Thermodynamics Theory + Questions.0001

8/18/2019 Thermodynamics Theory + Questions.0001

1/114

Basic ConceptsS K Mondal’s Chapter 1

1

1. Basic Concepts

Theory at a Glance (For GATE, IES & PSUs)

Intensive and Extensive PropertiesIntensive property: Whose value is independent of the size or extent i.e. mass of the system.

These are, e.g., pressure p and temperature T.

Extensive property: Whose value depends on the size or extent i.e. mass of the system (upper case

letters as the symbols). e.g., Volume, Mass (V, M). If mass is increased, the value of extensive

property also increases. e.g., volume V, internal energy U, enthalpy H, entropy S, etc.

Specific property: It is a special case of an intensive property. It is the value of an extensive

property per unit mass of system. (Lower case letters as symbols) eg: specific volume, density (v, ρ).

Thermodynamic System and Control Volume• In our study of thermodynamics, we will choose a small part of the universe to which we will

apply the laws of thermodynamics.

We call this subset a SYSTEM.

• The thermodynamic system is analogous to the free body diagram to which we apply the laws ofmechanics, (i.e. Newton’s Laws of Motion).

• The system is a macroscopically identifiable collection of matter on which we focusour attention (e.g., the water kettle or the aircraft engine).

SystemDefinition

• System: A quantity of matter in space which is analyzed during a problem.

• Surroundings: Everything external to the system.

• System Boundary: A separation present between system and surrounding.

Classification of the system boundary:-• Real solid boundary

• Imaginary boundary


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9/114

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10/114

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2 10V V

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11/114


11

( )

( )

( )

[ ]

2 2

1 1

2

3 3 2

2 11

3 3

10.

110

3

1 4.54.5 1.5 10

3 1.5

191.125 3.375 10 3

3

29.250 10.986 40.236

⎛ ⎞= = +⎜ ⎟⎝ ⎠

⎡ ⎤= − +

⎢ ⎥⎣ ⎦⎡ ⎤= − +⎢ ⎥⎣ ⎦

⎡ ⎤= − +⎢ ⎥⎣ ⎦= + =

∫ ∫V V

V V

Work done p dV V dV V

V V V ln

V

ln

ln

kJ

First Law of Thermodynamics:-

Q = W + ΔU2000 = 40.236 + ΔU∴ ΔU = 2000 – 40.236 = 1959.764 kJ

Example 2.

A fluid is contained in a cylinder piston arrangement that has a paddle that imparts work to the

fluid. The atmospheric pressure is 760 mm of Hg. The paddle makes 10,000 revolutions during

which the piston moves out 0.8m. The fluid exerts a torque of 1.275 N-m one the paddle. What is net

work transfer, if the diameter of the piston is 0.6m?

Solution:

Work done by the stirring device upon the system

W1 = 2πTN= 2π × 1.275 × 10000 N-m = 80kJThis is negative work for the system.

(Fig.)

Work done by the system upon the surroundings.

W2 = p.dV = p.(A × L)

= 101.325 × 4

π (0.6)2 × 0.80 = 22.9kJ

This is positive work for the system. Hence the net work transfer for the system.W = W1 + W2 = - 80 + 22.9 = - 57.l kJ .


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12

ASKED OBJECTIVE QUESTIONS (GATE, IES, IAS)

Previous 20-Years GATE Questions

GATE-1. List-I List II [GATE-1998]

A. Heat to work 1. Nozzle

B. Heat to lift weight 2. Endothermic chemical reaction

C. Heat to strain energy 3. Heat engine

D. Heat to electromagnetic energy 4. Hot air balloon/evaporation

5. Thermal radiation

6. Bimetallic strips

Codes: A B C D A B C D

(a) 3 4 6 5 (b) 3 4 5 6

(c) 3 6 4 2 (d) 1 2 3 4

Open and Closed systemsGATE-2. An isolated thermodynamic system executes a process, choose the correct

statement(s) form the following [GATE-1999]

(a) No heat is transferred

(b) No work is done

(c) No mass flows across the boundary of the system

(d) No chemical reaction takes place within the system

GATE-2a. Heat and work are [GATE-2011]

(a) intensive properties (b) extensive properties

(c) point functions (d) path functions

Quasi-Static ProcessGATE-3. A frictionless piston-cylinder device contains a gas initially at 0.8 MPa and

0.015 m3. It expands quasi-statically at constant temperature to a final volume

of 0.030 m3. The work output (in kJ/kg) during this process will be: [GATE-2009]

(a) 8.32 (b) 12.00 (c) 554.67 (d) 8320.00

Free Expansion with Zero Work TransferGATE-4. A balloon containing an ideal gas is initially kept in an evacuated and

insulated room. The balloon ruptures and the gas fills up the entire room.

Which one of the following statements is TRUE at the end of above process?(a) The internal energy of the gas decreases from its initial value, but the enthalpy

remains constant [GATE-2008]

(b) The internal energy of the gas increases from its initial value, but the enthalpy

remains constant

(c) Both internal energy and enthalpy of the gas remain constant

(d) Both internal energy and enthalpy of the gas increase


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13

GATE-5. Air is compressed adiabatically in a steady flow process with negligible change

in potential and kinetic energy. The Work done in the process is given by:

[GATE-1996, IAS-2000]

(a) – ∫Pdv (b) +∫Pdv (c) – ∫vdp (d) +∫vdp

pdV-work or Displacement WorkGATE-6. In a steady state steady flow process taking place in a device with a single inlet

and a single outlet, the work done per unit mass flow rate is given byoutlet

inlet

vdp=∫

, where v is the specific volume and p is the pressure. The

expression for w given above: [GATE-2008]

(a) Is valid only if the process is both reversible and adiabatic

(b) Is valid only if the process is both reversible and isothermal

(c) Is valid for any reversible process

(d) Is incorrect; it must be

outlet

inletvdpω = − ∫

GATE-7. A gas expands in a frictionless piston-cylinder arrangement. The expansion

process is very slow, and is resisted by an ambient pressure of 100 kPa. During

the expansion process, the pressure of the system (gas) remains constant at 300

kPa. The change in volume of the gas is 0.01 m3. The maximum amount of work

that could be utilized from the above process is: [GATE-2008]

(a) 0kJ (b) 1kJ (c) 2kJ (d) 3kJ

GATE-8. For reversible adiabatic compression in a steady flow process, the work

transfer per unit mass is: [GATE-1996]

( ) ( ) ( ) ( )a pdv b vdp c Tds d sdT ∫ ∫ ∫ ∫

Previous 20-Years IES Questions

IES-1. Which of the following are intensive properties? [IES-2005]

1. Kinetic Energy 2. Specific Enthalpy

3. Pressure 4. Entropy

Select the correct answer using the code given below:

(a) 1 and 3 (b) 2 and 3 (c) 1, 3 and 4 (d) 2 and 4

IES-2. Consider the following properties: [IES-2009]

1. Temperature 2. Viscosity

3. Specific entropy 4. Thermal conductivity

Which of the above properties of a system is/are intensive?

