chain. The energy of a rod is  when perpendicular to it. When the chain is in thermal equilibrium at temperature T , its average length is (a) Na / 2 (b) Na  (c) Na / 1  e 2 / k BT   (d) Na 1  e 2 / k BT  Ans. : (c) Solution: Let n1 no. of rods are parallel and n2 no. of rods are perpendicular. Energy of rod when it is perpendicular   Energy of rod when it is parallel is . e      e e   P     and P    e      e   e  e  e   e   n1ae   n2 ae  Nae  Na Average length  n1aP     n2 aP             e e e e 1  e 2  Since P    P  so n2  N , n1  0. H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 255  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics Q12. The excitations of a three-dimensional solid are bosonic in nature with their frequency  and wave-number k are related by   k 2 in the large wavelength limit. If the chemical potential is zero, the behavior of the specific heat of the system at low temperature is proportional to (a) T 1/ 2 (b) T (c) T 3 / 2 (d) T 3 Ans. : (c) Solution: If dispersion relation is   k s , At low temperature specific heat  T 3/ s Q13. Gas molecules of mass m are confined in a cylinder of radius R and height L (with R  L ) kept vertically in the Earth’s gravitational field. The average energy of the gas at low temperatures (such that mgL  k BT ) is given by (a) Nk B T / 2 (b) 3Nk B T / 2 (c) 2 Nk B T (d) 5 Nk B T / 2 Ans. : (d) 1 H h3  Solution: Z  e dpx dp y dpz dxdydz   px2   p 2y   p z2 L mgz  Z e e e dpz  dx dy  e 2 mk BT 2 mk BT 2 mk BT k BT dpx dp y dz    0   mgL    3 3 mgz  mk T  1  e B k T  mk T  2 L  2 Z   R2  B 2   e k BT dz  Z   R 2  B 2     2   0  2    mg   k T   B  ZN  Z N ,  ln z 5 Nk BT  E  k BT 2  , since mgL  k BT T 2 H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 256  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics NET/JRF (JUNE-2012) Q14. Consider a system of non-interacting particles in d dimensional obeying the dispersion relation   Ak s , where  is the energy, k is the wave vector; s is an integer and A is constant. The density of states, N    , is proportional to s d d s 1 1 1 1 (a)  d (b)  s (c)  s (d)  d Ans. : (b) Solution: We can solve this problem with intuition for example   Ak 2 1 3 1 Density of state in 3-dimensional N(ε)     2 2 2 1 Density of state in 2-dimensional N(ε)     0 2 1 1 1 Density of state in 1-dimensional N(ε)     2 2 d 1 Density of state in d-dimensional, where   Ak  N      s s Q15. The number of ways in which N identical bosons can be distributed in two energy levels, is N  N  1 N  N  1 (a) N  1 (b) (c) (d) N 2 2 Ans. : (a) Solution: Number of boson  N , Number of energy level  g N  g 1 So number of ways to distribute N boson into g level is, W  cN  N  1 since g  2. Q16. The free energy of the gas of N particles in a volume V and at a temperature T is  F  Nk B T ln a0V k B T  5/ 2  / N , where a 0 is a constant and k B denotes the Boltzmann constant. The internal energy of the gas is 3 5 (a) Nk B T (b) Nk B T 2 2  (c) Nk B T ln a 0V k B T  5/ 2  /N  3 2 Nk B T  (d) Nk B T ln a0V / k BT  5/ 2  Ans. : (b) H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 257  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics  Solution: F  Nk B T ln a0V k B T  5/ 2  / N , F  U  TS , U  F  TS  F   F   F  dF   SdT  PdV      S or S     U  F T   T V  T V  T V F  Nk B T ln C T  a 0Vk B5 / 2 5/ 2 where C  N  F   F     Nk B ln CT 5/ 2   Nk B T  C 5 3/ 2 T  T   Nk B T ln CT 5/ 2 5  Nk B T    T V CT  T V 5/ 2 2 2  F  5  F  5 T   F  Nk B T  U  F  T     Nk B T.  T V 2  T V 2 Q17. A system has two normal modes of vibration, with frequencies 1 and  2  21. What is the probability that at temperature T , the system has an energy less than 41 ? [In the following x  e   1 and Z is the partition function of the system.] (a) x 3 / 2 x  2 x 2  / Z (b) x 3 / 2 1  x  x 2  / Z (c) x 3 / 2 1  2 x 2  / Z (d) x 3 / 2 1  x  2 x 2  / Z Ans. : (d) Solution: There is two normal mode so there is two degree of freedom.  