UNIT – III
PART – A
1. State Boyle’s law.
Boyle’s law states. “The volume of a given mass of gas varies inversely as its absolute
pressure, when the temperature remains constant.
v
1
p
2. State Charle’s law.
Charle’s law states, “The volume of a given mass of a gas varies directly as its
absolute temperature, when the pressure remains constant.
VT
3. State Joule’s Law.
Joule’s law states, “The internal energy a given quantity of gas depends only on the
temperature”.
4. State Regnault’s law
Regnault’s law states that CP and Cv of a gas always remains constant.
5. State Avogadro’s law.
Avogadro’s law states. “Equal volumes of different perfect gases at the same
temperature and pressure, contain equal number of molecules”.
6. State Daltons law of partial pressure.
Dalton’s law of partial pressure states. “The total pressure of a mixture of gases is
equal to the sum of the partial pressures exerted by individual gases if each one of them
occupied separately in the total volume of the mixture at mixture temperature”.
P = P1 + P2 + P3 + …… Pk
7. Distinguish between ideal and real gas.
An ideal gas is one which strictly follows the gas laws under all conditions of
temperature and pressure.
In actual practice, there is no real gas which strictly follows the gas laws over the
entire range of temperature and pressure. However hydrogen, oxygen, nitrogen and air
behave as an ideal gas under certain temperature and pressure limits.
8. Define Joule – Thomson Co – efficient.
The Joule – Thomson co – efficient is defined as the change in temperature with
change in pressure keeping the enthalpy remains constant. It is denoted by the
 T 
 
 P h
9. Define co-efficient of volume expansion and isothermal compressibility.
Co-efficient of Volume expansion:
The co – efficient of volume expansions is defined as the change in volume with
change in temperature per unit volume keeping the pressure constant. It is denoted by 

1  v 
v  T p
Isothermal compressibility:
It is defined as the change in volume with change in pressure per unit volume
keeping the temperature constant. It is denoted by K
K
1  v 
v  P 
10. What is compressibility factor?
We know that, the perfect gas equation is pv = RT. But for real gas, a correction
factor has to be introduced in the perfect gas equation to take into account the deviation of
real gas from the perfect gas equation. The factor is known as compressibility factor (z) and
is defined by
Z
Pv
RT
11. State the assumptions made in kinetic theory gases.
1. There is no intermolecular force between particulars.
2. The volume of the molecules is negligible comparison with the gas
PART – B
1. A Mixture of 2 kg Oxygen (M=32kg/kgmol) and 2kfg argon (m=40 Kg/kmd) is present in an
insulated piston in cylinder arrangement at 100kpa, 300K. The piston now compresses the
mixture to half of its initial volume. Find the final pressure, temperature and piston work,
assume c, for oxygen and argon for oxygen and argon as 0.6618kj/kgk and 0.3122KJ/kgK
respectively
System : Closed
Process
: Adiabatic compression
Working fluid : Mixture of oxygen and argon.
1. Mo2 = 32kg/kg mol
2. Mar = 40kg/kg mol
3. T1 = 300t\k
Known 4. P1 = 100kpa
5. V2 = ½ V1
6. Cv02= 0.6618kj/kgk
7. Cv02= 0.3122kJ/kgK
Diagram:
2KgO2
2kg Ar
at 100 kpa, 300K
To find: 1. p2; 2. T2 ; 3.1W2
Analysis : a. p1 = V1 
MRT1
p1
Where m=mo2+mAr
=2+2
=4kg
 mo2 
 mAR 
R
 Ro2  
 RAR
mo

m
mo

m
 2
AR 
 2
AR 
=
2 8.314
2 8.314
x

x
2+2 32
2  2 40
=0.234
kj
kgk
V 1
 2  2 x0.23x300
100
3
=2.806m
(mo2Cvo2 )  (mr CvAR )
Cv 
(mo2  mAR )
Diagram:
2kg O2
+
2kg Ar
at 100kPa, 300K
Final State
V1 
V1
2
Initial State
2  0.6618  2  0.3122
22
kJ
 0.487
kgK
C p  Cv  R