(a) 1 only (b) 2 and 3 only (c) 2, 3 and 4 only (d) 1, 2, 3 and 4

IES-2a. Consider the following: [IES-2007, 2010]

1. Kinetic energy 2. Entropy


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15/114


15

(a) Open system [IES-2010]

(b) Steady flow diabatic system

(c) Closed system with a movable boundary

(d) Closed system with fixed boundary

IES-8. Which of the following is/are reversible process(es)? [IES-2005]

1. Isentropic expansion

2. Slow heating of water from a hot source

3. Constant pressure heating of an ideal gas from a constant temperature

source

4. Evaporation of a liquid at constant temperature

Select the correct answer using the code given below:

(a) 1 only (b) 1 and 2 (c) 2 and 3 (d) 1 and 4

IES-9. Assertion (A): In thermodynamic analysis, the concept of reversibility is that, a

reversible process is the most efficient process. [IES-2001]

Reason (R): The energy transfer as heat and work during the forward process

as always identically equal to the energy transfer is heat and work during thereversal or the process.

(a) Both A and R are individually true and R is the correct explanation of A

(b) Both A and R are individually true but R is NOT the correct explanation of A

(c) A is true but R is false

(d) A is false but R is true

IES-9a Which one of the following represents open thermodynamic system?

(a) Manual ice cream freezer (b) Centrifugal pump [IES-2011]

(c) Pressure cooker (d) Bomb calorimeter

IES-10. Ice kept in a well insulated thermo flask is an example of which system?

(a) Closed system (b) Isolated systems [IES-2009] (c) Open system (d) Non-flow adiabatic system

IES-10a Hot coffee stored in a well insulated thermos flask is an example of

(a) Isolated system (b) Closed system [IES-2010]

(c) Open system (d) Non-flow diabatic system

IES10b A thermodynamic system is considered to be an isolated one if [IES-2011]

(a) Mass transfer and entropy change are zero

(b) Entropy change and energy transfer are zero

(c) Energy transfer and mass transfer are zero

(d) Mass transfer and volume change are zero

IES-10c. Match List I with List II and select the correct answer using the code given

below the lists: [IES-2011]

List I

A. Interchange of matter is not possible in a

B. Any processes in which the system returns to

its original condition or state is called

C. Interchange of matter is possible in a

List II

1. Open system

2. System


16/114


16

D. The quantity of matter under consideration in

thermodynamics is called

3. Closed system

4. Cycle

Code: A B C D A B C D

(a) 2 1 4 3 (b) 3 1 4 2(c) 2 4 1 3 (d) 3 4 1 2

Zeroth Law of ThermodynamicsIES-11. Measurement of temperature is based on which law of thermodynamics?

[IES-2009]

(a) Zeroth law of thermodynamics (b) First law of thermodynamics

(c) Second law of thermodynamics (d) Third law of thermodynamics

IES-12. Consider the following statements: [IES-2003]

1. Zeroth law of thermodynamics is related to temperature

2. Entropy is related to first law of thermodynamics3. Internal energy of an ideal gas is a function of temperature and pressure

4. Van der Waals' equation is related to an ideal gas

Which of the above statements is/are correct?

(a) 1 only (b) 2, 3 and 4 (c) 1 and 3 (d) 2 and 4

IES-13. Zeroth Law of thermodynamics states that [IES-1996, 2010]

(a) Two thermodynamic systems are always in thermal equilibrium with each other.

(b) If two systems are in thermal equilibrium, then the third system will also be in

thermal equilibrium with each other.

(c) Two systems not in thermal equilibrium with a third system are also not in thermal

equilibrium with each other.

(d) When two systems are in thermal equilibrium with a third system, they are in

thermal equilibrium with each other.

International Temperature ScaleIES-14. Which one of the following correctly defines 1 K, as per the internationally

accepted definition of temperature scale? [IES-2004]

(a) 1/100th of the difference between normal boiling point and normal freezing point of

water

(b) 1/273.15th of the normal freezing point of water

(c) 100 times the difference between the triple point of water and the normal freezing

point of water

(d) 1/273.15th of the triple point of water

IES-15. In a new temperature scale say °ρ, the boiling and freezing points of water at

one atmosphere are 100°ρ and 300°ρ respectively. Correlate this scale with the

Centigrade scale. The reading of 0°ρ on the Centigrade scale is: [IES-2001]

(a) 0°C (b) 50°C (c) 100°C (d) 150°C


17/114


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18

Free Expansion with Zero Work TransferIES-21. Match items in List-I (Process) with those in List-II (Characteristic) and select

the correct answer using the codes given below the lists:

List-I List-II [IES-2001]

A. Throttling process 1. No work done

B. Isentropic process 2. No change in entropy

C. Free expansion 3. Constant internal energy

D. Isothermal process 4. Constant enthalpy


(a) 4 2 1 3 (b) 1 2 4 3

(c) 4 3 1 2 (d) 1 3 4 2

IES-22. The heat transfer, Q, the work done W and the change in internal energy U are

all zero in the case of [IES-1996](a) A rigid vessel containing steam at 150°C left in the atmosphere which is at 25°C.

(b) 1 kg of gas contained in an insulated cylinder expanding as the piston moves slowly

outwards.

(c) A rigid vessel containing ammonia gas connected through a valve to an evacuated

rigid vessel, the vessel, the valve and the connecting pipes being well insulated and

the valve being opened and after a time, conditions through the two vessels becoming

uniform.

(d) 1 kg of air flowing adiabatically from the atmosphere into a previously evacuated

bottle.

pdV-work or Displacement WorkIES-23. One kg of ice at 0°C is completely melted into water at 0°C at 1 bar pressure.

The latent heat of fusion of water is 333 kJ/kg and the densities of water and

ice at 0°C are 999.0 kg/m3 and 916.0 kg/m3, respectively. What are the

approximate values of the work done and energy transferred as heat for the

process, respectively? [IES-2007]

(a) –9.4 J and 333.0 kJ (b) 9.4 J and 333.0 kJ

(c) –333.0 kJ and –9.4 J (d) None of the above

IES-24. Which one of the following is the

correct sequence of the threeprocesses A, B and C in the

increasing order of the amount of

work done by a gas following ideal-

gas expansions by these processes?


