BASIC CONCEPTS OF THERMODYNAMICS 55

dharm
M-therm/th2-2.pm5

The work done during the process is given by

WpdV

V

V

=z

1

2

V 1 =^303015

1

..

p =. = 2 m

3

V 2 =

30 30

2 75

..

p =. = 0.4 m^3

∴ W = 10^5 30

2

04..

z V dVNm [Q 1 bar = 10

(^5) N/m (^2) ]
= 10^5 × 3.0 logeV^042. = 10^5 × 3.0(loge 0.4 – loge 2)
= – 3 × 10^5 loge (2/0.4) = – 3 × 10^5 × loge 5 = – 3 × 10^5 × 1.61 Nm
= – 4.83. × 10^5 Nm = – 4.83 × 10^5 J = – 483 kJ. (Ans.)
Example 2.20. To a closed system 150 kJ of work is supplied. If the initial volume is
0.6 m^3 and pressure of the system changes as p = 8 – 4V, where p is in bar and V is in m^3 ,
determine the final volume and pressure of the system.
Solution. Amount of work supplied to a closed system = 150 kJ
Initial volume = 0.6 m^3
Pressure-volume relationship, p = 8 – 4V
The work done during the process is given by
WpdV
V
V
=z
1
2
= 10^5 ()
.
84
06
2
z − VdV
V
= 10^5 842
2
06
V V
V
L −×
N
M
O
Q
P
.
= 10^5 [8(V 2 – 0.6) – 2(V 22 – 0.6^2 )]
= 10^5 [8V 2 – 4.8 – 2V 22 + 0.72]
= 10^5 [8V 2 – 2V 22 – 4.08] Nm or J
But this work is equal to – 150 × 10^3 J as this work is supplied to the system.
∴ – 150 × 10^3 = 10^5 [8V 2 – 2V 22 – 4.08]
or 2 V 22 – 8V 2 + 2.58 = 0
V 2 =
86442
4
8
4
±−××

2.58 ±6.585
= 0.354 m^3
Positive sign is incompatible with the present problem, therefore it is not considered.
∴ Final volume, V 2 = 0.354 m^3. (Ans.)
and, final pressure, p 2 = 8 – 4V = 8 – 4 × 0.354
= 6.584 bar = 6.584 × 105 N/m^2 or Pa. (Ans.)