The bulk modulus is a constant the describes how resistant a substance is to compression. It is defined as the ratio between pressure increase and the resulting decrease in a material's volume. Together with Young's modulus, the shear modulus, and Hooke's law, the bulk modulus describes a material's response to stress or strain.

Usually, bulk modulus is indicated by *K* or *B* in equations and tables. While it applies to uniform compression of any substance, it is most often used to describe the behavior of fluids.

It can be used to predict compression, calculate density, and indirectly indicate the types of chemical bonding within a substance. The bulk modulus is considered a descriptor of elastic properties because a compressed material returns to its original volume once the pressure is released.

The units for the bulk modulus are Pascals (Pa) or newtons per square meter (N/m^{2}) in the metric system or pounds per square inch (PSI) in the English system.

### Table of Fluid Bulk Modulus (K) Values

There are bulk modulus values for solids (e.g., 160 GPa for steel; 443 GPa for diamond; 50 MPa for solid helium) and gases (e.g., 101 kPa for air at constant temperature), but the most common tables list values for liquids. Here are representative values, in both English and metric units:

English Units(10 PSI)^{5} | SI Units(10 Pa)^{9} | |
---|---|---|

Acetone | 1.34 | 0.92 |

Benzene | 1.5 | 1.05 |

Carbon Tetrachloride | 1.91 | 1.32 |

Ethyl Alcohol | 1.54 | 1.06 |

Gasoline | 1.9 | 1.3 |

Glycerin | 6.31 | 4.35 |

ISO 32 Mineral Oil | 2.6 | 1.8 |

Kerosene | 1.9 | 1.3 |

Mercury | 41.4 | 28.5 |

Paraffin Oil | 2.41 | 1.66 |

Petrol | 1.55 - 2.16 | 1.07 - 1.49 |

Phosphate Ester | 4.4 | 3 |

SAE 30 Oil | 2.2 | 1.5 |

Seawater | 3.39 | 2.34 |

Sulfuric Acid | 4.3 | 3.0 |

Water | 3.12 | 2.15 |

Water - Glycol | 5 | 3.4 |

Water - Oil Emulsion | 3.3 | 2.3 |

The *K* value varies, depending on the state of matter of a sample, and in some cases, on the temperature. In liquids, the amount of dissolved gas greatly impacts the value. A high value of *K* indicates a material resists compression, while a low value indicates volume appreciably decreases under uniform pressure.

The reciprocal of the bulk modulus is compressibility, so a substance with a low bulk modulus has high compressibility.

Upon reviewing the table, you can see the liquid metal mercury is very nearly incompressible. This reflects the large atomic radius of mercury atoms compared with atoms in organic compounds and also the packing of the atoms. Because of hydrogen bonding, water also resists compression.

### Bulk Modulus Formulas

The bulk modulus of a material may be measured by powder diffraction, using x-rays, neutrons, or electrons targeting a powdered or microcrystalline sample. It may be calculated using the formula:

**Bulk Modulus ( K) = Volumetric stress / Volumetric strain**

This is the same as saying it equals the change in pressure divided by the change in volume divided by initial volume:

**Bulk Modulus ( K) = (p_{1} - p_{0}) / [(V_{1} - V_{0}) / V_{0}]**

Here, p_{0} and V_{0} are the initial pressure and volume, respectively, and p_{1} and V1 are the pressure and volume measured upon compression.

Bulk modulus elasticity may also be expressed in terms of pressure and density:

K = (p_{1} - p_{0}) / [(ρ_{1} - ρ_{0}) / ρ_{0}]

Here, ρ_{0} and ρ_{1} are the initial and final density values.

### Example Calculation

The bulk modulus may be used to calculate hydrostatic pressure and density of a liquid.

For example, consider seawater in the deepest point of the ocean, the Mariana Trench. The base of the trench is 10994 m below sea level.

The hydrostatic pressure in the Mariana Trench may be calculated as:

p_{1} = ρ*g*h

Where p_{1} is the pressure, ρ is the density of seawater at sea level, g is the acceleration of gravity, and h is the height (or depth) of the water column.

p_{1} = (1022 kg/m^{3})(9.81 m/s^{2})(10994 m)

p_{1} = 110 x 10^{6} Pa or 110 MPa

Knowing the pressure at sea level is 10^{5} Pa, the density of the water at the bottom of the trench may be calculated:

ρ_{1} = [(p_{1} - p)ρ + K*ρ) / K

ρ_{1} = [[(110 x 10^{6} Pa) - (1 x 10^{5} Pa)](1022 kg/m^{3})] + (2.34 x 10^{9} Pa)(1022 kg/m^{3})/(2.34 x 10^{9} Pa)

ρ_{1} = 1070 kg/m^{3}

What can you see from this? Despite the immense pressure on water at the bottom of the Mariana Trench, it isn't compressed very much!

### Sources

- De Jong, Maarten; Chen, Wei (2015). "Charting the complete elastic properties of inorganic crystalline compounds".
*Scientific Data*. 2: 150009. doi:10.1038/sdata.2015.9 - Gilman, J.J. (1969).
*Micromechanics of Flow in Solids*. New York: McGraw-Hill. - Kittel, Charles (2005).
*Introduction to Solid State Physics*(8th edition). ISBN 0-471-41526-X. - Thomas, Courtney H. (2013).
*Mechanical Behavior of Materials*(2nd edition). New Delhi: McGraw Hill Education (India). ISBN 1259027511.