Question
Download Solution PDFA closed gaseous system undergoes a reversible constant pressure process at 2 bar in which 100 kJ of heat is rejected and the volume changes from 0.2 m3 to 0.1 m3. The change in the internal energy of the system
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
By using the first law of thermodynamics
∆Q = ∆U + work done ----(1)
Where, ∆Q = change in heat, ∆U = change in internal energy of the system
Calculation:
P = 2 bar = 200 kPa, ∆Q = -100 kJ
For the constant pressure reversible process work done is given by
W.D. = P (V2 – V1 )
W.D. = 200 (0.1 - 0.2)
W.D. = -20 kJ
Using equation (1),
-100 = ∆U + (-20)
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