SECTION II – FIRST LAW APPLIED TO CLOSED SYSTEMS
0.2 kg of fluid, initially at a temperature of 165C, expands reversibly at a constant pressure of 7
bars until the volume is doubled. Find the final temperature, work done and heat transferred if the
fluid is steam with an initial quality (dryness fraction) of 70%.
T2 = ?
W12 = ?
Q12 = ?
Steam
P1 = P2 = 7 bars
T1 = 165C
x1 = 0.7
m = 0.2 kg
system boundary

Initial state is in the liquid-vapour saturation region since x1 = 0.7
v1  (1  x 1 ) v f1  x 1 v g1  (1  0.7)(0.001108)  0.7(0.2729)
v1  0.1914 m 3 / kg; 1  mv 1  mv 1  0.2(0.1914)  0,9373 m 3
u 1  (1  x 1 )u fg 1  x 1 u g1  u f1  x 1 u fg 1  696.44  0.7(1876.1)
u 1  2009.7 kJ / kg

Final state: P2 = 7 bars v2m v2 = 2v1 = 0,3727 m3/kg
v 2  v g P7 bars  0.2729 m 3 / kg  steam is a superheated vapour at the final state. Interpolation
is necessary!
T = 300V, P = 0.6 MPa  v = 0.4344 m3/kg, u = 2801.0 kJ/kg
T = 300C, P = 0.8 MPa  v = 0.3241 m3/kg, u = 2797.2 kJ/kg
v * 0.4344
(0.7  0.6)

 v*  0.37295 m 3 / kg
(0.3241  0.4344) (0.8  0.6)
u * 2801.0
(0.7  0.6)

 u*  2799.1 kJ / kg
(2797.2  2801.0) (0.8  0.6)
T  300C, P  0.7MPa 
T = 350C, P = 0.6 MPa  v = 0.4742 m3/kg, u = 2881.2 kJ/kg
T = 350C, P = 0.8 MPa  v = 0.3544 m3/kg, u = 2878.2 kJ/kg
v * *  0.4742
(0.7  0.6)

 v * *  0.4143 m 3 / kg
(0.3544  0.4742) (0.8  0.6)
u * *  2881.2
(0.7  0.6)

 u * *  2879.7 kJ / kg
(2878.2  2881.2) (0.8  0.6)
Section II – First Law Applied to Closed Systems
P  0.7 MPa , v  0.8328 m 3 / kg 
Page 5
T2  300
(0.3828  037925)

 T2  305.06C
(350  300) (0.4143  037925)
u 2  2799.1
(0.38281.37925)

 u 2  2807.3 kJ / kg
(2879.7  2799.1) (0.4143  0.37925)

Process in P-v and T-v diagrams
P
T
T = 305.06C


T = 305.05C
P1 = P2 = 7 bars


P1 = P2 = 71 bars


T = 165C
T1 = 165C


v
2

Work done: W12 = m  Pdv  mP (v 2  v1 )  0.2(700)(0.3828  0.1914)  W12  26.8 kJ

First Law: Q12 – W12 = U2 – U1
1
Q12 = W12 + (U2-U1) = W12 + m(u2-u1)
= mP(v2-v1) + m(u2–u1)
Q12 = 26.8 + 0.2(2807.3-2009.7) = 186.3 kJ

Alternatively, Q12 = mP(v2-v1) + m(u2-u1)
= m[(u2 + P2v2)-(u1+P1v1)]
Q12 = m(h2-h1)
h1  h f1  x1h fg 1  697.22  0.7(2066.3)  2143.6 kJ / kg
T = 300C, P = 0.6 MPa, h = 3061.6 kJ/kg
T = 300C, P = 0.8 MPa, h = 3056.5 kJ/kg
h * 3061.6
(0.7  0.6)

 h*  3059.1 kJ / kg (at P = 0.7 MPa, T = 300C)
(3056.5  3061.6) (0.8  0.6)
T = 350C, P = 0.6 MPa, h = 3165.7 kJ/kg
v
Section II – First Law Applied to Closed Systems
Page 6
T = 350C, P = 0.8 MPa, h = 3161.7 kJ/kg
h * *  3165.7
(0.7  0.6)

