Question
Download Solution PDFA stationary mass of gas is compressed without friction from an initial state of 0.3 m3 and 0.105 MPa to a final state of 0.15 m3 and 0.105 MPa the pressure remaining constant throughout the process. There is a transfer of 37.6 KJ of heat from the gas during the process. The internal energy of gas changes by
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Apply the first law of thermodynamics:
\( \Delta U = Q - W \)
Given:
- Initial volume: \( V_1 = 0.3~m^3 \)
- Final volume: \( V_2 = 0.15~m^3 \)
- Pressure: \( P = 0.105~MPa = 105~kPa \)
- Heat rejected: \( Q = -37.6~kJ \)
Step 1: Work done
\( W = P (V_2 - V_1) = 105 \times (-0.15) = -15.75~kJ \)
Step 2: Change in internal energy
\( \Delta U = -37.6 - (-15.75) = -21.85~kJ \)
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