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 work 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
CP 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
(c) 14.3
(b) 4.0
(d) 25.25
(Mark = 1)
Solution Heat supplied by heater in 20 minutes
 2 1000  20  60
 2400000 J
 2400 KJ
As all the electric energy has gone into heating the water
So Heat taken by water = 2400 KJ
T1
But heat taken by water   mcP dT
T2
 mcP T1  T2 
Mass of water
 40 103  
  1000kg / m 
1Litre  10 m 
3
3
 1000  40 10
 40kg
cP  Specific heat in kJ/kg K = 4.2.
3
3
40  4.2 T1  T2   240
2400
40  4.2
= 14.28 degree centigrade
T1  T2  
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.
1
(d) It is equivalent to 1st 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, since it is a relation among properties which are independent of the path.
Common data question
Nitrogen gas ( molecular weight 28 ) is enclosed in a cylinder by a piston, at the initial
condition of 2 bar, 298 k and 1 m3. In a particular process, the gas slowly expands under
isothermal condition, until the volume becomes 2 m3. 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
v2
Work done   Pdv
v1
In isothermal process T = Constant
For an ideal gas Pv = RT
Where R is constant
So Pv = Constant
Pv
1 1  P2 v2
P
P2v2
v
v2
Work done

v1
P2 v2
dv
v
v2
dv
v
v1
 P2v2 
 Pv
1 1 Loge
v2
v1
2
v2  2m3v1  1m3
 v2 
 Pv

1 1 Loge 
 v1 
 Pv
1 1 Loge 2
 2 105 1log e2
 2 105  0.6931
 138.629kJ
Entropy change of the system

d
T
{ for reversible process}
For isothermal process = du = 0
So from first law of thermodynamics
d  dw  du
d  dw
d  Pdv
S12  
d
Pdv

T
T
As T is constant
workdone
 S12 system 
T
138.629

298
 0.4651
Entropy change of surrounding 
HeatExchangewithSurrounding
SurroundingTemperature
WorkdoneonSurrounding
SurroundingTemperature
= - 0.4641

d  dw
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
(c) + 3000 KJ
(b) – 3000 KJ
(d) + 7000 KJ
(Mark – 1 )
Solution. (c) Using first law of thermodynamics.
3
  w  u
w  5000KJ
{work done is negative when it is done on the
System }
{Heat transferred is negative when it is taken from
the system or it is rejected}
  2000KJ
2000  5000  u
3000 KJ  u
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
(c) 10.35 MJ
A 450 k
(b) 30.95MJ
(d) 0.1 MJ
1250 k
B
Available energy
T
D
C
To
unavailable energy
S
Area under curve AB mcP 1250  450
 2000  0.5 1250  450 
 800, 000 KJ
 800MJ
1250
Entropy change from B to A

450
mc p dT
T
1250
450
1250
 2000  0.5 In
450
S B  A  1021.65 KJ / Kg .K
S B  A  mcP In
Area under C-D = S B  A  To
SB A  1021.65  303
309.56MJ
So Area ABCD = 800 – 309.56
4
= 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.
P
T
V
S
T
P
S
V
According to first Law of thermodymics, equal areas are enclosed by
(a) Figure 1 and 2
(c) Figure 1 and 4
(b) Figure 1 and 3
(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 process (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)
Possible processes
Constant pressure
P
Isothermal Compression
5
Isothermal Compression
Isothermal Expansion
P
Isothermal Compression
V
1
Isothermal Expansion
P
Constant Pressure Process
3
2
Isothermal Compression
V
Isothermal Compression
P
Constant Pressure
V
Isothermal Compression
6
9. Nitrogen at an initial state of 10 bar, 1 m3 and 300 K is expanded isothermally to a
final volume of 2 m3 , the P – v – T relation is
P