 0.487  0.234
 0.721
C
 p
Cv
0.721

0.487
 1.480
It is given that
V1
2
=1.4033
V2 
Since the process is adiabatic, assuming the process to be reversible
 1
T2  V1 
 
T1  V2 
 1
T2  V1 
 
T1  V2 
=300(2)0.48
=418.K
mRT2 
V2
4  0.234  418.4

1.403
p2 
 279.15kpa
1 W2 

p2V2  p1V1
  1
279.15 1.403  100  2.806
1.48  1
 231.3kJ
2. Consider an ideal gas at 300k and 0.86m3/kg. As a result of some disturbance that state
of the gas changes to 302K and 0.87m3.kg. Estimate the change in the pressure of the gas as
a result of these disturbances.
Let p= f(T.v)
 p 
 p 
dp =   T    dv
 T 
 v 
For an ideal gas
pv=RT
RT
v
 p  RdT
   2
v
 T v
Rtdv
 p 
and    2
v
 v T
RdT Rtdv
 dp 
 2
v
v
p
From the given data
dT=2K
dv=0.01m3/kg
Substituting these values we get
 0.287  2 0.287  301 0.01
dp  


0.865
 0.865
=-0.461kpa
3. Derive the equation.
  2v 
 CP 


T
 2
 p 

T
 T  p
and prove that Cp of an ideal gas is a function of T only.
From I law of thermodynamics.
Dq=dh-vdp
For an ideal gas undergoing constant pressure process
Dq=Cp dt
Or Tds = Cp dt
 s 
Cv  T  
 T v
Differentiating w.r.t. v at constant T we get
  s 
 Cv 
 v   T v  v 
 v

T
or
  s 
 Cv 
 v   T T  v 
 T

T
  p 
=  
T  T v
From Maxwell relation
 2 p 
 Cv 

T
 2
 v 

T
 T v
For an ideal gas
Pv=RT
RT
v
 p  R
  
 T v v
  p    R 

T  T  T  v 
0
p
 C 
or  v   0
 v T
Cv is a function of temperature alone.
4. Show that the relation p(v-b)=RT satisfies cyclic relation. where b and R are constant.
cyclic relation in terms of pa v and T can be expressed as
 p   v   T 
 v   T   p   1
 T   p  v
where,
  RT 
RT
 p 
 v   v  v  b    (v  b)2
 T

T
 T 
1
 p   p
 v  
 T 
 v
1

p  RT 
T  v  b v
1
=
R/v-b
v-b
=
R
 

  RT
RT
 u 
 T   T  p  b   Since v p  b 
 p

p 

=
R
P
RT  R   v  b 
 p   v   T  
         

2 
 v T  T  p  p v  (v  b)  p   R 
R
=
p(v-b)
=-1
5. Determine hfg for water at 1000C from the following data.
T(0C)
95
100
105
p(kpa)
84.55
101.35
120.82
Consider the clausius – clpeyron eqn
 p1  h fg  1 1 
 
  
 p2  R  T1 T2 
 1
h fg
1 
 120.82 
In 





 84.55   8.314    368  378 
 18.016 


0.357  (1.558 104 )h fg
h fg  2291kJ / kg
T1=95+273
=368K
T2=105+273
=378K
p1=84.55kpa
p2=120.82kpa
7. Determine the Joule Thomson coefficient for a van der walls gas given by the
a

equation.  p  2  (v  b)  RT
v 

Prove that for large volume (or low pressure), the inversion temperature is equal to za/br.
joule-Thomson Coefficient
 T 

 p 
 
=

1   T 
T    v 
Cp   p  p 
Differentiating
a

 p  v2  (v  b)  RTw.r.t.toT @constant, P we get


 2a  T  
a   T 

p


(
v

b
)
 3   R

v2   p  p

 v  p  p 
a 2a 2ab 
 v  
 T   p  v2  v2  v3   R
 p 

a 2ab 
 v  
 T   p  v2  v3   R
 p 




 T  1 
RT

 p  C 
 h p  p  a  2ab 
v2
v3 


a

p  2  (v  b) 


