2. Bulk modulus
Bulk modulus is defined by
\[
K=-\frac{\Delta P}{\Delta V/V_0}.
\]
The negative sign is included because a pressure increase usually causes a volume decrease:
\[
\Delta P>0,
\qquad
\Delta V<0.
\]
With the negative sign, \(K\) is positive for stable materials.
3. Compressibility
Compressibility is the reciprocal of bulk modulus:
\[
\beta=\frac{1}{K}.
\]
A large bulk modulus means low compressibility. A small bulk modulus means the material volume changes more easily.
4. Volume change formula
Starting from
\[
K=-\frac{\Delta P}{\Delta V/V_0},
\]
solve for volume change:
\[
\frac{\Delta V}{V_0}=-\frac{\Delta P}{K}.
\]
\[
\boxed{
\Delta V=-\frac{V_0\Delta P}{K}
}.
\]
For a pressure increase, this gives a negative \(\Delta V\), meaning the material volume decreases.
5. Pressure change formula
If volume change and bulk modulus are known, pressure change is
\[
\boxed{
\Delta P=-K\frac{\Delta V}{V_0}
}.
\]
A negative volume change produces a positive pressure increase.
6. Final volume
Final volume is found from
\[
V_f=V_0+\Delta V.
\]
For compression,
\[
V_f=V_0-|\Delta V|.
\]
For expansion,
\[
V_f=V_0+|\Delta V|.
\]
7. Units
Bulk modulus has the same units as pressure because volumetric strain is dimensionless:
\[
[K]=\mathrm{Pa}.
\]
Common units include
\[
1\ \mathrm{MPa}=10^6\ \mathrm{Pa},
\qquad
1\ \mathrm{GPa}=10^9\ \mathrm{Pa}.
\]
Compressibility has inverse pressure units:
\[
[\beta]=\mathrm{Pa^{-1}}.
\]
8. Typical bulk modulus values
| Material |
Approximate bulk modulus |
Meaning |
| Water |
\(2.2\ \mathrm{GPa}\) |
Low compressibility compared with gases |
| Seawater |
\(2.3\ \mathrm{GPa}\) |
Slightly less compressible than pure water |
| Hydraulic oil |
\(1{-}2\ \mathrm{GPa}\) |
Compressibility matters in hydraulic systems |
| Glass |
\(\sim35\ \mathrm{GPa}\) |
Much harder to compress than water |
| Aluminium |
\(\sim76\ \mathrm{GPa}\) |
High resistance to volume compression |
| Steel |
\(\sim160\ \mathrm{GPa}\) |
Very high bulk stiffness |
9. Pressure–volume graph
The bulk modulus relation can be written as
\[
\Delta P=-K\varepsilon_V.
\]
So a graph of \(\Delta P\) against \(\varepsilon_V=\Delta V/V_0\) is a straight line with slope \(-K\).
Compression appears in the quadrant where \(\Delta P>0\) and \(\Delta V/V_0<0\).
10. Liquids vs solids
Liquids usually have much lower bulk modulus than metals, so their volumes change more under pressure. Even so,
water and hydraulic oil are far less compressible than gases.
Solids such as steel and aluminium have high bulk modulus values, meaning very large pressures are needed to
produce even small fractional volume changes.
11. Worked example: water compression
Suppose water has
\[
V_0=2.0\ \mathrm{m^3},
\qquad
\Delta P=1.5\ \mathrm{MPa},
\qquad
K=2.2\ \mathrm{GPa}.
\]
Convert pressure and modulus:
\[
\Delta P=1.5\times10^6\ \mathrm{Pa},
\qquad
K=2.2\times10^9\ \mathrm{Pa}.
\]
Use
\[
\Delta V=-\frac{V_0\Delta P}{K}.
\]
\[
\Delta V
=
-\frac{(2.0)(1.5\times10^6)}{2.2\times10^9}.
\]
\[
\Delta V=-1.36\times10^{-3}\ \mathrm{m^3}.
\]
Since
\[
1\ \mathrm{m^3}=1000\ \mathrm{L},
\]
the volume decrease is
\[
|\Delta V|=1.36\ \mathrm{L}.
\]
The final volume is
\[
V_f=2.0-0.00136=1.99864\ \mathrm{m^3}.
\]
Key idea: bulk modulus measures resistance to volume change, while compressibility measures how much fractional
volume change occurs per unit pressure.
