A mixture of gas expands from 0.03 m3 to 0.06 m3 at a constant pressure of 1 MPa and absorbs 84 kJ of heat during the process. The change in internal energy of the mixture is
A. 30 kJ
B. 84 kJ
C. 54 kJ
D. 114 kJ
Answer: Option C
This Question Belongs to Mechanical Engineering >> Engineering Thermodynamics
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P1 = P2 = 1000 kPa
1Q2 = 84 kJ
1W2 = P1 (V2 – V1)
= 1000 (0.06 – 0.03) = 30 kJ
1Q2 = 1W2 + 1U2
U2 – U1= 1Q2 – 1W2 = 84 – 30 = 54 kJ