FIRST LAW OF THERMODYNAMICS 133dharm
M-therm/th4-2.pm5(ii) The work done per kg of fluid is given byWpdv
vv
=z
1(^2) = p(v
2 – v 1 ) = 1.6 × 10^5 (0.55 – 0.3) N-m
= 40 × 10^3 J = 40 kJ
∴ Work done = 40 kJ/kg. (Ans.)
(iii)Change in internal energy,
∆u = Q – W = 475.94 – 40 = 435.94 kJ/kg. (Ans.)
(iv)Change in enthalpy, (for non-flow process)
∆h = Q = 475.94 kJ/kg. (Ans.)
Example 4.19. 1 kg of gaseous CO 2 contained in a closed system undergoes a reversible
process at constant pressure. During this process 42 kJ of internal energy is decreased. Deter-
mine the work done during the process.
Take cp = 840 J/kg°C and cv = 600 J/kg°C.
Solution. Mass CO 2 , m = 1 kg
Decrease in internal energy, ∆u = – 42 kJ = – 42 × 10^3 J
Specific heat at constant pressure, cp = 840 J/kg°C
Specific heat at constant volume, cv = 600 J/kg°C
Let, initial temperature of CO 2 = T 1
Final temperature of CO 2 = T 2
Now change in internal energy,
∆U = m × cv(T 2 – T 1 )
- 42 × 10^3 = 1 × 600(T 2 – T 1 )
∴ T 2 – T 1 = – 42 10 600
×^3
= – 70°C
The heat supplied or rejected,
Q = mcp(T 2 – T 1 )
= 1 × 840 × (– 70) = – 58800 J or – 58.8 kJ
Applying first law to the process,
Q = ∆U + W - 58.8 = – 42 + W or W = – 16.8 kJ
∴ Work done during the process = – 16.8 kJ. (Ans.)
+Example 4.20. A fluid is contained in a cylinder by a spring-loaded, frictionless piston
so that the pressure in the fluid is a linear function of the volume (p = a + bV). The internal energy
of the fluid is given by the following equation
U = 42 + 3.6 pV
where U is in kJ, p in kPa, and V in cubic metre. If the fluid changes from an initial state of
190 kPa, 0.035 m^3 to a final state of 420 kPa, 0.07 m^3 , with no work other than that done on the
piston, find the direction and magnitude of the work and heat transfer.
Solution. Relation between pressure and volume, p = a + bV.
Equation of internal energy : U = 42 + 3.6pV
Initial pressure, p 1 = 190 kPa
Initial volume, V 1 = 0.035 m^3
Final pressure, p 2 = 420 kPa
Final volume, V 2 = 0.07 m^3