Library / Thermodynamics / THE-24
Thermodynamics
Fundamental thermodynamic relation (change in internal energy)
Change in internal energy of a simple compressible system for small reversible changes in entropy and volume.
- Ref
- THE-24
- Rev
- v1
- Field
- Thermodynamics
- Result
- dU [J]
For a simple compressible system, U changes by T dS (heat taken in reversibly) minus P dV (work done by the system). The relation involves only state variables, so it also holds for irreversible processes between the same equilibrium states. Use it for small changes; for a finite change, integrate.
Change in internal energy of a simple compressible system for small reversible changes in entropy and volume. The first term is the heat taken in, TdS, and the second is the work done by the system, PdV.
Calculator
- Heat term, T dS600 J
- Work term, P dV100 J
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- Heat contributes 600 J and expansion work removes 100 J, so the internal energy changes by 500 J.
- Setting dS = 0 (an adiabatic reversible change) gives dU = −P dV, the work done by the gas.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (J) | Local step | Change in result |
|---|---|---|---|---|---|---|
| P Pressure | 1 to 1e9 Pa | −999,400 | 600 | 1,000,000 | +10% | −2 % |
| dV Volume change | −1 to 1 m³ | −99,400 | 100,600 | 200,000 | +0.04 m³ | −800 % |
| dS Entropy change | −100 to 100 J/K | −30,100 | 29,900 | 60,000 | +4 J/K | +240 % |
| T Absolute temperature | 1 to 5,000 K | −98 | 9,900 | 9,998 | +10% | +12 % |
Sweep of Entropy change
| Entropy change (J/K) | Change in internal energy (J) |
|---|---|
| −100 | −30,100 |
| −75 | −22,600 |
| −50 | −15,100 |
| −25 | −7,600 |
| 0 | −100 |
| 25 | 7,400 |
| 50 | 14,900 |
| 75 | 22,400 |
| 100 | 29,900 |
Sweep one variable
Variables and how each one contributes
Change in internal energy
What the equation gives you. Everything else feeds it.
Absolute temperature
The absolute temperature, in K. Raises the heat term T dS and so the internal-energy change in proportion. Raising it by 10 % (to 330) moves the result by +12 %, from 500 to 560 J.
Entropy change
The entropy change, in J/K. Raises the internal-energy change through the heat term T dS. Raising it by 10 % (to 2.2) moves the result by +12 %, from 500 to 560 J.
Pressure
The pressure, in Pa. Lowers the internal-energy change through the work term P dV, if the volume expands. Raising it by 10 % (to 110,000) moves the result by −2 %, from 500 to 490 J.
Volume change
The volume change, in m³ (positive for expansion). Lowers the internal-energy change, because expansion does work on the surroundings. Raising it by 10 % (to 0.0011) moves the result by −2 %, from 500 to 490 J.
Worked example with the values above
- Heat term: T dS = 300 × 2 = 600 J.
- Work term: P dV = 100,000 × 0.001 = 100 J.
- dU = 600 − 100 = 500 J.
Assumptions and limits
- A simple compressible system in equilibrium, with only pressure–volume work.
- Small changes dS and dV, with T and P effectively constant over the step.
Common mistakes
- Using degrees Celsius for T.
- Using the work done on the system instead of by it (flipping the sign of the P dV term).
- Applying it to large changes without integrating.
Statistics and uncertainty
The two terms often nearly cancel, so the difference can be much less certain than either term: a 2 % error in each of two terms of similar size can give a much larger relative error in dU.
Where it is used
- Deriving the enthalpy, Helmholtz and Gibbs relations.
- Finding energy changes of a gas from entropy and volume changes.
Maths you need
- Entropy: dS = δQ/T
- First law: dU = δQ − δW