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01. THERMODYNAMICS 1. Basic Concepts & Zeroth Law Ans. (d) : Throttling Process–In this process, flow of fluid through some restricted or partially opened or...
01. THERMODYNAMICS 1. Basic Concepts & Zeroth Law Ans. (d) : Throttling Process–In this process, flow of fluid through some restricted or partially opened or of Thermodynamics porous body or very small opening. It is also known as 1. Zeroth law of thermodynamics forms the basis constant enthalpy process or isenthalpic process. of measurement of– Characteristics of throttling process– (a) Pressure (b) Temperature (i) No work transfer (c) Heat exchange (d) Work (ii) No heat transfer MRPL Tech. Asstt. Trainee 2021 (iii) It is irreversible process GSECL 23-02-2020 (Shift-I) (iv) It is isenthalpic process Mizoram PSC 2019 (Paper-2) 3. Which of the following is the extensive NPCIL 07.11.2019 (2.00-4.00 pm) property of thermodynamic system? SSC JE 27-09-2019 (Shift-1) (a) Pressure (b) Volume HPCL JE 07-11-2019 (Shift-2) (c) Temperature (d) Density GPSC (Mech) 21.07.2019 SSC JE 22-03-2021 Shift-II Nagaland PSC 2018 Paper-I RRB JE CBT-II 31.08.2019 IInd Shift VIZAG JET 28-10-2018 WBPSC 2018, 2000 SSC JE 25. 1. 2018 (3.15 pm), Vizag JET 2017 Nagaland PSC 2018 Paper-I GPSC AMVI (ME) 24-12-2016, WBPSC 2016 Karnataka PSC JE 09-09-2017 GPSC AMVI (Auto) 24-12-2016 Vizag JET 2017 (MP Sub Engineer 5 April 2016 Evening) UPSSSC JE 2015 ISRO ISAC 13-11-2016, KPCL JE 2016 RRB JE [Exam Date : 04-01-2015 (Yellow Paper)] (M.P. Sub Engineer 2015) SSC JE 2013 GSSB ITI Supervision Inst. (Mech) 18-01-2015 ESE-1999 RRB SSE [Exam Date : 01-09-2015 (Shift-I)] RRB JE [Exam Date : 26-08-2015 (Shift-III)] Ans : (b) Volume is a extensive property. RRB SSE [Exam Date : 02-09-2015 (Shift-II)] Properties can be classified on the basis of size/extent Mizoram PSC 2014 (Paper-I) of the system RRB JE [Exam Date : 14-12-2014 (05 Yellow Paper)] RRB JE [Exam Date : 14-12-2014 (01 Red Paper)] UPRVUNL JE 2014, MPSC AMVI 2013 GPSC Motor Vehicle (Auto) 02-12-2012 SSC JE 2008, ESE-1996 Ans : (b) Zeroth law of thermodynamics forms the basis of measurement of temperature. This law states that when two bodies are in thermal equilibrium with a third body, they are also in thermal equilibrium with each other. First law of thermodynamics defines internal energy. 4. Heat and work are : Second law of thermodynamics defines entropy of the (a) Intensive properties system. (b) Extensive properties (c) Point functions 2. During throttling process- (d) Path function (a) Internal energy does not change Kerala PSC AM 09.08.2021 (b) Pressure does not change UPCL AE 29.08.2021 (c) Entropy does not change ISRO LPSC 11-05-2014 (d) Enthalpy does not change (MP Engineer JE Paper II 2011) MRPL Tech. Asstt. Trainee 2021 SSC JE 2021, 2010 NPCIL 07.11.2019 (9.00-11.00 am) GSSSB (SI) 14-07-2019 SSC JE 27-09-2019 (Shift-2) ISRO MCF 23-06-2019 HPCL JE 07-11-2019 (Shift-1) Ans. (d) : Heat–Energy that is transferred from one SSC JE 22 1. 2018 (3.15 pm) body to another as the result of a difference in ISRO VSSC 06-08-2017 temperature. SSC JE 4 March 2017 Shift-I, SSC JE 2013 GPSC Motor Vehicle (Auto) 02-12-2012 Heat flows from the hotter body to the colder body. MPSC AMVI 2011 Temperature–Temperature is a physical quantity that (MP Engineer JE Paper II 2011) expresses hot and cold. Thermodynamics 11 YCT Heat and work are path function. These are denoted Ans : (b) Thermodynamic System–It is a region in by dynamic path. So it is inexact and imperfect space upon which the study/analysis is focused. differential. Boundary–It is the separation between system and surrounding. 2 2 W W12 ; W W2 W1 1 1 2 2 1 Q Q12 ; 1 Q Q2 Q1 Q2 –B–1 Q2 – C–1 ; W2 –B–1 W2 – C – 1 Types of System– (1) Open System–In this type of system both mass and 5. Control volume in a thermodynamic system energy transfer take place. refers to– Ex.- Turbine, pump, piston-cylinder arrangement with (a) A specified mass in fluid flow valve etc. (b) Mass that moves across the boundary Most of engineering devices are open system. (c) Fixed region in space for thermodynamic study (2) Closed System–A system is said to be a close (d) Transfer of energy across the boundary system, there is no mass transfer but energy transfer SSC JE 2021 RRB JE CBT-II 31.08.2019 IInd Shift takes place. Ex.-Rigid closed container with conductive Nagaland PSC 2018 Paper-I walls, piston cylinder arrangement without valve. SSC JE 29. 1. 2018 (3.15 pm) (3) Isolated System–There is no mass as well as energy SSC JE 3 March 2017 Shift-II transfer takes place. RRB JE [Exam Date : 27-08-2015 (Shift-I)] Ex.-Universe, hot fluid in perfect insulated ESE-2010 flask/container. Ans : (c) Control volume in a thermodynamic system 7. Which of the following images indicates single refers to fixed region in space for thermodynamic study. riveted joints? * This system has no any definite boundary i.e. it has an imagninary boundary. * Turbine impeller, pump, compressor etc. are held in (a) the category of control volume system. (b) 6. A closed thermodynamic system is one in (c) which: (a) There is no energy or mass transfer across the boundary (b) There is no mass transfer but energy transfer exists (c) There is no energy transfer but there is mass (d) transfer (d) Both energy and mass transfer exist UPSSSC JE 19.12.2021 SSC JE 28-10-2020 (3 to 5 pm) Ans. (b) : GPSC AMVI 2020 Nagaland PSC 2018 Paper-I NPCIL 03.06.2018 SSC JE 29. 1. 2018 (10.15 am) Double riveted lap joint SSC JE 4 March 2017 Shift-II TSPSC AE 2015 Mizoram PSC 2014 (Paper-I) UPRVUNL JE 2014 RRB JE [Exam Date : 14-12-2014 (05 Yellow Paper)] Single riveted lap joint GPSC Motor Vehicle (Auto) 02-12-2012 (ESE-2010, 1999) Thermodynamics 12 YCT Double riveted butt joint 11. Internal Energy is a function of only (a) Pressure (b) Temperature (absolute) (c) Volume (d) Pressure & Temperature RRB JE CBT-II 31.08.2019 IInd Shift Nagaland NPCIL 07.11.2019 (9.00-11.00 am) Double riveted lap joint HPCL JE 07-11-2019 (Shift-1) PSC (JE) 2017 (Paper-2) RRB JE [Exam Date : 27-08-2015 (Shift-III)] 8. Fill in the blanks with the appropriate answer: GSSB ITI Supervision Inst. (Mech) 18-01-2015 A.............. system is exemplified by a jet RRB JE [Exam Date : 04-01-2015 (Yellow Paper)] engine. UPSSSC JE 2015 (a) Isotherm JK SSB JE (103) 2014 (b) Control volume system Mizoram PSC 2014 (Paper-I) (c) constant-mass systems SC JE 2012 (d) Isobaric GPSC Motor Vehicle (Mech) 02-12-2012 UPSSSC JE 19.12.2021 MP Engineer JE Paper II 2011 Ans. (c) : Constant mass system is exemplified by a jet Ans. (b) : Internal energy represents the total