fiziks Institute for NET/JRF, GATE, IIT JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Kinetic Theory, Thermodynamics OBJECTIVE QUESTIONS IIT-JAM-2005

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1 Institute for NE/JRF, GAE, II JAM, JES, IFR and GRE in HYSIAL SIENES Kinetic heory, hermodynamics OBJEIE QUESIONS II-JAM Q. he molar specific heat of a gas as given from the kinetic theory is R. If it is not specified whether it is or, one could conclude that the molecules of the gas (a) are definitely monatomic (b) are definitely rigid diatomic are definitely non-rigid diatomic (d) can be monatomic or rigid diatomic Ans. : (d) 5R Solution: If molecule is mono atomic then p = and if molecule is rigid diatomic then 5R =. Q. he value of entropy at absolute zero of temperature would be (a) zero for all the materials (b) finite for all the materials zero for some materials and non-zero for others (d) unpredictable for any material Solution: If system will achieve absolute zero then it is perfectly ordered system then entropy will be zero. Q. Which of the following statements is INORRE? (a) Indistinguishable particles obey Maxwell-Boltzmann statistics (b) All particles of an ideal Bose gas occupy a single energy state at = 0 he integral spin particles obey Bose-Einstein statistics (d) rotons obey Fermi-Dirac statistics Solution: Distinguishable particles obey Maxwell-Boltzmann statistics. Near II, Hauz Khas, New Delhi 6 hone: / B/6, Jia Sarai, Near II Hauz Khas, New Delhi 6 fiziks.physics@gmail.com 9

2 Institute for NE/JRF, GAE, II JAM, JES, IFR and GRE in HYSIAL SIENES II-JAM-006 Q4. A solid melts into a liquid via first order phase transition. he relationship between the pressure and the temperature of the phase transition is = - + 0, where 0 is a constant. he entropy change associated with the phase transition is.0 J mole - K -. he d lausius-lapeyron equation for the latent heat is L = Δv. Here Δ v = v liquid vsolid d is the change in molar volume at the phase transition. he correct statement relating the values of the volumes is (a) v = v (b) v v liquid solid liquid = solid v liquid = v solid (d) v quid = v solid + Ans. : Solution: Since = + 0 d = d d dl It is given L = Δv L= Δ v = Δ v d d - - Since ds =.0 Jmole K.0 J mole - K - dq mdl ds = = d = = Δv Δ v= II-JAM-007 Q5. Experimental measurements of heat capacity per mole of Aluminum at low temperatures show that the data can be fitted to the formula, = a + b, where a = JK - mole -, b =.48 x 0-5 JK -4 mole - and is the temperature in in Kelvin. he entropy of a mole of Aluminum at such temperatures is given by the formula b a b (a) a + + c, where c > 0 is a constant (b) + + c, where c > 0 is a constant 4 b a + (d) a b + 4 Near II, Hauz Khas, New Delhi 6 hone: / B/6, Jia Sarai, Near II Hauz Khas, New Delhi 6 fiziks.physics@gmail.com 9

3 Institute for NE/JRF, GAE, II JAM, JES, IFR and GRE in HYSIAL SIENES Solution: cd v ds = s= a + b d b = + + a c II-JAM-008 Q6. he chemical potential of an ideal Bose gas at any temperature is (a) necessarily negative (b) either zero or negative necessarily positive (d) either zero or positive Ans. : (b) Solution: he chemical potential of an ideal Bose gas at any temperature is either zero or negative.(zero in case of photon ). Q7. A thermodynamic system is maintained at constant temperature and pressure. In thermodynamic equilibrium, its (a) Gibbs free energy is minimum (b) enthalpy is maximum Helmholtz free energy is minimum (d) internal energy is zero Solution: A thermodynamic system is maintained at constant temperature and pressure can be defined by Gibbs energy dg = Sd + d 0 i.e. Gibbs free energy is minimum. II-JAM-009 Q8. A box containing moles of a diatomic ideal gas at temperature 0 is connected to another identical box containing moles of a monatomic ideal gas at temperature 5 0. here are no thermal losses and the heat capacity of the boxes is negligible. Find the final temperature of the mixture of gases (ignore the vibrational degrees of freedom for the diatomic molecules). (a) 0 (b) (d) 0 Ans. : Solution: Internal energy of the system remains conserve i.e. Umonoatomic + Udiatomic = Umixture Near II, Hauz Khas, New Delhi 6 hone: / B/6, Jia Sarai, Near II Hauz Khas, New Delhi 6 fiziks.physics@gmail.com 94