19/114


19

[IES-2006](a) A – B – C (b) B – A – C (c) A – C – B (d) C – A – B

IES-25. An ideal gas undergoes an

isothermal expansion from

state R to state S in a turbineas shown in the diagram given

below:

The area of shaded region is

1000 Nm. What is the amount is

turbine work done during the

process?

(a) 14,000 Nm (b) 12,000 Nm

(c) 11,000 Nm (d) 10,000 Nm [IES-2004]

IES-26. Assertion (A): The area 'under' curve on pv plane, pdv∫

represents the work of

reversible non-flow process. [IES-1992]

Reason (R): The area 'under' the curve T –s plane Tds∫

represents heat of any

reversible process.

(a) Both A and R are individually true and R is the correct explanation of A




IES-27. If pdv∫

and vdp∫

for a thermodynamic system of an Ideal gas on valuation

give same quantity (positive/negative) during a process, then the process

undergone by the system is: [IES-2003]

(a) Isomeric (b) Isentropic (c) Isobaric (d) Isothermal

IES-28. Which one of the following expresses the reversible work done by the system

(steady flow) between states 1 and 2? [IES-2008]2 2 2 2

1 1 1 1

(a) (b) (c) (d) pdv vdp pdv vdp− −∫ ∫ ∫ ∫

Heat Transfer-A Path FunctionIES-29. Assertion (A): The change in heat and work cannot be expressed as difference

between the end states. [IES-1999]

Reason (R): Heat and work are both exact differentials.(a) Both A and R are individually true and R is the correct explanation of A





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20

Previous 20-Years IAS Questions

Thermodynamic System and Control VolumeIAS-1. The following are examples of some intensive and extensive properties:

1. Pressure 2. Temperature [IAS-1995]

3. Volume 4. Velocity

5. Electric charge 6. Magnetisation

7. Viscosity 8. Potential energy

Which one of the following sets gives the correct combination of intensive and

extensive properties?

Intensive Extensive

(a) 1, 2, 3, 4 5, 6, 7, 8

(b) 1, 3, 5, 7 2, 4, 6, 8

(c) 1, 2, 4, 7 3, 5, 6, 8(d) 2, 3, 6, 8 1, 4, 5, 7

Zeroth Law of ThermodynamicsIAS-2. Match List-I with List-II and select the correct answer using the codes given

below the lists: [IAS-2004]

List-I List-II

A. Reversible cycle 1. Measurement of temperature

B. Mechanical work 2. Clapeyron equation

C. Zeroth Law 3. Clausius Theorem

D. Heat 4. High grade energy

5. 3rd

law of thermodynamics6. Inexact differential


(a) 3 4 1 6 (b) 2 6 1 3

(c) 3 1 5 6 (d) 1 4 5 2

IAS-3. Match List-I with List-II and select the correct answer: [IAS-2000]

List-I List-II

A. The entropy of a pure crystalline 1. First law of thermodynamics

substance is zero at absolute zero

temperature

B. Spontaneous processes occur 2. Second law of thermodynamics

in a certain direction

C. If two bodies are in thermal 3. Third law of thermodynamics

equilibrium with a third body,

then they are also in thermal

equilibrium with each other

D. The law of conservation of energy 4. Zeroth law of thermodynamics.


(a) 2 3 4 1 (b) 3 2 1 4


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21

(c) 3 2 4 1 (d) 2 3 1 4

International Temperature Scale

IAS-4. A new temperature scale in degrees N is to be defined. The boiling andfreezing on this scale are 400°N and 100°N respectively. What will be the

reading on new scale corresponding to 60°C? [IAS-1995]

(a) 120°N (b) 180°N (c) 220°N (d) 280°N

Free Expansion with Zero Work TransferIAS-5. In free expansion of a gas between two equilibrium states, the work transfer

involved [IAS-2001]

(a) Can be calculated by joining the two states on p-v coordinates by any path and

estimating the area below

(b) Can be calculated by joining the two states by a quasi-static path and then finding

the area below(c) Is zero

(d) Is equal to heat generated by friction during expansion.

IAS-6. Work done in a free expansion process is: [IAS-2002]

(a) Positive (b) Negative (c) Zero (d) Maximum

IAS-7. In the temperature-entropy diagram

of a vapour shown in the given figure,

the thermodynamic process shown by

the dotted line AB represents

(a) Hyperbolic expansion

(b) Free expansion

(c) Constant volume expansion(d) Polytropic expansion

[IAS-1995]

IAS-8. If pdv∫

and vdp∫

for a thermodynamic system of an Ideal gas on valuation

give same quantity (positive/negative) during a process, then the process

undergone by the system is: [IAS-1997, IES-2003]

(a) Isomeric (b) Isentropic (c) Isobaric (d) Isothermal

IAS-9. For the expression pdv∫

to represent the work, which of the following

conditions should apply? [IAS-2002]

(a) The system is closed one and process takes place in non-flow system(b) The process is non-quasi static

(c) The boundary of the system should not move in order that work may be transferred

(d) If the system is open one, it should be non-reversible


22/114


23/114


23

Answers with Explanation (Objective)

Previous 20-Years GATE Answers

GATE-1. Ans. (a)

GATE-2. Ans. (a, b, c) For an isolated system no mass and energy transfer through the system.

0, 0, 0 or ConstantdQ dW dE E = = ∴ = =

GATE-2a. Ans. (d)

GATE-3. Ans. (a) Iso-thermal work done (W) = 211

lnV

RT V

⎛ ⎞⎜ ⎟⎝ ⎠

21 1

1ln

0.030800 0.015 ln

0.015

8.32kJ/kg

V P V V

⎛ ⎞= ⎜ ⎟⎝ ⎠

⎛ ⎞= × × ⎜ ⎟

⎝ ⎠=

GATE-4. Ans. (c) It is free expansion. Since vacuum does not offer any resistance, there is no work

transfer involved in free expansion.

Here,

2

1

0δω =∫ and Q1-2=0 therefore Q1-2 = U Δ + W1-2 so, U Δ = 0

GATE-5. Ans. (c) For closed system W = pdv+∫ , for steady flow W = vdp−∫ GATE-6. (c)

GATE-7. Ans. (b) W = Resistance pressure.Δ V = 1 × Δ V = 100 × 0.1 kJ = 1kJGATE-8. Ans. (b) W vdp= −∫

Previous 20-Years IES Answers

IES-1. Ans. (b)

IES-2. Ans. (d) Intensive property: Whose value is independent of the size or extent i.e. mass of

the system.