energy engine. associated with molecules. We know that, for an ideal 9. Which of the following statements about an gas– ideal gas is false? U = mCvT (a) The total volume of the individual molecule dU = mCvdT is lesser than the volume that the gas The above equation is valid for an ideal gas occupies undergoing any process. (b) Motion is frictionless * According to Joule's law, internal energy of an ideal (c) Follow Charle's law gas is a function of temperature only. (d) Presence of intermolecular forces and interactions 12. The area under the T-S diagram curve at any UPSSSC JE 19.12.2021 thermodynamic process represents : Ans. (d) : Ideal gas has no molecular force of attraction (a) heat rejected only or repulsion or there are absence of intermolecular (b) heat absorbs only forces and interactions. (c) work done during the process The total volume of the individual molecule is lesser (d) heat absorbed and rejected (or negligible) than the volume that the gas occupies. GSECL 23-02-2020 (Shift-2) It's motion is frictionless. RRB SSE [Exam Date : 03-09-2015 (Shift-III)] It follows all gas laws– Ex. Charle's law, Boyle's law, (M.P. Sub Engineer 2015) Gay-Lussac law. UPRVUNL JE 2014 10. Enthalpy is: JK SSB JE (103) 2014 (a) internal energy / pressure volume product (MP Engineer JE Paper II 2011) (b) internal energy – pressure volume product Ans : (d) (c) internal energy pressure volume product (d) internal energy + pressure volume product SSC JE 27-10-2020 (Shift-1) HPCL JE 20-04-2019 SSC JE 27-09-2019 (Shift-1) SSC JE 22 1. 2018 (10.15 am) SSC JE 29. 1. 2018 (10.15 am) RRB SSE [Exam Date : 01-09-2015 (Shift-II)] The area under the T-S diagram curve at any SSC JE 2015 thermodynamic process represents heat absorbed RRB JE [Exam Date : 14-12-2014 (04 Green Paper)] and rejected. Ans. (d) : Enthalpy is the addition of internal energy i.e. Q T.dS and product of pressure volume. H= U + PV The area under the P-V diagram curve at any thermodynamic process represents work done Internal External energy during the process. energy (Flow work) w P.dV Enthalpy is the compound property. For ideal gas– 13. Isolated system indicates– (a) Mass of substance cross the boundary dh CpdT For unit mass (b) Energy of substance cross the boundary dH mCp dT For total mass (c) Both mass and energy of substance cross the boundary Thermodynamics 13 YCT (d) Both mass and energy of substance does not Ans. (c) : Internal energy and enthalpy for an ideal gas cross the boundary is the function of temperature. SSC JE 28-10-2020 (3 to 5 pm) * Change in internal energy– Nagaland PSC 2018 Paper-I dU mC v dT ISRO IPRC 10-12-2016 RRB JE [Exam Date : 28-08-2015 (Shift-I)] * Change in enthalpy– RRB SSE [Exam Date : 02-09-2015 (Shift-III)] dH mCpdT (ESE-2011) Ans : (d) Isolated system indicates that both mass and * In an isothermal process the temperature is constant. energy of substance does not cross the boundary. * So, the internal energy in isothermal process remains Example: Thermos Flask, Casserole, Thermodynamic constant. universe. 16. Change in internal energy in a closed system is Note–Systems are classified on the basis of mass flow equal to heat transferred if the reversible and energy flow across the boundary. process takes place at constant : (a) Pressure (b) Temperature System Mass Energy (c) Volume (d) Internal energy Closed Not cross Cross Nagaland PSC 2018 Paper-I Open Cross Cross SSC JE 27. 1. 2018 (3.15 pm) SSC JE 29. 1. 2018 (3.15 pm) Isolated Not cross Not cross ISRO VSSC 08-02-2015 14. A process in which no heat crosses the JK SSB JE (103) 2014 boundary of the system is called : (ESE-2008, 2005) (a) Ideal process (b) Adiabatic process Ans : (c) First law of thermodynamics for closed (c) Isothermal process (d) Isobaric process system– DSSSB JE 16-11-2019 (12.30-2.30) Q dU W SSC JE 23. 1. 2018 (3.15 pm) KPCL JE 2016 RRB SSE [Exam Date : 01-09-2015 (Shift-I)] UPSSSC JE 2015 SSC JE 2014 (MORNING) MPSC AMVI 2013 Ans. (b) A process in which no heat crosses the boundary of the system is called adiabatic process. Q = dU if W = 0 i.e. Q 0 Now W for closed system– Q dU W W = PdV If, V = Constant W dU........ (For adiabatic Q = 0) Then, PdV = 0, W = 0 Note–Reversible adiabatic process is also called isentropic Internal energy is equal to heat transferred for closed process. system. 17. Under certain polytropic process, value of n = 1....... process : (a) Adiabatic (b) Reversible (c) Irreversible (d) Isothermal ISRO VSSC 01-07-2018 (MP Sub Engineer 5th April 2016 Morning UPSSSC JE 2016 ISRO LPSC 11-05-2014 RRB-JE 30.08.2019, Ist Shift 15. In an isothermal process, the internal energy NPCIL 07.11.2019 (2.00-4.00 pm) (a) Increases (b) Decreases Ans. (d) (c) Remains constant (d) None of the above PVn =P Vn GPSC AMVI 2020 1 1 2 2 SSC JE 1 March 2017 Shift-II then n = 1 SSC JE 4 March 2017 Shift-I P V = constant UPRVUNL JE 2016 (Shift-II) 1 1 WBPSC 2016 So, it is Isothermal process ISRO VSSC 08-02-2015 18. In a free expansion process UPSSSC JE 2015 (a) the work done is zero SSC JE 2014 (MORNING) (b) the heat transfer is zero WBPSC 2001 (c) the work done and heat transfer both are zero Thermodynamics 14 YCT (d) the work done is zero, but the heat transfer Ans. (d) : Thermodynamic Property–All measurable increases characterstics of the system known as properties. For (MP Sub Engineer 4th April 2016 Evening) example pressure, temperature, volume, enthalpy, (MP Engineer JE Paper II 2011) entropy, density etc. SSC JE 2010 These are two types– (ESE-2012) (i) Intensive/Intrinsic Properties–All the ISRO LPSC 11-05-2014 thermodynamic properties which independent of mass RRB JE [Exam Date : 27-08-2015 (Shift-III)] (size) of system. Ans. (c) In a free expansion process the work done and Example–Pressure, temperature, thermal conductivity, heat transfer both are zero. density and all the specific properties. Free expansion process–A free expansion occurs when (ii) Extensive/Extrinsic Properties– a fluid is allowed to expand suddenly into a vacuum All the thermodynamic properties which are depend on chamber through an orifice of large dimensions. In this extent or mass of the system. Example–Volume, mass, process, no heat is supplied or rejected and no external energy, enthalpy etc. work is done. 22. Which of the following are intensive properties? In free expansion process– (1) Kinetic Energy Q1-2 = 0, W1-2 = 0 and du = 0 (2) Specific Enthalpy P, V, T = C (ideal gas), h = C, (3) Pressure But dS (4) Entropy Select the correct answer using the code given 19. Which of the following is NOT an extensive below property? (a) 1 and 3 (b) 2 and 3 (a) Enthalpy (b) Entropy (c) 1, 3 and 4 (d) 2 and 4 (c) Specific enthalpy (d) None of these (ESE-2005, 2007) MRPL Tech. Asstt. Trainee 2021 SSC JE 29. 