COMPRESSIBILITY: SOLIDS, LIQUIDS, GASES, EXAMPLES - PHYSICAL - 2025

The compressibility of a substance or material is the change in volume that it experiences when it is subjected to a change in pressure. In general, the volume decreases when pressure is applied to a system or object. However, sometimes the opposite occurs: a change in pressure can produce 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 with increasing frequency of collisions, repulsive forces take place.

A submarine experiences compression forces while submerged. Source: pixabay.com.

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

A solid has its molecules very close together, and when trying to bring them closer together, repulsive forces appear that make the task difficult. Therefore it is said that solids are not very compressible. In the molecules of liquids 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. A very large pressure variation would be required to achieve an appreciable volume change under so-called normal pressure and temperature conditions. On the other hand, gases, as their molecules are widely spaced, are easily compressed and decompressed.

Solid compressibility

When an object is immersed 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:

The force exerted by the fluid on the submerged object is perpendicular to the surface. Source: Wikimedia Commons.

Pressure is defined as force per unit area, which will cause a volume change ΔV proportional to the initial volume of the object V _o. This change in volume 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 proportionality constant, which is called the volumetric modulus of the material:

B = -Stress / Strain

B = -ΔP / (ΔV / V _o)

Since ΔV / V _o 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 modulus is known as compressibility and is denoted by the letter k. Thus:

Here k is the negative of the fractional change in volume per increase in pressure. Its units in the International System are the inverses of Pa, that is, m ² / N.

The equation for B or for k if you prefer, is applicable to both solids and liquids. The volumetric modulus concept is rarely applied to gases. A simple model is explained below to quantify the volume decrease 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:

Solved exercises-examples

-Solved exercise 1

A solid brass sphere whose volume is 0.8 m ³ is dropped into the ocean to a depth where the hydrostatic pressure is 20 M Pa greater than at the surface. How will the volume of the sphere change? It is known that the modulus of compressibility of brass is 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 / V _o)

Thus:

ΔV = -5.71.10 ^-4 x 0.8 m ³ = -4.57 x 10 ^-4 m ³

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

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

-Solved exercise 2

By putting your ear against the railroad tracks, you can tell when one of these vehicles is approaching in the distance. How long does it take for sound to travel on a steel rail if the train is 1 km away?

Data

Steel density = 7.8 x 10 ³ kg / m3

Steel compressibility modulus = 2.0 x 10 ¹¹ Pa.

Solution

The modulus of compressibility B calculated above applies to liquids as well, although a great deal of effort is generally 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 ° C and one atmosphere 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 verified immediately with numerical application.

-Solved exercise 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 find out what volume n moles of a given gas have when it is kept confined to a pressure P and a temperature T, we use the equation of state. In the equation of state for an ideal gas, where intermolecular forces are not taken into account, the simplest model states that:

_Ideal PV = n. R. T

Where R is the ideal gas constant.

Changes in gas volume can take place at constant pressure or constant temperature. For example, keeping the temperature constant, the isothermal compressibility Κ _T is:

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

Therefore B _T the isothermal compressibility modulus is:

And also the adiabatic B _adiabatic compressibility modulus is important, for which there is no incoming or outgoing heat flow.

B _adiabatic = γp

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

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

Data

The adiabatic compressibility modulus of air is 1.42 × 10 ⁵ Pa

The density of air is 1,225 kg / m ³ (at atmospheric pressure and 15 ºC)

Solution

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

Where Z is the gas compressibility coefficient, 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 gaseous molecules collide more frequently and the repulsive forces between them increase. This can lead to an increase in volume in the real gas, whereby Z> 1.

In contrast, at lower pressures, the molecules are free to move and attractive forces predominate. In this case, Z <1.

For the simple case of 1 mole of gas n = 1, if the same pressure and temperature conditions are maintained, by dividing the previous equations term by term, we obtain:

-Solved exercise 5

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

a) The compressibility factor.

b) The molar volume of the real gas.

c) What types of forces predominate: attractive or repulsive?

Solution

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

V _real = 0.88 V _ideal

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 supplied data:

R = 0.082 L.atm / mol.K

The molar volume is calculated by solving and substituting values:

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

References

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