4 Institute for NE/JRF, GAE, II JAM, JES, IFR and GRE in HYSIAL SIENES U monotomic = n, U diatomic = n 5R R v =, v =, n = n =, 0 =, = 50 n + n Let the common temperature of mixture is and specific heat is = n + n and number of moles of mixture is n= n+ n, then n + n = n =.5 0 Q9. Isothermal compressibility κ of a substance is defined asκ Ans. : Solution: moles of an ideal gas will be n (a) (b) = nr and k = =. II-JAM-00 = (d) n. Its value for n Q0. A gas of molecules each having mass m is in thermal equilibrium at a temperature. Let v x, v y, v z be the artesian components of velocity, v, of a molecules. he mean value of ( v α v + β v ) x y z is k (a) ( α + β ) B m k + (b) ( α + β ) B m ( k ) B m β α k (d) ( α + β ) B m Solution: ( ) v αv + βv = v + α v + β v αv v + βv v βαv v x y z x y y x y z x y z ( ) v αv + βv = v + α v + β v α v v + β v v βα v v x y z x y y x y z x y z Near II, Hauz Khas, New Delhi 6 hone: / B/6, Jia Sarai, Near II Hauz Khas, New Delhi 6 fiziks.physics@gmail.com 95

5 Institute for NE/JRF, GAE, II JAM, JES, IFR and GRE in HYSIAL SIENES k B vx = vy = vz = and vx = vy = vz = 0 m k B ( vx αvy + βvz) = ( + α + β ) m Q. A trapped air bubble of volume 0 is released from a depth h measured from the water surface in a large water tank. he volume of the bubble grows to 0 as it reaches just below the surface. he temperature of the water and the pressure above the surface of water (0 5 N/m ) remain constant throughout the process. If the density of water is 000 kg/ m and the acceleration due to gravity is 0m/ s, then the depth h is (a) m (b) 0 m 50 m (d) 00 m Ans. : (b) Solution: At depth h pressure = 0 + ρgh and volume = 0 where At surface pressure = 0 and volume = 0 ( ρ ) + gh = = 0 N / m, g = 0 m/ s h= 0m ρ = 000 kg / m II-JAM-0 Q. onsider free expansion of one mole of an ideal gas in an adiabatic container from volume to. he entropy change of the gas, calculated by considering a reversible process between the original state( ) to the final state ( ) where is the, temperature of the system is denoted by Δ S. he corresponding change in the entropy of the surrounding is Δ S. Which of the following combinations is correct? (a) Δ S = Rln ( / ), Δ S = Rln ( / ) (b) Δ S = Rln ( / ), Δ S = Rln ( / ) ( ) Δ S = Rln /, Δ S = 0 (d) ( ) Δ S = Rln /, Δ S = 0, Near II, Hauz Khas, New Delhi 6 hone: / B/6, Jia Sarai, Near II Hauz Khas, New Delhi 6 fiziks.physics@gmail.com 96

6 Institute for NE/JRF, GAE, II JAM, JES, IFR and GRE in HYSIAL SIENES Ans. : Solution: Free expansion is irreversible process when gas expand to which can be explained by choosing any path between two state (because entropy is state function). So one can choose reversible isothermal process. So Δ S = Rln. Hence it is free expansion so entropy of surrounding is Δ S = 0. Q. A gas of molecular mass m is at temperature. If the gas obeys Maxwell-Boltzmann velocity distribution, the average speed of molecules is given by (a) k B k (b) B m m k B πm (d) 8k B Ans. : (d) II-JAM-0 Q4. For a liquid to vapour phase transition at tr, which of the following plots between specific Gibbs free energy g and temperature is correct? (a) g (b) g πm Liquid apour Liquid apour tr tr g (d) g Liquid apour Liquid apour tr tr Near II, Hauz Khas, New Delhi 6 hone: / B/6, Jia Sarai, Near II Hauz Khas, New Delhi 6 fiziks.physics@gmail.com 97

