1
Basic Thermodynamics
Syllabus:

Zeroth, first and second law of thermodynamics, thermodynamic system and
processes, car not cycle, inerversibility and availability, behavior of ideal gas and real
gases, properties of pure substance, calculation of wo
rk and heat in ideal processes,
analysis of thermodynamic, cycles related to energy conversion.
Gate
→ 2003
1.
A 2 kw, 40 liters water heater is switched on for 20 minutes. The heat capacity
for water is 4.2 kj/kg k. Assuming all the electrical energy has gone in to
hearting the water, increase of the water temperature in degree
centigrade is
(a) 2.7
(b) 4.0
(Mark = 1)
(c) 14.3
(d) 25.25
Solution
Heat supplied by heater in 20 minutes
As all the electric energy has gone into heating the water
So Heat taken by water = 2400 KJ
But heat ta
ken by water
Mass of water
Specific heat in kJ/kg K = 4.2.
= 14.28 degree centigrade
2.
Considering the relationship Tds = Pdv between the entropy (s), internal energy
(u), pressure (P), temperature (T) and Volume (v) which of the following
statements is correct ?
(a)
It is applicable only for a reversible process.
(b)
For an irreversible process Tds
> du + Pdv.
(c)
It is valid only for an ideal gas.
2
(d)
It is equivalent to 1
st
Law, for a reversible process.
( Mark
–
2)
Solution. (c) Tds = du + Pdv
Is valid only for an ideal gas both for reversible and irreversible process undergone by a
closed
system, sin
ce it is a relation among properties which are independent of the path.
Common data question
Nitrogen g
a
s
( molecular weight 28 ) is enclosed in a cylinder by a piston, at the initial
condition
of 2 bar, 298 k and 1 m
3
. In a
particular process, the gas
slowly expands under
isothermal condition, until the volume becomes 2 m
3
. Heat exchange occurs with the
atmosphere at 298 k during this process.
3.
The work interaction for the nitrogen gas is
(a) 200 KJ
(b) 138

6 KJ
(c) 2 KJ
(d)

200KJ
(Mark
–
2)
4.
The entropy change for the universe during the process in KJ/K is
(a) 0.4652
(b) 0.0067
(c) 0
(d)

0.6711
(Mark

2 )
Solution
Work done
In isothermal process T = Constant
For an ideal gas Pv = RT
Where R is
constant
So Pv = Constant
Work done
3
Entropy change of the system
{ for reversible proc
ess}
For isothermal process = du = 0
So from first law of
thermodynamics
As T is constant
Entropy change of surrounding
=

0.4641
So entropy change of universe is zero.
Gate 2004
5.
A gas contained in a cylinder is compressed, the work required for compression
being 5000 KJ. During the process, the heat interaction of 2000 KJ causes the
surrounding to be heated
. The change in internal energy of the gas during the
process is
(a)
–
7000 KJ
(b)
–
3000 KJ
(c) + 3000 KJ
(d) + 7000 KJ
(Mark
–
1 )
Solution.
(c) Us
ing first law of thermodynamics.
4
{
work done is negative when it is done on the
S
ystem
}
{
Heat transferred is negative when it is taken from
the system or it is rejected}
6.
A Steel billet of 2000 kg mass is to be cooled
from 1250 k to 450 k . The heat
released during process is to be used as a source of energy. The ambient
temperature is 303 k and specific heat of steel is 0.5 kJ/kg K. The available
energy of this billet is.
(a) 490.44 MJ
(b) 30.95MJ
(c) 10.35 MJ
(
d) 0.1 MJ
Area under curve AB
Entropy change from B to A
Area under C

D
=
So Area ABCD = 800
–
309.56
5
= 490MJ.
Gate 2005
7.
The following four figures have been drawn to represent a fictitious
thermodynamic cycle, on the P
–
v and T
–
S planes.
According to first Law of thermodymics, equal areas are e
nclosed by
(a) Figure 1 and 2
(b) Figure 1 and 3
(c) Figure 1 and 4
(d) Figure 2 and 3
8.
A Reversible thermodynamic cycle containing only three processes and producing
work is to be constructed. The constraints are
(i)
There must be one isothermal proce
ss (ii) There must be one isentropic
process (iii) The maximum and minimum cycle pressures and clearance
volume are fined (iv) Polytrophic processes are not allowed. The number of
possible cycle are
(a) 1
(b) 2
(c) 3
(d) 4
(Mark
–
2)
Poss
ible processes
Constant pressure
Isothermal Compression
Isothermal Compression
P
6
V
Isothermal Expansion
Isothermal Compression
P
Isothermal Compression
Constant Pressure Process
2
V
Isothermal Expansion
P
3
1
V
Isothermal
Compression
Isothermal Compression
P
Constant Pressure
7
9.
Nitrogen at an initial state of 10 bar, 1 m
3
and 300 K is expanded isothermally to a
final volume of 2 m
3
, the P
–
v
–
T relation is
Where a > 0. The fi
nal pressure
(a) Will be slightly less than 5 bar.
(b) Will be slightly more less than 5 bar.
(c) Will be exactly 5 bar.
(d) Can not be ascertained in the absence of the value of a.
(Mark
–
2)
Solution. (b)
As proc
ess is isothermal
–
RT = constant
So
is slightly more than 5 bar as a is positive
.
The following table of properties was printed out for saturated liquid
and saturated
v
apor of ammonia. The title for only the first two columns are available. All that we
know that the other columns (columns 3 to 8 ) contain data on specific properties,
namely, internal energy (KJ/kg ),enthalpy (KJ/kg ) and entropy (KJ/kg)
t
0
P(KPa)