Specific property: It is a special case of an intensive property. It is the value of anextensive property per unit mass of system (Lower case letters as symbols) e.g., specific

volume, density (v, ρ).

IES-2a. Ans. (c) Kinetic energy21 mv

2 depends on mass, Entropy kJ/k depends on mass so

Entropy is extensive property but specific entropy kJ/kg K is an intensive property.


24/114


24

IES-3. Ans. (a) Extensive property is dependent on mass of system. Thus volume is extensive

property.

IES-4. Ans. (a) Extensive property: Whose value depends on the size or extent i.e. mass of the

system (upper case letters as the symbols) e.g., Volume, Mass (V, M). If mass is

increased, the value of extensive property also increases.

IES-4a Ans. (d) The properties like temperature, viscosity which are Independent of the MASS of

the system are called Intensive property

IES-5. Ans. (d)

• But remember 100% heat can’t be convertible to work but 100% work can beconverted to heat. It depends on second law of thermodynamics.

• A thermodynamic system is defined as a definite quantity of matter or a region inspace upon which attention is focused in the analysis of a problem.

• The system is a macroscopically identifiable collection of matter on which we focusour attention

IES-5a Ans. (d)

IES-6. Ans. (b) In closed thermodynamic system, there is no mass transfer but energy transferexists.

IES-7. Ans. (c)

IES-8. Ans. (d) Isentropic means reversible adiabatic. Heat transfer in any finite temp difference is

irreversible.

IES-9. Ans. (a) The energy transfer as heat and work during the forward process as always

identically equal to the energy transfer is heat and work during the reversal or the

process is the correct reason for maximum efficiency because it is conservative system.

IES-9a. Ans. (b)

IES-10. Ans. (b) Isolated System - in which there is no interaction between system and the

surroundings. It is of fixed mass and energy, and hence there is no mass and energy

transfer across the system boundary.

IES-10a Ans. (a)IES-10b. Ans. (c)

IES-10c. Ans. (d)

IES-11. Ans. (a) All temperature measurements are based on Zeroth law of thermodynamics.

IES-12. Ans. (a) Entropy - related to second law of thermodynamics.

Internal Energy (u) = f (T) only (for an ideal gas)

Van der Wall's equation related to => real gas.

IES-13. Ans. (d)

IES-14. Ans. (d)

IES-15.Ans. (d)0 300 0

150 C100 300 100 0

C C

− −= ⇒ = °

− −


25/114


25

IES-16. Ans. (b) Both A and R are correct but R is not correct explanation for A. Temperature is

independent of thermometric property of fluid.

IES-17. Ans. (d)

IES-18. Ans. (d) But it will occur at absolute zero temperature.

IES-19. Ans. (c) Time Constants: The time constant is the amount of time required for athermocouple to indicated 63.2% of step change in temperature of a surrounding media.

Some of the factors influencing the measured time constant are sheath wall thickness,

degree of insulation compaction, and distance of junction from the welded cap on an

ungrounded thermocouple. In addition, the velocity of a gas past the thermocouple probe

greatly influences the time constant measurement. In general, time constants for

measurement of gas can be estimated to be ten times as long as those for measurement

of liquid. The time constant also varies inversely proportional to the square root of the

velocity of the media.

IES-20. Ans. (c)

IES-20a Ans. (d)

IES-21. Ans. (a)

IES-22. Ans. (c) In example of (c), it is a case of free expansion heat transfer, work done, and

changes in internal energy are all zero.

IES-23. Ans. (a) Work done (W) = P Δ V = 100× (V2 – V1) = 100×2 1

m m⎛ ⎞−⎜ ⎟ρ ρ⎝ ⎠

= 100 kPa× 1 1999 916

⎛ ⎞−⎜ ⎟

⎝ ⎠ = –9.1 J

IES-24. Ans. (d) 4 (2 1) 4kJ= = × − =∫ AW pdV 1

3 (7 4) 4.5kJ2

1 (12 9) 3kJ

= = × × − =

= = × − =

∫

∫

B

C

W pdV

W pdV

IES-25. Ans. (c) Turbine work = area under curve R–S

( )

( )

3

5

1 bar 0.2 0.1 m 1000 Nm

10 0.2 0.1 Nm 1000Nm 11000Nm

pdv= = × − +

= × − + =

∫

IES-26. Ans. (b)

IES-27. Ans. (d) Isothermal work is minimum of any process.

0[ is onstant]

pv mRT

pdv vdp T c

pdv vdp

=

+ =

= −∫ ∫∵

IES-28. Ans. (b) For steady flow process, reversible work given by

2

1 vdp−∫ .IES-29. Ans. (c) A is true because change in heat and work are path functions and thus can't be

expressed simply as difference between the end states. R is false because both work and

heat are inexact differentials.


26/114


26

Previous 20-Years IAS Answers

IAS-1. Ans. (c) Intensive properties, i.e. independent of mass are pressure, temperature, velocity

and viscosity. Extensive properties, i.e. dependent on mass of system are volume, electriccharge, magnetisation, and potential energy. Thus correct choice is (c).

IAS-2. Ans. (a)

IAS-3. Ans. (c)

IAS-4. Ans. (d) The boiling and freezing points on new scale are 400° N and 100°N i.e. range is

300°N corresponding to 100°C. Thus conversion equation is

°N = 100 + 3 × °C = 100+ 3 × 60 = 100 + 180 = 280 °N

IAS-5. Ans. (c)

IAS-6. Ans. (c) Since vacuum does not offer any resistance, there is no work transfer involved in

free expansion.

IAS-7. Ans. (b)

IAS-8. Ans. (d) Isothermal work is minimum of any process.

IAS-9. Ans. (a)IAS-10. Ans. (c) For closed system W = pdv+∫ , for steady flow W = vdp−∫ IAS-12. Ans. (b) Start with D. ∫PdV only valid for quasi-static path so choice (c) & (d) out.

Automatically C-4 then eye on A and B. Bottle filling of gas is variable flow so A-2.

IAS-13. Ans. (c) W = pdv∫ where pressure (p) is an intensive property and volume (v) is anextensive property

IAS-14. Ans. (a) Pressure is intensive property but such option is not there.


27/114

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28/114

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29/114

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30/114

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31/114

First Law of ThermodynamicsS K Mondal’s Chapter 2

32

u = specific internal energy, kJ/kg

dv = change in specific volume, m3/kg.

