ME8391 Engineering Thermodynamics MCQ

Multiple Choice Questions 108 Pages

Anna University

Mechanical Engineerng

ASM

Contributed by

Anees Singh Mall

Vel Tech High Tech Dr.Rangarajan Dr.Sakunthala
Engineering College
Department of Mechanical Engineering
Multiple Choice Questions
III SEMESTER
ME8391-Engineering Thermodynamics
(Regulation – 2017)
Academic Year 2020 – 2021
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9
01.
THERMODYNAMICS

1.

Basic Concepts & Zeroth
Law of Thermodynamics

1. The energy of the isolated system is always a
constant, which is given by:
(a) Zeroth law of thermodynamics
(b) First law of thermodynamics
(c) Second law of thermodynamics
(d) Third law of thermodynamics
(e) Law of stable equilibrium
CGPSC Polytechnic Lecturer 2017

Ans. (b) : Isolated System–An isolated system is a
thermodynamic system that cannot exchange either
energy or mass outside the boundaries of the system.
So, ∆E = 0, ∆m = 0
So from the first law of thermodynamics energy of the
isolated system will be remain constant.
2. A series of operations, which takes place in a
certain order and restore the initial conditions
at the end, is known as
(a) Reversible cycle
(b) Irreversible cycle
(c) Thermodynamic cycle
(d) None of these
Vizag Steel (MT) 2017
UPPSC AE 12.04.2016 Paper-II
Ans. (c) : Thermodynamic cycle– A series of
operation which takes place in a certain order and
restore the initial conditions at the end, is known as
thermodynamic cycle.
3. Heat and work are :

(a) Intensive properties
(b) Extensive properties
(c) Point functions
(d) Path functions
OPSC AEE 2019 PAPER - II
UPRVUNL AE 2014

UP Jal Nigam AE 2016
Nagaland PSC CTSE 2017 Paper-2
UKPSC AE 2007 Paper -II

Ans.

(d) : Heat and work are path function similarities
between heat and work :
(i) Both are recognized at the boundary of the system,
as they cross the boundary phenomena.
(ii) System possesses - energy, but neither heat and work.
(iii) Both are associated with process not state. Heat and
work have no meaning at a state.
(iv) Both are path functions.
• Path function- Magnitude depends on the path
followed during the process as well as the end states.
• Point function- Magnitude depends on state only
and not a how the system approaches that state.
4. What are the properties of a thermodynamic
system whose value for the entire system is
equal to the sum of their values for individual
parts of the system?
(a) Thermodynamic properties
(b) Extensive properties
(c) Intensive properties
(d) Specific properties
TNPSC AE 2018

Ans. (b) : Extensive properties- Extensive properties
of matter that changes as the amount of matter changes
Example- Volume, enthalpy, entropy etc.
Intensive properties- An intensive properties is a bulk
property, meaning that it is a local physical property of
a system that does not depends on the system size or the
amount of material in the system.
Example- Pressure, temperature, refractive index,
density and hardness of object.
5. Consider the following properties:
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
Gujarat PSC AE 2019

Ans : (d) : Properties of intensive are as follows-
• Temperature
• Viscosity
• Specific entropy
• Thermal conductivity
• Specific volume
• Specific enthalpy
6. An open system
(a) is a specified region where transfers of
energy and / or mass take place
(b) is a region of constant mass and only energy
is allowed to cross the boundaries
(c) cannot transfer either energy or mass to or
from the surroundings
(d) has an enthalpy transfer across its boundaries
and the mass within the system is not
necessarily constant
BPSC Poly. Lect. 2016
HPPSC Lect. 2016
Ans : (a)
system mass
transfer
Energy
transfer
open system
closed system
Isolated system
√

×
×
√

√
×
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7. Match the following :
1. Closed system a. Increase in static pressure
2. Open system b. Increase in kinetic energy
3. Pump c. Heat, mass and work
interact
4. Turbine d. Heat and work interact
5. Nozzle e. Delivers work
(a) 1–d, 2–c, 3–a, 4–e, 5–b
(b) 1–c, 2–d, 3–b, 4–e, 5–a
(c) 1–c, 2–a, 3–d, 4–e, 5–b
(d) 1–d, 2–c, 3–e, 4–a, 5–b
OPSC Civil Services Pre. 2011

