Estimate compressibility at constant temperature using trusted inputs. Switch methods, convert units, and review sensitivity. Results display clearly for faster decisions today.
Isothermal compressibility is defined as: κT = - (1/V) (∂V/∂P)T.
For two measured states at the same temperature, the calculator uses a finite-difference approximation: κT ≈ - (1/Vref) (ΔV/ΔP). You may choose Vref as the average volume, V₁, or V₂.
If you provide bulk modulus, it uses KT = 1/κT. For an ideal gas at constant temperature, it uses κT = 1/P.
| Material | P₁ (Pa) | V₁ (m³) | P₂ (Pa) | V₂ (m³) | Approx. κT (1/Pa) |
|---|---|---|---|---|---|
| Water (near room conditions) | 1.0e5 | 1.0000 | 2.0e5 | 0.99995 | ~4.5e-10 |
| Light oil (illustrative) | 1.0e5 | 1.0000 | 5.0e5 | 0.9996 | ~1.0e-9 |
| Ideal gas (illustrative) | 1.0e5 | — | — | — | 1.0e-5 |
Isothermal compressibility, κT, quantifies how strongly a material volume responds to pressure at fixed temperature. It is essential in pump selection, hydraulic stiffness, liquid storage, and high‑pressure process design where small volume changes can produce large pressure swings.
The thermodynamic definition is κT = −(1/V)(∂V/∂P)T. The minus sign makes κT positive for normal materials because volume decreases as pressure increases. The inverse quantity is the isothermal bulk modulus, KT = 1/κT.
When you have two measurements at the same temperature, the calculator estimates κT using κT ≈ −(1/Vref)(ΔV/ΔP). This approach is practical for lab data, piston‑cylinder tests, or tabulated P–V pairs from material datasheets.
Using Vref as the average of V₁ and V₂ is a robust default for small steps. If the dataset is anchored to a known operating state, choose V₁ or V₂ as the reference to match that state. Large pressure steps can hide curvature in V(P), so smaller steps improve fidelity.
κT is commonly reported in 1/Pa, 1/MPa, or 1/bar. This calculator converts inputs to SI internally and then displays κT in multiple units, alongside KT in Pa, MPa, and GPa. Scientific notation is used automatically for very small values.
Near room temperature, many liquids have κT on the order of 10−10 to 10−9 1/Pa, corresponding to KT around 1–3 GPa. Water is often cited near ~4.5×10−10 1/Pa, while light oils can be somewhat more compressible. For an ideal gas, κT = 1/P, giving ~1.0×10−5 1/Pa at 1 atm.
Bulk modulus sets pressure‑volume stiffness: larger KT means a “stiffer” fluid or solid under compression. In many fluids, acoustic behavior is tied to compressibility; lower compressibility generally supports higher wave speeds, although the exact relationship depends on the thermodynamic path and material model.
Ensure both states are truly isothermal: temperature drift can dominate small volume changes. Use consistent units, verify that ΔP is not near zero, and prefer precise volume measurements (calibrated displacement or density methods). If κT becomes non‑positive, recheck point ordering and data quality.
1) What does a larger κT mean?
A larger κT means the material compresses more for a given pressure change at constant temperature, so its volume is more sensitive to pressure and its bulk modulus is lower.
2) Why can κT appear negative in my result?
Negative values usually indicate inconsistent inputs, reversed state order, or data outside normal behavior. Check that P₂ and V₂ correspond to the same sample and temperature, and that volume decreases as pressure increases.
3) Which method should I choose?
Use the two‑state method when you have measured P–V pairs. Use bulk modulus if KT is known from a datasheet. Use the ideal‑gas method only for gases close to ideal behavior.
4) What pressure step is best for finite differences?
Smaller pressure steps reduce curvature errors and better approximate the derivative. Choose a step large enough to exceed measurement noise but small enough to stay within the same operating regime.
5) Is κT the same as compressibility used in acoustics?
It is related but not always identical. Acoustic analysis often uses an adiabatic path, while κT is isothermal. Many liquids show similar magnitudes, but the thermodynamic path can matter.
6) Can I use density instead of volume?
Yes, if you can convert consistently. Since V ∝ 1/ρ for a fixed mass, compressibility can be expressed using density changes with pressure. Ensure the same mass basis and temperature for both states.
7) What units are recommended for reporting?
For engineering, 1/MPa or 1/bar is often readable, while 1/Pa is standard in SI. This tool reports several options so you can match your report format or material datasheet.