1
v 

  
C p  p  a  2ab 

v2
v3 

2a 3ab 

bp   2 
1 
v
v
= 

C p  p  a  2ab 
v2
v3 

For large value of v at low-pressure.
2a 

bv  
 T 
1 
v

 p  = C 
p
 h
p 



=
=
When
1 
2a 
b  

Cp 
pv 
1 
2a 
b 

Cp 
RT 
 T 
 p   0
 h
2a
RTinversion
2a
or Tinversion 
bR
then b=
8. Determine the change of internal energy, enthalpy and entropy for an isothermal process
when the gas obeys Vander Wall’s equation.
System
To obtain
: A van der Waal’s gas under going a process
: An expression for each of the following
1)u2-u1
2)h2-h1
3)s2-s1
We know that from Eqn. 9.25
 u 
 p 
 v   T  T   p
 T
 v
For Van der Waals gas
a

 p  v2  (v  b)  RT


p
RT a

v  b v2
R
 p 
 T   v  b
 v
RT
 u 
Therefore   
p
 v T v  b
a

= p  2   p
v 

a
 p 
 T   v2
 v
a
duT  2 dVT
v
1 1
(u2  u1 )T  a   
 V1 V2 
1 1
(u2  u1 )  a   
 V1 V2 
We know that
U2=h2-p2v2
U1=h1-p1v1
There fore
H2 –h1 = (u2-u1)+(p2v2-p1v1)
1 1
( p2v2  p1v1 )  a   
 V1 V2 
Equation 9.34 gives
ds  Cv
dT
T
 p 
   dv
 T v
But from Equations 9.25
1
 p   u 
 T    v   p  T
 v   p

It has already been proved that for Vander Waals gas
a
 u 
 v   v2
 T
a 1
 p  
Therefore     p  2 
v T
 T v 
Substituting the above expression in the expression for ds we get
ds =Cv
dT 
a 1
 p     dv
T 
v2  T
dT RT 1

  dv
T v b T
dT
dv
=Cv
R
T
v b
=Cv
Let us Consider an is other malo process
V b 
(ss  s1 )T  RIn  2

 V1  b 
9. Prove that Cp-Cv=R for an ideal gas.
System : Ideal gas undergoing process
To prove
Proof
: Cp-Cv=R
: From Equation 9.37
 v   p 
C p  Cv  T    
 T  p  T v
For an ideal gas
pv=RT
R
 v 
 T   p
 p
R
 v 
 T   V
 p
Therefore
R R
C p  Cv  T  
p V
R
RT
PV
R
10. Calculate the specific volume of dry saturated steam at a pressure of 147 kPa at which
the values of temperature T and latent heat L are 110.790C and 2223.3KJ/kg respectively.
Further, saturation temperature of steam at pressure of 157 kPa is 112.74 0C . Neglect the
specific volume of water.
From Clausius – Clapeyron equation
j fg
dp