Ans. (a) :
1. Closed system Heat and work interact
2. Open system Heat, mass and work interact
3. Pump Increase in static pressure
4. Turbine Delivers work
5. Nozzle Increase in kinetic energy
8. The law which provides the basis of
temperature measurement is:
(a) Third law of thermodynamics
(b) Zeroth law of thermodynamics
(c) First law of thermodynamics
(d) Second law of thermodynamics
TNPSC 2019
UPRVUNL AE 2014, 2016

Ans. (b) : 1. Zeroth law of thermodynamics provides
the basis of temperature measurement.
2. First law of thermodynamics provides conservation
law of energy.
3. Second law of thermodynamics provides the basic
concept of entropy.
9. Which one of the following is the extensive
property of the system?
(a) Volume (b) Pressure
(c) Temperature (d) Density
OPSC AEE 2019 PAPER - II
UPRVUNL AE 2014

Ans : (a) : These properties are dependent on mass e.g.
volume, energy, Heat capacity (C
v
, C
p
), enthalpy,
entropy.
10. Pressure exerted by a gas in a closed container
is:
(a) Weak function of Density and Temperature
(b) Weak function of Density and Volume
(c) Strong function of Density and Temperature
(d) Strong function of Density and Volume
OPSC AEE 2019 PAPER - II

Ans : (c) : Pressure exerted by a gas in a closed
container is strong function of density and temperature.
11. Which of the following items is not a path
function?
(a) Heat
(b) Work
(c) Kinetic energy
(d) Thermal conductivity
BPSC AE 2012 Paper - V

Ans : (d) : Thermal conductivity is not a path functions.
12. Which one of the properties given below is an
intensive property of the system
(a) composition (b) volume
(c) kinetic energy (d) entropy
TSPSC AEE 2015

Ans. (a) : Volume, Kinetic energy and entropy are the
example of extensive property where as composition is
the example of intensive property.
13. Which pair of the following alternatives is
correctly matched ?
List – I List – II
(a) Heat – Point function
(b) Energy – Path function
(c) Entropy – Second law of
thermodynamics
(d) Gibbs function – Path function
UKPSC AE 2012 Paper–II

Ans. (c) : Entropy – Second law of
thermodynamics
14. Thermodynamic work is the product of
(a) Two intensive
(b) Two extensive properties
(c) An intensive property and change in an
extensive property
(d) An extensive property and change in an
intensive property
UPPSC AE 12.04.2016 Paper-II
Ans : (c) For quasistatic process work done is
calculated by
P.dV
∫

Pressure = Intensive property.
Volume (dV) = change in an extensive property.
Thermodynamic work is the product of An intensive
property and change in an extensive property.
15. Bi-Metallic strips made of two different
materials bend during a rise is temperature
because of
(a) Differences in coefficient of linear expansion
(b) Differences in elastic properties
(c) Differences in thermal conductivities
(d) Difference in stress
TNPSC AE 2014

Ans. (a) : Bi-Metallic strips made of two different
materials bend during a rise is temperature because of
differences in coefficient of linear expansion.
16. The gauge pressure in a truck tire before and
after the journey was recorded as 200 kPa and
220 kPa respectively at the location where
atmospheric pressure and temperature was
100 kPa and 27ºC respectively. How much the
rise in temperature of tire air after trip?
(a) 20ºC (b) 27ºC
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(c) 47ºC (d) Insufficient data
OPSC AEE 2019 PAPER - II

Ans : (c) : Given
P
atm
= 100 kPa, T
atm
= 27 ºC = 300 K
P
gauge1
= 200 kPa, P
gauge2
= 220 kPa
[P
1
]
Abs
= 300 kPa, [P
2
]
Abs
= 320 kPa
then,
V
1
= V
2
1 1
2 2
P T
P T
=

1 2
2
1
300 320
300
T P
T
P
×
×
= =

2
320 47º
T K C
= =

17. The condition for the reversibility of a cycle is
(a) the pressure and temperature of the working
substance must not differ, appreciably, from
those of the surroundings at any stage in the
process
(b) all the processes, taking place in the cycle of
operation, must be extremely slow
(c) the working parts of the engine must be
friction free
(d) all of the above
Gujarat PSC AE 2019

Ans : (d) :
The condition for the reversibility of a cycle
is

The pressure and temperature of the working
substance must not differ, appreciably from those in the
process.

