Question
Download Solution PDFA gas is compressed in a frictionless piston from an initial state of y m3 and 1 MPa to a final state of 0.2 m3 and 1 MPa. There is a transfer of 40 kJ of heat from the gas and a drop of 20 kJ in internal energy. What is the initial state volume of the gas?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
According to the first law of thermodynamics
δQ = dU + δW
Isobaric work done
δW = PdV = P (Vfinal - Vinitial)
Calculation:
Initial condition ⇒ P1 = 1 MPa, V1 = y m3
Final condition ⇒ P2 = 1 MPa, V2 = 0.2 m3
Heat Transfer = -40 kJ (from the gas)
Change in Internal energy (u2 – u1) = -20 kJ (drop)
According to first law of thermodynamics
δQ = dU + δW
-40 = -20 + δW
⇒ δW = -20 kJ ---(I)
Since, the process is isobaric (as pressure remains same)
So, isobaric work done δW = PdV = P (Vfinal - Vinitial)
δW = P (Vfinal - Vinitial) = -20 kJ
1000 kPa × (0.2 – y) m3 = -20 kJ
\(0.2 - y = \frac{{ - 20}}{{1000}} = - 0.02 \Rightarrow y = 0.22\)
∴Initial volume (y) = 0.22 m3Last updated on May 28, 2025
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