dT T V fg
(Or )Vg 
h fg dp

T dT

h fg T

T P

2223.3
112.74  110.79

 273  110.79 157 147
 1.133 / Kg
11. Prove that constant pressure lines in the wet region of a mollier diagram are straight and
not parallel and that the slope of a constant pressure line in the superheat region increases
with temperature.
Let h=f(s.p)
 h 
 h 
dh    ds    dp
 s  p
 p s
Also dh=Tds+vdp
Comparing the coefficients of dh, we get
 h 
 s   T
 p
This relation give the slop of the isobars in the Mollier diagram as shown below.
h
P=C
dh
ds
s
 In the wet region for a given pressure temperature remains constant and hence it is
a straight line.
 Higher the pressure, higher will be the saturation temperature and higher will be the
slope. there fore isobars slope upward more steely as the pressure increases.
 As temperature increases beyond the saturation line, that is, in the superheated
vapour region, these lines bend slightly upward in that region.
13. AVessel of volume 0.28 m3 contain 10 kg of air at 302K. Determine the pressure exerted
by the air using
(1) Perfect gas equation
(2) Vander Waals equation
(3) Generalised compressibility chart.
[Tale Critical temperature of air is 132.8 K; Critical pressure of air is 37.7 bar]
Given data:
Volume, v = 0.28 m3
Mass, m = 10 kg
Temperature T = 302.8K
Critical Temperature (Tc) = 132.8K
Critical Pressure (pc) = 37.7 bar
= 37.7  100 kN/m2
To find:
Pressure (p)
Solution:
1. Perfect gas equation:
pv = mRT
mRT
 p=
v
10  0.287  302
 p=
[
0.28
= 3095.5 N/m2
p = 3095.5 kPa
[
R for air is0.287 kJ/kgk]
1 N/m2  1 pascal]
2. Vander Waals equation:
q

....(A)
 p  v2  (v-b) = RT


27R2 (Tc )2 27  (0.287)2  (132.8)2
a=

64pc
64p
[Critical Pressure, pc = 37.7 bar = 37.7  100 kN/m2]
[
1 bar = 100 kN/m2 ]
27  (0.287)2 (132.8)2
a
64  (37.7)  (100)
a = 0.162.
We know that
b
RTc 0.287  132.8

8pc 8  37.7  100
b = 1.26  10-3
Substituting a, b values in Vander Waals Equation
0.162 

-3
(A)   p 
 v-1.26  10  0.287  302
2
v


Where
V - Specific volume
Volume v 0.28
v=
 
mass
m 10

v = 0.028 m3/kg
Substituting Specific volume


0.162 
  p+
0.028-1.26  10-3  0.287  302
2 
(0.028)


 p = 3034.7 kN/m2


3. Generalized compressibility chart:
TR =
T
302

 2.27K
Tc 132.8
TR  2.27 K
vR 
vPc 0.028  37.7  100

RTc
0.287  132.8
vR  2.76 m3 / kg
Reduced temperature is 2.27 K and reduced specific volume is 2.76 m 3/kg both are
intersects at one point. Mark this point on compressibility chart. From chart, corresponding
(z) value is 0.99. We know that,
pv
RT
p  0.028
 0.99 =
0.287  302
 p = 3064.5 kN/m2
Compressibility factor (z) =
Results :
1. Pressure p (By perfect gas equation) = 3095.5 kN/m2
2. Pressure p (By Vander Waals equation) = 3034.7 kN/m2
3. Pressure p (By compressibility chart) = 3064.5 kN/m2
15. Compute the specific volume of steam at 0.75 bar and 570 K using Vander Waals
equation. Take critical temperature of steam is 647.3 K and Critical pressure is 220.9 bar.
Given data:
Pressure, p = 0.75 bar = 0.75  100kN/m2 = 0.75  100 kPa
[ 1 bar = 100 kN/m2  100kPa]
Temperature, T = 570 K
Critical Temperature, Tc = 647.3 K
Critical Pressure, pc = 220.9  100 kN/m2
To find
Specific volume (v)
Solution :
We know that Vander Waals equation
a

 p  v 2  (v - b) = RT


27R2 (Tc )2
Where, a =
64pc
Universal Gas Constant
8.314

kJ/ kgK
Molecular weight of Steam
18
R  0.462 kJ/kgK
Where, R =
27  (0.462)2  (647.3)2
64  (220.9)  100
a = 1.70
RTc 0.462  647.3
We know, b =

8pc 8  220.9  100
a =
b = 1.69  10-3
Substituting a, b and pressure and temperature values in Vander Waals equation.
1.70 

  0.75  100  2   v  1.69  103  0.462  570
v 

1.70 

  75  2  (v - 1.69  10-3) = 263.34
v 


By trial and error method, we get
Specific volume v = 3.58m3/kg
Result :