1  1 Energy of harmonic oscillator is E   n1  1   n2   2.  2  2  1  1 E   n1  1   n2   21 where n1  0,1,2,3.... and n 2  0,1,2,3....  2  2 31 51 Ground state energy E  , first excited state energy E . Second excited state 2 2 71 energy E  which is doubly degenerate state so g  2 , other state have more 2 energy than 41. 3  1 5  1 7  1 PE  41   e  2 e  2  2e  2  x 3 2 1  x  2 x  2 where x  e   1. Z Z Q18. Bose condensation occurs in liquid He 4 kept at ambient pressure at 2.17 K. At which temperature will Bose condensation occur in He 4 in gaseous state, the density of which is 1000 times smaller than that of liquid He 4 ? (Assume that it is a perfect Bose gas.) (a) 2.17 mK (b) 21.7 mK (c) 21.7  K (d) 2.17  K H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 258  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics Ans. : (b) 2  N 3 Solution: For bosons T    V  Q19. Consider black body radiation contained in a cavity whose walls are at temperature T. The radiation is in equilibrium with the walls of the cavity. If the temperature of the walls is increased to 2T and the radiation is allowed to come to equilibrium at the new temperature, the entropy of the radiation increases by a factor of (a) 2 (b) 4 (c) 8 (d) 16 Ans. : (c)  8 5 k B4T 4  F   32 5 k B4  3 Solution: For Black Body, energy is given by F  V , S       VT. 45 2C 3  T V  45 C  3 3  S  T 3 , If temperate increase from T to 2T then entropy will incase S to 8S. NET/JRF (DEC-2012) Q20. The entropy of a system,  S  , is related to the accessible phase space volume  by S  k B ln E , N , V  where E , N and V are the energy, number of particles and volume respectively. From this one can conclude that  (a) does not change during evolution to equilibrium (b) oscillates during evolution to equilibrium (c) is a maximum at equilibrium (d) is a minimum at equilibrium Ans. : (c) Solution: Entropy is maximum at equilibrium. Q21. Let W be the work done in a quasistatic reversible thermodynamic process. Which of the following statements about W is correct? (a) W is a perfect differential if the process is isothermal (b) W is a perfect differential if the process is adiabatic (c) W is always a perfect differential (d) W cannot be a perfect differential H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 259  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics Ans. : (b) Solution: Work done is perfect differential in adiabatic process. Q22. The free energy difference between the superconducting and the normal states of a  material is given by F  f S  f N      where  is an order parameter and 2 4 2  and  are constants s.t.   0 in Normal and   0 in the super conducting state, while   0 always, minimum value of F is 2 2 3 2 5 2 (a)  (b)  (c)  (d)   2 2 2 Ans. : (b)  F 4 Solution: F        2    2 4 3 2  2  2   2   0    3 2   2  2 2 Putting the value, F     2  F   2  min 2 Q23. A given quantity of gas is taken from the state A C reversibly, by two P A paths, A C directly and A B C as shown in the figure. During the process A C the work done by the gas is 100 J and the heat absorbed is 150 J. If during the process A B C the work done by the B C gas is 30 J , the heat absorbed is V (a) 20 J (b) 80 J (c) 220 J (d) 280 J Ans. : (b) Solution: During path AC , dU  dQ  dW  150  100  50 J Since, internal energy is point function, so dU will same in all path In path ABC , dQ  dU  dW  50  30  80 J. H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 260  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics Q24. Consider a one-dimensional Ising model with N spins, at very low temperatures when almost all spins are aligned parallel to each other. There will be a few spin flips with each flip costing an energy 2 J. In a