The **compressibility** of a substance or material is the change in volume that it experiences when it is subjected to a change in pressure. Usually the volume decreases when applying a pressure on a system or object. However, sometimes the opposite occurs: a change in pressure can cause an explosion in which the system increases in volume, or when a phase change occurs.

In some chemical reactions this can happen and in gases as well, since by increasing the frequency of collisions, repulsive forces take place.

When imagining how easy or difficult it can be to compress an object, consider the three states that matter is normally in: solid, liquid, and gas. In each of them the molecules keep certain distances from each other. The stronger the bonds that unite the molecules of the substance that composes the object and the closer they are, the more difficult it will be to cause a deformation.

A solid has its molecules very close, and when trying to bring them closer together, repulsive forces appear that make the task difficult. For this reason it is said that solids are not very compressible. In liquid molecules there is more space, so their compressibility is greater, but even so the change in volume usually requires large forces.

So solids and liquids are hardly compressible. It would take a very large pressure variation to achieve an appreciable volume change under so-called normal conditions of pressure and temperature. On the other hand, gases, since their molecules are widely spaced, are easily compressed and decompressed.

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**solid compressibility**

When an object is submerged in a fluid, for example, it exerts pressure on the object in all directions. In this way we can think that the volume of the object will decrease, although in most cases this will not be appreciable.

The situation can be seen in the following figure:

Pressure is defined as force per unit area, which will cause a change in volume ΔV proportional to the initial volume of the object Vo. This volume change will depend on its qualities.

Hooke’s law states that the deformation experienced by an object is proportional to the stress applied to it:

Stress ∝ Strain

The volumetric deformation experienced by a body is quantified by B the required constant of proportionality, which is called the *bulk modulus of material*:

B= -Effort/unit strain

B = -ΔP/ (ΔV/Vo)

Since ΔV/Vo is a dimensionless quantity, since it is the quotient between two volumes, the volumetric module has the same units of pressure, which in the International System are Pascals (Pa).

The negative sign indicates the expected reduction in volume, when the object is compressed enough, that is, the pressure increases.

**-Compressibility of a material**

The inverse or reciprocal value of the volumetric module is known as *compressibility *and is denoted by the letter* k. *Therefore:

Here *what* is the negative of the fractional change in volume per increase in pressure. Its units in the International System are the inverse of Pa, that is, m2 /N.

The equation for B or for k if you prefer, is applicable to both solids and liquids. The concept of bulk modulus is rarely applied to gases. A simple model is explained below to quantify the decrease in volume that a real gas can experience.

**The speed of sound and the modulus of compressibility**

An interesting application is the speed of sound in a medium, which depends on its compressibility modulus:

Where c is the speed of sound, B is the volumetric modulus and Ρ is the density of the medium.

**Solved exercises-examples**

**-Exercise solved 1**

A solid brass sphere whose volume is 0.8 m3 is dropped into the ocean to a depth at which the hydrostatic pressure is 20 M Pa greater than at the surface. What change will the volume of the sphere experience? The modulus of compressibility of brass is known to be *B. *= 35,000 MPa,

**Solution**

1 M Pa = 1 Mega Pascal = 1. 10 6 Pa

The pressure variation with respect to the surface is DP = 20 x 10 6 Pa. Applying the equation given for B, we have:

B = -ΔP/ (ΔV/Vo)

Therefore:

So:

ΔV = -5.71.10 -4 x 0.8 m3 = -4.57 x 10-4 m3

The volume difference can have a negative sign when the final volume is less than the initial volume, so this result agrees with all the assumptions we have made so far.

Such a high modulus of compressibility indicates that a large change in pressure is required for the object to experience an appreciable decrease in volume.

**-Exercise solved 2**

Putting your ear against the train tracks, you know when one of these vehicles is approaching in the distance. How long does it take for sound to travel along a steel rail if the train is 1 km away?

**Data**

Steel density = 7.8 x 10 3 kg/m3

Modulus of compressibility of steel = 2.0 x 10 11 Pa.