Specific heat at constant volumeThe specific heat of a substance at constant volume C

v is defined as the rate of change of

specific internal energy with respect to temperature when the volume is held constant, i.e.,

∂⎛ ⎞= ⎜ ⎟∂⎝ ⎠v

v

uC

T

For a constant volume process

( )2

1

.

T

vv

T

u C dT Δ = ∫

The first law may be written for a closed stationary system composed of a unit mass of a pure

substance.

Q = Δu + Wor d Q = du + d W

For a process in the absence of work other than pdv work

d W = pdvTherefore d Q = du + pdv

Therefore, when the volume is held constant

( ) ( )

( )2

1

v v

v

u

.

T

v

T

Q

Q C dT

= Δ

= ∫

Since u, T and v are properties, Cv is a property of the system. The product m⋅

Cv is called the

heat capacity at constant volume (J/K).

Specific heat at constant pressureThe specific heat at constant pressure pC is defined as the rate of change of specific enthalpy

with respect to temperature when the pressure is held constant.

PC P

h

T

∂⎛ ⎞= ⎜ ⎟∂⎝ ⎠

For a constant pressure process

( )2

1

.

T

P P

T

h C dT Δ = ∫

The first law for a closed stationary system of unit mass

dQ = du + pdv

Again, h = u + pv

Therefore dh = du + pdv + vdp

= d Q + vdp

Therefore dQ = dh – vdp

Therefore ( dQ )P = dh


32/114


33

or ( ) ( h) p pQ = Δ

∴ Form abow equations

( )2

1

P .

T

P

T

Q C dT = ∫

pC is a property of the system, just like Cv. The heat capacity at constant pressure is equal to m

pC (J/K).

Application of First Law to Steady Flow Process S.F.E.E

S.F.E.E. per unit mass basis

(i)

+ + + = + + +2 2

1 2

1 1 2 2

C Cd Q d Wh g z h g z

2 d m 2 d m [h, W, Q should be in J/Kg and C in m/s and g in m/s2]

(ii)+ + + = + + +

2 2

1 1 2 21 2

gZ Q gZ Wh h

2000 1000 dm 2000 1000 dm

C d C d

[h, W, Q should be in KJ/Kg and C in m/s and g in m/s2]

S.F.E.E. per unit time basis

⎛ ⎞+ + +⎜ ⎟τ⎝ ⎠

⎛ ⎞= + + +⎜ ⎟

τ⎝ ⎠

211 1 1

2

22 2 2

w z2

w z2

x

C dQ h g d

dW C h g

d

Where, w = mass flow rate (kg/s)

Steady Flow Process Involving Two Fluid Streams at the Inlet and Exit of the Control

Volume


33/114

S

M

W

E

S

Th

en

N

A

dr

be

K Mon

ss balance

ere v = spec

ergy balan

me examp

e following

gineering sy

zzle and D

ozzle is a d

p, whereas

ow a nozzle

Firdal’s

1 1

1

A C

v

ific volume (

ce

⎛ +⎜

⎝

⎛ = ⎜

⎝

1 1

3 3

w h

w h

e of steady

xamples ill

tems.

ffuser:

evice which

a diffuser in

which is ins

1

2

C h +

st La

+1

2 2

2

w

A C

v

m3/kg)

+

+

2

11

2

33

2

2

C Z g

C Z g

flow proc

strate the a

increases t

creases the

lated. The s2

1

Q Z g

dm+ +

d

of T

34

=

=

2

3 3

3

w w

A C

v

⎛ +⎟ ⎜

⎝

⎞ ⎛ +⎟ ⎜

⎝ ⎠

2 2

4 4

w h

w h

sses:-

pplications o

e velocity

pressure of

teady flow e2

2

22

C h= + +

ig.

herm

+

+

3 4

4

4

w

A C

v

+

+ +

2

22

2

44

2

2

C Z g

C Z

f the steady

r K.E. of a

fluid at th

ergy equati

2

x W

g dm

+ d

odyn

4

⎞+⎟

τ ⎠

⎞+⎟ τ ⎠

x

dQ

d

dW

d

flow energy

fluid at the

expense of

on of the con

micsCh

equation in

expense of

its K.E. Fig

trol surface

pter 2

some of the

its pressure

ure show in

gives


34/114


35

Here 0; 0,x dW dQ

dm dm= = and the change in potential energy is zero. The equation reduces to

2 2

1 21 2

2 2

C C h h+ = + (a)

The continuity equation gives1 1 2 2

1 2

A A w = =

C C

v v (b)

When the inlet velocity or the ‘velocity of approach’ V1 is small compared to the exit velocity V2,

Equation (a) becomes2

2

1 2

2 1 2

2

2( ) /

C h h

or C h h m s

= +

= −

where (h1 – h2) is in J/kg.

Equations (a) and (b) hold good for a diffuser as well.

Throttling Device:

When a fluid flows through a constricted passage, like a partially opened value, an orifice, or a

porous plug, there is an appreciable drop in pressure, and the flow is said to be throttled. Figure

shown in below, the process of throttling by a prettily opened value on a fluid flowing in an insulated

pipe. In the steady-flow energy equation-

0, 0x W dQ

dm dm= =

d

And the changes in P. E. are very small and ignored. Thus, the S.F.E.E. reduces to2 2

1 21 2

2 2

C C h h+ = +

(Fig.- Flow Through a Valve)

Often the pipe velocities in throttling are so low that the K. E. terms are also negligible. So

1 2h h=

or the enthalpy of the fluid before throttling is equal to the enthalpy of the fluid after throttling.

Turbine and Compressor:

Turbines and engines give positive power output, whereas compressors and pumps require power

input.

For a turbine (Fig. below) which is well insulated, the flow velocities are often small, and the K.E.

terms can be neglected. The S.F.E.E. then becomes


35/114


36

(Fig.-. Flow through a Turbine)

1 2

1 2

x

x

dW h h

dm

W or h h

m

= +

= −

The enthalpy of the fluid increase by the amount of work input.

Heat Exchanger:

A heat exchanger is a device in which heat is transferred from one fluid to another, Figure shown in

below a steam condenser where steam condense outside the tubes and cooling water flows through

the tubes. The S.F.E.E for the C.S. gives

c 1 s 2 c 3 s 4

s 2 4 c 3 1

w w w w

, w ( ) w ( )

h h h h

or h h h h

+ = +

− = −

Here the K.E. and P.E. terms are considered small, there is no external work done, and energy

exchange in the form of heat is confined only between the two fluids, i.e. there is no external heat

interaction or heat loss.

