Target of 1000 likes to release next booklet for channel members , keep sharing and subscribing INSP . “ Physics will never be the same again ” Theory Notes On “ ENTROPY ” INSP WEB PORTAL : www.inspedu.in YouTube : INDIAN SCHOOL OF PHYSICS Telegram channel ENTROPY INSP BOOKS FOR JEE & OLYMPIAD YOUTUBE : INDIAN SCHOOL OF PHYSICS [ INSP ] INSP WEB PORTAL ( www.inspedu.in) ENTROPY JEE ADV 2023 [EXTRA SYLLABUS ] FOUNDATION OLYMPIAD Improve your JEE mains and JEE ADV score by practicing more than 10K+ specially curated physics problems on INSP web portal [INSP][www.inspedu.in] YT : Indian school of physics NITIN SACHAN ENTROPY (En → Inside; Trope → Transformation): It is denoted by S in the name of Sadi Carnot. → The second law leads to the concept of entropy. → It is the property of a state of a system. → For example, an engine will lose some energy to the surrounding. This puzzle of loss leads to entropy. From the Carnot engine (reversible engine) Heat Engine (Reversible) Q1 T1 = Q 2 T2 Q1 Q 2 = T1 T2 Q1 Q2 +− = 0 T1 T2 w.r.t heat engine, the algebraic sum of Q a cyclic process for a reversible heat engine is zero. T Mathematically, The ratio Q R =0 T Q = entropy (S) T [INSP][www.inspedu.in] YT : Indian school of physics NITIN SACHAN It can also be proved that a reversible cyclic engine operates in a cycle that does not take the heat at a constant temperature. The source is also not at a constant temperature. Because if the heat engine (system) runs not at a constant temperature, the surrounding temperature also varies. In this case, we can derive the same equation as above by dividing the given cycle into an infinite number of Carnot cycles, as shown in the figure. P V Let us consider a thermodynamical plane P – V. In this case; we can divide this thermodynamically reversible cycle into an infinite number of small elemental Carnot's cycles consisting of reversible isothermal and reversible adiabatic processes, as shown in the figure. For each elemental cycle Q1 Q 2 = T1 T2 Similarly Q1 Q2 = T1 T2 Q1 Q2 = T1 T2 By adding all the equations, we get [INSP][www.inspedu.in] YT : Indian school of physics NITIN SACHAN Q1 Q 2 = T1 T2 Q1 Q2 − = 0 T2 cycle T1 Hence Q1 Q 2 +− cycle T1 T2 Q R = T = 0 Q R = 0 (for an infinite number of cycles T → ) Now for a reversible heat engine Q R = 0 …. (*) T Some portions QR are positive, and some QR are negative and Q R = exact differential. So, it is defined as a new quantity S where T dS = 0 b dQ R = dS = Sb − Sa T a S= QR = entropy of the system T [INSP][www.inspedu.in] YT : Indian school of physics NITIN SACHAN Clausius Inequality Entropy for a reversible process: P C 2 b a 1 V Along the path 1 − a − 2 − b − 1 (Re versible) , we can write Q =0 T Q Q + =0 T 1a 2 T 2b1 1 − a − 2 − c −1 (Reversible) Q Q + =0 T 1a 2 T 2c1 … (1) Q Q + = T 1a 2 T 2c1 … (2) Q Q + T 1a 2 T 2c1 Q Q + =0 T 1b2 T 2c1 Q = 0 (for a reversible cycle) T Q Property T Re v = Entropy [INSP][www.inspedu.in] YT : Indian school of physics NITIN SACHAN Entropy for Irreversible Process P c 2 b a 1 V For the cycle 1 − a − 2 − b − 1 (Re versible process) Q Q + =0 T 1a 2 T 2b1 … (1) For the cycle 1 − a − 2 − c − 1 (Irr e versible process) Q Q − + 0 T 2b1 T 2c1 ….(2) From (1) and (2) Q Q T T 2c1 2b1 Q dS T 2C1 …(3) From (1) and (3), For any general process dS Q T ….