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Formula of Bulk Modulus

Fluids and properties of matter formulasClass 11Fluids and solidsUpdated 1 Nov 2026

K=−ΔPΔV/V=−V ΔPΔVK = -\frac{\Delta P}{\Delta V / V} = -\frac{V\,\Delta P}{\Delta V}
SI unit
Pa (N/m²)
Dimensions
[M¹ L⁻¹ T⁻²]
where
  • K bulk modulus (also B) Pa
  • ΔP increase in pressure on the body Pa
  • V original volume m³
  • ΔV change in volume (negative when compressed) m³

The formula of bulk modulus is K=−ΔPΔV/VK = -\frac{\Delta P}{\Delta V / V}: the increase in pressure on a body divided by the fractional change in its volume. The minus sign is there because more pressure makes the volume smaller, so ΔV is negative and K comes out positive.

Bulk modulus measures how hard it is to squeeze something into a smaller space. Water has K ≈ 2.2 × 10⁹ Pa, so even a hundred times atmospheric pressure shrinks it by less than half a percent. Air at room conditions has a K more than ten thousand times lower. Unlike Young's modulus, bulk modulus applies to solids, liquids and gases alike, and it is grouped with the other elastic moduli in the fluids and solids summary.

Quick facts

  • Formula

    K=−ΔPΔV/VK = -\frac{\Delta P}{\Delta V / V}

  • SI unit

    pascal (Pa)

  • Dimensional formula

    [M¹ L⁻¹ T⁻²]

  • Compressibility

    β=1K\beta = \frac{1}{K}, unit Pa⁻¹

  • Water

    about 2.2 × 10⁹ Pa

  • Ideal gas

    K = P (isothermal), K = γP (adiabatic)

Symbols in the formula of bulk modulus

Quantities in K = −ΔP / (ΔV/V)
SymbolMeaningSI unit
KKBulk modulusPa
ΔP\Delta PExtra pressure applied evenly on all sides (hydraulic stress)Pa
VVVolume before the changem³
ΔV\Delta VNew volume minus old volumem³
ΔVV\frac{\Delta V}{V}Volume strainnone

The top of the fraction is a stress (the hydraulic kind, equal to the pressure rise) and the bottom is a volume strain. So bulk modulus is stress over strain, exactly like the other moduli. Many calculators, including the one above, ask for the size of the volume decrease, which removes the minus sign: K = P V / ΔV with both P and ΔV taken as positive.

Worked examples

Example 1: compressing an oil sample

A 2.0 L sample of hydraulic oil is put under an extra pressure of 3.0 × 10⁶ Pa. Its volume falls by 4.0 cm³. Find the bulk modulus.

Convert: V = 2.0 × 10⁻³ m³, ΔV = −4.0 × 10⁻⁶ m³.

K=−3.0×106×2.0×10−3−4.0×10−6=6.0×1034.0×10−6=1.5×109 PaK = -\frac{3.0\times10^6 \times 2.0\times10^{-3}}{-4.0\times10^{-6}} = \frac{6.0\times10^3}{4.0\times10^{-6}} = 1.5\times10^9\ \text{Pa}

Example 2: water deep in the sea

Water (K = 2.2 × 10⁹ Pa) is under an extra pressure of 1.0 × 10⁷ Pa, about what you find 1 km down in the ocean. By what fraction does its volume shrink?

ΔVV=−ΔPK=−1.0×1072.2×109=−4.5×10−3\frac{\Delta V}{V} = -\frac{\Delta P}{K} = -\frac{1.0\times10^7}{2.2\times10^9} = -4.5\times10^{-3}

A drop of about 0.45 percent. A litre of water would lose roughly 4.5 cm³.

Example 3: compressibility

Find the compressibility of water.

β=1K=12.2×109=4.5×10−10 Pa−1\beta = \frac{1}{K} = \frac{1}{2.2\times10^9} = 4.5\times10^{-10}\ \text{Pa}^{-1}

Watch the sign. If you put ΔV in as a positive number, drop the minus sign from the formula too, or K will come out negative.

Bulk modulus of gases and the speed of sound

For an ideal gas compressed slowly at constant temperature, PVPV stays constant and the bulk modulus equals the pressure itself, K = P. Compressed quickly with no heat flow (adiabatic), PVγPV^\gamma stays constant and K = γP. Sound waves are fast compressions, so the adiabatic value applies. The speed of sound in any fluid is

v=Kρv = \sqrt{\frac{K}{\rho}}

For water, v=2.2×109/1000≈1.5×103 m/sv = \sqrt{2.2\times10^9 / 1000} \approx 1.5\times10^3\ \text{m/s}, several times faster than in air. More on this on the speed of sound formula page.

Typical values and comparison

Approximate bulk moduli
SubstanceK in Pa
Steel1.6 × 10¹¹
Glassabout 3.7 × 10¹⁰
Water2.2 × 10⁹
Ethanolabout 0.9 × 10⁹
Air (adiabatic, 1 atm)about 1.4 × 10⁵

Solids are the hardest to compress, liquids next and gases by far the easiest. Comparing K with Young's modulus and the modulus of rigidity shows which deformation a material resists most.

Common mistakes

  • Dropping the minus sign without making ΔV positive. That gives a negative bulk modulus, which is impossible for a stable material.
  • Using the final volume for V. Use the original volume.
  • Mixing litres and m³. 1 L = 10⁻³ m³ and 1 cm³ = 10⁻⁶ m³. Since V and ΔV appear as a ratio, any matching units work.
  • Using total pressure. ΔP is the change in pressure, not the final value.

Frequently asked questions

What is the formula of bulk modulus?

K = −ΔP / (ΔV / V), the pressure increase divided by the volume strain. The minus sign makes K positive, since the volume decreases.

Which SI unit does bulk modulus have?

The pascal (Pa), the same as pressure, because volume strain has no unit.

What is compressibility?

Compressibility is the reciprocal of bulk modulus, 1/K. A large compressibility means the material squeezes easily; gases have the largest values.

Where does the minus sign in the bulk modulus formula come from?

An increase in pressure (positive ΔP) causes a decrease in volume (negative ΔV). The minus sign cancels this so the modulus is a positive number.