Fig. -

Figure (shows in below) a steam desuperheater where the temperature of the superheated steam is

reduced by spraying water. If w1, w2, and w3 are the mass flow rates of the injected water, of the

steam entering, and of the steam leaving, respectively, and h1, h2, and h3 are the corresponding

enthalpies, and if K.E. and P.E. terms are neglected as before, the S.F.E.E. becomes

1 1 2 2 3 3w h w h w h+ =

and the mass balance gives

w1 + w2 = w3


36/114

S

Th

pr

un

(1)

(2)

(b)

(c)

K Mon

e above law

ctical situa

ity.

Work dev

(a) Water

In this ca

1 1 1v +z g p

(b) Steam

In this ca

(hW =

Work abs

(a) Centrifu

The system

In this sy

1 1 1v +z g p

Centrifug2

11

h2

C + +

Blowers –

Firdal’s

is also call

ions as wor

loping sys

turbines

e Q = 0 and2

12

z g2

C + =

or gas turbi

e generally

)2

11 2 – h

C ⎛ + ⎜

⎝

rbing syst

al water pu

is shown in

tem Q = 0 a

2W z g+ = +

l compress2

2 –2

C W Q =

In this case

st La

ed as stea

developing

ems

ΔU = 0 and

2 2p v W+

nes

ΔZ can be as2

2 Q2

C ⎞−+ Δ⎟

⎠

ms

p

the Figure b

Fig.

nd ΔU = 0; t2

22 2

p v 2

C +

or – In this

2h +

we have Δ z

of T

37

y flow en

system and

equation bec

sumed to be

elow

e energy eq

system Δz =

= 0,1 1v p =

herm

rgy equati

work absor

omes

zero and the

uation now

0 and the eq

2 2p v and Q

odyn

on. This ca

tion system

equation be

ecomes,

uation beco

= 0; now th

micsCh

n be applie

. Let the m

comes

es,

energy sim

pter 2

to various

ss flow rate

lifies to


37/114

S

(d)

(e)

(3)

(a)

(b)

(c)

(vi

M

ca

un

wisu

K Mon

1u +W

Fans – In

and hence t

2

2

2

C W =

Reciproca

equation ap

or

Non-work

Steam boi

equation for

Steam conare very sm

(h1 – h2) and

heat lost by

Steam noz

In this syst

possible hea

The en

i) Unstead

ny flow pro

be analyze

der non-stea

hin the conface, as giv

Firdal’s

2

22

u as2

C = +

ans the tem

e energy eq

ing compr

lied to a rec

(1

2

h – h

h

Q

W Q

=

= +

developin

ler – In thi

a boiler bec

denser – Ill. Under st

this heat is

steam will b

le:

m we can a

t loss also ze

rgy equatio

1

2

h

or C

+

Flow Ana

esses, such

by the con

dy state con

rol volume in below:

st La

2 1 C C

perature ris

ation for fa

ssor – In a

iprocating c

)2

1

–

– h

W

and absor

system we

mes Q = (h2

this systeady conditi

also equal t

equal to he

ssume ΔZ aro.

for this cas2

1 2

2

2

1 1

2 2

2(

C C h

C h

= +

+

ysis:-

s filling up

rol volume t

itions (Figu

accumulat

of T

38

e is very sm

s becomes,

reciprocatin

mpressor is

ing syste

neglect ΔZ, – h1)

the work dns the chan

o the chang

at gained by

d W to be

e becomes.

2)h

and evacuati

echnique. Co

re-shown in

d is equal to

herm

all and heat

compresso

s

KE and W

ne is zero age in enthal

in enthalp

the cooling

ero and hea

ng gas cylin

nsider a dev

below). The

the net rate

odyn

loss is negl

ΔKE and Δ

(i.e.) ΔZ = Δ

nd we can ay is equal t

of cooling

ater.

t transfer w

ers, are not

ice through

rate at whic

of mass flo

micsCh

cted (i.e.) Δ

E are negli

E = W = 0;

lso assumeheat lost b

ater circula

hich is noth

steady, Suc

hich a flui

the mass o

across the

pter 2

h = 0, q = 0

ibly energy

the energy

Z and ΔKEsteam. Q =

ted (i.e.) the

ing but any

processes

is flowing

fluid

ontrol


38/114

S

W

Th

ac

Ra

W

Fo

an

Or

K Mon

ere vm is t

Over a

vmΔ =

e rate of acc

oss the cont

te of energyv

dE=

d

⎛ = ⎜

⎝ v

E

τ

ere m is th

11

dE

d

h +2

∴

⎛ ⎜⎝

v

C

τ

llowing Figu

vEΔ =

Equati

d the equati

vdE

dτ =

dEv =

Firdal’s

1v

dmw

dτ = −

e mass of fl

y finite peri

1 2m – mΔ Δ

umulation o

rol surface. I

increase = R2

11 1

2

h + +2

mC + +

2

⎛ ⎜⎝

C w

mass of flui

11

d mU +

d

dm+Z g

dτ

⎛ ⎜⎝

⎞⎟

⎠

τ

re shows all

1h⎛

− + ⎜⎝

∫W

n (A) is gen

vdE = 0dτ

n reduces

dQ dW

d dτ τ −

Q – dW or

st La

12

dm d

dτ = −

id within th

od of time

f energy wit

f Ev is the e

ate of energ

1

v

dQg +

dτ

gZ

⎞−⎟

⎠

⎞⎟

⎠

w

d in the con2

v

2

+ mgZ

dQh +

dτ

⎞⎟

⎠

⎛ − ⎜

⎝

these energ2

11+Z g dm

2

⎞⎟ ⎠

C

eral energy

or a closed s

dQ = dE +

of T

39

ig.

2m

τ

e control vol

in the cont

ergy of fluid

inflow – Ra2

22 2 2h + +Z

2

⎛ ⎜⎝

C

rol volume a

2

2 22

dm+Z g

2 d

⎞⎟

⎠ τ

flux quanti2

21 2h +

2

⎛ − ⎜

⎝ ∫

C

quation. Fo

ystem w1 =

W

herm

ume at any i

ol volume is

within the

te of energydW

gd

⎞−⎟

⎠ τ

t any instan

dW

d−

τ

ties. For any

2 2+Z g dm ⎞⎟ ⎠

Fig.

steady flow

, w2 = 0, the

odyn

nstant.

equal to th

ontrol volu

outflow.

equati

t

(equ

time interv

,

n from equa

micsCh

net rate of

e at any ins

...on A

)........tion B

l, equation

tion (A),

pter 2

energy flow

tant,

B) becomes


39/114

SFl

Ex

Va

tec

th

an

su

p

Sy

w

W

be

vol

U

K Monw Processes

ample of a

riable flow p

hnique, as il

Consid beginning

d gas flows i

ply to the p

P P P,T , v , h ,

stem Tech

ich would e

Energy

1E m=

ere ( 2m –

2E

E

=

Δ =

The P.E.

n omitted.