(4) The above equation is said to be Clausius Inequality. For internal reversible cycle Q dS = T int ernally reversible [INSP][www.inspedu.in] YT : Indian school of physics NITIN SACHAN Q dS [General path] T Hence, Q dS = + ( S)generation T dS is a state function. ( S)generation Is a path function SOME IMPORTANT POINTS 1. A process is reversible if it is both internally as well as externally reversible 2. Internally reversible means no entropy is generated within the system, whereas externally reversible means no entropy is generated outside the system. 3. Entropy is the measure of unavailable energy 4. Though entropy is a property (state function), entropy generation is not a property & is a path function. The entropy of a system can change in the following three ways. 1. Heat interaction → External Interaction 2. Mass interaction 3. Entropy generation → Internal Irreversibility *Work, kinetic energy & potential energy are available energies & hence no entropy is associated with energy interactions in these forms. We know that, dS = [INSP][www.inspedu.in] Q + ( S)gen T YT : Indian school of physics NITIN SACHAN Case (1): Internally reversible ( S)gen = 0 (i) Heat is added to the system Q > 0 (positive) Hence, dS >0 (positive) (ii) Heat rejected by the system, Q → negative dS = Q → negative T (iii) Adiabatic system Q →zero dS = Q = zero T Case (2): Internally Irreversible ( S)gen 0 dS = (i) Q + ( S)gen T If some heat is added to the system, then Q →positive We know that dS = Q + ( S)gen T Q 0 and (S)gen 0 (always) dS 0 T (ii) If heat is rejected by the system, then Q →negative , (S)gen 0 (always) dS = Q + ( S)gen T dS can be positive, negative or zero (iii) For adiabatic system, Q = 0 dS = Q + ( S)gen T Q = 0 dS = (S) gen T For a reversible adiabatic process (S)gen = 0 dS = 0 For an irreversible adiabatic process (S)gen is positive (always) dS > 0 Therefore dS 0 [INSP][www.inspedu.in] YT : Indian school of physics NITIN SACHAN ENTROPY CHANGE OF THE UNIVERSE We know that dS = Q + ( S)generation T The universe is an isolated system, So Q it is zero. o Q dS = + ( S)generation 0 T i.e., according to the second law, all the processes are possible for which dS 0 . This is an increase in the entropy principle. Now we consider two systems A and B interact with each other (in everything) Moreover, the two systems do not interact with anything else. Hence the combined system itself is considered an isolated system. Considering the number of systems like the above, we can call it the universe (in general). Universe A B C Z Within the universe, one interacts with the other. However, one universe does not interact with the other. (S)universe 0 (S)universe = (S)A + (S)B + (S)C + ... In the case of A and B systems, if A is a system and B is surrounding (S) universe = (S)system + (S)surrounding In irreversible heat transfer [INSP][www.inspedu.in] YT : Indian school of physics NITIN SACHAN Let us consider the irreversible heat transfer process The thermal capacity of reservoir = c = T1 > T 2 Thermal heat capacity of the reservoir = Thermal heat capacity of the reservoir = (S) A = − Q T1 ( S) B = + Q T2 1 1 (S)universe = Q − + 0 ( T1 T2 T1 >T2 ) Ex1: Consider two bodies, A and B, of finite heat capacities having temperatures T1 and T2, respectively; when heat transfer (Q) occurs between A and B, the temperature of A decreases, and that of B increases. Finally, the process is naturally stopped and will be at a mean temperature. If we consider the heat capacities of A and B are identical. Then mean temperature = T1 + T2 (the process is natural or irreversible) 2 [INSP][www.inspedu.in] YT : Indian school of physics NITIN SACHAN (S)universe = (S)A + (S)B SA = T T Q f cp dT = = cp ln f 0 ( T T1 T T1 Tf T1 ) (reversible heat transfer) SA 0 (obviously, it loses heat) Similarly T T Q f cp dT SB = = = cp ln f 0 T T2 T T2 SB 0 (S)universe = Cp ln Tf2 T1T2 (S)universe = Cp ln (T1 + T2 ) 2 0 T1T2 ( T1T2 T1 + T2 ) ( GM AM) 2 If a reversible heat engine connects two bodies, then the final temperature of the combined system Tf =? Suniverse = ? Engine Taken internal irreversibility = 0 for A, B. The entire system (A, B, H.E.R.) is reversible if the heat engine is reversible. i.e., (S)universe = (S) A + (S) HE + (S) B R [INSP][www.inspedu.in] YT : Indian school of physics NITIN SACHAN A heat engine is reversible or irreversible; the entropy change = 0 (always) ( Entropy is a property, and the heat engine operates in a thermodynamic cycle) SA , SB they operate on a reversible process so that SA + SB = 0 (summation = 0) (S)universe = 0 (S) universe = C p ln Tf T − C p ln f T1 T2 = Cp ln Tf 2 =0 T1T2 Tf 2 = 1 Tf = T1T2 T1T2 Conclusion: All natural processes in this universe, considered an isolated system, occur in such a way (S)universe 0 . This is the direction of constraint, as mentioned earlier. A directional constraint for any natural process occurs in such a way that it makes a permanent change in the surroundings, i.e., the algebraic sum of the entropy change of the process and the entropy change of the interacting system and surroundings must be greater than zero. If the processes are natural, i.e., the entropy of the universe monotonically increases (i.e., always increases). This is the directional; constraint (in general). [INSP][www.inspedu.in] YT : Indian school of physics NITIN SACHAN Ex2: The coefficient of performance of The heat pump is 50% of the theoretical maximum. It maintains a house at 200C, which leaks energy of 0.6kW per degree of temperature difference to the ambient. For a maximum of 1.0kW, power is ignored. Find the minimum outside temperature for which the heat pump is a good heat source. Sol: Surrounding temperature Ta =? Leaking energy by house = Q.L. To maintain the house at 2930K QH (taken from pump) = QL (leaked energy) (to maintain a steady state) i.e., QH = QL = ( 0.6)( TH − Ta ) kW Q.H. = (coefficient of performance)Heat pump W = ( c.o.p )HP ( W = 1) But (c.o.p)HP = (0.5) (c.o.p)ideal (c.o.p) = QH QH => (c.o.p) = W QH − Qa ( c.o.p )ideal = ( TH − Ta ) 2 T TH H =>QH = 0.5 TH − Ta TH − Ta = 0.6 (T.H. – Ta) 0.5 T = H 0.6 T.H. = 293 K Ta = 277.4 K 40 C [INSP][www.inspedu.in] YT : Indian school of physics NITIN SACHAN Ex3: We wish to produce refrigeration at -300C. A reservoir is available at 2000C, and the ambient temperature is 300CThus the work is done by a cyclic heat engine [C.H.E.] operating between the 2000C reservoir and the surroundings. This work will be used to drive the refrigerator. What will be the ratio of the heat transferred from the 2000C reservoir to the heat transferred from the –300C reservoir? Assuming all processes are reversible. Sol: Drives Refrigerator Heat Engine W Q1 =? QL From heat engine T W = Q1 = Q1 1 − a TH From refrigerator c.o.p = QL W W= Q2 QL TL (reversible heat pump) = = c.o.p Q H − Q L Ta − TL T −T QL L = Q L a 1/ c.o.p TL T −T T L Q1 1 − a = Q L a T T H L Q1 = 0.687 QL W= [INSP][www.inspedu.in] YT : Indian school of physics NITIN SACHAN T- S Diagram dS = Q + ( S)generation T If a process is internally reversible, then ( S)generation it is zero. dS = Q T TdS = Q or Q = TdS T 2 P 3 = 1 V 4 S Area under the T-S diagram gives heat transfer for an internally reversible process when projected on the entropy axis. rev = 1 − QL T ; rev = 1 − L QH TH If arbitrary = reversible [Re versible] If arbitrary reversible [Irreversible] If arbitrary reversible [Im possible] [INSP][www.inspedu.in] YT : Indian school of physics NITIN SACHAN CLAUSIUS INEQUALITY The value of the cyclic integral Q / T is always less than or equal to zero. Q 0 T This inequality is valid for all cycles, reversible or irreversible. For Reversible cycle Heat Energy rev Q Q Q =+ H − L T TH TL Q H TH Q H Q L = = Q L TL TH TL Q = zero T rev Note If a device is shown exchanging heat with more than two reservoirs, then Clausius inequality is advisable. Q 0 T [INSP][www.inspedu.in] YT : Indian school of physics NITIN SACHAN PROBLEMS 1. An ideal gas goes from an initial state ( V1 ,T1 ) to a final state ( V2 ,T2 ) by an unspecified process. The entropy change is (A) nCp ln Ans: Sol: T2 T1 (B) nC V ln T2 T1 (C) nCp ln B From the first law of thermodynamics, dQ = dU + dW => dQ dT PdV = nC v + T T T T1 T2 (D) nC V ln T1 T2 dQ = nCv dT + PdV … (1) The entropy change is f dQ i T f dT = nC v + i T S = … (2) f i P dV T … (3) From the ideal gas equation P nR = T V Substituting this into the second term of equation (3), we have f dT + nR i T T V = nC v ln 2 + nR ln 2 T1 V1 S = nC v f i dV V We have calculated S without consideration of the path followed. Since entropy is a state variable, the entropy change will be the same regardless of the process between the two states. For an isothermal process, T2 = T1 ln T2 = ln (1) = 0 T1 => S = nR ln V2 V1 For an isochoric process, V2 = ln (1) = 0 V1 T S = nC v ln 2 T1 ln [INSP][www.inspedu.in] YT : Indian school of physics NITIN SACHAN 2. Figure shows a container with a membrane partition. The left-hand side of the container, of volume V1 , is filled with an ideal gas; the right-hand side is initially vacuumed. The total volume of the container is V2 . The membrane is punctured and free expansion of gas, which is an irreversible process. The entire system is thermally insulated from its surroundings. Vacuum Gas Membrane The entropy change associated with this free expansion of the ideal gas is (A) nR ln Ans: Sol: V1 V2 (B) 2nR ln V1 V2 (C) nR ln V2 V1 (D) 2nR ln V2 V1 C From the first law of thermodynamics, Q = U + W => In free expansion, W = 0 The temperature of an ideal gas is unchanged during the free expansion; internal energy is a function of temperature only; hence. U = 0 for isolated system; there is no heat transfer to or from the surroundings; Q = 0. dQ = 0 It seems that S = 2 1 dQ is zero, but this is not the case. T We know that S = nCp ln T2 V + nR ln 2 T1 V1 This expression applies to any reversible or irreversible process between the same initial and final states. Since the temperature of the gas is constant (free expansion), S = nR ln V2 V1 Since the final volume Vf is greater than the initial volume, the entropy change is positive; entropy has increased. This is one of the examples in which dS 0 but dQ = 0 [INSP][www.inspedu.in] YT : Indian school of physics NITIN SACHAN 3. Figure shows a well-insulated container with a thin membrane partition. Each of the two chambers has two different liquids. The membrane is punctured, and liquids are allowed to mix. Consider m1 = m2 and C1 = C2 . The entropy change associated with this