configuration with r spin flips, the energy of the system is E   NJ  2rJ and the number of configuration is N C r ; r varies from 0 to N. The partition function is N N N  J   J   J  (a)   (b) e  NJ / k BT (c)  sinh  (d)  cosh   k BT   k B T   k B T  Ans. : (d) Solution: Let us consider only three energy levels, E r  2 J  2rJ i.e. E 0  2 J , E1  0 and E2  2J. Q2  Ce 2 0   E0  2C1e  E1  2C 2 e  E2   e  2J  2e 0  e  2 J    e J  e J  2 2  4 4 2 Cr r 0 2  e J  e J  Q2     cosh  J 2  cosh  J 2  Q N  cosh  J  N.  2  Q25. Consider a system of three spins S1, S2 and S3 each of which can take values +1 and -1. The energy of the system is given by E   J S1 S 2  S 2 S 3  S 3 S1  where J is a positive constant. The minimum energy and the corresponding number of spin configuration are, respectively, (a) J and 1 (b) 3 J and 1 (c) 3 J and 2 (d) 6 J and 2 Ans. : (c) Solution: If we take S1  S2  S3  1 i.e.    S1 S2 S3 Then energy, E   J 1  1  1  1  1  1  3J Again S1  S2  S3  1 , then    Energy  E   3J So, minimum energy is  3J  and there are two spin configuration. If we take    S1 S2 S3 Then we get Maximum energy E  J. H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 261  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics NET/JRF (JUNE-2013) Q26. Ten grams of ice at 00 C is added to a beaker containing 30 grams of water at 250 C. What is the final temperature of the system when it comes to thermal equilibrium? (The specific heat of water is 1 cal / gm / 0 C and latent heat of melting of ice is 80 cal / gm ) (a) 00 C (b) 7.50 C (c) 12.50 C (d) 1.250 C Ans. : (a) Solution: The amount of heat required to melt the ice of mass 10 gm at 00 C is Q  m  L  10  80  800Cal , where L is the latent heat of melting of ice and m is the mass of the ice. The amount of heat available in water of mass 30 gm at 250 C is Q  m  Cv  T  30 1 25  750Cal Since the heat available is less than the heat required to melt the ice therefore ice will not melt as a result the temperature of the system will be at 00 C only. Q27. A vessel has two compartments of volume V1 and V2 , containing an ideal gas at pressures p1 and p 2 , and temperatures T1 and T2 respectively. If the wall separating the compartments is removed, the resulting equilibrium temperature will be p1T1  p 2T2 V1T1  V2T2 p1V1  p 2V2 (d) T1T2  1/ 2 (a) (b) (c) p1  p 2 V1  V2  p1V1 / T1    p 2V2 / T2  Ans. : (c) p1V1 p2V2 Solution: V  V1  V2 , n  n1  n2   , U1  U 2  U , n1CvT1  n2CvT2  nCvT , T1 T2 p1V1  p2V2 n1T1  n2T2  nT  T  p1V1 p2V2  T1 T2 H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 262  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics Q28. For temperature T1  T2 , the qualitative temperature dependence of the probability distribution F v  of the speed v of a molecule in three dimensions is correctly represented by the following figure: T2 T1 F(v) F(v) (a) T1 (b) T2 v v T1 T2 T1 F(v) F(v) (c) (d) T2 Ans. : (a) v v Solution: Area under the F  v  is conserve and the mean velocity shift towards right for higher temperature. Q29. A system of non-interacting spin- 1/ 2 charged particles are placed in an external magnetic field. At low temperature T , the leading behavior of the excess energy above the ground state energy, depends on T as: ( c is a constant) (a) cT (b) cT 3 (c) e  c / T (d) c (is independent of T ) Ans. : (c)  kTB H  H  B   H  e  e kT  Solution: U    B H tanh B    B H   B H BH  kT  e kT  e  kT    Excess energy from the ground level  kTB H  H  B    B H  H  B    H  B  e  e   (  H )   H 1   e  e    H  e    B H  B H  kT kT kT kT 2   BH  B B   B  H  H  B  B H B H   e kT  e kT     e kT e  B kT   e kT  e kT       C At low temperature, the lower value, U  e T , where C   B H. H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 263  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics Q30. Consider a system of two