Specific volume v = 3.58 m3/kg.
16. An ideal gas mixture consisting of 3 kg of air and 7 kg of nitrogen at a temperature of
25C occupies a volume of 1m3. Determine the specific enthalpy, the specific internal energy
and specific entropy of the mixture. Assume that air and nitrogen are ideal gases. Take R for
air is 0.287 kJ/kgK and for nitrogen is 0.297 kJ/kgK. The other properties of air and nitrogen
are given as under.
Name of the gas
Nitrogen
Air
Properties
H, kJ/kg
309.64
298.52
Given data :
For air
Mass, ma = 3 kg
Temperature, T = 25C + 273 = 298 K
Volume, v = 1m3
Gas constant, R = 0.287 kJ/kgK
Enthalpy, h = 298.52 kJ/kgK
Internal energy, U = 212.90 kJ/kg
Entropy, S = 2.35 kJ/kgK
For Nitrogen
Mass, mn = 7 kg
Temperature, T = 298 K
Volume, v = 1m3
R = 0.297 kJ/kgK
h = 309.64 kJ/kg
U = 221.11 kJ/kg
S = 6.46 kJ/kgK
To find :
1.
2.
3.
4.
5.
Specific volume (v) of the mixture
Pressure (p) of the mixture
Specific Enthalpy (h) of the mixture
Specific Internal energy (U) of the mixture
Specific Entropy (S) of the mixture
Solution:
U, kJ/kg
221.11
212.90
S, kJ/kg
6.46
2.35
Specific Volume of air, v a 
volume v
1


mass
ma 3
va  0.333m3 / kg
Specific Volume of nitrogen, vn 
volume 1 1


mass
mn 7
vn  0.142m3 / kg
Pressure of air,
maRaTa
 pv = mRT
v
3  0.287  298
pa 
1
pa  265.5 kN/m2
Pa 
Pressure of nitrogen,
mnRnTn 7  0.297  298

v
1
2
pn  619.5 kN/ m
pn 
For mixture
Total mass, m = ma  mn  3  7  10kg
m = 10 kg
Temperature, T = 298 K
Volume,
v = 1m3
v
1

m 10
= 0.1m3 / kg
1. Specific Volume of mixture, vm 
2. Pressure of mixture, p = pa  pn  265.5  619.5
= 876.07 kN/m2
maha  mnhn
m
3  298.52+ 7  309.64
=
10
= 306.34 kJ/kg
3. Specific Enthalpy of mixture, h =
maUa  mnUn
m
3  212.90+7  221.11
=
10
= 218.64 kJ/kg
4. Internal energy of mixture, U =
maSa  mnsn
m
3  2.35+7  6.46
=
10
= 5.227 kJ/kg.
5. Specific Entropy of mixture, S =
Results:
1.
2.
3.
4.
5.
Specific Volume of the mixture, vm = 0.1m3/kg
Pressure of the mixture, p = 876.07 kN/m2
Specific Enthalpy of the mixture, h = 306.34 J/kg
Specific Internal energy of the mixture, U = 218.64 kJ/kg
Specific Entropy of the mixture, S = 5.227 kJ/kg.
17. A perfect gas of 0.25 kg has a pressure of 298 Kpa, a temperature of 80C, and a volume
of 0.08m3. The gas undergoes an irreversible adiabatic process to a final pressure of 350 kPa
and find volume of 0.10 m3, work done on the gas is 25 kJ. Find cp, cv.
Given data :
m
p1
T1
v1
p2
v2
W
= 0.25 kg
= 298 Kpa
= 80C + 273 = 353 K
= 0.08m3
= 350 Kpa
= 0.10m3
= -25 kJ
[Work done on the gas in Negative valve]
To find:
Cp and Cv
Solution:
We know that,
Perfect gas equation
P1v1 = mRT1
 R=
p1v1 298  0.08