The working parts of the engine must be friction free.
18. Which of the following is an example of
irreversible process?
(a) Polytropic expansion of fluid
(b) Unrestricted expansion of gases
(c) Isothermal expansion
(d) Electrolysis
JPSC AE 2013 Ist Paper

Ans. (b) :

Irreversible process-
In an irreversible process, there is
a loss of heat due to friction, radiation and conduction.
In an actual practice, most of the processes are
irreversible to some degree The main causes for the
irreversibility may be.
(i) Mechanical and fluid friction
(ii) Unrestricted expansion
(iii) Heat transfer with a finite temperature difference.
19. Which of the following is considered as
thermodynamic potential?
(a) Temperature (b) Internal energy
(c) Enthalpy (d) Entropy
JPSC AE 2013 Ist Paper

Ans. (c) :
Enthalpy is considered as thermodynamic
potential.
20. Ratio of absolute temperature
2
1
T
T
at two
different states in an adiabatic process is :
(a)
1
1
2
1
P
P
γ−
 
 
 
(b)
( )
1
2
1
P
P
γ−
γ
 
 
 

(c)
1
2
1
P
P
γ
γ−
 
 
 
(d)
1
1
1
2
P
P
γ−
 
 
 

OPSC Civil Services Pre. 2011
UKPSC AE-2013, Paper-II
Ans. (b) :

Ratio of absolute temperature
2
1
T
T
at two different states
in an adiabatic process is :
1
2 2
1 1
T P
T P
γ−
γ
 
=
 
 

21. Pressure of 1 (one) atmosphere is equivalent to:
(a) 0.101325 MPa (b) 750 mm of mercury
(c) 1 bar (d) 0.101325 kPa
OPSC Civil Services Pre. 2011

Ans. (a) :

1 atmosphere pressure = 101325 Pa
= 1.01325 bar
= 760 mm of mercury
= 760 Torr
= 0.101325 MPa
= 14.6959 Pound-force per
square inch
22. A steel ball of mass 1 kg and specific heat
0.4 kJ/kg
°
C is at a temperature of 60
°
C. It is
dropped into 1 kg water at 20
°
C. The final
steady state temperature of water is :
(a) 23.5 °C (b) 30 °C
(c) 35 °C (d) 40 °C
RPSC Vice Principal ITI 2018

Ans. (a) :

m
b
= 1 kg
c
b
= 0.4 kJ/kgk

1
b
T
= 60°C
m
w
= 1 kg

1
w
T
= 20°C
Let the final steady state temperature of ball and water
be T°C.

1 1
b b b w w w
m c (T T) m c (T T )
− = −

1 × 0.4 (60°–T) = 1 × 4.18 (T–20°)
T = 23.5°C
23. A 120-V electric resistance heater draws 10 A.
It operates for 10 min in a rigid volume.
Calculate the work done on the air in the
volume.
(a) 720000 kJ (b) 720 kJ
(c) 12000 J (d) 12 kJ
RPSC Vice Principal ITI 2018

Ans. (b) :

W = VIT = 120 V × 10 A × (10 × 60) s
= 72 × 10
4
J = 720 kJ.
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24. Air is compressed in a cylinder such that the
volume changes from 0.2 to 0.02 m
3
. The initial
pressure is 200 kPa. If the pressure is constant,
the approximate work is
(a) –36 kJ (b) –40 kJ
(c) –46 kJ (d) –52 kJ
RPSC Vice Principal ITI 2018

Ans. (a) :

Work done, If the pressure is constant,
w =
2
1
PdV
∫

= P(V
2
– V
1
)
= 200 × (0.02 – 0.2)
= –36 kJ
25
.
Match List 1 with List 2 and choose the correct
answer from the code-

List-I
(laws of
thermodynamics)
List-II
(Defines)
A.
First

(i)
Absolute zero
temperature

B
.
Second

(ii)
Internal Energy

C
.
Zeroth

(iii)
Temperature

D
.
Third
(iv)
Entropy
1. (A) (B) (C) (D)
i ii iii iv
2. (A) (B) (C) (D)
iii iv ii i
3. (A) (B) (C) (D)
iv ii i iii
4. (A) (B) (C) (D)
ii iv iii i
RPSC INSP. OF FACTORIES AND BOILER 2016