1. 2018 (3.15 pm) SSC JE 23. 1. 2018 (10.15 am) Mizoram PSC 2018 (Paper-1) SSC JE 25. 1. 2018 (3.15 pm) RRB SSE [Exam Date : 02-09-2015 (Shift-II)] SSC JE 29. 1. 2018 (3.15 pm) Ans : (b) Specific enthalpy, pressure are intensive SSC JE 2013 properties. Ans. (c) Properties may be of two types – Properties can be classified on the basis of size/extent Intensive properties– Intensive properties are of the system– independent on the mass of the system, e.g. pressure, temperature, specific enthalpy, specific volume, specific energy, density etc. Extensive properties– Extensive properties are dependent upon the mass of the system e.g. volume, energy, enthalpy, entropy etc. 20. Of the following, 'path function' quantity is (a) Work done (b) pressure (c) Enthalpy (d) Temperature HPCL JE 07-11-2019 (Shift-1) NPCIL 07.11.2019 (9.00-11.00 am) Note–Each specific extensive property are intensive. Mizoram PSC 2015 (Paper-1) Example: Specific volume, specific entropy, specific Mizoram PSC 2014 (Paper-I) enthalpy etc. SSC JE 2012 23. Which of the following is not a property of Ans. (a) Heat and work done are the path system? function. (a) Temperature (b) Pressure Heat and work are not thermodynamic properties (c) Specific volume (d) Heat because these are not point function. Its value depends GPSC AMVI 2020 SSC JE 2012 on the path of the process in the system. GPSC Motor Vehicle (Auto) 02-12-2012 21. Which of the following is an intensive (MP Engineer JE Paper II 2011) property? Ans. (d) Heat is a transient energy and not a property of (a) Temperature the system. It is a path function. (b) Pressure This energy is due to temperature difference. (c) Volume 24. Throttling is........ process : (d) Both temperature and pressure (a) Reversible (b) Irreversible NPCIL 07.11.2019 (2.00-4.00 pm) (c) Adiabatic (d) Isothermal ISRO MCF 23-06-2019 ISRO SDSC 08-04-2018 SSC JE 27. 1. 2018 (3.15 pm) SSC JE 25. 1. 2018 (3.15 pm) Karnataka PSC JE 09-09-2017 SSC JE 3 March 2017 Shift-I RRB JE [Exam Date : 14-12-2014 (04 Green Paper)] UPSSSC JE 2016 Thermodynamics 15 YCT Ans. (b) : Throttling is an irreversible process. Ans. (c) : Given, T1 = 27°C = 273 + 27 = 300 K A throttling process is defined as a process in which there T2 = ? is no change in enthalpy from state 1 to state 2[h1= h2]. V1 = V There is no work done ( W= 0) V2 = 2V Enthalpy will remain constant. P = Constant 25. Temperature of a gas is produced due to According to Charles's law– (a) its heating value VT (b) kinetic energy of molecules (c) repulsion of molecules V1 T1 (d) attraction of molecules V2 T2 SSC JE 4 March 2017 Shift-I SSC JE 3 March 2017 Shift-II V GPSC AMVI (ME) 24-12-2016 T2 2 T1.....(i) UPRVUNL JE 2015 V1 WBPSC 2004 Put the value in equation (i) Ans. (b) : Temperature of a gas is produced due to 2V kinetic energy of molecules. T2 300 600K V The average kinetic energy of gas particles is proportional to the absolute temperature of the gas, and T2 600 273 327C all gasses at the same temperature have the same average kinetic energy. 28. Which of the following is not an extensive 3 property : ke = KT (a) entropy 2 (b) enthalpy ke T (c) internal energy 26. What is the temperature at which a system goes under a reversible isothermal process (d) density without heat transfer? RSMSSB JEN 13.12.2020 (Degree) (a) Absolute zero temperature SSC JE 4 March 2017 Shift-II (b) Critical temperature (M.P. Sub Engineer 2015) (c) Reversible temperature Ans. (d) Extensive property:- A property whose value (d) Boiling temperature depends upon mass, shape, size of the system is known SSC JE 23. 1. 2018 (3.15 pm) as extensive property. (MP Sub Engineer 5th April 2016 Morning) UPSSSC JE 2015 Ex.– Volume, Mass, Entropy, Enthalpy, Kinetic energy, MPSC AMVI 2011 Potential energy etc. Ans. (a) According to the definition of Kelvin scale– Intensive property:- A property whose value does not Q T depend upon the mass, shape and size of the system, called intensive property. Q t Tt Ex.:- Density, Pressure, Temperature, Specific volume, Q Q T Tt 273.16 Specific entropy etc. Qt Qt 29. The ratio of two specific heats of air is equal : From the above equation, it is clear that the heat (a) 0. 17 (b) 0. 24 transferred isothermally between two adiabatic bodies (c) 0.1 (d) 1.41 decreases if temperature decreases. RSMSSB JEN 26.12.2020 (Degree) Here Qt is the thermometric property. Therefore, from ISRO VSSC 01-07-2018 above equation if Q is having smaller value then T also SSC JE 2007 lowers. The smallest possible value of Q is zero and corresponding T is zero. Ans. : (d) The ratio of two specific heats of air i.e. Thus, the temperature at which a system undergoes a specific heat at constant pressure (C P) to specific heat at reversible isothermal process without transfer of heat is constant volume (C V) is– called as absolute zero temperature. CP 1.41 27. A perfect gas at 27ºC is heated at constant CV pressure till its volume is doubled. The final Note :- for CO2 = 1.28, for steam = 1.33 temperature is (a) 54ºC (b) 108ºC 30. Density of water is maximum at : (c) 327ºC (d) 600ºC (a) 00C (b) 00K 0 ISRO VSSC 01-07-2018 (c) 4 C (d) 1000C RRB JE [Exam Date : 21-12-2014 (07 Red Paper)] RSMSSB JEN 26.12.2020 (Degree) RRB JE [Exam Date : 14-12-2014 (02 Red Paper)] SSC JE 2014 (Morning) SSC JE 4 March 2017 Shift-I SSC JE 2014 (Evening) Thermodynamics 16 YCT Ans. (c) : Steam formation diagram Pump/Compressor Nozzle/Diffuser Isolated system– A. Thermos Flask/Casserole b. Thermodynamic universe 34. Constant volume process is also called as (a) Isobaric (b) Isochoric (c) Isentropic (d) Adiabatic SSC JE 2021 ISRO MCF 23-06-2019 SSC JE 27-09-2019 (Shift-1) Fig show that volume of water is minimum and density KPCL JE 2016 is maximum at 4°C temperature. Ans. (b) : Constant volume (V = C) process is also 31. What is the work done for a constant volume called isochoric or isometric process. process? (a) 0.5 (b) 0 (c) –1 (d) 1 SSC JE 27-10-2020 (Shift-1) ISRO VSSC 01-07-2018 SSC JE 27-10-2020 (Shift-3) Ans. (b) : As per question, Work is zero for isochoric process in close system. dV = 0 i.e V = Constant V1 = V and V2 = V In open system, work is (W) = VdP (non zero) So, W1–2 = P(V2 – V1) (For closed system) 35. An Ideal gas at 27ºC is heated at constant = P(V – V) pressure till its volume becomes three times. =0 What would be the temperature of gas? 32. When there is no heat exchange between two (a) 81ºC (b) 627ºC bodies that are at the same temperature, they (c) 543ºC (d) 327ºC are said to be in Nagaland PSC 2018 Paper-I (a) thermal equilibrium SSC JE 4 March 2017 Shift-II (b) energy equilibrium RRB JE [Exam Date : 29-08-2015 (Shift-III)] (c) heat equilibrium TSPSC AE 2015 (d) electrical equilibrium Ans. (b) : Given, SSC JE 27-09-2019 (Shift-1) T1 = 27ºC = 27 + 273 = 300K RRB JE [Exam Date : 14-12-2014 (03 Green Paper)] P = C GPSC (Mech) 21.07.2019 V2 = 3V1 Ans. (a) : When there is no heat exchange between two VT bodies that are at the same temperature they are said to V1 T1 V be in thermal equilibrium. T2 T1 2 = 300 × 3 = 900 K Electrical equilibrium–When potential difference is V2 T2 V1 zero. T 2 = 900 – 273 = 627ºC 33. Which one of the following represents open 36. Total heat of a substance is also known as : thermodynamic system? (a) internal energy (b) entropy (a) Manual ice cream freezer (c) latent heat (d) enthalpy (b) Centrifugal pump GSECL 23-02-2020 (Shift-I) (c) Pressure cooker SSC JE 1 March 2017 Shift-II Mizoram PSC 2015 (Paper-1) (d) Bomb calorimeter Nagaland PSC 2018 Paper-I Ans. (d) : Total heat of a substance is also known as ISRO IPRC 10-12-2016 enthalpy. RRB JE [Exam Date : 29-08-2015 (Shift-II)] Specific. enthalpy (h) = u + pv J/kg (ESE-2011) and enthalpy (H) = U + PV Joule Ans : (b) Closed system– Enthalpy is a compound property, meant for solving A. Piston-Cylinder Arrangement without valve certain type of thermodynamic problem (Generally flow B. Bomb Calorimeter process). C. Motor Car battery H U PV for total mass (m) D. Pressure cooker h u pv for unit mass Open system– A. Motor car engine (I.C. engine) B. Any wheel (Water turbine/Gas turbine/Steam Turbine) C. Boiler/Condenser Thermodynamics 17 YCT 37. The universal gas constant in S.I. unit is equal Ans. (a) : At 0 K (–273 degree Celsius) the particles to stop moving and all disorder disappears. Thus nothing (a) 8314 (b) 831.4 can be colder than absolute zero on the Kelvin scale. (c) 8480 (d) 848 41. In a thermodynamic system, a process in which SSC JE 23. 1. 2018 (10.15 am) volume remains constant is called.............. WBPSC 2009 process. GPSC Motor Vehicle (Auto) 02-12-2012 (a) isobaric (b) isometric Ans. (a) : The universal gas constant in S.I. unit is (c) adiabatic (d) isentropic equal to 8314 J/kg-mole K or 8.314 kJ/kg-mole K. MPSC AMVI 2013 The universal gas constant (generally denoted by SSC JE 24. 1. 2018 (10.15 am) Ru) of gas is the product of the gas constant (R) and the RRB JE [Exam Date : 14-12-2014 (03 Green Paper)] molecular mass (M) of the gas. Ans. (b) : In a thermodynamic system, a process in R u MR which volume remains constant is called isometric process or isochoric process. Where - For closed system, M = Molecular mass of gas expressed in kg-mole At V=C R = Gas constant W = 0 38. The process in which no heat enters or leaves This process is governed by Gay-Lussac Law. the system is called– 42. The heat absorbed or rejected during a (a) Isobaric (b) Isothermal polytropic process is equal to : (c) Isentropic (d) Isochoric 1/ 2 SSC JE 2014 (Evening) n (A) workdone BSSC WRD JE 2016 1 RRB JE [Exam Date : 27-08-2015 (Shift-III)] 1/ 2 Ans : (c) The process in which no heat enters or n (B) workdone leaves the system is called Isentropic process or n 1 adiabatic process. n The temperature of the gas changes, as the work is (C) workdone done at the cost of internal energy. 1 2 In this process, the change in internal energy is equal n to the work done. (D) workdone 1 Q du W Q 0 (a) (A) only (b) (B) only du W (c) (C) only (d) (D) only (ESE-2005, 2002) Note- Reversible adiabatic process is called isentropic SSC JE 25. 1. 2018 (3.15 pm) process. Ans. (c) Polytropic process– It is most commonly used 39. Molar specific heat of an ideal gas depend on process in practice. This thermodynamics process is (a) Its pressure said to be governed by the law PVn = Const. where 'n' is (b) Its temperature the index which can vary from– (c) Both its pressure and temperature 1 Cv Mizoram PSC 2016 (Paper-2) 1 1 Ans. (a) : Irreversible process, cannot lead to the C v Cp initial state of the system and the surroundings without external inputs. dT dT This process can not be drawn in a firm line. So, dS v dS p Isenthalpic process is an irreversible adiabatic So, on T - S diagram slope of constant volume is more process. than slope of constant pressure. 272. Which of the following is Extensive property ? 269. An adiabatic process is one in which (a) Entropy (b) Kinetic Energy (a) the temperature of the gas changes (c) Internal Energy (d) All of the above (b) no heat enters or leaves the gas GPSC AMVI (Auto) 24-12-2016 (c) the change in internal energy is equal to the Ans. (d) : Extensive Property– The property of the mechanical work done system, whose value for entire system is equal to the (d) All of the above sum of their values for the individual parts of the system MP Sub Engineer 5 April 2016 Evening are called extensive properties. Ans. (d) Some facts about adiabatic process– i.e. P = P1 + P2 + P3 +......... No heat enters or leaves to or from the working Ex.– Entropy, kinetic energy, internal energy enthalpy substance. etc. The temperature of the gas changes. 273. An aircraft engine is an example of: The change in internal energy is equal to the work (a) closed system (b) open system done. (c) isolated system (d) quasi equilibrium In adiabatic process– UPRVUNL JE 2016 (Shift-II) Q = 0 Ans. (b) : An aircraft engine is an example of open Q dU W system. An open system is defined as in which the mass and heat energy can be transfer to its dU W surroundings. Open system is also called a control It is expressed by the relation, PV = constant. volume system. Where is the isentropic index and it's value is– Example - Air compressor, Turbine etc. 2 274. Consider the following: For monoatomic gas, 1 1.67 3 1. Temperature 2 2. Enthalpy For diatomic gas, 1 1.4 3. Internal energy 5 4. Specific entropy 2 Which of these are extensive properties? For polyatomic gas, 1 1.33 6 (a) 1,2,3,4 (b) 2,4 270. A process, which can be reversed bringing both (c) 2,3 (d) 3,4 the state of the system and the surroundings to UPRVUNL JE 2016 (Shift-II) the initial condition without any other inputs is Ans. (c) : Extensive Properties - The extensive (a) Irreversible process (b) Reversible process properties are defined as the properties which depend (c) Constant process (d) None of these on the mass or the amount of matter present in the Mizoram PSC 2016 (Paper-2) system. Ans. (b) : Reversible process is defined as the Example - Volume, mass, internal energy etc. process, which can be reversed bringing both the state Intensive Properties - An intensive property of a of the system and the surroundings to the