 P  2  v  RT Where a > 0. The final pressure
v 

(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)
a 

Solution. (b)  P  2  V  RT
V 

As process is isothermal – RT = constant


a 
a 
 P1  2  V1   P2  2  V2
V1 
V2 


P1  10bar
V2  2m3
V1  1m3
T1  300 K
2
a 
a

10  1   P2  
1 
4

a
10  a  2 P2 
2
a
2 P2  10 
2
P2  5  a
So P2 is slightly more than 5 bar as a is positive.
The following table of properties was printed out for saturated liquid and saturated
vapor 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)
t0
P(KPa)
-20
190.2
88.76
0.3657
89.05
5.6155
1299.5
0
429.6
179.69
0.7114
180.36
5.3309
1318
1418
1442.2
7
20
587.5
272.89
1.0408
274.3
5.0860
1332.2
40
1554.9
368.74
1.3574
371.43
4.8662
1341.0
1460.2
1470.2
20. The specific enthalpy data are in columns
(a) 3 and 7
(b) 3 and 8
(c) 5 and 7
(d) 5 and 8
10. when saturated liquid at 400C is throttled to – 200C the quality atenit will be
(a) 0.189
(c) 0.231
(b) 0.212
(d) 0.788
10. (d) We know that h = u + PV
h = enthalpy u = internal energy.
So 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 400C = 371.43 KJ/kg.
Enthalpy of vapor at – 200C = ( 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.
Specific volume
Temperature
In oC
45
342.24
PSat (bar)
Saturated
Liquid
0.09593
150
Enthalpy (KJ/kg)
Saturated
Vapour
Saturated
Liquid
Saturated
Vapour
0.001010
15.26
188.45
2394.8
0.001658
0.010337
1610.5
2610.5
8
Specific enthalpy of water in KJ/kg at 150 bar and 450C is
(a) 203.6
(b) 200.53
(c) 196.38
(d) 188.45
Solution. (b) Specific enthalpy at 150 and 450C
 188.45 
150  0.09593 105  0.001010
103
 203.59 KJ / kg.
13. Match items from groups I, II, III, IV and V.
Group II
Group I
E – Heat
F work
When added to
the system is
G Positive
H Positive
Group III
Group IV
Group V
Differential
Function
Phenomenon
I Enact
K Path
M Transient
J in enact
L Point
N Boundary
(d) Heat when added to the system is positive and is in enact differential because it is
path dependent and is boundary phenomenon work when added to the system is
negative and is in enact differential because it is path dependent and is transient
phenomenon.
Statement for linked answer question 14 and 15.
A football was inflated to a gauge pressure of 1 bar when the ambient temperature was
150C. When the game started next day, the air temperature at the stadium was 50C.
Assume that the volume of foot ball remains constant 2500 cm3.
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
9
(a) 2.23 bar
(c) 1.07 bar
(b) 1.94 bar
(d) 1 bar
Solution. 14 (d) Gauge pressure with in foot ball Pt  Patm  Pguage
= 1.01 + 1
= 2.01 bat
Temperature when total pressure is 2.01 bar = 150C or 2880k.
Volume of air with in foot ball = 2500 cm3.
Since the volume remains same
PV
PV
1 1
 2 2
T1
T2
V1  V2
Next day temperature T2  5o C or 2780k.
P1 P2

T1 T2
2.01 P2

288 278
P2  1.94 bar
P2  Ptotal  Patm  P2 gauge  1.94
P2 gauge  1.94  1.01
= 0.93 bar.
Heat lost =  mCV dT
PV
Rm  1 1
T1
2.01
m
2500
100 
3
105
288  287
 0.000017447 105 / 287
1.7447kg

287
 0.00607kg
mCV dT  0.00607  0.72  15  5
Heat lost  0.04376 KJ
 43.76 J
15 (c) Pgauge  1bar
{ final pressure}
Pgauge  1  1.01  2.01
{ final pressure}
10
P2 P1

T2 T1
2.01 P1

278 288
P1  2.08bar
Pgauge  1.07bar
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)   du  dw
(b) Tds  du  Pdv
(d)   du  Pdv
(c) Tds  du  dw
(d)   du  Pdv is true for reversible process and closed system.
37. Water has a critical specific volume of 0.003155 m3/kg. A closed and rigid steel
tank of volume .025m3 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)
(b)
(c)
(d)
Will rise.
Will fall.
Will remain constant.
May rise or fall depending on the amount of heat transferred.
Solution. Critical volume = 0.003155m3 / kg.
0.025
 0.0025m3 / kg.
Specific volume of mixture 
10
Which is less than critical volume
As it is a constant volume process
Critical Volume
v
Saturated Liquid
Saturated Volume
v
v
P
V
11
40. Which combination of the following statements is correct ?
P : A gas cools upon expansion only.
12