Ans : (4)

(A) First→(ii) Internal Energy
(B) Second→(iv) Entropy
(C) Zeroth→ (iii) Temperature
(D) Third → (i) Absolute zero Temperature
26. Which of the following processes is irreversible
process
(a) Isothermal (b) Adiabatic
(c) Throttling (d) All of the above
Vizag Steel (MT) 2017

Ans. (c) :

Throttling process in which

No change in enthalpy from state one to state two (h
1

= h
2
)

No work is done (W = 0)

Process called isenthalpic

Process is adiabatic (Q = 0)
27. A cylinder contains 5m
3
of ideal gas at a
pressure of 1 bar. This gas is compressed in a
reversible isothermal process till its pressure
increases to 5 bar.
(a) 804.7 (b) 953.2
(c) 981.7 (d) 1012.2
Vizag Steel (MT) 2017

Ans. (a) :

Volume (v
1
) = 5.0m
3

Initial pressure (P
1
) = 1 bar = 100 kPa
Final Pressure (P
2
) = 5 bar = 500 kPa
We know that,
Work done (w)
2
1
P
P
VdP...........(i)
=
∫

Since, process is isothermal
Therefore, PV = mRT = constant.....(i)
Now equations (i) and (ii)

2
1
P
2
1
P
P
c
w dP c
P P
 
= =
 
 
∫
ln

2
1 1 1 1 2 2
1
P
w P V ( P V P V )
P
 
= =
 
 
∵

500
w (100 5)
100
 
= × ×
 
 
ln

= 804.7 kJ
28. If 'h' refers t heat, T refers to temperature,
then in the throttling process,
(a)
2
1 2
h h
=
(b)
1 2
h h
=

(c)
1 2
fg
s
h
h h
T
= +
(d)
2 1
fg
s
h
h h
T
= +

TNPSC 2019

Ans. (b) :
The throttling process is a constant enthalpy
process.
h
1
= h
2

29. Which of the following processes are
thermodynamically reversible?
(a) Throttling
(b) Free expansion
(c) Constant volume and constant pressure
(d) Isothermal and adiabatic
TNPSC 2019

Ans. (d) :
Throttling, Free expansion, Constant volume
and constant pressure processes and adiabatic process
are thermodynamically irreversible whereas isothermal
and adiabatic processes are thermodynamically
reversible.
30. The most efficient method of compressing air is
to compress it
(a) adiabatically (b) isentopically
(c) isothermally (d) isochorically
TNPSC 2019

Ans. (c) :
The work of compression or the steady flow
work input to the gas.
1
2
c 1 1
1
P
W P V 1
1 P
γ−
γ
 
 
γ
 
= −
 
 
γ −
 
 
 

For reversible adiabatic compression.
Similarly, for reversible polytropic compression
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n 1
n
2
n 1 1
1
Pn
W P V 1
n 1 P
−
 
 
 
 
 
 
= −
 
 
−
 
 
 

For reversible isothermal compression of an ideal gas
W
t
= P
1
V
1

n
ℓ
(P
2
/P
1
).
So, for same pressure ratio P
2
/P
1
the isothermal
compression needs the minimum work, whereas
adiabatic compression needs the maximum work, while
the polytropic compression work lies between iso-
thermal and adiabatic.
31. The internal energy of an ideal gas is function
of
(a) pressure only
(b) absolute temperature only
(c) pressure and volume
(d) pressure, volume and temperature
Gujarat PSC AE 2019

TSPSC AEE 2015
UKPSC AE 2007 Paper -II
OPSC AEE 2015 PAPER - II

Ans. (b) :
The internal energy of an ideal gas is function
of absolute temperature only.
U =
f
(T only)
32. Work output from a system is at the expense of
internal energy is a non flow process carried
out
(a) at constant pressure
(b) at constant volume
(c) adiabatically
(d) polytropically
TSPSC AEE 2015

Ans. (c) :
Work output from a system is at the expense
of internal energy is a non flow process carried out
adiabatically.
33. The absolute zero pressure will be
(a) When the molecular momentum of the system
becomes zero
(b) at sea level
(c) at the temperature of –273K
(d) at the centre of the earth
APPSC AEE 2016

Ans. (a) :
Generally, absolute zero pressure is the point
where there exist a minimum temperature i.e. zero. That
can be possible only when molecular momentum of
system become zero. There should not be any motion of
particles so there is no collision of particles, kinetic
energy nullifies and the temperature becomes zero.
34. An ideal gas can be taken from point K to point
N in three different paths:
K
→
L
→
N, K
→
N, K
→
M
→
N. Which of the
following is a true statement?