Ising spins S1 and S 2 taking values  1 with interaction energy given by    JS1 S 2 , when it is in thermal equilibrium at temperature T. For large T , the average energy of the system varies as C / k B T , with C given by (a)  2 J 2 (b)  J 2 (c) J 2 (d) 4 J Ans. : (b) Solution: The interaction energy is given by E   J S1 S 2 where S1 and S 2 taking values  1. Possible values of the Energy of the system are E1   J 11   J , E 2   J  1 1   J E3   J 1  1   J , E 4   J  1  1   J E  r  J  J  E g e r r kT 2 Je J kT  2 Je  J kT  kTJ  e  e kT  J  1  kT   J   1   kT    U  r   J E J J  J J   J   J   r   e kT  e  kT   gr e r kT 2e kT  2e kT   1   kT  1     kT  J2 J  U   C   J 2 (For large T ,  1 ) kT kT Q31. Consider two different systems each with three identical non-interacting particles. Both have single particle states with energies  0 ,3 0 and 5 0 ,  0  0. One system is populated 1 by spin  fermions and the other by bosons. What is the value of E F  E B where E F 2 and EB are the ground state energies of the fermionic and bosonic systems respectively? (a) 6 0 (b) 2 0 (c) 4 0 (d)  0 Ans. : (b) Solution: Energy of Fermion = 2 1 0  3 0  5 0 Energy of boson = 3 1 0  3 0 E F  E B = 5 0  3 0  2 0 H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 264  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics NET/JRF (DEC-2013) 1 Q32. Three identical spin- fermions are to be distributed in two non-degenerate distinct 2 energy levels. The number of ways this can be done is (a) 8 (b) 4 (c) 3 (d) 2 Ans. : (b) Solution: Total number of degeneracy g  (Number of energy state (n))  (Number of degeneracy due to spin ( 2 s  1 )) 1 1 n  2, s  , g  2  (2.  1)  4 2 2 Number of particle, N  3. So number of ways, g cN  4 c3  4 Q33. Consider the melting transition of ice into water at constant pressure. Which of the following thermodynamic quantities does not exhibit a discontinuous change across the phase transition? (a) Internal energy (b) Helmholtz free energy (c) Gibbs free energy (d) Entropy Ans. : (c) Solution: Ice to water: 1st order phase transition. So Gibbs free energy is continuous, so it doesn’t exhibit discontinuous change. Q34. Two different thermodynamic systems are described by the following equations of state: 1 3RN 1 1 5RN 2  1  1 and 2    2  where T 1, 2  , N 1, 2  and U 1, 2  are respectively, the T 2U T 2U temperatures, the mole numbers and the internal energies of the two systems, and R is the gas constant. Let U tot denote the total energy when these two systems are put in U 1 contact and attain thermal equilibrium. The ratio is U tot 5 N 2  3N 1 N 1 N 2  (a) (b) (c) 1 (d) 1 3N 1  5 N 2  3N 1  5 N 2  N  N 2  N  N 2  Ans. : (b) 1 3RN 1 1 5RN 2  Solution:  and  T 1 2U 1 T 2  2U 2  3 5 Now U tot  U (1)  U  2  RN 1T 1  RN  2T  2 2 2 H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 265  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics 1 3RN 1T 1  U 1 2  3 N 1T 1    U tot 1 3N 1T 1  5 RN  2T  2  3N 1T 1  5 N  2T  2 2  3N 1 At thermal equilibrium T 1  T  2 , thus 3N 1  5 N 2  Q35. The speed v of the molecules of mass m of an ideal gas obeys Maxwell’s velocity distribution law at an equilibrium temperature T. Let  vx , v y , vz  denote the components   2 of the velocity and k B the Boltzmann constant. The average value of  vx   v y , where  and  are constants, is  (a)  2   2 k B T / m    (b)  2   2 k B T / m (c)     k B T / m (d)     k B T / m 2 2 Ans. : (b) Solution: Ideal gas obeys Maxwell velocity distribution law at equilibrium temperature. Then   2 average value of  vx   v y  v  2 Now x  v y   2 v x2   2 v y2  2  v x v y k BT v x  0, v y  0 and vx2   v y2  vz2 m  v  2 Then x  v y   2 v x2   2 v y2  2 v x v y  v    2 B   2   2  2 k BT kT kT x  v y  2 m m m Q36. The entropy S of a thermodynamic system as a function of energy S C E is given by the following graph. The