7 Institute for NE/JRF, GAE, II JAM, JES, IFR and GRE in HYSIAL SIENES Q5. A tiny dust particle of mass kg is floating in air at 00K. Ignoring gravity, its rms speed (in µm/s) due to random collisions with air molecules will be closest to (a) 0. (b) 0 (d) 00 Ans. : Solution: v rms k = = = 0 0 m/ s m.4 0 Q6. When the temperature of a blackbody is doubled, the maximum value of its spectral energy density, with respect to that at initial temperature, would become (a) times (b) 8 times 6 times (d) times U Solution: U = U = U = 64U 4 4 U II-JAM-0 Q7. A blackbody at temperature emits radiation at a peak wavelength λ. If the temperature of the blackbody becomes 4, the new peak wavelength is (a) λ 56 Ans. : (d) (b) λ 64 λ 6 λmax λ λ Solution: From wein Law λmax = λmax λmax = = = 4 4 Q8. Let N MB, N BE, N FD denote the number of ways in which two particles can be distributed in two energy states according to Maxwell-Boltzmann, Bose-Einstein and Fermi-Dirac statistics respectively. hen N : N : N is MB BE (a) 4 : : (b) 4 : : 4 : : (d) 4 : : Solution: N = g = n = FD (d) λ 4 Near II, Hauz Khas, New Delhi 6 hone: / B/6, Jia Sarai, Near II Hauz Khas, New Delhi 6 fiziks.physics@gmail.com 98

8 Institute for NE/JRF, GAE, II JAM, JES, IFR and GRE in HYSIAL SIENES For Maxwell W N n = g = = n 4 For boson N = g = n = ; W n+ g + = = = ng For Fermion N = g = n = ; g W = = = gg n Q9. wo thermally isolated identical systems have heat capacities which vary as v = β (where β > 0 ). Initially one system is at 00 K and the other at 400 K. he systems are then brought into thermal contact and the combined system is allowed to reach thermal equilibrium. he final temperature of the combined system is. Ans. : Solution: here is not a unique value of temperature rather range of temperature he maximum temperature when work done is zero so dq+ dq = 0 4 max max max ( 00) ( 400) 4 4 mβ d + mβ d = 0 = max ( 00) + ( 400) 4 4 = max = 60K he minimum temperature system is when process is reversible so change in entropy of system is zero Δ S +Δ = min S0 0 mβ ( ) ( ) min mβ d + d = 0 min = = 5K min So 5K 60K Near II, Hauz Khas, New Delhi 6 hone: / B/6, Jia Sarai, Near II Hauz Khas, New Delhi 6 fiziks.physics@gmail.com 99

9 Institute for NE/JRF, GAE, II JAM, JES, IFR and GRE in HYSIAL SIENES II-JAM-04 Q0. In - dimension, an ensemble of N classical particles has energy of the form x E = + kx. he average internal energy of the system at temperature is m (a) Nk B (b) Nk B Nk B (d) Nk B Ans. : (d) Q. A solid metallic cube of heat capacity S is at temperature 00 K. It is brought in contact with a reservoir at 600 K. If the heat transfer takes place only between the reservoir and the cube, the entropy change of the universe after reaching the thermal equilibrium is (a) 0.69 S (b) 0.54 S 0.7 S (d) 0.9 S Ans. : (d) Q. A real gas has specific volume v at temperature. Its coefficient of volume expansion and isothermal compressibility are α and k, respectively. Its molar specific heat at constant pressure p and molar specific heat at constant volume v are related as (a) = R (b) p v + p = v vα + k Ans. : p vα = v + (d) p = v k Q. At atmospheric pressure ( = 0 5 a), aluminium melts at 550 K. As it melts, its density decreases from 0 kg/m to.9 0 kg/m. Latent heat of fusion of aluminium is 4 0 J/kg. he melting point of aluminium at a pressure of 0 7 a is closest to (a) 55. K (b) 55.6 K K (d) K Ans. : (b) Near II, Hauz Khas, New Delhi 6 hone: / B/6, Jia Sarai, Near II Hauz Khas, New Delhi 6 fiziks.physics@gmail.com 00