(a) The same work is done during each process.
(b) The same amount of heat is added to the gas
during each process
(c) The same change in internal energy during
each process
(d) The same entropy generated during each
process
APPSC AEE 2016

Ans. (c) :
From this diagram we get that the same
change in internal energy during each process because
of internal energy. It is the property of the system (i.e
point function) whereas work and heat for each process
will be different because of area under P–V and T–S
diagram will be different for each process (i.e path
function).

35. Which of the following is not a property of the
system?
(a) Temperature (b) Pressure
(c) Volume (d) Heat
TNPSC AE 2017
UKPSC AE 2012 Paper–II

Ans. (d) :
Heat is not a property of the system it is the
form of energy whereas temperature, pressure and
volume are the property of the system.
36. In reference to Thermodynamic equilibrium, it
is required to have,
(a) Mechanical Equilibrium
(b) Chemical Equilibrium
(c) Thermal Equilibrium
(d) Mechanical, Chemical and Thermal
Equilibrium

TNPSC AE 2017

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Ans. (d) :
In reference to Thermodynamic equilibrium,
it is required to have, Mechanical, Chemical and
Thermal Equilibrium.
For Mechanical equilibrium
0
∆ ≈
P

For Thermal equilibrium
0
∆ ≈
T

For Chemical equilibrium no change in concentration of
reactants and products.
37. If the value of n is zero in the equation PV
n
=
C, then the process is called
(a) constant volume process
(b) constant pressure process
(c) idiabatic process
(d) isothermal process
TNPSC AE 2018

Ans. (b) :
If the value of n is zero in the equation pV
n
=
c, then the process is called constant pressure process.
38. In a free expansion of a gas between two
equilibrium states, work transfer involved.
(a) can be calculated by joining the two states on
p-v coordinates by any path and estimating
area below
(b) can be calculated by joining 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
TNPSC AE 2018

Ans. (c) :
In a free expansion of a gas between two
equilibrium states, work transfer involved is zero.
39. The internal energy of a substance depends on
(a) temperature (b) pressure
(c) entropy (d) enthalpy
TNPSC AE 2018

Ans. (a) :
The internal energy of a substance depends
on temperature.
U = mc
V
∆T
40. The temperature at which the volume of gas
becomes zero is called
(a) absolute scale of temperature
(b) absolute zero temperature
(c) absolute temperature
(d) dew point temperature
TNPSC AE 2018

Ans. (b) :
The temperature at which the volume of gas
becomes zero is called absolute zero temperature.
At absolute zero temperature (– 273.15 K),
momentum of gas molecules becomes zero.
41. A fan consumes 20 W of electric power and
discharges air from a ventilated room at 0.25
kg/s. The maximum air outlet velocity is nearly
(a) 4.7 m/s (b) 8.7 m/s
(c) 10.2 m/s (d) 12.7 m/s
UPSC JWM 2017

Ans. (d) :
Fan convert electrical energy into mechanical
energy.
( )
2
1
Power Mass flow rate of air × discharge v
elocity
2
= ×

2
1
P mv
2
=
ɺ

2P 2 20
v 12.65m / s
m 0.25
×
= = =
ɺ

Maximum air outlet velocity
12.7m / s
≃

42. In a reversible adiabatic process the ratio
(T
1
/T
2
) is equal to-
(a)
1
1
2
p
p
γ
γ
−
 
 
 
(b)
1
1
2
v
v
γ
γ
−
 
 
 

(c)
( )
1
2
1 2
v v
γ
γ
−
(d)
2
1
v
v
γ
 
 
 

RPSC AE 2018
UKPSC AE 2012 Paper–II

Ans. (a) :
We know that, Polytropic process
PV
n
= constant ...(1)
If process is reversible adiabatic then η = γ = 1.4
For ideal gas

constant
PV
T
=

then

2 1 2
1 1 2
V P T
V T P
 
= ×
 
 
...(2)