initial system that does not depend on the system size or condition without any other inputs. amount of matter in the system. Thermodynamics 51 YCT Example - Temperature, pressure, density, specific Ans : (d) A frictionless quasi static process is called a entropy etc. reversible process. Isentropic process is a reversible Note- All the specific properties are intensive process, so frictionless adiabatic process is always properties. isentropic. e.g - Specific entropy, specific internal energy etc. The process which is not reversible is known as 275. The air is expanding from a very minute hole irreversible. of cycle tube in an evacuated room. The work All actual process are irreversible process. done during process is: 279. Which of the cycles below is not a reversible (a) negative (b) Positive cycle: (c) zero (d) minimum (a) Carnot (b) Ericsson UPRVUNL JE 2016 (Shift-II) (c) Joule (d) None of the above Ans. (c) : The air is expanding from a very minute RSMSSB JEN 16.10.2016 (Degree) hole of cycle tube in an evacuated room. The work Ans. (d) : All the given cycles are reversible cycle. done during process is zero. Because in throttling process work done is also zero. 280. Range of temperature measurement of a * At constant volume process, the work done will be resistance thermometer is ________ zero (for closed system). (a) 50º F to 200º F (b) 100º F to 400º F 276. The general law of expansion or compression is (c) 200º F to 800º F (d) 400º F to 1800º F PVn = C. The process is said to be hyperbolic, if Karnataka PSC RTO (Mech.) 10.07.2016 n is equal to Ans. (d) : Range of temperature measurement of a (a) (b) 1 resistance thermometer is 418ºF to 1832ºF. Resistance (c) 0 (d) thermometer is an electrical transducer. ISRO IPRC 10-12-2016 281. 200W electric bulb was switched on in a 5×6×6 Ans. (b) : The general law of expansion or compression meter size thermally insulated room having a is PVn = C. The process is said to be hyperbolic, if n is temperature of 500C. The room temperature at equal to 1. the end of 24 hour will be: Hyperbolic process also known as isothermal process. (a) 181.50C (b) 161.50C 0 It is the process where the gas is heated or expanded (c) 145 C (d) 1350C and contracted in such a way that the product of its UPRVUNL JE 2016 (Shift-II) pressure and volume remains constant. Ans. (b) : air = 1.2 kg/m3 277. Area under process line in P-V (Pressure- (Cv)air = 0.718 kJ/kgK Volume) and T-S (Temperature-Entropy) 200 Power (P) = 200 W = kW diagrams give.............and.............respectively. 1000 (a) Q and W (Heat and Work) time (t) = 24 hour = 24 × 3600 sec. (b) W and Q (Work and Heat) T1 = 500C (c) U, W (Internal energy and Work) W (d) H, W (Enthalpy and Work) Pt = mCvT P t ISRO IPRC 10-12-2016 Ans. (b) : m VCv T = Pt and V 200 1.2×5×6×6×0.718 (T2–50) ×24×3600 1000 155.088 (T2 – 50) = 17280 17280 (T2 – 50) T2 = 111.42 + 50 For close system– 155.088 W P. V2 V1 Q T. S2 S1 T2 = 161.42 161.50C W P.dV Q T.dS 282. Which of the following conversion formula is correct for temperature units So, Area under process line in P-V diagram gives work C F 32 R 492 and area under process line in T-S diagram gives heat. (a) 100 180 180 278. Choose the correct statement : C F 32 R 460 (a) An adiabatic process is always reversible (b) (b) An isentropic process is never reversible 100 180 180 (c) An adiabatic process is always isentropic C F 32 R (c) (d) A frictionless adiabatic process is always 100 180 80 isentropic (d) None of the above DMRC 2016 (Paper-I) ISRO IPRC 10-12-2016 Thermodynamics 52 YCT Ans. (a) : The correct conversion formula for 287. An ideal gas at 2270C is heated at constant temperature units is– pressure till volume becomes three times. The C F 32 K 273 R 492 temperature of gas will then be: (a) 6810C (b) 9270C 100 180 100 180 (c) 1227 C 0 (d) 10270C Scale Absolute Freezing Boiling UPRVUNL JE 2016 (Shift-II) zero point of point of Ans. (c) : Charles's law states that, at constant temp. water water pressure, the volume V of a gas directly proportional to Kelvin 0K 273 K 373 K its absolute temperature T. Celcius –2730C 00C 1000C V = Constant Fahrenheit –4600F 320F 2120F T 0 0 Rankine 0R 492 R 6720R V1 V2 283. Heat transfer occurs due to...............difference T1 T2 (a) Thermal conductivity (b) Temperature (c) Specific heat (d) Heat flux Given - T1 = 227 + 273 = 500 K V1 = V, V2 = 3V ISRO IPRC 28-08-2016 V 3V Ans. (b) : According to second law of thermodynamics, heat transfer or heat flow takes place from a body at a 500 T2 higher temperature to a body at a lower temperature. T2 = 1500 K So, heat transfer occurs due to temperature difference. T2 = 1500 – 273 = 12270C If T1 = T2 No heat transfer. 288. The relation between specific heat, CP and CV is 284. The reading of temperature on Celsius scale is given by : 400C. What is equivalent reading of (a) Cv/Cp = (b) Cp–Cv = R/J temperature on Fahrenheit scale? (c) (Cp–Cv)J=R (d) Cv–Cp = R/J (a) 1140F (b) 1040F UPSSSC JE 2016 0 (c) 110 F (d) 1340F Ans. (b) : The relation between specific heat Cp and Cv UPRVUNL JE 2016 (Shift-II) is, Cp – Cv = R/J Ans. (b) : C = 400C, F=? Here, Cp = Specific heat at constant pressure C F 32 Cv = Specific heat at constant volume For air, 5 9 = 1.4 40 F 32 Cp = 1.008 kJ/kg–K 5 9 Cv = 0.728 kJ/kg–K F = 72 + 32 = 1040F R = 0.287 kJ/kg–K 285. What is the gas constant (R) ? 289. In a throttling process the enthalpy of the (a) It is work per degree per mass substance (b) It is work per mole (a) Remains constant (b) Increases (c) It is work measured as Joule (c) Decreases (d) None of the above (d) It is change in temperature per mole RSMSSB JEN 16.10.2016 (Degree) KPCL JE 2016 Ans. (a) : A throttling process is defined as a process in Ans. (a) : Gas Constant (R) :– Gas constant (R) is which there is no change in enthalpy from state one to work per degree per mass state two (h 1 = h2) means enthalpy remains constant. Unit of gas constant (R), in S.I. units is N-m/kg.K or The throttling process is very fast and there is no J/kg. K moving part. So the process is adiabatic (i.e. no heat Value of gas constant for air is 287 J/kg.K transfer) and has no work transfer. 