free mixing of liquids is Ans: Sol: T1 + T2 2 (A) mC ln / T1T2 2 T1 − T2 2 (B) mC ln / T1T2 2 T1 + T2 2 (C) mC ln / T1T2 4 T1 − T2 2 (D) mC ln / T1T2 4 A We first compute the final temperature of the mixture Q net = 0 m1C1 ( Tf − T1 ) + m2C2 ( Tf − T2 ) = 0 m C T + m 2 C 2 T2 Tf = 1 1 1 m1C1 + m 2 C 2 Change in entropy of liquid 1, f dQ i T Tf dT = m1C1 T1 T Since Tf T1 this entropy change is negative, S1 = The entropy change of liquid 2, S2 = m 2 C 2 ln Tf T2 Since Tf T2 this entropy change is positive, the total entropy change is the sum of the individual entropy changes. Stotal = S1 + S2 T T = m1C1 ln f + m 2C 2 ln f Tl T2 The first term is negative, and the second is positive, but total entropy change is positive. Let us see it in a particular case. m1 = m2 , C1 = C2 [INSP][www.inspedu.in] YT : Indian school of physics NITIN SACHAN T + T2 Tf = 1 2 Stotal Tf2 = mC ln T1T2 T1 + T2 2 = mC ln / T1T2 2 T +T Since 1 2 the arithmetic -mean of two temperatures T1 T2 2 GM => Stotal 0 T1T2 is the geometric mean, and AM > This proves that total entropy change will be positive in this irreversible mixing process. 4. Heat Q is transferred from a hot reservoir at temperature T2 to a cold reservoir at temperature T1 . The entropy change is T2 + T1 T1T2 (A) Q Ams: Sol: (B) Q T2 − T1 2T1T2 (C) Q T2 + T1 2T1T2 (D) Q T2 − T1 T1T2 D The process of heat transfer is irreversible. Entropy change of hot reservoir S = f i Q dQ =− T T2 Entropy change of cold reservoir, S = f i Q dQ =+ T T1 The total entropy change of both reservoirs is Stotal = Q Q Q − T1 T2 T2 − T1 T1T2 Since T2 T1 the total entropy change is positive. [INSP][www.inspedu.in] YT : Indian school of physics NITIN SACHAN 5. Ans: Sol: An electric current of 10 amp is maintained for 1 sec. In a resistor of 25 ohms, the resistor's temperature is kept constant at 27°C. What are the resistor's entropy change and the entropy change of the universe? (A) 0 and 8.33 J/K respectively (B) 8.33 J/K and 0 respectively (C) 4.33 J/K and 0 respectively (D) 0 and 4.33 J/K respectively A We know that heat dissipated in a resistor is given by 2 Q = i 2 Rt = (10 ) 25 1 = 2500 joules As the resistor is maintained at 27 o C (300 K), the heat Q is not absorbed by the resistor but taken away by the surroundings. Hence the entropy – change of the resistor ( S)resistor = Q =0 T The entropy change of the surroundings is ( S)surroundings = 6. Ans: Sol: Q 2500 = T 300 = 8.33 joules / K This is also the change of entropy of the universe. Find the change in entropy when 10kg of ice 2 o C is changed to water at 2 o C . (Latent heat of ice is 80 k. cal. Per kg.). (A) 1.001 kcal/ K (B) 2.001 kcal/ K (C) 3.001 kcal/ K (D) 4.001 kcal/ K C Change of entropy T L + 2.302 ms log10 2 T T 1 275 80 = + 2.303 1 1 log10 273 273 = 0.2930 + 0.007137 = 0.3001 cal./K = ( S2 − S1 ) = change in entropy for 10kg (=10,000 gm) = 10,000 0.3001 = 3001cal / o K = 3.001 k cal / o K = 3.001 k cal / o K 7. Ans: Sol: Find the change in entropy when 0.1 kg. of ice −10o C is completely converted into steam at 100o C , given specific heat of ice = 2100 J kg −1K −1 . The ice has Latent heat of fusion = 80 cal. gm −1 and latent of steam = 540 cal. gm −1 . (A) 147 cal / C (B) 167 cal / C (C) 187 cal / C (D) 207 cal / C D Change of entropy T L 2 1 + T1 T1 T L +2.3026 ms log10 2 + 2 T1 T2 = 2.3026 ms log10 [INSP][www.inspedu.in] YT : Indian school of physics NITIN SACHAN 273 80 + = 2.3026 1 0.5 log10 273 263 373 540 + +2.3026 