Now, the

ume. Then t

ing the firs

Q = ΔE

= 2m

Firdal’s

variable fl

rocesses ma

lustrated be

r a processhe bottle co

nto the bottl

ipeline is ve

P Pu and v .

ique: Assu

entually en

of the gas b

(1 1 2u m –+

)1 is the m

2 2

2 1

u

– E m=

erms are ne

e is a chan

he work don

W =

=

for the proc

+ W

2 1 1u –m u –

st La

w proble

be analyze

low.

in which a gtains gas of

e till the ma

y large so t

e an envelo

er the bottle

efore filling.

)2

1m2

P P

C u

⎛ +⎜

⎝ ss of gas in

(2 1 1u – m u

glected. The

ge in the vo

e

( 2p – V p V

( p 0 –

p ⎡⎣

( 2 – m – m

ess

)2 1m – m⎡⎢⎣

of T

40

:

d either by t

s bottle is fi mass m1 at

s of gas in t

at the state

pe (which is

, as shown i

⎞⎟

⎠

he pipeline

)2

1 1m – u

2

P C ⎛

⎜⎝

gas in the b

lume of gas

)

)2 1 – m v

)1 Pp v p

(2

PP

+ u2

⎤−⎥

⎦

herm

e system te

lled from a ptate p1, T1,

e bottle is

of gas in the

extensible)

Figure abo

nd tube wh

P u

⎞+ ⎟

⎠

ttle is not i

because of

⎤⎦

)2 1 – m P P p v

odyn

chnique or t

ipeline (Fig1, h1 and u1.

2 at state p2 pipeline is c

f gas in the

e.

ich would en

motion, an

the collapse

micsCh

e control vo

re shown inThe valve is

, T2, v2, h2 a

onstant at

pipeline and

ter the bottl

so the K.E.

of the enve

pter 2

lume

below). Inopened

d u2. The

the tube

.

terms have

lope to zero


40/114

S

w

CFi

wr

Si

No

DLe

ap

As

A

or

or

w

qu

K Mon

= 2m

ich gives th

ntrol Voluure above,

itten on a ti

ce hP and C

w

v 2

2 2

E U –

m u –

=

=

ischar t us conside

plying first l

suming K.E.

ain

ich shows

asi-static.

For ch

Firdal’s

2 1 1u – m u –

energy bal

e Techniqpplying the

e rate basi

vdE dQ

dτ dτ= +

are consta

vE Q ⎛

Δ = + ⎜⎝

1 2 2

1 1 P

m u –

u – h

=

⎛

⎜⎝

ing a a tank dis

aw to the co

and P.E. of

d(mmdu+ udm

( )

=

= =

+

= −

= −

+ == 0

dm du

m pv

V vm c

vdm mdv

dm dv

m v

du dv

pv v

d u pv

dQ

hat the pro

rging the ta

st La

)2 1m – m2

C ⎛ ⎜⎝

nce for the

ue: Assumeenergy equa

-2

PP

dh +

2 dτ

C ⎛ ⎞⎜ ⎟⎝ ⎠t, the equat

(2

PP 2+ m

2

⎞⎟ ⎠

C

(

1 1

2

P

2

m u

m2

C ⎞

−⎟ ⎠

d Chaharging a fl

trol volume,

= V dU d

the fluid to

u) = hdm= udm+ pv

=

.

0

0

nst

cess is adia

nk

of T

41

P+ h

⎞⎟ ⎠

rocess.

a control voltion in this

ion is integr

)1m

)1

ging auid into a s

⎛ + +⎜

⎝ h

e small and

m

batic and

herm

ume boundecase, the foll

ted to give f

Tankpply line (F

⎞+ ⎟

⎠

2

o

gz2

C

dQ = 0

Chargi

odyn

d by a controwing energ

or the Total

igure). Since

out

ut

dm

ng and Dis

micsCh

l surface asbalance m

process

x dW = 0 a

charging a

pter 2

shown iny be

d dmin = 0,

Tank


41/114


42

( ) = Δ = −

= −

∫ 2 2 1 1in2 2 1 1

V

p p

hdm U m u m u

m h m u m u

where the subscript p refers to the constant state of the fluid in the pipeline. If the tank is initially

empty, m1 = 0.

= 2 2 p pm h m u

Since

=

=2

2

p

p

m m

h u

If the fluid is an ideal gas, the temperature of the gas in the tank after it is charged is given by

=

= γ2

2

p p vc T c T

or T T

PROBLEMS & SOLUTIONSExample 1

The work and heat transfer per degree of temperature change for a closed system is given by

1 1/ ; /

30 10

dW dQ kJ C kJ C

dT dT ° ° = =

Calculate the change in internal energy as its temperature increases from 125ºC to 245ºC.

Solution:

( ) ( )

( )

2

1

2

1

2 1

30

1 1245 125

30 30 30

10

1245 125 12

10 10

T

T

T

T

dT dW

dT W T T

dT dQ

dT Q kJ

=

= = − = −

=

= = − =

∫

∫

Applying First Law of Thermodynamics

Q = W + ΔUΔU = Q – W = 12 – 4 = 8kJ.

Example 2

Air expands from 3 bar to 1 bar in a nozzle. The initial velocity is 90 m/s. the initial temperature is

150ºC. Calculate the velocity of air at the exit of the nozzle.

Solution:

The system in question is an open one. First Law of Thermodynamics for an open system gives2 2

1 21 1 1 2 2 2

2 2

C C w h Z g Q w h Z g W

⎡ ⎤ ⎡ ⎤+ + + = + + +⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦

Since the flow is assumed to be steady.

1 2w w=

Flow in a nozzle is adiabatic flow.


42/114


43/114


44

( )

( )

( )

i i e e 2 2 1 1

e

1 2

2

0

0

0 2

2 0

2 0

Q m h = m h m u – m u W

Here Q 0, W 0, m 0 no mass leaving from control vol.

