FIRST LAW OF THERMODYNAMICS 121
dharm
M-therm/th4-2.pm5
= 0.105(0.20 – 0.40) MJ = – 21 kJ [Q 1 MJ = 10^3 kJ]
Substituting this value of W1–2 in equation (i), we get
- 42.5 = (U 2 – U 1 ) – 21
∴ U 2 – U 1 = – 42.5 + 21 = – 21.5 kJ
Hence ‘decrease’ in internal energy = 21.5 kJ. (Ans.)
Example 4.5. A container is divided into compartments by a partition. The container is
completely insulated so that there is no heat transfer. One portion contains gas at temperature T 1
and pressure p 1 while the other portion also has the same gas but at temperature T 2 and pressure
p 2.
How will the First Law of Thermodynamics conclude the result if partition is removed?
Solution. Refer Fig. 4.13.
According to First Law of Thermodynamics,
δQ = δU + δW
When partition removed, δQ = 0
δW = 0
∴δU = 0.
Gas
T, p 11
Gas
T, p 22
Insulated walls
Partition
Fig. 4.13
Conclusion. There is conservation of internal energy.
Example 4.6. Air enters a compressor at 10^5 Pa and 25°C having volume of 1.8 m^3 /kg and
is compressed to 5 × 105 Pa isothermally.
Determine : (i) Work done ;
(ii)Change in internal energy ; and
(iii)Heat transferred.
Solution. Initial pressure of air, p 1 = 10^5 Pa
Initial temperature of air, T 1 = 25 + 273 = 298 K
Final pressure of air, p 2 = 5 × 10^5 Pa
Final temperature of air, T 2 = T 1 = 298 K (isothermal process)
Since, it is a closed steady state process, we can write down the first law of thermodynamics
as,
Q = (u 2 – u 1 ) + W ......per kg