1 1log10 373 273 273 + 0.2930 = 2.3026 0.5 log10 263 +2.3026 1 1 0.1335 +1.447 = 2.3026 0.5 0.0162 + 0.2930 +2.3026 0.1335 + 1.447 273 + 0.2930 = 2.3026 0.5 log10 263 +2.3026 1 1 0.1335 +1.447 = 0.01865 + 0.2930 + 0.3119 + 1.447 = 2.07065cal. / o K Change in entropy for 0.1 kg. ( = 100gm) = 100 2.07065 = 207 cal. / o K 8. Ans: Sol: Calculate the increase in helium's standard entropy when heated from 298K to 1000K at constant pressure. (A) 25.16 Jk −1 mol−1 (B) 18.8 Jk −1 mol−1 (C) 12.9 Jk −1 mol−1 (D) 8.17 Jk −1 mol−1 A The change of entropy is given by dQ where dQ = 1.Cp .dT T C dT dS = p T T2 T dT = Cp log e 2 or S2 − S1 = Cp T1 T T1 T = 2.303Cp log10 2 T1 1000 5 C = 5 R = 2.303 R log10 p 2 2 298 1000 5 = 2.303 8.31 log10 2 298 Solving we get, S2 − S1 = 25.16J K −1mol−1 dS = [INSP][www.inspedu.in] YT : Indian school of physics NITIN SACHAN 9. Ans: Sol: Calculate the change in entropy when 1kg of ice −10o C is wholly converted into steam 100o C . ( ( Cp )ice = 2100 J / kg.K . Latent heat of fusion of ice = 336000 joules/kg, ( Cp )water = 4200J / kg / K Latent heat of steam = 22.68 105 joule /kg.) (A) 7946.38 J / K (B) 8166.38 J / K (C) 8699.38 J / K (D) 8978.38 J / K C Consider the change of entropy as follows: (i) Change is the entropy of 1kg of ice at - 10o C in converting into ice at 0 o C is given by 273 = mCp loge ( T2 / T1 ) = 1 2100 2.303 log10 263 = 2100 2.303 0.0162 = 78.38 Joule /K (ii) Change of entropy when 1kg of ice 0 o C is converted into the water at 0 o C is given by mL 1 336000 = = 1.23 103 Joule / K T 273 (iii) Change of entropy when 1kg of water 0 o C is converted into the water it 100o C is given T 373 = mCp log e 2 = 1 4200 2.303log10 273 T 1 = 4200 2.303 0.1355 = 1311 Joule / K (iv) Change of entropy when 1kg of water 100o C is converted into steam 100o C is given by mL = = (1 22.68 105 / 273) T = 6.08 103 joule / K The total change of entropy is given by 78.38 + 1.23 103 + 1311 + 6.08 103 joule/K = 8699.38 Joule / K . 10. Ans: Silver's molar specific heat capacity at constant pressure in the range 40 K to 120K varies as Cp = 0.076T - 0.0026 T2 − 0.15cal / ( mole − K ) . 5 moles of silver are heated from 40K to 120 K, calculate the change in entropy. (A) 11.25 cal/K (B) 16.25 cal/K (C) 21.25 cal/K (D) 26.25 cal/K C Sol: S = 120 120 40 T 40 S = 5 = 5 Cp dT 120 40 0.076T − 0.00026 T 2 − 0.15 dT T 0.076 − 0.00026T − ( 0.15 / T ) dT = 5 0.076 (120 − 40 ) − 0.00013 (1202 − 402 ) − 0.15 2.3026 log10 (120 / 40 ) = 50.076 80 − 0.0013 12800 − 0.15 2.3026 0.4771 = 21.25 cal/K [INSP][www.inspedu.in] YT : Indian school of physics NITIN SACHAN 11. A mass m of a liquid at temperature Tx K is mixed with an equal mass m of the same liquid at temperature Ty K . The entropy change of the universe will mClog e ( Tx + Ty ) / ( Tx Ty ) be. Find + ? Ans: Sol: (A) 2 (B) 4 (C) 6 (D) 8 B When equal masses at temperature Tx and Ty are mixed, let the temperature of the mixture be T. Then { Heat lost = heat gained} m C ( Tx − T ) = mC ( T − Ty ) Tx + Ty 2 Temperature of mixture T = Entropy changes when the temperature of m units of the liquid changes from Tx T. S1 = mCloge ( T / Tx ) Similarly, entropy changes when the temperature of m units of the liquid change from Ty S2 = m C log e ( T / Ty ) A total change of entropy of the universe S = S1 + S2 = mC log e ( T / Tx ) + mC log e ( T / Ty ) = mC log e T − log e Tx + log e T − log e Ty = mC 2 log e T − log e Tx − log e Ty T = mC log e y T T x y 2 Tx + Ty ) ( T = ( Tx + Ty ) / 2 = mC log e 4Tx Ty T + Ty = mC log e x 2 Tx Ty = 2mClog e ( Tx + Ty ) / 2 Tx Ty 12. Ans: 1 gm mole of an ideal gas expands isothermally from V0 to 4V0 . Find the change in its entropy in terms of gas constant R (A) 1.128 R J/K (B) 1.218 R J/K (C) 1.386 R J/K (D) 1.428 R J/K C Sol: W = V2 V1 PdV = V2 V1 RT dV dT ( PV = RT ) = RTloge ( V2 / V1 ) = 2.3026 RTlog10 ( V2 / V1 ) W = 2.3026 R log10 (4) ] Change in entropy = T [INSP][www.inspedu.in] YT : Indian school of physics NITIN SACHAN 13. The formula gives the variation of specific heat of a substance with temperature Cp = a + bT + cT2 . The change in entropy as the system goes from T1 to T2 is T2 1 2 2 − b ( T2 − T1 ) + ( T2 − T1 ) C T1 (A) 2.3026a log10 T2 1 2 2 + b ( T2 − T1 ) − ( T2 − T1 ) C T1 (B) 2.3026a log10 T2 1 2 2 − b ( T2 − T1 ) − ( T2 − T1 ) C T1 (C) 2.3026a log10 T2 1 2 2 + b ( T2 − T1 ) + ( T2 − T1 ) T C 1 (D) 2.3026a log10 Ans: D [Hint: dQ = CP dT and dS = CP T2 S = CP T1 dT T dT T T2 a = + b + cT dT T T1 = a loge T + bT +12 cT2 14. T2 T1 Calculate the change in entropy when heat flows between two bodies at temperatures T1 T2 and until both reach the standard temperature T0. (The heat capacities of the bodies are C1 and C2 .) T0 T0 + C 2 log10 T1 T2 (A) 2.306 C1 log10 T T (B) 2.306 C1 log10 0 − C2 log10 0 T1 T2 T0 T0 + C 2 log10 T1 T2 (C) 2.306 2C1 log10 T T (D) 2.306 C1 log10 0 + 2C2 log10 0 T1 T2 Ans: A Hint : C1 ( T1 − T0 ) = C2 ( T0 − T2 ) C T +C T 1 1 2 2 C + C S1 = C1 loge ( T0 / T4 ) and S2 = C2 loge ( T0 / T2 ) 1 2 T T S = 2.306 C1 log10 0 + C2 log10 0 T1 T2 T = [INSP][www.inspedu.in] YT : Indian school of physics NITIN SACHAN 0.2 kg ice 0C is in 0.5kg water 20C . Calculate the change in entropy of the ice as it melts 0C . Lf= 3.33 105 J / kg Ans: +244.0 J/K Sol: Convert all temperatures into Kelvin. f dQ m L Smelt = Sf − Si = = ice f i T T 15. = 16. Sol: 17. Sol: 18. ( 0.2 kg ) ( 3.33 105 J / kg ) 273K = +244.0 J / K 0.1 kg ice 0C is in 0.5 kg water 20C . Calculate the entropy change of the ice-water mixture as it moved to the final temperature. f dQ f mice water c w dT f dT Sice water = Sf − Si = = = mice water c w i T i i T T T 276.4K = mice water c w ln f = ( 0.1kg )( 4186J / kg K ) ln = +5.2J / K 273K Ti Suppose 0.1 kg ice 0C is in 0.5 kg water at 20C . Find the entropy change of the water as it melted the ice. f m f dT dQ water c w dT = = m water c w i T i i T T T 277.1K = mwater c w ln f = ( 0.5kg )( 4186J / kg K ) ln = −116.8J / K 293K Ti Swater melt = Sf − Si = f Suppose 0.1 kg ice at 0C is in 0.5 kg water at 20C . Find the entropy change of the water as it cooled down to the final temperature. Ans: Sol: 19. f m f dT c dT dQ = water w = m water c w i T i i T T T 276.4K = mwater c w ln f = ( 0.5kg )( 4186J / kg K ) ln = −5.3J / K T 277.1K i Suppose 0.1 kg ice at 0C is in 0.5 kg water at 20C . Determine the change in entropy of the ice/water system in thermal equilibrium. Swater cooled = Sf − Si = Ans: Sol: Stotal = Smelt + Sice water = (+122.0 + 5.2 = +5.1 J / K [INSP][www.inspedu.in] f + Swater melt + Swater cooled + – 116.8 + – 5.3) J/K YT : Indian school of physics NITIN SACHAN 20. One mole of an ideal, monatomic gas is initially at point A with pressure ( pA ) , volume ( VA ) , and temperature ( TA = TH ) . After that, it is allowed to expand isothermally to point B Calculate the change in entropy of the gas SAB going from A to B. A Q AB p B T H V Sol: 21. Eint = Q − W = 0 QAB = WAB = nRTH ln ( VB / VA ) B dQ 1 B Q AB SAB = = dQ = A T TH A TH SAB = nR ln ( VB / VA ) 0 1 mole of an ideal, monatomic gas is initially at point A with pressure ( pA ) , volume ( VA ) , and temperature ( TA = TH ) . Then, the gas expands isothermally to point B and cools isochronally to point C. What is the change in entropy of the gas SBC going from B to C? A B T H p C TL QBC V Ans: Sol: Eint = QBC dEint = dQBC = 3 / 2nR dT C dQ TL 3 / 2 nR dT SBC = = = 3 / 2 nR ln ( TL / TH ) B T TH T 3 3 SBC = nR ln ( TL / TH ) = − nR ln ( TH / TL ) 0 2 2 22. 1 mole of an ideal, monatomic gas is initially at point A with pressure ( pA ) , volume ( VA ) , and temperature ( TA = TH ) . After that, it is allowed to expand isothermally to point B, then cool isochronally to point C, and then contract isothermally to point D Find the entropy change of the gas SCD going from C to D. [INSP][www.inspedu.in] YT : Indian school of physics NITIN SACHAN A p B T H D QCD C TL V Sol: 23. Eint = Q − W = 0 QCD = WCD = nRTL ln ( VA / VB ) D dQ Q 1 D SCD = = dQ = CD C T TL C TL SCD = nR ln ( VA / VB ) = −nR ln ( VB / VA ) 0 1 mole of an ideal, monatomic gas is initially at point A with pressure ( pA ) , volume ( VA ) , and temperature ( TA = TH ) . Then, it expands isothermally to point B, cools isochronally to point C, contracts isothermally to point D, and warms up isochronally back to the initial point A.Calculate the change in entropy of the gas SDA going from D to A. A p Q DA B D C TH TL V Sol: SDA 24. Eint = QDA dQDA = 3 / 2nR dT A dQ TH 3 / 2 nR dT SDA = = = 3 / 2 nR ln ( TH / TL ) D T TL T 3 = nR ln ( TH / TL ) 0 2 1 mole of an ideal, monatomic gas is initially at point A with pressure ( pA ) , volume ( VA ) , and temperature ( TA = TH ) . It then expands isothermally to point B, cools isochronally to point C, contracts isothermally to point D, and warms up isochronally back to the initial point A. Calculate the net entropy change in one cycle. [INSP][www.inspedu.in] YT : Indian school of physics NITIN SACHAN A p B D C TH TL V Sol: 25. Easy way : S cycle = 0 ! 1 mole of an ideal, monatomic gas is initially at point A with pressure ( pA ) , volume ( VA ) , and temperature ( TA = TH ) . It then expands isothermally to point B, cools isochronally to point C, contracts isothermally to point D, and warms up isochronally back to the initial point A. Find the net entropy change in one cycle. A p B D C TH TL V Ans: Sol: Harder way : Snet = SAB + SBC + SCD + SDA V 3 T V 3 T = nR ln B − nR ln H − nR ln B + nR ln H => Snet = Scycle = 0 ! VA 2 TL VA 2 TL 26. A Carnot engine that operates between the temperatures TH = 1700K and TL = 600K , the engine performs 1200 J of work each cycle TH QH QL TL What is the efficiency of this engine? [INSP][www.inspedu.in] YT : Indian school of physics NITIN SACHAN TL 300 = 1− = 0.647 TH 850 A Carnot engine that operates between the temperatures TH = 1700K and TL = 600K performs 1200 J of work each cycle. = 1− 27. TH QH QL TL Calculate the heat extracted from the high-temperature reservoir each cycle. W W 1200 J = = 0.647 Q H = = = 1854 J QH 0.647 28. A Carnot engine that operates between the temperatures TH = 1700K and TL = 600K , the engine performs 1200 J of work each cycle TH QH QL TL Calculate the heat expelled into the low-temperature reservoir each cycle. QL = QH − W = 1854J − 1200J = 654J [INSP][www.inspedu.in] YT : Indian school of physics NITIN SACHAN [INSP][www.inspedu.in] YT : Indian school of physics NITIN SACHAN