286. Which of the following variables controls the 290. If Cp is the specific heat at constant pressure physical properties of a perfect gas ? and Cv is the specific heat at constant volume (a) pressure (b) temperature for air, then _____. (c) volume (d) all of the above (a) Cp < Cv (b) Cp = Cv GPSC AMVI (Auto) 24-12-2016 (c) Cp > Cv (d) There is no such relation Karnataka PSC (MVI) 26.06.2016 Ans. (d) : * Pressure, temp., volume variables control the physical properties of perfect gas. Ans. (c) : The ratio of specific heat constant pressure to * Perfect gas has no molecular force. the specific heat at constant volume is always greater * Does not change its phase during process. than one. * Obey's a set of common governing equations – Cp (γ) = 1.4 = (i) PV = RT (ii) u = f(T) (iii) h = f(T) air Cv * CP and CV remains constant with temperature Cp > C v variation. Thermodynamics 53 YCT 291. The amount of heat required to raise the Ans. (a) : Solid and liquid have only one specific heat, temperature of 1 kg of water through 10C, is while gases have two specific heats. called In the case of solid and liquids a small change in (a) specific heat at constant volume temperature causes a negligible change in the volume (b) specific heat at constant pressure and pressure, hence the external work performed is (c) kilo calorie negligible. In such cases, all the heat is utilised for (d) none of these raising the temperature. Thus there is only one value of RRB JE [Exam Date : 04-01-2015 (Red Paper)] specific heat for solids and liquids. Ans : (c) The amount of heat required to rise the Cp = Cv + R temperature of 1 Kg. of water through 1ºC, is called 1 kilo calorie. PV = mRT (1000 calorie = 1 kilo calorie) PV P R= = 292. The quantity of heat required to melt one kg of mT ρT ice from and at 00 C is If solid isincomprissible (a) 1000 J (b) 100 kJ (c) 1000 kJ (d) 335 kJ ρ=0 GSSB ITI Supervision Inst. (RAC) 18-01-2015 R=0 Ans. (d) : The quantity of heat required to melt one kg Cp = Cv of ice from at 0oC is 335 kJ-which is called the latent heat of melting. 296. Which of the following is the property of a At 0 C,0 L.H.Ice = 80 k cal/kg system : (a) Pressure and temp = 80 4.18 = 334.4 (b) Internal energy 335kJ (c) Volume and density 293. The latent heat of Ice is (d) All of the above (a) 100 kJ/kg (b) 1000 kJ/kg ISRO VSSC 08-02-2015 (c) 335 kJ/kg (d) 200 kJ/kg Ans : (d) In the given option all are properties of a GSSB ITI Supervision Inst. (RAC) 18-01-2015 system. Ans. (c) : 335 kJ/kg of energy are required to melt 1kg Properties of system are two types– of ice at 00C. (1) Extensive Property–It depends upon the mass of * Liquid water has 335 kJ/kg more energy than ice at system. the same temperature. Example–Volume, kinetic energy, potential energy etc. * This energy is released when the liquid water Intensive Property–It does not depend on the mass of system. subsequently freezes and it is called the latent heat Example–Density, pressure, temperature, dryness of fusion. fraction, specific enthalpy, specific entropy etc. 294. Specific heat is the amount of heat required to 297. If a process can be stopped at any stage and raise the temperature reversed so that the system and surroundings (a) By unit degree of a substance are exactly restored to their initial states, it is (b) By unit degree of an unit mass known as : (c) Of a unit mass by 100 (a) Adiabatic process (d) None of the above (b) Isothermal process GSSB ITI Supervision Inst. (Mech) 18-01-2015 (c) Ideal process Ans. (b) : Specific heat - (d) Frictionless process * Specific heat is the quantity of heat essential to raise RRB SSE [Exam Date : 03-09-2015 (Shift-II)] the unit degree temperature of an unit mass. Ans (c) : If a process can be stopped at any stage and Q = mCt reversed so that the system and surrounding are exactly 0 if t = 1 C, m = 1 kg restored to their initial stage is known as ideal process. All the isothermal process, adiabatic process, isobaric then, C Q process, isochoric process are ideal process. This is not where, C = specific heat actual process. Unit kJ/kg-C = kJ/kg-K 298. Ice kept in a well-insulated thermoflask is an 295. Solid and liquid have example of which system? (a) One value of specific heat (a) Closed system (b) Two value of specific heat (b) Isolated system (c) Three value of specific heat (c) Open system (d) No value of specific heat (d) Non-flow adiabatic system UPSSSC JE 2015 RRB JE [Exam Date : 29-08-2015 (Shift-I)] Thermodynamics 54 YCT Ans : (b) Ice-kept in a well insulated thermoflask is So slope of constant volume line on temperature an example of isolated system, because in well insulated T entropy diagram is given by. thermo flask, mass and energy do not cross the CV boundary. Note–Slope of constant pressure line on temperature Note– T System Mass Energy entropy diagram is given by. CP Close do not cross Cross Open Cross Corss 302. Which of the following is an extensive property? (a) Pressure (b) Heat Capacity Isolated do not cross do not cross (c) Temperature (d) Specific Volume 299. All of the following are intensive properties RRB JE [Exam Date : 26-08-2015 (Shift-I)] EXCEPT? Ans : (b) Heat capacity is an extensive property. (a) Mass (b) Density (c) Pressure (d) Temperature Heat capacity = m C RRB JE [Exam Date : 27-08-2015 (Shift-I)] Ans : (a) Mass is a extensive property. Properties can be classified on the basis of size/extent of the system. 303. Assertion (A) : The change in heat and work cannot be express as difference between the end states. Reason (R) : Heat and work both are exact differential. 300. A system and its environment put together (a) Both A and R are true and R is correct constitute– explanation of A (a) An adiabatic system (b) Both A and R are true But R is not a correct (b) An isolated system explanation of A (c) A segregated system (c) A is true but R is false (d) A homogeneous system (d) A is false but R is true RRB JE [Exam Date : 26-08-2015 (Shift-II)] UPRVUNL JE 2015 Ans : (b) A system and its environment put together Ans. (c) : constitute an Universe. Universe is Isolated system. Heat and work both are path function. So the change Isolated system is a system of fixed mass and no heat in heat and work cannot be express as difference or work cross its boundary. between the end states. 301. Slope of constant volume line on temperature i.e. Q Q1 – Q2 & W W1 – W2 entropy diagram is given by– Properties are point function and are exact or perfect (a) Cp/T (b) T/Cp differentials. Ex.– Enthalpy, entropy, internal energy (c) Cv/T (d) T/Cv etc. RRB JE [Exam Date : 26-08-2015 (Shift-II)] 304. Heat interaction between a system and its Ans : (d) surroundings (a) Represents energy in transit (b) Does not depend on the choice of the system (c) Can be identified after the completion of process (d) Is a property of the system and its differential is exact We know that– UPRVUNL JE 2015 Q = T.dS Ans. (a) : Heat interaction between a system and its mCVdT =T.dS [Q = mCVdT for constant volume] surroundings represents energy in transit. Slope on T-S diagram– 305. Spot the odd one out. (a) Specific enthalpy (b) Kinetic Energy dT T dT T ...... for unit mass (c) Work (d) Pressure dS v mC v dS v Cv UPRVUNL JE 2015 Thermodynamics 55 YCT Ans. (c) : Specific enthalpy, kinetic energy and Ans. (d) : * The ratio of CP to CV of a perfect gas is pressure are point function. called isentropic index. Work is a path function which depends on the end Cp states, as well as the path followed. 