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Engineering Thermodynamics
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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

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Correct Answer: Option C
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1 Comment
Zoo Topia
Zoo Topia 6 years ago
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