The bulk modulus (K) of a substance measures the substance's resistance to uniform compression. It is defined as the pressure increase needed to cause a given relative decrease in volume. Its base unit is the pascal.
As an example, suppose an iron cannon ball with bulk modulus 160 GPa is to be reduced in volume by 0.5%. This requires a pressure increase of 0.005×160 GPa = 0.8 GPa (116,000 psi).
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The bulk modulus K can be formally defined by the equation:
where P is pressure, V is volume, and ∂P/∂V denotes the partial derivative of pressure with respect to volume. The inverse of the bulk modulus gives a substance's compressibility.
Other moduli describe the material's response (strain) to other kinds of stress: the shear modulus describes the response to shear, and Young's modulus describes the response to linear strain. For a fluid, only the bulk modulus is meaningful. For an anisotropic solid such as wood or paper, these three moduli do not contain enough information to describe its behaviour, and one must use the full generalized Hooke's law.
Strictly speaking, the bulk modulus is a thermodynamic quantity, and it is necessary to specify how the temperature varies in order to specify a bulk modulus: constanttemperature (K_{T}), constantentropy (adiabatic K_{S}), and other variations are possible. In practice, such distinctions are usually only relevant for gases.
For a gas, the adiabatic bulk modulus K_{S} is approximately given by
and the isothermal bulk modulus K_{T} is approximately given by
where
In a fluid, the bulk modulus K and the density ρ determine the speed of sound c (pressure waves), according to the formula
Solids can also sustain transverse waves: for these materials one additional elastic modulus, for example the shear modulus, is needed to determine wave speeds.
For crystalline solids with a symmetry lower than cubic the bulk modulus is not the same in all directions and needs to be described with a tensor with more than one independent value. It is possible to study the tensor elements using powder diffraction under applied pressure.
Material  Bulk modulus in GPa  Bulk modulus in psi 

Glass (see also diagram below table)  35 to 55  5.8×10^{6} 
Steel  160  23×10^{6} 
Diamond^{[1]}  442  64×10^{6} 
Water  2.2×10^{9} Pa (value increases at higher pressures) 
Air  1.42×10^{5} Pa (adiabatic bulk modulus) 
Air  1.01×10^{5} Pa (constant temperature bulk modulus) 
Solid helium  5×10^{7} Pa (approximate) 

Conversion formulas  

Homogeneous isotropic linear elastic materials have their elastic properties uniquely determined by any two moduli among these, thus given any two, any other of the elastic moduli can be calculated according to these formulas.  
The bulk modulus (K) of a substance measures the substance's resistance to uniform compression. It is defined as the pressure increase needed to cause a given relative decrease in volume. Its base unit is the pascal.
As an example, suppose an iron cannon ball with bulk modulus 160 GPa is to be reduced in volume by 0.5%. This requires a pressure increase of 0.005×160 GPa = 0.8 GPa (116,000 psi).
Contents 
The bulk modulus K can be formally defined by the equation:
where P is pressure, V is volume, and ∂P/∂V denotes the partial derivative of pressure with respect to volume. The inverse of the bulk modulus gives a substance's compressibility.
Other moduli describe the material's response (strain) to other kinds of stress: the shear modulus describes the response to shear, and Young's modulus describes the response to linear strain. For a fluid, only the bulk modulus is meaningful. For an anisotropic solid such as wood or paper, these three moduli do not contain enough information to describe its behaviour, and one must use the full generalized Hooke's law.
Strictly speaking, the bulk modulus is a thermodynamic quantity, and it is necessary to specify how the temperature varies in order to specify a bulk modulus: constanttemperature (isothermal $K\_T$), constantentropy (adiabatic $K\_S$), and other variations are possible. In practice, such distinctions are usually only relevant for gases.
For a gas, the adiabatic bulk modulus $K\_S$ is approximately given by
K_S=\gamma\, P
and the isothermal bulk modulus $K\_T$ is approximately given by
K_T=P\,
where
In a fluid, the bulk modulus K and the density ρ determine the speed of sound c (pressure waves), according to the NewtonLaplace formula
Solids can also sustain transverse waves: for these materials one additional elastic modulus, for example the shear modulus, is needed to determine wave speeds.
It is possible to measure the bulk modulus using powder diffraction under applied pressure.
Material  Bulk modulus in GPa  Bulk modulus in psi 

Glass (see also diagram below table)  35 to 55  5.8×10^{6} 
Steel  160  23×10^{6} 
Diamond^{[1]}  442  64×10^{6} 
Water  2.2×10^{9} Pa (value increases at higher pressures) 
Air  1.42×10^{5} Pa (adiabatic bulk modulus) 
Air  1.01×10^{5} Pa (constant temperature bulk modulus) 
Solid helium  5×10^{7} Pa (approximate) 