306. Enthalpy of formation is defined as enthalpy of Cv compounds at : * The heat ratio or ratio of specific heat capacities (a) 25º C and 10 atmospheres (Cp/Cv) is also known as the adiabatic index. (b) 25º C and 1 atmospheres 310. An adiabatic wall is one which– (c) 0º C and 1 atmosphere (a) prevents thermal interaction (d) 100º C and 1 atmosphere (b) permits thermal interaction RRB SSE [Exam Date : 02-09-2015 (Shift-III)] (c) encourage thermal interaction Ans : (b) (d) discourage thermal interaction Enthalpy of formation is defined as enthalpy of RRB JE [Exam Date : 29-08-2015 (Shift-III)] compounds at 25ºC and 1 atmospheres. The enthalpy of formation is the standard reaction Ans : (a) Adiabatic wall : enthalpy for the formation of compound from its An adiabatic wall is an insulating wall that does not allow the flow of heat energy between the system and elements (atoms or molecules) in their most stable its surrounding or between two system. reference states at the choosen temperature (298.15 K or 25ºC) and at 1 bar pressure. Diathermic wall :- 307. 1 m3 of air at 10 kg/cm2 is allowed to expand A diathermic wall is a conducting wall which allows freely to a volume to 10 m3. The work done will heat energy to flow between the system and its be surroundings or between two system. (a) Zero (b) +ve 311. Work done during a process can be determined (c) –ve (d) 105 kgm by P.dV when the process is : UPSSSC JE 2015 (a) Isentropic (b) Isothermal Ans. (a) : Free expansion is a process where no (c) Adiabatic (d) Quasi-static external pressure is applied. Therefore – p=0 JK SSB JE (103) 2014 (ESE-13) Work done is given by, W = pdv Ans. (d) : Work done during a process can be = 0 dv = 0 determined by P.dV when the process is quasi-static. Quasi-static process is a thermodynamic process that 308. A perfect gas at 270ºC is heated at constant pressure till its volume is doubled. The final happens slowly enough for the system to remain in temperature is : internal equilibrium. (a) 54ºC (b) 327ºC All reversible processes are quasi-static. (c) 108ºC (d) 600ºC 312. Two block which are at different states are ISRO VSSC 08-02-2015 brought into contact with each other and Ans : (*) Given, that– allowed to reach a final state to thermal T1 = 270º C = 270 + 273 = 543º K equilibrium. The final temperature attained is V1 = V specified by the V2 = 2V (a) Zeroth Law of Thermodynamics T2 = ? (b) First Law of Thermodynamics P = Constant (c) 2nd Law of Thermodynamics Then by Charles's law– (d) 3rd Law of Thermodynamics V T at P = Constant. RRB JE [Exam Date : 21-12-2014 (09 Yellow Paper)] V1 T1 Ans : (a) Zeroth law of thermodynamics– i.e V2 T2 This law states that when two bodies are in thermal TV 2V equilibrium with a third body, they are also in thermal T2 1 2 543 equilibrium with each other. V1 V T2 = 1086 K or T2 813º C Note–According to commission 'No answer' is correct. 309. The ratio of Cp to Cv of a perfect gas is called: (a) isothermal index (b) adiabatic index (c) ideal gas index (d) isentropic index Mizoram PSC 2015 (Paper-1) Thermodynamics 56 YCT 313. 1000C is equal to............. 0F. 316. Which gas among the following has the highest (a) 540 (b) 273 value of adiabatic index? (c) 212 (d) 132 (a) Oxygen (b) Methane ISRO LPSC 11-05-2014 (c) Helium (d) Nitrogen Ans : (c) Given– SSC JE 2014 (Evening) °C = 100° °F = ? Ans. : (c) Adiabatic index is the ratio of specific Temperature scale equation– heats. C F 32 K 273.15 Value of '' for different gases – 100 180 100 Gases Value of C F 32 100 F 32 (i) Monoatomic gases – 5/3 So, (ii) Diatomic gases – 7/5 5 9 5 9 900 = 5F – 160 (iii) Polyatomic gases – 4/3 5F = 1060 (iv) Air – 1.41 1060 (v) Methane – 1.29 F 212 (vii) N2, O2 – 1.41 5 In Above list, the value of '' for Helium So, 100C 212F (Monoatomic) is maximum. 314. The temperature at which the volume of a gas 317. The specific heat of the gas in an isothermal becomes zero is called: process is (a) Absolute scale temperature (a) Zero (b) Infinite (b) Absolute zero temperature (c) Negative (d) Remains constant (c) Absolute temperature RRB JE [Exam Date : 21-12-2014 (09 Yellow Paper)] (d) None of these RRB JE [Exam Date : 14-12-2014 (06 Yellow Paper)] Ans. : (b) Specific heat–Specific heat is the amount of heat required to raise temperature of unit mass through Ans : (b) A little consideration will show, that the unit degree. 1 volume of gas goes on decreasing by of its Q m C T 273th original volume for every 1ºC decrease in temperature. Q 1 Specific heat C= 1ºC Volume m T 273th According to question– It is thus obvious, that at temperature of –273ºC, the T = T2 – T1 = 0 volume of the gas would become zero. The temperature at which the volume of a gas become then, Q C zero is called absolute zero temperature. 0 –273ºC = 0K. 318. The amount of heat required to raise the 315. The boiling and freezing points for water are temperature of the unit mass of gas by one marked on a temperature scale P as 130o P and degree in a closed container, is called: o -20 P respectively. What will be the reading on (a) specific heat at constant volume (b) specific heat at constant pressure this scale corresponding to 60oC on Celsius (c) kilo Joule scale? (d) calorific value (a) 60oP (b) 70oP ISRO LPSC 11-05-2014 (c) 90oP (d) 110oP Ans : (a) The amount of heat required to raise the SSC JE 2014 (MORNING) temperature of unit mass of gas by one degree in a Ans. : (b) Given that – closed container is called specific heat at constant Temperature on P–scale volume. 