10 Institute for NE/JRF, GAE, II JAM, JES, IFR and GRE in HYSIAL SIENES II-JAM-05 Q4. A system consists of N number of particles, N >>. Each particle can have only one of the two energies E or + ε ( ε ) E > 0. If the system is in equilibrium at a temperature, the average number of particles with energy E is N N (a) (b) / k e ε + Ans. : (d) Solution: ( E E ) ( ε ) E+ E ε k k k N = Ne = e N = Ne N / k e ε + (d) k Ne ε / Q5. A rigid and thermally isolated tank is divided into two compartments of equal volume, separated by a thin membrane. One compartment contains one mole of an ideal gas A and the other compartment contains one mole of a different ideal gas B. he two gases are in thermal equilibrium at a temperature. If the membrane ruptures, the two gases mix. Assume that the gases are chemically inert. he change in the total entropy of the gases on mixing is (a) 0 (b) R ln ln (d) R ln Ans. : (d) Solution: For A, number of microstate after mixing is For A, number of microstate before mixing is A B Δ S = Rln Rln = Rln A Similarly, for B Δ S = Rln Δ S =Δ S +Δ S = Rln B Q6. A rigid triangular molecule consists of three non-collinear atoms joined by rigid rods. Ans. : A he constant pressure molar specific heat ( p ) of an ideal gas consisting of such molecules is (a) 6 R (b) 5 R 4 R (d) R Solution: D.O.F = 6 6R U U = = = R = + R= 4R B Near II, Hauz Khas, New Delhi 6 hone: / B/6, Jia Sarai, Near II Hauz Khas, New Delhi 6 fiziks.physics@gmail.com 0

11 Institute for NE/JRF, GAE, II JAM, JES, IFR and GRE in HYSIAL SIENES SEION B: MSQ Q7. As shown in the diagram AB and D are two isotherms at temperatures and, respectively ( ) >. A and BD are two reversible adiabats. In this arnot cycle, which of the following statements are true? Q Q (a) = (b) he entropy of the source decreases he entropy of the system increases Q D (d) Work done by the system W = Q Q, (b) and (d) Q8. In the thermodynamic cycle shown in the figure, one mole of a monatomic ideal gas is taken through a cycle. AB is a reversible isothermal expansion at a temperature of 800 K in which the volume of the gas is doubled. B is an isobaric contraction to the original volume in which the temperature is reduced to00 K. A is a constant volume process in which the pressure and temperature return to their initial values. he A net amount of heat (in Joules) absorbed by the gas in one complete cycle is. Ans. : 45 Solution: rocess A B is isothermal expansion A A = 800 K, A, A and B = 800 K, B = A, B = rocess B is isobaric B olume A = B =, = A, = 00K A is Isochoric B Δ Q = nra ln = 460J A ressure A Q B Near II, Hauz Khas, New Delhi 6 hone: / B/6, Jia Sarai, Near II Hauz Khas, New Delhi 6 fiziks.physics@gmail.com 0

12 Institute for NE/JRF, GAE, II JAM, JES, IFR and GRE in HYSIAL SIENES nγrδ γ Δ Q = nδ = = R γ ( γ ) R Δ Q = ( ) ( ) γ ( γ ) ( ) R = 500 = 694J otal heat exchange is Q+ Q + Q = 45 = 044J Q9. One gram of ice at 0 o is melted and heated to water at 9 o. Assume that the specific heat remains constant over the entire process. he latent heat of fusion of ice is 80 alories/gm. he entropy change in the process (in alories per degree) is. Ans. : 0.9 ML 80 0 d 0 Solution: Δ S = =, Δ S = M 7 Δ S =.ln 7 7 Δ S =Δ S+ΔS 80 0 Δ S = +.ln = = II-JAM-06 Q0. A spherical closed container with smooth inner wall contains a monoatomic ideal gas. If the collisions between the wall and the atoms are elastic, then the Maxwell speeddistribution function dnv dv for the atoms is best represented by: (a) (b) dn v dv 0 v dn v dv 0 v dn v dv (d) dn v dv Ans. : 0 v 0 v Near II, Hauz Khas, New Delhi 6 hone: / B/6, Jia Sarai, Near II Hauz Khas, New Delhi 6 fiziks.physics@gmail.com 0