20
190.2
88.76
0.3657
89.05
5.6155
1299.5
1418
0
429.6
179.69
0.7114
180.36
5.3309
1318
1442.2
8
20
587.5
272.89
1.0408
274.3
5.0860
1332.2
1460.2
40
1554.9
368.74
1.3574
371.43
4.8662
1341.0
1470.2
20. The specific enthalpy data are in colu
mns
(a) 3 and 7
(b) 3 and 8
(c) 5 and 7
(d) 5 and 8
10.
when saturated liquid at 40
0
C is throttled to
–
20
0
C the quality atenit will be
(a) 0.189
(b) 0.212
(c) 0.231
(d) 0.788
10. (d) We know that h = u + PV
h = enthalpy u = internal energy.
S
o all the value in column 5 are greater than
corresponding value in column 3 so
column 5 represent
specific enthalpy of saturate liquid and similarly all the values of
column 8 are greater than corresponding values of column 7.50 the column 8
represent
specific enthalpy of saturated steam.
11.
Enthalpy of saturated liquid at 40
0
C = 371.43 KJ/kg.
Enthalpy of vapor at
–
20
0
C = ( let x be dry ness fraction)
= 89.05+x (1418
–
89

05).
During throttling enthalpy remains constant
371.43 = 89.05 + x (1418
–
89.
05)
282.38 = x (1418
–
89.05)
x = 0.212.
Gate → 2006
12.
Given below is an extract from steam tables.
Temperature
In
o
C
P
Sat
(bar)
Specific volume
Enthalpy (KJ/kg)
Saturated
Liquid
Saturated
Vapour
Saturated
Liquid
Saturated
Vapour
45
0.09593
0.001010
15.26
188.45
2394.8
342.24
150
0.001
658
0.010337
1610.5
2610.5
9
Specific enthalpy of water in KJ/kg at 150 bar and 45
0
C is
(a) 203.6
(b) 200.53
(c) 196.38
(d) 188.45
Solution. (b) Specific enthalpy at 150 and 45
0
C
13.
Match items from groups I, II, III
, IV a
nd V.
Group I
Group II
Group III
Group IV
Group V
When added to
the system is
Differential
Function
Phenomenon
E
–
eeat
G P潳itive
I
Enact
K⁐ath
M⁔ransie湴
F⁷潲k
e P潳itive
Jn enact
L P潩nt
N⁂潵n摡ry
(搩 eeat when a摤d搠d漠ohe system
†
is 灯pitive an搠is in enact 摩ffere湴ial 扥cause it is
灡th 摥灥n摥nt an搠is 扯bn摡ry 灨en潭en潮 w潲欠when a摤d搠d漠† he system is
negative n搠is in enact 摩fferential 扥cause it is 灡th 摥灥n摥nt an搠is transient
灨en潭enon.
Statement f潲 lin步搠 ans
wer 煵esti潮 ㄴn搠ㄵ⸠
A f潯o扡ll was inflate搠 t漠aauge 灲essure 潦 ㄠ扡r⁷hen the am扩ent tem灥rature was
ㄵ
0
C. When the game started next day, the air temperature at the stadium was 5
0
C.
Assume that the volume of foot ball remains constant 2500 cm
3
.
14.
The amount of heat lost by the air in the football and the gauge pressure of air in
the football at the stadium respectively equal.
(a) 30.6J, 1.94 bar
(b) 21.8J, 0.93 bar
(c) 61.1J, 1.94 bar
d) 43.7J, 0.93 bar
15.
Gauge pressure of air to which the
ball must have been originally inflated so that
it would be equal 1 bar gauge at the stadium is
10
(a) 2.23 bar
(b) 1.94 bar
(c) 1.07 bar
(d) 1 bar
Solution. 14 (d) Gau
ge pressure with in foot ball
= 1.01 + 1
= 2.01
bat
Temperature when total pressure is 2.01 bar = 15
0
C or 288
0
k.
Volume of air with in foot ball = 2500 cm
3
.
Since the volume remains same
Next day temperature
or 278
0
k.
1.94 bar
= 0.93 bar.
Heat lost =
Heat lost
15 (c)
{ final pressure}
{ final pressure}
11
Gate 2007
16.
(d) Which of the following relationships is valid only for reversible processes
under gone by a
closed system of simple compressible substance ( neglect
changes in kinetic and potential energy ) ?
(a)
(b)
(c)
(d)
(d)
is true for reversible process and closed system.
37. Water has a critical specific volume of 0.003155 m
3
/kg. A closed and rigid steel
tank of volume .025m
3
contains a mixture of water and steam at 0.1 MPa. The mass
of the mixture is 10 kg. The
tank is now slowly heated. The liquid level inside the
tank.
(a)
Will rise.
(b)
Will fall.
(c)
Will remain constant.
(d)
May rise or fall depending on the amount of heat transferred.
Solution. Critical volume = 0.003155m
3
/ kg.
Specific volume of mixture
Which is less than critical volume
As it is a constant volume process
P
V
Critical Volume
v
Saturated Volume
v
Saturated Liquid
v
12
40. Which combination of the following statements is correct ?
P : A gas cools upon expansion only.
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