The entropy change of a calorically perfect ideal gas can be calculated from two equilibrium temperatures and pressures.
Difficulty
Intermediate
Read time
4 min
Prerequisites
Consistent SI units, algebra, and the physical assumptions listed below
Source
Teng-Jui Lin — Equation SheetsAdapted from Teng-Jui Lin, Equation Sheets — Mechanical Engineering, under CC BY 4.0.
Adapted notation and added original explanations, variables, assumptions, applications, and a worked example. Corrected the pressure-term sign: ds = c_p dT/T − (∂v/∂T)_P dP for an ideal gas, then integrated with constant c_p on a mass basis.
The entropy change of a calorically perfect ideal gas can be calculated from two equilibrium temperatures and pressures. Entropy is a state property, so this expression applies even if the actual path between those states is irreversible.
| Symbol | Meaning | Units |
|---|---|---|
| Final minus initial specific entropy | ||
| Constant mass-specific heat at constant pressure | ||
| Specific gas constant | ||
| Initial absolute temperature | ||
| Final absolute temperature | ||
| Initial absolute pressure | ||
| Final absolute pressure |
Check whether two ideal-gas states could be connected isentropically.
Calculate the state-property part of an entropy balance.
An ideal gas changes from 300 K, 100000 Pa to 400 K, 200000 Pa. Use c_p = 1005 J/(kg·K) and R_s = 287 J/(kg·K).
Adapted from Teng-Jui Lin, Equation Sheets — Mechanical Engineering, under CC BY 4.0.
Adapted notation and added original explanations, variables, assumptions, applications, and a worked example. Corrected the pressure-term sign: ds = c_p dT/T − (∂v/∂T)_P dP for an ideal gas, then integrated with constant c_p on a mass basis.
Assumptions
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