TITLE.PM5

120 ENGINEERING THERMODYNAMICS

dharm

M-therm/th4-2.pm5

Temperature before compression = 40°C or 313 K
Temperature after compression = 160°C or 433 K
The work done during the compression process, W = – 30 kJ
According to first law of thermodynamics,
Q = ∆U + W = (U 2 – U 1 ) + W
= mcv (T 2 – T 1 ) + W
= 0.3 × 0.75(433 – 313) – 30 = – 3 kJ
Hence, heat ‘rejected’ during the process = 3 kJ. (Ans.)
Note. Work, W has been taken –ve because it has been supplied from outside.
Example 4.4. When a stationary mass of gas was compressed without friction at constant
pressure its initial state of 0.4 m^3 and 0.105 MPa was found to change to final state of 0.20 m^3 and
0.105 MPa. There was a transfer of 42.5 kJ of heat from the gas during the process.
How much did the internal energy of the gas change?
Solution.

Boundary

Weight

Piston

Gas Q = 42.5 kJ

Fig. 4.12
Initial state
Pressure of gas, p 1 = 0.105 MPa
Volume of gas, V 1 = 0.4 m^3
Final state
Pressure of gas, p 2 = 0.105 MPa
Volume of gas, V 2 = 0.20 m^3
Process used : Constant pressure
Heat transferred, Q = – 42.5 kJ
(–ve sign indicates that heat is rejected)
Change in internal energy, ∆∆∆∆∆U = U 2 – U 1 :
First law for a stationary system in a process gives
Q = ∆U + W
or Q1–2 = (U 2 – U 1 ) + W1–2 ...(i)

Here WpdV

V

V

12

1

2

− =z = p(V 2 – V 1 )