0 evacuated

0.718 0.287 273

0.718 25 0.287 298

103.48 /

103.48 /

0.71

i

i

i i i i

m m m

h u

h u pv u T T

u

u kJ kg u

or u u kJ kg

u u

+ + +

= = =

= ∴ =

∴ =

= + = + + +

= + × + ×

= + =

− =

= +2

2 0

2

8

103.48144.2

0.718 0.718

T

u uT C

−= = = °

Example 4

A system whose mass is 4.5 kg undergoes a process and the temperature changes from 50° C to

100°C. Assume that the specific heat of the system is a function of temperature only. Calculate theheat transfer during the process for the following relation ship.

801.25 /

160n

c kJ kg C t

= + °+

[t is in oC]

Solution:

[ ] ( )

[ ] ( ) ( )( )

100 100

1 2

50 50

100 100

50 50

100100

50

50

804.5 1.25

160

4.5 1.250.0125 2.0

14.5 1.25 ln 0.0125 2.0

0.01251

4.5 1.25 50 1.25 2.0 0.625 20.0125

4

⎛ ⎞= = +⎜ ⎟+⎝ ⎠

⎧ ⎫⎪ ⎪= +⎨ ⎬

+⎪ ⎪⎩ ⎭

⎧ ⎫⎪ ⎪⎡ ⎤= + +⎨ ⎬⎢ ⎥⎣ ⎦⎪ ⎪⎩ ⎭

⎧ ⎫⎡ ⎤= × + + − +⎨ ⎬⎢ ⎥⎣ ⎦⎩ ⎭

=

∫ ∫

∫ ∫

nQ mc dt dt

t

dtdt

t

t t

ln ln

1 3.25.5 62.5 358

0.0125 2.625

⎧ ⎫+ =⎨ ⎬⎩ ⎭

ln kJ


44/114

S

A

CT

st

ar

no

G

G

G

K Mon

SKED

pplicati

mmon De inlet an

am for a

e as indi

tations are

TE-1. If

th

(a)

TE-2. As

wa

en

(a)

TE-3. Th

th

Ac

(a)

Firdal’s

OBJE

revio

on of

ata for Qd the outl

adiabatic

ated in t

as usually

ass flow

turbine (i

12.157

ume the a

ter at the

rgy effects

0.293

followin

rmodynam

ording to

Figures 1an

st La

TIVE

s 20-

irst La

estionst conditio

steam tu

he figure.

followed.

ate of stea

MW) is:

ove turbi

inlet to th

, the specif

(

four fig

ic cycle, on

he first la

2 (b) Fig

of T

45

UEST

ears

to St

1 and Qns of

bine

The

m through

(b) 12.941

e to be pa

pump is

c work (in

b) 0.35 1

ures have

the p-v an

of thermo

ures 1and 3

herm

IONS (

GAT

ady Fl

:

the turbin

(c

t of a simp

1000 kg/m3.

kJ/kg) sup

(c)

been dr

T-s plane

dynamics,

(c) Figure

odyn

ATE

Que

w Pro

e is 20 kg/

) 168.001

le Rankine

Ignoring

lied to the

2.930

wn to r

.

qual areas

1and 4

micsCh

, IES,

tions

ess S.

[G

the powe

[G

(d)

cycle. The

inetic an

pump is:

[ (d)

present a

[

are enclos

(d) Figures

pter 2

IAS)

.E.E

TE-2009]

output of

TE-2009]

168.785

density of

potential

ATE-2009]3.510

fictitious

ATE-2005]

ed by

2 and 3


45/114


46

Internal Energy – A Property of SystemGATE-4. A gas contained in a cylinder is compressed, the work required for

compression being 5000 kJ. During the process, heat interaction of 2000 kJ

causes the surroundings to the heated. The change in internal energy of the

gas during the process is: [GATE-2004](a) – 7000 kJ (b) – 3000 kJ (c) + 3000 kJ (d) + 7000 kJ

GATE-4a. The contents of a well-insulated tank are heated by a resistor of in

which 10 A current is flowing. Consider the tank along with its contents

as a thermodynamic system. The work done by the system and the heat

transfer to the system are positive. The rates of heat (Q), work (W) and

change in internal energy during the process in kW are [GATE-2011]

(a) Q = 0, W = –2.3, = +2.3 (b) Q = +2.3, W = 0, = +2.3

(c) Q = –2.3, W = 0, = –2.3 (d) Q = 0, W = +2.3, = –2.3

Discharging and Charging a TankGATE-5. A rigid, insulated tank is initially

evacuated. The tank is connected with a

supply line through which air (assumed to

be ideal gas with constant specific heats)

passes at I MPa, 350°C. A valve connected

with the supply line is opened and the tank

is charged with air until the final pressure

inside the tank reaches I MPa. The final

temperature inside the tank

(A) Is greater than 350°C

(B) Is less than 350°C

(C) Is equal to 350°C

(D) May be greater than, less than, or equal to

350°C, depending on the volume of the tank

Previous 20-Years IES Questions

First Law of Thermodynamics

IES-1. Which one of the following sets of thermodynamic laws/relations is directlyinvolved in determining the final properties during an adiabatic mixing

process? [IES-2000]

(a) The first and second laws of thermodynamics

(b) The second law of thermodynamics and steady flow relations

(c) Perfect gas relationship and steady flow relations

(d) The first law of thermodynamics and perfect gas relationship

23Ω

( U)

Δ U Δ UΔ U Δ U


46/114


47

IES-2. Two blocks which are at different states are brought into contact with each

other and allowed to reach a final state of thermal equilibrium. The final

temperature attained is specified by the [IES-1998]

(a) Zeroth law of thermodynamics (b) First law of thermodynamics

(c) Second law of thermodynamics (d) Third law of thermodynamics

IES-3. For a closed system, the difference between the heat added to the system and

the work done by the system is equal to the change in [IES-1992]

(a) Enthalpy (b) Entropy

(c) Temperature (d) Internal energy

IES-4. An ideal cycle is shown in the figure. Its

thermal efficiency is given by

3 3

1 1

2 2

1 1

1 11

(a)1 (b) 11 1

γ

⎛ ⎞ ⎛ ⎞− −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

− −⎛ ⎞ ⎛ ⎞− −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

v v

v v

p p

p p

( )

( )

( )

( )3 1 3 11 1

2 1 1 2 1 1

1(c)1 (b) 1γ

γ

− −− −

− −

v v v v p p

p p v p p v

[IES-1998]

IES-5. Which one of the following is correct? [IES-2007]

The cyclic integral of )( W Q δ δ − for a process is:(a) Positive (b) Negative (c) Zero (d) Unpredictable

IES-6. A c

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