 0.270
mT1 0.25  353
Characteristic gas constant, R = 0.270 kJ/kgK
Similarly
P2v2 = mRT2
 T2 
p2v 2
350  0.10

mR 0.25  0.270
T2  518.5K
We know that
Heat transfer, Q = W + U
Q  W  mCv  T2  T1  U= m Cv  T2  T1 
 Q = -25 + 0.25  Cv (518.5  353)
For adiabatic process, Q = 0
 0 = -25 + 0.25  Cv (518.5  353)
 Cv  0.604 kJ/kgK
We know that, R = Cp  Cv
0.270 = Cp  0.604
 Cp  0.874 kJ/ kgK
Results:
Cv = 0.604 kJ/kgK
Cp = 0.874 kj/kgK

1   v 
 T 

T    v 

 P h CP   T p

 
From equation, we can determine the Joule-Thomson coefficient () in terms of
measurable properties such as pressure (p), temperature (T), specific volume (v) and C p.
Let, Enthalpy is a function of pressure and temperature.
i.e. h = f(p, T)
 h 
 h 
 dh=  dp    dT
 T p
 p T
For Throttling process, Enthalpy remains contact
H=C
 dh = 0
Substitute dh value in equation.
 h 
 h 
 O =   dp    dT
 T p
 p T
 h 
 h 
   dp    dT  0
 T p
 p T
Divided by dT
 h   p   h 
        0
 p T  T h  T p
 h   p 
 h 
       
 T p
 p T  T h
 h   1 
 h 
       
 T p
 p T   



 T  
=  
 p h 
 h 
1  h 
     
  p T
 p p
 h 
1  h 
 Cp       
  p T
 p p
 h 
The property  Cp    is known as constant temperature coefficient.
 p T
18. Prove that internal energy of an ideal gas is a function of temperature alone.
Solution:
We know that, ideal gas Equation
Pv = RT
RT
p
v
...(A)
We know that, Internal energy Equation
 p 
dU  Cv dT  T   dv  pdv
 T 
Divided by dv
[From Equation (30)]
 U 
 T 
 p 
    Cv    T    p
 v T
 v T
 T v

 U 
 p 
    T  p 
 v T
 T v


 T 
Cv    0
 v T

  RT 
 U 
   T 
p
T  v 
 v T
[From Equation (A)]
R
 U 
    T p
v
 v T


=p-p
p=
RT 
v 
 U 
   0
 v T
If the temperature remains constant, there is no change in internal energy with
volume. Hence internal energy is a function of temperature alone.
19. Prove that specific heat at constant volume (Cv) of a Vander Waals gas is a function of
temperature alone?
Solution:
We know that, Vander Waals equation
a

 p  v 2  (v - b)= RT


a
RT

 p + 2  
v  (v  b)

 p=
RT
a
 2
(v-b) v
R
 p 
   
 T v v  b
  2p 
 2 0
 T v
We know that,
 2p 
 Cv 

T
 2
 v 

T
 T 

 Cv 

 v   0

T


2p

0

T2

Thus if the temperature remains constant, there is no change in specific heat with
specific volume. Hence specific heat of constant volume is function of temperature alone.
20. Find the value of (h)T for a fluid that obeys the equation of state
P
RT a

v v2
Solution:
We know that,
 U 
 p 
    T   p
 v T
 T 
  p 

 dU = T    p dv
  T v

We know that
P
RT a
 (Given)
v v2
 P
RT a
 2
v
v
RT
a
P  2
v
v
 RT

a
 
 p  dv   2  dv
 v

v 
........(A)
RT
a
 (dU)T   2  dv [From equation (A) and P =
]
v
v 
 U =
v2

v
v1
dv
2
v2
 -1
 U = a  
 v  v1
1 1
 U = a 
 .......(B)
 v1 v 2 
We know that,
h  U  P2v 2  P1v1 
h 
a a
  P2v 2  p1v1
v1 v 2
a a  RT a  2  RT a 
 
 v  
 2  v1
v1 v 2  v 2 v 22 
 v1 v1 


P=
a a a a
 h =       
 v1 v 2   v1 v 2 
=
2a 2a

v1 v 2
1 1
 (h)r = 2a   .
 v1 v 2 
RT a

 2 (Given)
v v