 h * *  3163.7 kJ / kg (at P = 0.7 MPa, T = 350C)
(3161.7  3165.7) (0.8  0.6)
h 2  3059.1
(0.3828  0.37925)

 h 2  3969.7 kJ / kg
(3163.7  3059.1) (0.4143  0.37925)
Q12 = m(h2-h1) = 0.2 (3069.7 – 2143.6)  Q12 = 185.2 kJ
The cylinder shown below is fitted with a piston that is restrained by a spring so arranged that for
zero volume in the cylinder the spring is fully extended. The spring force is proportional to the
spring displacement and the weight of the piston is negligible. The enclosed volume in the cylinder
is 120 litres when the piston encounters the stops. The cylinder contains 4 kg of water initially at
350 KPa, 1% quality, and the water is then heated until it exists as a saturated vapour. Show this
process on a P-  diagram and determine:
(a) The final pressure in the cylinder
(b) The work done by the water during the process
(c) Heat added during the process
(d) Change in internal energy
P1 = 350 kPa
x1= 0.01
m = 4 kg
T1 = Tsat = 138.88C
system boundary
v1 = *1-0.01)(0.001079) + 0.0l(0.5243)
= 0.00631121 m3/kg
 2  120 litres  120 x 1000 x 10 6  0.12 m 3

Process in the P-  diagram.
P


T = 155.5C const

P1 =350 MPa = 0.35 MPa


1 = 0.025 m3 1 = 0.12 m3

Section II – First Law Applied to Closed Systems

Page 7
Consider a free body diagram of the piston
Fs
y+
Patm

P
Wp (here negligible!)
For any equilibrium position of the piston,
PA – PatmA – Fs = 0  P = Patm + Fs/A
Fs = ksy  P = Patm + ks y/A
  = Ay  P = Patm + ks   /A2

Process 1  2: P1 = Patm +
k s 1  0
(350  101)A 2
249 2

k


A
s
2
2
0.025

A
ks = 9960 A2 KN/m2
P2 = Patm + ks(  2-0)/A2 = 101 + 9960A2
(0.12)
A2
P2 = 1296.2 KPa
(a) Process 2  3: v = const  v3 = vg = 0.12/4 = 0.03 m3/kg
Interpolate for P3 with v3 = 0.03 m3/kg between P = 6 MPa, vg = 0.03244 m3/kg and P = 7
MPa, vg = 0.02737 m3/kg in the saturated water Pressure table.
P3  6
(0.03  0.03244)

 P3  6.481 MPa
(7  6) (0.02737  0.03244)
(b) W13  W12  W23
3
W23   Pd; d  0 sin ce 2  3
2
(since  2  3 )

(P1  P2 )
(350  1296.2)
( 2 1 ) 
(0.12  0.025)
2
2
W13 = 78.2 kJ
Section II – First Law Applied to Closed Systems
Page 8
(c) Heat added during process
First Law: Q13 – W13 = U3 –U1
Q13 = W13 + (U3-U1) = 78.2 + 7926.8 = 8005.0 kJ
(d) Change in the internal energy
U = U3 – U1 = m (U3 – U1)
U1  U f1  X1 U fg 1  583.95  0.01(1965.0)  603.6 kJ / kg
U3 = Ug @ P = 6.481 MPa; interpolation in the saturated water Pressure table.
U 3  2589.7
(6.481  6)

 U 3  2585.3 kJ / kg
(2580.5  2589.7)
(7  6)
U = 4(2585.3 – 603.6) = 7926.8 kJ
Air at 1 atm and 20C occupies on initial volume of 1000 cm3 in a cylinder. The air is confined by
a piston which has a constant restraining force so that the gas pressure always remain constant.
Heat is added to the air until its temperature reaches 260C. Calculate the heat added, the work the
air does on the piston, and the change in internal energy of the air.

