A 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

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ESE Mechanical 2015 Paper 1: Official Paper
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  1. -100 kJ
  2. -80 kJ
  3. -60 kJ
  4. -40 kJ

Answer (Detailed Solution Below)

Option 2 : -80 kJ
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Detailed Solution

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Concept:

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:

Given:  

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)

∆U = -80 kJ
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