temperatures of the phases B A A, B and C denoted by T A , TB and TC respectively. E Satisfy the following inequalities: (a) TC  TB  T A (b) T A  TC  TB (c) TB  TC  T A (d) TB  T A  TC Ans. : (c) Solution: Temperatures of phase are: TA , TB , TC H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 266  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics  dS  1 S Since,   C  dE  T B dS A Hence, will be slope, then it will be zero for B - phase dE E So TB   and in C and A phases, internal energy of C phase is more, so TC  T A Now TB  TC  T A Q37. A system of N classical non-interacting particles, each of mass m , is at a temperature T and is confined by the external potential V r   Ar 2 (where A is a constant) in three 1 2 dimensions. The internal energy of the system is A  k BT  (c) N 2mA k B T 3 ln  3/ 2 (a) 3 Nk B T (b) Nk B T (d) N 2 m  m  Ans. : (a) Solution: V r   1 2 1 2   Ar  A x 2  y 2  z 2 it is harmonic oscillator. 2 3N 1  kT  So its partition function will be z N    N     ln Z N Internal energy, U  kT 2  3 NkT T Q38. A Carnot cycle operates as a heat engine between two bodies of equal heat capacity until their temperatures become equal. If the initial temperatures of the bodies are T1 and T2 , respectively and T1  T2 , then their common final temperature is (a) T12 / T2 (b) T22 / T1 (c) T1T2 (d) 1 T1  T2  2 Ans. : (c) Solution: For heat Carnot engine the change in entropy for source and sink TF dT  T TF dT T  dS1   and dS 2    log  F   log  F  T1 T   T1 T2 T  T2  T T S  dS1  dS 2  log F  log F. T1 T2 T  2 Since, Carnot engine is reversible in nature, so log F  0  TF  T1T2 T1T2 H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 267  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics NET/JRF (JUNE-2014) Q39. Which of the graphs below gives the correct qualitative behaviour of the energy density E r   of blackbody radiation of wavelength  at two temperatures T1 and T2 T1  T2  ? (a) (b) T2 T2 Er   Er   T1   (c) (d) T2 T2 Er   Er   T1 T1 Ans. : (c)   Q40. A system can have three energy levels: E  0,  . The level E  0 is doubly degenerate, while the others are non-degenerate. The average energy at inverse temperature  is  e   e       (a)   tanh   (d)   tanh   (b) 1  e  e    (c) 0  2  Ans. : (d) Solution: E  0,   , E  0 doubly degenerate z   gi e   Ei  2  e   0  e    e   z  2  e   e   ln z  ln 2  e   e       Now E    ln  z      ln 2  e   e      1  2 e  e       e    e                    e 2  e 2   e e E              tanh     2          2    e 2  e 2     e 2  e 2             H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 268  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics Q41. The free energy F of a system depends on a thermodynamic variable  as F  a 2  b 6 with a, b  0. The value of  , when the system is in thermodynamic equilibrium, is (b)  a / 6b  (c)  a / 3b  (d)  a / b  1/ 4 1/ 4 1/ 4 (a) zero Ans. : (c) Solution: Frequency F  a 2  b 6 , a, b  0 2 F F F is equilibrium i.e.  0 , now  2a  6b 5  2  F 1/ 4 a  a   0  2a  6b 5   4      3b  3b  Q42. For a particular thermodynamic system the entropy S is related to the internal energy U and volume V by S  cU 3 / 4V 1 / 4 where c is a constant. The Gibbs potential G  U  TS  PV for this system is 3PU cU US (a) (b) (c) zero (d) 4T 3 4V Ans. : (c) Solution: S  cU 3/ 4V 1/ 4 , dU  TdS  PdV  S  1  S  1 c  3 1/ 4 1/ 4 4 U 1/ 4         U V  T   U V T  U V T 4 3c V 1/ 4 1 5 / 4  U  SV U S V 5 / 4 1/ 4 4     P    P  U  V  S c 3 c 3 4 U 1/4 S V 5/4 1/4 4 1 G U   cU V 3/4 1/4  U V  U  U  U  0 3c V 1/4 c 3 3 3 Q43. The pressure of a non-relativistic free Fermi gas in three-dimensions depends, at T  0 , on the density of fermions n as (a) n 5 / 3 (b) n1 / 3 (c) n 2 / 3 (d) n 4 / 3 Ans. : (a) H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 269  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics 2 Solution: Pressure P  nEF , EF  n 2 / 3 , at T  0 3 2 2 P  n  n 2 / 3  n 5 / 3  P  n5 / 3 3 3 Q44. The vander Waals’ equation of state for a gas is given by  a   P  2 V  b   RT  V  where P, V and T represent the pressure, volume and temperature respectively, and a and b are constant parameters. At the critical point, where all the roots of the above cubic equation are degenerate, the volume is given by a a 8a (a) (b) (c) (d) 3b 9b 27b 2 27bR Ans. : (d)  a   P   2 P  Solution:  P  2  V  b   RT , for critical volume    0,  2   0  V   V   V  a ab PV   Pb  2  RT V V P a 2ab 2 P 2a 6ab 2a 6ab  0  P  2  3  0,  0  3  4  0  3  4  Vc  3b V V V V 2 V V V V H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 270  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics NET/JRF (DEC-2014) Q45. The pressure P of a fluid is related to its number density  by the equation of state P  a  b 2 where a and b are constants. If the initial volume of the fluid is V0 , the work done on the system when it is compressed, so as to increase the number density from an initial value of  0 to 2  0 is (a) a 0V0 (b) a  b 0  0V0  3a 7  b  (c)   0   0V0 (d) a ln 2  b 0  0V0  2 3  Ans. : (d) n n2 n Solution: P  a  b 2  P  a b 2   V V V V2 dV V2 dV n n W   P  dV  an  bn 2  , where V1  , V2  V1 V V1 V 2 0 2 0  W   n  a ln 2  b 0    0V0  a ln 2  b0  ,  n  0V0 Work done on the system  W   a ln 2  b0  0V0 Q46. An ideal Bose gas is confined inside a container that is connected to a particle reservoir. Each particle can occupy a discrete set of single-particle quantum states. If the probability that a particular quantum state is unoccupied is 0.1 , then the average number of bosons in that state is (a) 8 (b) 9 (c) 10 (d) 11 Ans. : (b) Q47. In low density oxygen gas at low temperature, only the translational and rotational modes of the molecules are excited. The specific heat per molecule of the gas is 1 3 5 (a) kB (b) k B (c) kB (d) kB 2 2 2 Ans. : (d) Solution: Total D.O.F. = 3 transition + 2 rotation i.e. f  5 k B T 5k B T U 5 U f   CV   kB 2 2 T 2 H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 271  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics Q48. When a gas expands adiabatically from volume V1 to V2 by a quasi-static reversible process, it cools from temperature T1 to T2. If now the same process is carried out adiabatically and irreversibly, and T2 is the temperature of the gas when it has equilibrated, then V V  T2V1 (a) T2  T2 (b) T2  T2 (c) T2  T2  2 1  (d) T2   V2  V2 Ans. : (b) Q49. A random walker takes a step of unit length in the positive direction with probability 2 / 3 and a step of unit length in the negative direction with probability 1 / 3. The mean displacement of the walker after n steps is (a) n / 3 (b) n / 8 (c) 2n / 3 (d) 0 Ans. : (a) 2 1 Solution: P  1   P  1  3 3 2 1 1 n For one step  1    , for n step  3 3 3 3 Q50. A collection N of non-interacting spins S i , i  1, 2,....., N , S i  1 is kept in an external magnetic field B at a temperature T. The Hamiltonian of the system is B H   B i S i. What should be the minimum value of for which the mean value k BT 1 Si  ? 3 1 1 (a) N ln 2 (b) 2 ln 2 (c) ln 2 (d) N ln 2 2 2 Ans. : (c) B B  e kT e kT Solution: P  Si  1  B B , P  Si  1  B B   e kT e kT e kT  e kT B B   1e kT e kT  B  Si   Si    tanh   B B  kT  e kT  e kT B For N particle Si   N tanh kT S 1  B  1 B 1 According to question, i    tanh     ln 2 N 3  kT  3 kT 2 H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 272  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics NET/JRF (JUNE-2015) Q51. A system of N non-interacting classical particles, each of mass m is in a two dimensional harmonic potential of the form V  r    x 2  y 2 where  is a positive    1  constant. The canonical partition function of the system at temperature T is    :  k BT  N N    2    2m  2N    N  2m 2  (a)    (b)   (c)   (d)  2   2m        2m      Ans. (d)  Solution: V  r    x 2  y 2   px2  p 2y  x 2  y 2 1     z1  2 h  e 2 mkT dpx  e 2 mkT dp y  e   kT dx  e  kT dy 2 mkT 2 mkT 1 1 1   z1  2  2 h2 h 2 2  2  kT kT N  2 2 m   2 2 m  z1   2   kT   z N   2 2  2  h   h   Q52. A system of N distinguishable particles, each of which can be in one of the two energy levels 0 and  , has a total energy n  , where n is an integer. The entropy of the system is proportional to  N !  