1 1 2 2
P V = P V
γ γ

1
2 1
1 2
V P
V P
γ
   
=
   
   

From equation (2)

1
1 2 1
2 1 2
P T P
P T P
γ
     
× =
     
     

1
1 1
2 2
T P
T P
γ
γ
 
−
 
 
   
=
   
   

43. An isolated system-
(a) is a specified region where transfer of energy
and/or mass takes place
(b) is a region of constant mass and only energy
is allowed to cross the boundaries
(c) cannot transfer either energy or mass to or
from the surroundings
(d) is one in which mass within the system is not
necessarily constant
RPSC AE 2018
TSPSC AEE 2015

Ans. (c) :

Isolated system
—An isolated system cannot
transfer either energy or mass to or from the
surrounding ∆m = 0, ∆E = 0 for isolated system.
Example
—Universe, thermal flask bottle etc.
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15

44. In the polytropic process equation PV
n
=
constant if n is infinitely large, the process is
termed as-
(a) Constant volume (b) Constant pressure
(c) Constant temperature (d) Adiabatic
RPSC AE 2018
TNPSC AE 2018
UKPSC AE 2007 Paper -II

Ans. (a) :
In the polytropic process
PV
n
= constant
If
n
= ∞

then
PV
∞
= constant.
Then this polytropic process is termed as costant
volume process.
If n = 1
PV = constant (Isothermal process)
If n = 0
PV
0
= constant
Constant pressure process
If n = 1.4
PV
1.4
= constant (adiabatic process)
Process

Value of 'n' index

Constant pressure

0

Isothermal

1

Adiabatic

1.4

Constant volume

∞

45. The instrument which measures the
temperature of the source without direct
contact is
(a) Bi-metallic cut-out
(b) Vapour pressure thermometer
(c) Pyrometer
(d) Thin film thermometer
TNPSC AE 2014

Ans. (c) :

Pyrometer–
Pyrometer is a device use for
measuring relatively high temperature, such as are
encountered in furnace. Most pyrometer works by
measuring radiation from the body whose temperature
is to be measured. Radiation devices have the advantage
of not touch the material being measured.
46. Work done for an isothermal process is
(a)
1 1 2 2
1
PV PV
n
−
−
(b)
2 1
( )
P V V
−

(c)
1 1 1 2
ln( / )
PV P P
(d)
1 1 1 2
ln( / )
PV V V

TNPSC AE 2013

Ans. (c) :
1
isot 1 1
2
P
W P V ln
P
 
=
 
 

1
2 2
2
P
P V ln
P
 
=
 
 

2
1 1
1
V
P V ln
V
 
=
 
 

Work done for polytropic process

1 1 2 2
poly
P V P V
W
n 1
−
=
−

Work done for constant volume process for closed
system.
W
V=C
= zero
work done for constant pressure process for closed
system
W
P=C
= P (V
2
– V
1
)
47. Which of the following is correct statement for
'Energy (E)'?
(a) 'Energy (E)' is not the property of system
(b) 'Energy (E)' is path function
(c) Change in energy between two states of a
system is different for different path followed
(d)
0
dE
=
∫


UPRVUNL AE 2016

Ans. (d) : • Energy E is the property of system and it is
the point function.
• Change in energy between two states of a
system is same for different path followed.
• For point function

0
dE
=
∫


48. Internal energy of system containing perfect
gas depends on
(a) Pressure only
(b) Temperature only
(c) Pressure and temperature
(d) Pressure temperature and specific heat
HPPSC AE 2018

Ans. (b) : Internal energy of a system containing perfect
gas (ideal gas) depends on temperature only.
U = f(T) [For ideal gas only]
49. Which of the following equations is incorrect?
(where V,P,T and Q are volume, pressure,
temperature and heat transfer respectively)
(a)
0
=
∫

dV
(b)
0
=
∫

dP

(c)
0
=
∫

dT
(d)
0
=
∫

dQ

RPSC LECTURER 16.01.2016

Ans. (d) :

0
Q
≠
∫


0
V
=
∫


0
P
=
∫


0
T
=
∫


Because V,P,T are properties of the system whereas Q
and work (W) not a properties of the system.
Cyclic integration of the system properties will be equal
to zero.
50. Which of the following statements is correct for
"Energy"
(a) It is a point function
(b) It is a path function
(c) It is not a conserved quantity
(d) It can be measured by thermometer
RPSC LECTURER 16.01.2016