13 Institute for NE/JRF, GAE, II JAM, JES, IFR and GRE in HYSIAL SIENES dn Solution: mv v exp dv k Q. For an ideal gas, which one of the following S - diagram is valid? (a) Isochor S Isobar (b) Isobar S Isochor Isochor (d) Isochor Isobar Isobar S S S Solution: = = isochoric and isobaric plot. > from the slope of Sdiagram one can plot Q. If U, F, H and G represent internal energy, Helmholtz free energy enthalpy and Gibbs free energy respectively, then which one of the following is a correct thermodynamic relation? (a) du = d ds (b) dh = d + ds df = d + Sd (d) dg = d + Sd Ans. : (b) Solution: H = U + dh = du + d + d ds d + d + d dh = d + ds S Near II, Hauz Khas, New Delhi 6 hone: / B/6, Jia Sarai, Near II Hauz Khas, New Delhi 6 fiziks.physics@gmail.com 04

14 Institute for NE/JRF, GAE, II JAM, JES, IFR and GRE in HYSIAL SIENES Q. One mole of an ideal gas with average molecular speed v 0 is kept in a container of fixed volume. If the temperature of the gas is increased such that the average speed gets doubled, then (a) the mean free path of the gas molecule will increase (b) the mean free path of the gas molecule will not change the mean free path of the gas molecule will decrease (d) the collision frequency of the gas molecule with wall of the container remains unchanged. Ans. : (b) Solution: For fixed volume if temperature is increased then pressure is also increased by same amount so mean free path with not change k λ = π d p Q4. he diagram below shows three possible paths for an ideal gas to reach the final state f from an initial state i. Which among f (a) he work done by the gas is maximum along path -. (b) Minimum change in the internal energy occurs along path -. Maximum heat transfer is for path - (d) Heat transfer is path independent Solution: work done is area under curve so it is maximum in path Hence change in internal energy is same in all path so heat exchange is also maximum in path the following statements ( s ) is (are) correct? Q5. A cylinder contains 6 g of O. he work done when the gas is compressed to 75% of the original volume at constant temperature of [Universal gas constant R 8. J / ( mole K) 6 n= = 0.5, = 00k Near II, Hauz Khas, New Delhi 6 hone: / = ] 0 7 is. J. 8 B/6, Jia Sarai, Near II Hauz Khas, New Delhi 6 fiziks.physics@gmail.com 05 i

15 Institute for NE/JRF, GAE, II JAM, JES, IFR and GRE in HYSIAL SIENES Ans. : d 75 Solution: W = d nr = nr ln = ln.75 = 58J 00 Q6. An aluminum plate of mass 0. kg at Ans. : is immersed in 0.5 litre of water at 0 0 kept inside an insulating container and is then removed. If the temperature of the water is found to be 0, then the temperature of the aluminum plate is 0 (he specific heat of water and aluminum are 400 J / kg respectively, the density of water is 000 kg / m ) Solution: M asa( a f ai ) MwSw( w f wi) K and 900 J / kg = ( 68) = ( 96 9) ( af 68) = 450 ( af 68) = = = = 67.6 af = 94.6 Q7. If there is a 0% decrease in the atmospheric pressure at a hill compared to the pressure Ans. : at sea level, then the change in the boiling point of water is. 0 (ake latent heat of vaporisation of water as 70 kj / kg and the change in the specific volume at the boiling point to be. m / kg) af K Solution: 5 v v 5 0 d d d L = = = = = d v v L 70 0 Near II, Hauz Khas, New Delhi 6 hone: / B/6, Jia Sarai, Near II Hauz Khas, New Delhi 6 fiziks.physics@gmail.com 06