**Solution**

**liquid compressibility**

The modulus of compressibility B calculated above applies to liquids as well, although great effort is usually required to produce an appreciable decrease in volume. But fluids can expand or contract as they heat up or cool down, and equally if they are depressurized or pressurized.

For water under standard conditions of pressure and temperature (0 ºC and one atmosphere of pressure approximately or 100 kPa), the volumetric modulus is 2100 MPa. That is, about 21,000 times atmospheric pressure.

For this reason, in most applications, liquids are usually considered incompressible. This can be checked immediately with numerical application.

**-Exercise solved 3**

Find the fractional decrease in the volume of water when it is subjected to a pressure of 15 MPa.

**Solution**

**Compressibility in gases **

** **Gases, as explained above, work a little differently.

To know what volume they have *no* moles of a given gas when confined to a pressure *P* already a temperature *you*, the equation of state is used. In the equation of state for an ideal gas, where intermolecular forces are not taken into account, the simplest model shows that:

** ***P.Videal = n. R.T*

where R is the ideal gas constant.

Changes in the volume of the gas can be carried out at constant pressure or at constant temperature. For example, holding the temperature constant, the isothermal compressibility ΚT is:

Instead of the symbol «delta» that was used before when defining the concept for solids, for a gas it is described with a derivative, in this case a partial derivative with respect to P, keeping T constant.

Therefore *BT* the isothermal modulus of compressibility is:

*BT = p*

And Badiabatic adiabatic modulus of compressibility is also important, for which there is no incoming or outgoing heat flow.

Badiabatic = *γp*

where γ is the adiabatic coefficient. With this coefficient you can calculate the speed of sound in air:

**-Exercise solved 4**

Applying the above equation, find the speed of sound in air.

**Data**

The adiabatic modulus of compressibility of air is 1.42×105 Pa

The density of air is 1.225 kg/m3 (at atmospheric pressure and 15 ºC)

**Solution**

**The compressibility factor Z**

Instead of working with the modulus of compressibility, such as unit volume change per pressure change, it may be interesting to *compressibility factor of a real gas*a different but illustrative concept of how the real gas compares to the ideal gas:

*Q. Vreal = ZR T*

Where Z is the coefficient of compressibility of the gas, which depends on the conditions in which it is found, being generally a function of both the pressure P and the temperature T, and can be expressed as:

** ***Z = f(P,T)*

In the case of an ideal gas Z = 1. For real gases the Z value almost always increases with pressure and decreases with temperature.

As the pressure increases, the gas molecules collide more frequently and the repulsive forces between them increase. This can lead to an increase in volume of the real gas, so Z > 1.

On the other hand, at lower pressures, the molecules are free to move and the attractive forces predominate. In such a case, Z < 1.

For the simple case of 1 mol of gas n = 1, if the same pressure and temperature conditions are maintained, dividing the above equations term by term gives:

It is concluded that:

*Vreal = Z Videal*

**-Exercise solved 5**

There is a real gas at 250 ºK and 15 atm pressure, which has a molar volume 12% less than that calculated by the ideal gas equation of state. If the pressure and temperature are held constant, find:

a) The compressibility factor.

b) The molar volume of the real gas.

c) What kind of forces dominate: attractive or repulsive?

**Solution**

a) If the actual volume is 12% less than the ideal, it means that:

Vreal = 0.88 Videal

Therefore for 1 mole of gas, the compressibility factor is:

Z = 0.88

b) Choosing the ideal gas constant with the appropriate units for the data supplied:

R = 0.082 L.atm/mol.K

The molar volume is calculated by isolating and substituting values:

c) Attractive forces predominate, since Z is less than 1.

**References**

Atkins, P. 2008. Physical Chemistry. Panamerican Medical Editorial. 10 – 15. Giancoli, D. 2006. Physics: Principles with Applications. 6th. Ed Prentice Hall. 242 – 243 and 314-15 Mott, R. 2006. Fluid Mechanics. Pearson Education.13-14. Rex, A. 2011. Fundamentals of Physics. Pearson Education. 242-243. Tipler, P. (2006) Physics for Science and Technology. 5th Ed. Volume 1. Editorial Reverté. 542.