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16

Ans. (a) :

Note:-Option (a) is given by RPSC.
51. The heat capacity of the substance is given by
expression:
(a) Heat capacity = (Mass) + (specific heat)
(b) Heat capacity = (Mass) – (specific heat)
(c) Heat capacity = (Mass)
×
(specific heat)
(d) Heat capacity = (Mass) / (specific heat)
(e) Heat capacity = (Mass)
2
+ (specific heat)
2

(CGPCS Polytechnic Lecturer 2017)

Ans. (c) : Heat Capacity–Heat capacity or thermal
capacity is a physical property of matter, defined as the
amount of heat to be supplied to a given mass of a
material to produce a unit change in its temperature.
Heat capacity is an extensive property. It is denoted by
C
C = m
×
c (J/k)
Where
m
→
mass of the system
c
→
Specific heat
52. Which of the following is NOT an intensive
thermodynamic property?
(a) Temperature (b) Pressure
(c) Energy (d) Specific volume
(e) Specific energy
(CGPCS Polytechnic Lecturer 2017)

Ans. (c) : Intensive Properties–An intensive property
is a property of matter that depends only on the type of
matter is a sample and not depends on the amount of
mass.
Example–Pressure, temperature, density, viscosity
specific enthalpy, specific entropy, specific volume etc.
Extensive Properties– An extensive property is a
property that depends on the amount of matter in a
sample.
Example– Volume, enthalpy, entropy, mass energy etc.
53. A piston cylinder arrangement has air at 600
kPa, 290 K and volume of 0.01 m
3
. During a
constant pressure process, if it gives 54 kJ of
work, the final volume must be
(a) 0.10 m
3
(b) 0.05 m
3

(c) 0.01 m
3
(d) 0.15 m
3

JPSC AE PRE 2019

Ans. (a) : Given,
P
1
= 600 kPa
T
1
= 290 K
W = 54 kJ
V
1
= 0.01 m
3

V
2
= ?
W = PdV
= P(V
2
– V
1
)
54 × 10
3
= 600 × 10
3
(V
2
– 0.01)
V
2
= 0.1 m
3

54. A polytropic process with n = –1, initiates with
P = V = 0 and ends with P = 600 kPa and V =
0.01 m
3
. The work done is
(a) 2 kJ (b) 3 kJ
(c) 4 kJ (d) 6 kJ
JPSC AE PRE 2019

Ans. (b) : Given,
n = –1
P
1
= 0
V
1
= 0
P
2
= 600 kPa
V
2
= 0.01 m
3

1 1 2 2
1
PV PV
W
n
−
=
−

0 600 0.01
1 1
− ×
=
− −

= 3 kJ
55. Work done in a free expansion process is
(a) Positive (b) Negative
(c) Zero (d) Maximum
JPSC AE PRE 2019
TNPSC AE 2018
UKPSC AE 2007 Paper -II

Ans. (c) :
From first law of thermodynamics
δ
Q = dU +
δ
W,
since, for free expansion (T
1
= T
2
)
δ
Q = 0 and dU = 0
So,
δ
W = 0
56. For an ideal gas, enthalpy is represented by
(a) H = U – RT (b) H = U + RT
(c) H = RT – U (d) H = –(U + RT)
JPSC AE PRE 2019

Ans. (b) : H = U + PV
PV = mRT
for unity mass, m = 1
H = U + mRT
H = U + RT
57. Which one of the following represents the
energy in storage?
(a) Work (b) Heat
(c) Energy (d) Internal energy
JPSC AE PRE 2019

Ans. (d) : Internal energy—Internal energy of a
system is the sum of potential energy and kinetic energy
of that system.
58. Heat transfer takes place as per
(a) Zeroth Law of Thermodynamics
(b) First Law of Thermodynamics
(c) Second Law of Thermodynamics
(d) Kirchhoff's Law
SJVN ET 2013
Nagaland PSC CTSE 2017 Paper-2

Ans. (c) : Heat transfer takes place as per second law of
thermodynamics.
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17