Conversion formulas  

Homogeneous isotropic linear elastic materials have their elastic properties uniquely determined by any two moduli among these, thus given any two, any other of the elastic moduli can be calculated according to these formulas.  
$(\backslash lambda,\backslash ,G)$  $(E,\backslash ,G)$  $(K,\backslash ,\backslash lambda)$  $(K,\backslash ,G)$  $(\backslash lambda,\backslash ,\backslash nu)$  $(G,\backslash ,\backslash nu)$  $(E,\backslash ,\backslash nu)$  $(K,\backslash ,\; \backslash nu)$  $(K,\backslash ,E)$  $(M,\backslash ,G)$  
$K=\backslash ,$  $\backslash lambda+\; \backslash tfrac\{2G\}\{3\}$  $\backslash tfrac\{EG\}\{3(3GE)\}$  $\backslash tfrac\{\backslash lambda(1+\backslash nu)\}\{3\backslash nu\}$  $\backslash tfrac\{2G(1+\backslash nu)\}\{3(12\backslash nu)\}$  $\backslash tfrac\{E\}\{3(12\backslash nu)\}$  $M\; \; \backslash tfrac\{4G\}\{3\}$  
$E=\backslash ,$  $\backslash tfrac\{G(3\backslash lambda\; +\; 2G)\}\{\backslash lambda\; +\; G\}$  $\backslash tfrac\{9K(K\backslash lambda)\}\{3K\backslash lambda\}$  $\backslash tfrac\{9KG\}\{3K+G\}$  $\backslash tfrac\{\backslash lambda(1+\backslash nu)(12\backslash nu)\}\{\backslash nu\}$  $2G(1+\backslash nu)\backslash ,$  $3K(12\backslash nu)\backslash ,$  $\backslash tfrac\{G(3M4G)\}\{MG\}$  
$\backslash lambda=\backslash ,$  $\backslash tfrac\{G(E2G)\}\{3GE\}$  $K\backslash tfrac\{2G\}\{3\}$  $\backslash tfrac\{2\; G\; \backslash nu\}\{12\backslash nu\}$  $\backslash tfrac\{E\backslash nu\}\{(1+\backslash nu)(12\backslash nu)\}$  $\backslash tfrac\{3K\backslash nu\}\{1+\backslash nu\}$  $\backslash tfrac\{3K(3KE)\}\{9KE\}$  $M\; \; 2G\backslash ,$  
$G=\backslash ,$  $\backslash tfrac\{3(K\backslash lambda)\}\{2\}$  $\backslash tfrac\{\backslash lambda(12\backslash nu)\}\{2\backslash nu\}$  $\backslash tfrac\{E\}\{2(1+\backslash nu)\}$  $\backslash tfrac\{3K(12\backslash nu)\}\{2(1+\backslash nu)\}$  $\backslash tfrac\{3KE\}\{9KE\}$  
$\backslash nu=\backslash ,$  $\backslash tfrac\{\backslash lambda\}\{2(\backslash lambda\; +\; G)\}$  $\backslash tfrac\{E\}\{2G\}1$  $\backslash tfrac\{\backslash lambda\}\{3K\backslash lambda\}$  $\backslash tfrac\{3K2G\}\{2(3K+G)\}$  $\backslash tfrac\{3KE\}\{6K\}$  $\backslash tfrac\{M\; \; 2G\}\{2M\; \; 2G\}$  
$M=\backslash ,$  $\backslash lambda+2G\backslash ,$  $\backslash tfrac\{G(4GE)\}\{3GE\}$  $3K2\backslash lambda\backslash ,$  $K+\backslash tfrac\{4G\}\{3\}$  $\backslash tfrac\{\backslash lambda(1\backslash nu)\}\{\backslash nu\}$  $\backslash tfrac\{2G(1\backslash nu)\}\{12\backslash nu\}$  $\backslash tfrac\{E(1\backslash nu)\}\{(1+\backslash nu)(12\backslash nu)\}$  $\backslash tfrac\{3K(1\backslash nu)\}\{1+\backslash nu\}$  $\backslash tfrac\{3K(3K+E)\}\{9KE\}$ 
The bulk modulus (letter K) is a measure of the resistance of a substance to compression on all sides. Gases are easily compressed, while solids and liquid are only compressed with difficulty. It is measured by a measuring instrument when another object is placed under pressure.