0 Tb = 130 P Hint Closecontainer = Constant volume T0 = –200P 0 & Tc = 60 C There are two types of specific heat of gas– T 0 T p 20 1. Constant volume specific heat (Cv) c 2. Constant pressure specific heat (Cp) 100 0 130 20 R 60 0 Tp 20 Cp Cv (From mayer equation) J 100 150 90 = TP + 20 Cp 0 Cp Cv TP 70 P Cv i.e temperature on P–scale is 700P. = Adiabatic index & = 1.4 for air Thermodynamics 57 YCT 319. The pressure exerted on the walls of a Ans. (d) : Intensive property- Independent of mass. container by a gas is due to the fact that Gas (1) Pressure molecules: (2) Temperature (a) Stick to the walls of the container (3) density (b) Lose their kinetic energy (4) All specific properties (c) Get accelerated towards the wall 323. The valve of Universal Gas constant is : (d) Change their momentum due to collision (a) 8.314 kJ/kg mole °K with the wall (b) 8.314 Joule/kg mole °K RRB JE [Exam Date : 14-12-2014 (01 Red Paper)] (c) 0.8314 kJ/kg mole °K Ans. (d) : (d) 83.14 Joule/kg mole °K RPSC MVSI 2013 Ans. (a) : Universal gas constant R = 8.314 kJ/kg mole °K 324. The fixed point/points for celcius temperature scale is/are : (a) Ice point as 0oC (b) Steam point as 100oC (c) Both ice and steam points as 0oC & 100oC respectively (d) Triple point of water as 0.01oC DMRC 21-07-2013 320. For a particular ideal gas, the value of R is Ans : (c) The fixed points for celcius temperature scale 0.280 kJ/kgK and the value of is 1.375. The are both ice and steam points as 0ºC and 100ºC value of Cp and Cv are, respectively, in kJ/ kg K : respectively. (a) 1.111, 0.66 (b) 1.2, 0.70 C F 32 K 273.15 (c) 1.25, 0.8 (d) 1.0267, 0.7467 SSC JE 2014 (Evening) 5 9 5 Ans. : (d) Here, CP – CV = R ºC = Temperature in degree celsius. CP – CV = 0.280 kJ/kg K........ (i) F = Temperature in Fahrenheit CP K = Temperature in kelvin. Note–Triple point of the water is the point where all CV the phases coexist in equilibrium together. CP For water, its value is 273.16 K or 0.01ºC. 1.375.......(ii) CV Pressure = 0.61 kPa from equation (i) & (ii) 325. A non flow quasi-static (reversible) process occurs for which CV 0.7467 P = (–3V + 16) bar, where V is volume in m3. CP 1.0267 What is work done when V changes from 2 to 6 m3? 321. In a steady flow process, the value of: (a) heat transfer is constant (a) 16 × 105 J (b) 16.5 × 105 J (c) 16 × 103 J (d) 16.5 × 102 J (b) work transfer is constant (c) mass flow at inlet and outlet is same (ESE-2013) 6 (d) all of these V2 5 RRB JE [Exam Date : 14-12-2014 (04 Green Paper)] Ans. (a) : W V1 PdV (3V 16) 10 dV 2 Ans. (d) : Steady Flow Process–A steady flow process 6 is one in which matter and energy flow in and out of 3V 2 5 open system does not experience any change in the 16V 10 2 2 mass and energy of the system. Ex.: Pipes, nozzles, diffusers and pumps. 3 In this process– 2 2 2 6 2 16 6 2 10 5 Heat transfer is constant = 16 × 105 J Work transfer is constant 326. A quasi-static process : Mass flow at inlet and outlet is same. (A) is a reversible process 322. Which one of the following is an intensive (B) need not be a reversible process property of a thermodynamic system? (C) proceeds very slowly (a) Volume (b) Mass (D) occurs when the driving force is finite but (c) Energy (d) Specific volume very small : RPSC MVSI 2013 (a) (A) only is correct Thermodynamics 58 YCT (b) (A), (C) and (D) are all correct Ans. (d) : First law of thermodynamics throws the light (c) (C) and (D) are only correct on the concept of internal energy. (d) (C) only is correct * Second law– Indicates the limit of converting heat GPSC Motor Vehicle (Mech.) 02-12-2012 into work and introduces the principle of increase of Ans. (b) : A Quasi-Static Process– entropy. (i) Is a reversible process. * Third law – Defines the absolute zero of entropy. (ii) Proceeds very slowly. 331. The polytropic index of expansion 'n' in the (iii) Occurs when the driving force is finite but very equation PVn = C for constant volume process small. is– 327. Temperature and pressure of perfect gas are (a) 1 (b) 1.4 (a) Intensive property (c) (d) 0 (b) Extensive property (MP Engineer JE Paper II 2011) (c) Both intensive and extensive property Ans. (c) The polytropic index of expansion 'n' in the (d) None of the above equation PVn = C, for constant volume or isochoric GPSC Motor Vehicle (Auto) 02-12-2012 process is . Ans. (a) : Temperature and pressure of perfect gas are intensive properties. 332. Work transfer between the system and the 328. Which aspect does not pertain to a free surroundings expansion process? (a) Is a point function (a) Pressure remains constant (b) Is always given by P.dV (b) No change in the temperature of the system (c) Is a function of pressure only (c) No gain or loss of heat (d) Depends on the path followed by the system (d) Work done is Zero (ESE-2011) SSC JE 2012 Ans. (d) : Work transfer is transient in nature, inexact Ans. (a) A free expansion process occurs when a differential and boundary phenomenon. So work is a fluid is allowed to expand suddenly into a vacuum path function. chamber through an orifice of large dimensions. For quasi-static (reversible) process, work done is In this process, no heat is supplied or rejected and calculated by P.dV. no external work is done. In free expansion process– Pr essure (P) , Volume (V) , Q 0, W 0 dU 0, U C, T C, h C 329. The ratio of specific heats of gas at constant pressure and at constant volume : Conditions for applying above equation– (a) varies with temperature (i) System must be closed system (b) varies with pressure (ii) Process must be reversible process (c) is always constant (iii) Work should cross the boundary (d) none of these 333. Match List-I with List-II and select the correct TNPSC 2012 (Paper-4) answer using the code given below the lists: Ans. (c) : for Ideal gas, the ratio of specific heat of gas List-I List-II at constant pressure (CP) and constant volume (CV) is A. n = always constant. B. n = 1.4 330. The first law of thermodynamics throws the C. n = 1.0 light on the concept of : D. n = 0 (a) Entropy (b) Temperature (c) Strain (d) Internal Energy Arunachal Pradesh PSC Boiler Inspector 2012 Thermodynamics 59 YCT Codes: Ans. (b) : Throttling Process–When a perfect gas is A B C D expanded through an aperture of minute dimensions, (a) 4 3 2 1 such as a narrow throat or a slightly opened valve, the (b) 1 3 2 4 process is termed as throttling process. (c) 4 2 3 1 During this process, no heat is supplied or rejected (d) 1 2 3 4 and also no external work is done. (ESE-2011) In throttling process– Ans. (a) : h = 0, du = 0, Q = 0, W1–2 = 0 Pressure, Volume 337. Heat addition to a system at constant volume will increase its (a) Enthalpy (b) Internal energy (c) Work done (d) Specific volume WBPSC 2009 Ans. (b) : 334. The work done in the expansion of a gas from volume V1 to V2 under constant pressure 'p' is equal to : (a) zero (b) P(V2 – V1) (c) P(V2 + V1) (d) P(V2 V1) For close system at constant volume, SSC JE 2010 Work done (W) = PdV = 0 Q = du + W Ans. : (b) The work done in the expansion of a gas from volume V1 to V2 under constant pressure P is Q du equal to P(V2 – V1). So, heat addition to a system at constant volume will 2 increase its internal energy. W1 2 P.dV Note-Heat addition to a system at constant volume also 1 increases pressure, temperature and entropy. W12 P V2 V1 338. Consider the following properties: 1. Temperature 2. Viscosity