System
Air
P1 = P2 = 1 atm
1 = 1000 cm3
T1 = 20C
W12
Q12
T2 = 260C

1 atm = 101.3 KPa
20C = 293 K
260C = 533 K
1000 cm3 = 10-3 m3
Q12 = ?
W12 = ?
U = ?
system boundary
Can air be treated as an ideal gas under the given conditions?
PR1  PR 2  0.1013 / 3.76  0.027

TR1  293 / 133  2.203; TR 2  533 / 133  4.008

 Air can be treated as an ideal gas
Z1  Z 2  1
See p.807 of text 
Section II – First Law Applied to Closed Systems
Page 9

Mass in the system

P11 101.3 10 3

 1.205  10 3 kg
RT1 0.287  293
Volume at the final state
mRT 2 1.205  10 3  0.287  533
P2  2  mRT 2   2 

P2
101.3
P11  mRT1  m 
 2  1.82 10 3 m 3

Work done by the air on the piston
W12   Pd  P2  1   101.3(1.82  1)  10 3  W12  0.083 kJ  83 J
2
1

Internal energy at the initial and final states
U 1  206.91
(293  290)

 U 1  209.06 kJ / kg
210.49  206.91 (295  290)
U 2  381.84
(533  530)

 U 2  384.09 kJ / kg
(389.34  381.84) (540  530)
U = m(U2-U1) = 1.205 x 10-3 (384.09-209.06) = 0.215 kJ

Heat transfer
First Law: Q12 – W12 = U  Q12=W12 + U
Q12 = (0.083 + 0.215) kJ = 0.298 kJ = 298 J

Process in the P-  diagram
P
P1 = P2




T = 533 K
T = 293 K
1000 cm3
1820 cm3

Section II – First Law Applied to Closed Systems
Page 10
Air in a closed vessel of fixed volume, 0.14 m3, exerts a pressure of 10 bars at 250C. If the vessel
is cooled so that the pressure falls to 3.5 bars, determine the final temperature and heat transferred.

System
T2 = ?
Q12 = ?
Air
 = 0.14 m3
P1 = 10 bars
P2 = 3.5 bars
T1 = 250C
Assumptions
 System is stationary
 Potential energy datum is at the system
level
system boundary
Q12

Can air be treated as an ideal gas under the given conditions?
Pr1 
P1
P
1
0.35

 0.266, Pr2  2 
 0.093
Pcrit 3.76
Pcrit 3.76
Tr1 
T1
523

 3.932
Tcrit 133
Z1  0.95  1; Pr2 is small  Z 2  1
 Air can be treated as an ideal gas under the given conditions.

Mass of air in the system
P1  mRT1  m

P1 1000  0.14

 0.933 kg
RT1 0.287  523
Final temperature
P2 
350  0.14

MR 0.933  0.287
T2  183 K  90C
P2   mRT 2  T2 

Heat transferred
First Law: Q12 – W12 = U2 – U1
2
W12   Pd but 1  2    d  0  W12  0
1
U 2  U1  mC vave (T2  T1 ); Tave 
Tave  353 K
T1  T2 (523  183)

K
2
2
Section II – First Law Applied to Closed Systems
Page 11
C vave  0.721
(353  350)

 C vave  0.721 kJ / kg
(0.726  0.721) (400  350)
(U2 – U1) = 0.933(0.721)(183-523) = –228.7 kJ = Q12
indicates heat transfer from
the system (i.e. cooling)
0.9 kg of air, initially at a pressure of 15 bars and a temperature of 250C expands reversibly and
polytropically to 1.5 bars. Find the final temperature, work done and heat transferred if the index of
expansion is 1.25.

System
Air
m = 0.9 kg
P1 = 15 bars
P2 = 1.5 bars
T1 = 250C
T2 = ?
W12 = ?
Q12 = ?
Assumptions
• System is stationary
• Potential energy datum is at the
level of the system.
system boundary

Can air be treated as an ideal gas under the given conditions?
Pr 41 
Tr1 
P1
P
1.5
0.15

 0.399, Pr2  2 
 0.040
Pcrit 3.76
Pcrit 3.76
T1
523

 3.93 (see page 807, Fig . A  30a , Text )
Tcrit 133
Z1 = 0.975  1; since Pr2 is so small, Z2  1 for all Tr values  air can be treated as an ideal gas
under the given conditions.

