In some chemical reactions, this can happen and also in gases, because as the frequency of collisions increases, repulsive forces are taking place.
When imagining how easy or difficult it can be to compress an object, you need to consider the three states that matter normally is: solid, liquid, and gas. In each of them, the molecules maintain 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 is to cause 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, solids are said to be poorly compressible. In liquid molecules, there is more space, so their compressibility is greater, but even so the change in volume usually requires large forces.
Therefore, solids and liquids are hardly compressible. A very large pressure variation would be needed to obtain a noticeable change in volume under so-called normal pressure and temperature conditions. On the other hand, gases, like widely spaced molecules, are easily compressed and decompressed.
Compressibility of solids
When an object is immersed in a fluid, for example, it presses 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 is not 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 change in volume ΔV proportional to the initial volume of the object V o . This volume change will depend on your qualities.
Hooke’s law states that the deformation experienced by an object is proportional to the effort applied to it:
Effort ∝ Deformation
The volumetric deformation experienced by a body is quantified by B, the necessary constant of proportionality, which is called the volumetric modulus of the material :
B = – Unit deformation / deformation
B = -ΔP / (ΔV / V o )
As ΔV / V or is a dimensionless quantity, since it is the ratio between two volumes, the volumetric module has the same pressure units, which in the International System are Pascals (Pa).
The minus sign indicates the expected reduction in volume when the object is sufficiently compressed, ie the pressure increases.
-Compressibility of a material
The inverse or reciprocal value of the volumetric modulus is known as compressibility and is indicated by the letter k. Therefore:
Here k is the negative of the fractional change in volume due to an increase in pressure. Units in the International System are the inverse of Pa, which is m 2 / N.
The equation for B or for k, if preferred, is applicable to both solids and liquids. The volumetric module concept is rarely applied to gases. A simple model for quantifying the decrease in volume that a real gas can experience is explained below.
The speed of sound and the compressibility module
An interesting application is the speed of sound in a medium, which depends on its compressibility module:
Exercise -Fixed 1
A sphere of solid brass whose volume is 0.8 m 3 is launched into the ocean at a depth where the hydrostatic pressure is 20 M Pa greater than at the surface. What change will the volume of the sphere experience? It is known that the compressibility modulus of brass is B = 35,000 MPa,
1 M Pa = 1 Mega pascal = 1. 10 6 Pa
The pressure variation relative to the surface is DP = 20 x 10 6 Pa. Applying the equation given to B, you have:
B = -ΔP / (ΔV / V o )
ΔV = -5.71.10 -4 x 0.8 m 3 = -4.57 x 10 -4 m 3
The difference in volume can have a negative sign when the final volume is less than the initial volume, therefore, this result is in accordance with all the assumptions we made so far.
Such a high compressibility modulus indicates that a large change in pressure is needed for the object to experience a considerable decrease in volume.
– Exercise solved 2
Placing the ear on the train tracks is known when one of these vehicles approaches from a distance. How long does sound take when traveling on a steel rail if the train is 1 km away.
Steel density = 7.8 x 10 3 kg / m3
Steel compressibility module = 2.0 x 10 11 Pa.
The compressibility modulus B calculated above is also applied to liquids, although it usually takes a great deal of effort to produce an appreciable decrease in volume. But fluids can expand or contract as they heat or cool, and also if they are depressurized or pressurized.
For water under standard conditions of pressure and temperature (0°C and an atmosphere pressure of approximately 100 kPa), the volumetric module is 2100 MPa. That’s about 21000 times atmospheric pressure.
Therefore, in most applications, liquids are generally considered to be incompressible. This can be verified immediately with numerical application.
– Exercise solved 3
Find the fractional decrease in water volume when subjected to a pressure of 15 MPa.
Compressibility in gases
Gases, as explained above, work a little differently.
To find out what volume they have n moles of a given gas when it is kept confined to a pressure P and a temperature T , 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 states that:
ideal PV = n. RT
Where R is the ideal gas constant.
Changes in gas volume can be carried out at constant pressure or at 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 of solids, a gas is described with a derivative, in this case a partial derivative with respect to P, keeping T constant.
Therefore B t isothermal compressibility modulus is:
And the adiabatic B- adiabatic compressibility modulus is also important, for which there is no incoming or outgoing heat flow.
Adiabatic B = γp
Where γ is the adiabatic coefficient. With this coefficient, you can calculate the speed of sound in air:
Applying the above equation, find the speed of sound in air.
The adiabatic air compressibility modulus is 1.42 × 10 5 Pa
The density of the air is 1,225 kg / m 3 (at atmospheric pressure and 15 °C)
Instead of working with the compressibility module, as a unit volume change per pressure change, the compressibility factor of a real gas can be interesting, a different but illustrative concept of how real gas compares to ideal gas.
P.V real = ZRT
Where Z is the compressibility coefficient of the gas, which depends on the conditions in which it is found, and is generally a function of pressure P and temperature T, which can be expressed as:
Z = f (P, T)
In the case of an ideal gas Z = 1. For real gases, the value of Z almost always increases with pressure and decreases with temperature.
As the pressure increases, the gas molecules collide more often and the repulsive forces between them increase. This can lead to an increase in actual gas volume, whereby Z> 1.
On the other hand, at lower pressures, molecules are free to move and attractive forces predominate. In this 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, we obtain:
real V = ideal ZV
– Exercise solved 5
There is a real gas at pressures of 250 ºK and 15 atm, with a molar volume 12% smaller 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 real gas.
c) What kind of forces predominate: attractive or repulsive?
a) If the actual volume is 12% less than the ideal, it means that:
Real V = 0.88 Ideal V
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 provided:
R = 0.082 L.atm / mol.K
Molar volume is calculated by clearing and replacing values:
c) Attractive forces predominate, since Z is less than 1.