59. 2 kg of substance receives 500 kJ and
undergoes a temperature change from 100
o
C to
200
o
C. Then average specific heat of substance
during the process will be:
(a) 5 kJ/kg
o
K (b) 2.5 kJ/kg
o
K
(c) 10 kJ/kg
o
K (d) 25 kJ/kg
o
K
SJVN ET 2013

Ans. (b) : m = 2kg
Q = 500 kJ
T
1
= 100
o
C
T
2
= 200
o
C
Q = mc
∆
T
500 = 2 × c (200 – 100)

o
c 2.5 kJ / kg K
=

60. The work done during an isothermal process
is:
(a)
2
1 1 e
1
v
P V log
v
 
 
 

(b)
1
1 2 e
2
v
P V log
v
 
 
 

(c)
2
2 2 e
1
P
P V log
P
 
 
 

(d)
2 2 1 1
P V P V
n 1
−
−

TRB Polytechnic Lecturer 2017

Ans. (a) : W
isothermal
=
2
1 1 e
1
V
P V log
V
 
 
 

61. Thermocouples are generally used for
measuring temperature:
(a) 500
o
C (b) 1000
o
C
(c) 1500
o
C (d) 2000
o
C
SJVN ET 2013
TNPSC AE 2018

Ans. (c) :
Thermocouples are generally used for
measuring temperature up to 1500
o
C
62. Which of the following is not a point function
of the system.
(a) Temperature (b) Pressure
(c) Specific volume (d) Heat
TRB Polytechnic Lecturer 2017

Ans. (d) :
Heat is path function whereas temperature,
pressure and specific volume are point function.
63. The enthalpy of an ideal gas is a function of
(a) Pressure only (b) Volume only
(c) Temperature only (d) None of these
Nagaland PSC CTSE 2017 Paper-2

Ans. (c) :
Enthalpy is a function of temperature only.
64. 1ºC is equal to
(a) 273.15 K (b) 274.15 K
(c) 283.15 K (d) 263.15 K
Nagaland PSC CTSE 2017 Paper-2

Ans. (b) :
1ºC is equal to Kelvin, 274.15 K.
65. Identify open system and closed system from
the following
(a) Blood circulation and respiration in human
body
(b) Fuel system and radiator in cars
(c) Air compressor and boiler
(d) Shell and tube heat exchanger and Blower
Nagaland PSC CTSE 2017 Paper-2

Ans. (c) :
Air compressor is an open system and boiler
comes under closed system (boiler having closed
chamber).
66. Compression efficiency is compared against
(a) Ideal compression
(b) Adiabatic compression
(c) Isentropic compression
(d) Isothermal compression
Nagaland PSC CTSE 2017 Paper-2

Ans. (d) :
Compression efficiency is compared against
isothermal compression.
67. Which thermometer is independent of the
substance or material used in its construction?
(a) mercury thermometer
(b) alcohol thermometer
(c) ideal gas thermometer
(d) resistance thermometer
Nagaland PSC CTSE 2017 Paper-2

Ans. (c) :
Ideal gas thermometer in independent of the
substance or material used in its construction.
68. The pressure of air in an automobile tyre at
temperature of 27
o
C is 1.75 bar (gauge). Due to
running the temperature of air in the rises to
87
o
C. What will be the gauge pressure during
this running? [P
atm
= 1.01 bar, volume of tyre is
assumed constant]
(a) 2.302 bar (b) 2.914 bar
(c) 1.677 bar (d) 3.180 bar
SJVN ET 2019

Ans. (a) :
P
abs
= P
atm
+ P
gauge

= 1.01 + 1.75
= 2.76 bar
1 1 2 2
1 2
P V P V
T T
∴ =

( )
abs 2
2
abs
1
P T
2.76 360
P
T 300
×
×
= =

(
)
2
abs
P 3.312 1.01
= −

2.302 bar
=

69. A non-flow reversible process takes place
according to
3
15
V m ,
P
=
where P is in bar.
What will be the work done if pressure changes
from 1 bar to 10 bar? [Given. ln (10) = 2.3025]
(a) 3.453 MN-m, expansion
(b) 3.453 N-m, compression
(c) 3.453 MN-m compression
(d) 3.453 N-m, expansion
SJVN ET 2019

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ME8391 Engineering Thermodynamics MCQ

ME8391 Engineering Thermodynamics MCQ

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