Properties at the initial state
mRT1 0.9  0.287  523

 0.09 m 3
P1
1500

0.09
v1  1 
 0.1 m 3 / kg
m
0.9
u 1  374.36
(523  520)

 u 1  376.60 kJ / kg (see Table A  17, page 788 of Text )
(381.84  374.36) (530  520)
1 
Section II – First Law Applied to Closed Systems

Page 12
Properties at the final state
P
Polytropic Process: P v  P2 v  v 2   1
 P2
n
1 1
 15 
v2  

 1 .5 



1/ n
v1
1 / 1.25
(0.1)  0.631 m 3 / kg
Ideal gas: P2 v 2  RT2  T2 
T2 
n
2
P2 v 2
P2
150  0.631
 329.8 K  56.8C
0.287
u 2  232.02
(329.8  325)

 u 2  235.47 kJ / kg
(235.61  232.02)
(330  325)

Work done by the air
2
2
W12   Pd  m  Pdv  m
1
1
W12  (0.9)

P2 v 2  P1 v1
(1  n )
[150(0.631)  1500(0.1)]
 199.3 kJ
(1  1.25)
Heat transferred
First Law: Q12 – W12 = U2 – U1
Q12 = W12 + m(U2 – U1) = 199.3 + 0.9(235.47 – 376.60)
Q12 = 72.3 kJ
Alternatively, we can write u 2  u 1  c vave (T2  T1 )
c vave  0.726
(0.733  0.726)

(426.4  400)
 c vave  0.730 kJ / kg K (see Table A  2, page 764 of Text )
(450  400)
T1  T2 523  329.8

 426.4 K
2
2
Q12 = 199.3 + 0.9(-141.04) = 72.4 kJ
NOTE: Tave 
Section II – First Law Applied to Closed Systems

Page 13
Process in the P-  diagram
P

P1 = 1.5 MPa
Pvn = const
T = 533 K = const
P2 = 0.15 MPa


T = 329.8 K = const
const

Polytropic processes involving ideal gases

Process that obeys Pvn = const is a polytropic process.
- n: polytropic index of expansion or compression
- polytropic processes not limited to ideal gases

If n = 0, Pvn = const becomes P = const  ISOBARIC PROCESS

(const )1 / 2
If n = , v =
becomes v = const  ISOCHORIC PROCESS
P1 / 2

If n = 1, Pvn = const becomes Pv = const and for an ideal gas, Pv = RT  T = const 
ISOTHERMAL PROCESS

If n = k, where k = cp/cv  ADIABATIC PROCESS (NO HEAT TRANSFER)
(for ideal gas)

When n > 0 (i.e. positive), P and v cannot increase or decrease simultaneously i.e. P
corresponds to v and vice versa. Compression and expansion processes fall into quadrants 2
and 4 respectively with n positive and an initial state at A.
n=k
n
P


A

n=0
n=1


v
Section II – First Law Applied to Closed Systems
Page 14

It is possible, physically, for processes to fall into quadrants 1 and 3. n will then be negative and
P and v will increase or decrease simultaneously. However, such processes do not often occur
in practical systems.

One example of a polytropic
process with n < 0 is the
expansion of a gas behind a
piston that is attached to a
linear spring.
Patm
P
GAS
x
k (spring constant)
Heat
A (Surface Area of piston)
Free-body diagram of piston.
FP
FPatm
Fs
At any intermediate equilibrium state (recall process is reversible and therefore quasistatic), Fx = 0
 Fp – Fs - FPatm = 0  Fp = FPatm + Fs
Fp = P.A, FPatm = PatmA, Fs = kX
x
A
x  P and x  v
- P and v increase simultaneously  n < 0
P = Patm + k
- Process will fall in quadrant 1 starting at A

Relationships between P, v and T for polytropic processes involving ideal gases
n
P 
v
P2  v1 
    1   2 
- Pv = const  P v  P2 v 
v 2  P1 
P1  v 2 
n
1 1
n
n
2
- Pv = RT  P1v1 = RT1, P2v2 = RT2 
n
 T2   P2
P2  T2  n 1
  
    
P1  T1 
 T1   P1
n
 v1 
T 
    2 
 v2 
 T1 
n
n 1
 v1
 
 v2



P2
P1
 v2

 v1
 T2
 
 T1
n 1
n
  T2
  
  T1
1
 n 1
T2  v1 

  

T
1
 v2 

n 1
1/ n