For an internally reversible process on a closed stationary system of a simple compressible substance:Question 4Select one:a.TdS = dU – PdVb.TdS > dU +PdVc.TdS = dh + VdPd.TdS = dU + PdV
Question
For an internally reversible process on a closed stationary system of a simple compressible substance:Question 4Select one:a.TdS = dU – PdVb.TdS > dU +PdVc.TdS = dh + VdPd.TdS = dU + PdV
Solution
The correct answer is:
d. TdS = dU + PdV
This is the fundamental thermodynamic relation for a closed system undergoing an internally reversible process. It states that the change in entropy (dS) times temperature (T) is equal to the change in internal energy (dU) plus the pressure (P) times the change in volume (dV).
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