N!  (a) N ln n (b) n ln N (d) ln   n ! N  n  !  (c) ln    n!    Ans. : (d) Solution: No of ways for above configuration is  N Cn N N W   Entropy=k ln n N n n N n H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 273  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics Q53. The condition for the liquid and vapour phases of a fluid to be in equilibrium is given by dP Q the approximate equation  1 (Clausius-Clayperon equation) where vvap is the dT Tvvap volume per particle in the vapour phase, and Q1 is the latent heat, which may be taken to be a constant. If the vapour obeys ideal gas law, which of the following plots is correct? ln P ln P (a) (b) (c) ln P (d) ln P O T O T O T O T Ans. (c) dP Q RT dP Ql P dP Ql dT C Solution:  l , vap  dT Tvap P   dT RT 2  P  R T 2  ln P   T  Q54. Consider three Ising spins at the vertices of a triangle which interact with each other with a ferromagnetic Ising interaction of strength J. The partition function of the system at  1  temperature T is given by    :  k BT  (a) 2e3 J  6e   J (b) 2e 3 J  6e  J (c) 2e3 J  6e 3 J  3e  J  3e   J (d)  2 cosh  J  3 Ans. (b) Solution: H  J  S1S 2  S1S3  S 2 S3  S1 S2 S3 E 1 1 1 3J 1 1 1  1 1 1   J 1 1 1  1 11  1 1 1  J 1 1 1 1 1 1 3J  z  2e3 J  6e  J H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 274  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics Q55. A large number N of Brownian particles in one dimension start their diffusive motion from the origin at time t  0. The diffusion coefficient is D. The number of particles crossing a point at a distance L from the origin, per unit time, depends on L and time t as  L2 4 Dt  L2 4 Dt N NL N (a) e  4 Dt  (b) e L2 (c) e  4 Dt  (d) Ne L2 4 Dt 4 Dt 16 Dt 3 Ans. (a) Solution: From Einstein Smoluchowski theory dx   x2  p  x  dx  exp   4 Dt  4 Dt  N   L2  Number of particle passing from point L at origin .exp   4 Dt  4 Dt   Q56. An ideal Bose gas in d -dimensions obeys the dispersion relation  k  Ak s , where A   and s are constants. For Bose-Einstein condensation to occur, the occupancy of excited states d s   s Ne  c  d 0 e      1  where c is a constant, should remain finite even for   0. This can happen if d 1 1 d 1 d 1 d (a)  (b)   (c) 1 (d)  1 s 4 4 s 2 s 2 s Ans. (c) d s   s Solution: Ne  c  d 0 e      1 B.E. condensation is possible in 3-D d s 1 d 3 1 For materlistic particle g  2     s 2 s 2 d  s d For massless particle g  2  2 3 s s d In both cases  1 s H.No. 40-D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi-110016 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com | Email: [email protected] 275  fiziks Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics NET/JRF (DEC-2015) Q57. The heat capacity of the interior of a refrigerator is 4.2 kJ / K. The minimum work that must be done to lower the internal temperature from 18o C to 17o C , when the outside temperature is 27o C will be (a) 2.20 kJ (b) 0.80 kJ (c) 0.30 kJ (d) 0.14 kJ Ans. : (b) Q58. For a system of independent non interacting one-dimensional oscillators, the value of the free energy per oscillator, in the limit T 0 , is 1 3 (a)  (b)  (c)  (d) 0 2 2 Ans. : (a) N         Solution: For the given system Z N   2sinh   F  kT ln Z N  NkT ln  2sinh    2kT    2kT           2 e 2 kT  e 2 kT                  NkT ln  NkT ln  e 1  e    NkT ln e  NkT ln 1  e kT  2 kT kT 2 kT