TITLE.PM5

SECOND LAW OF THERMODYNAMICS AND ENTROPY 269

dharm

/M-therm/th5-3.pm5

Solution. Refer Fig. 5.32.
Initial pressure, p 1 = 1.05 bar = 1.05 × 10^5 N/m^2
Initial volume, V 1 = 0.04 m^3
Temperature, T 1 = 15 + 273 = 288 K
Final pressure, p 2 = 4.8 bar = 4.8 × 10^5 N/m^2
Final temperature, T 2 = T 1 = 288 K.
The process is shown on a p-v and a T-s diagram in Figs. 5.32 (a) and 5.32 (b) respectively.
The shaded area in Fig. 5.32 (a) represents work input, and the shaded area on Fig. 5.32 (b)
represents heat rejected.
Characteristic gas constant,

R R

M

=Universal gas constant,

Molecular weight,

(^0) =^8314
28
= 297 Nm/kg K
(a)
(b)
Fig. 5.32