ORÍGENES Y EVOLUCIÓN DE LA EXCLUSIÓN SOCIAL

(The relationship between pressure, volume and temperature in gases.)

A gas can be thought of as behaving in the same way as a collection of rigid, perfectly elastic spheres in constant motion that interact with each other and with the walls of the container.

Using this concept it is possible to deduce equations for behaviour, and this idealized representation (known as a perfect gas) results in the perfect gas equation of state, which holds true over a wide range of pressures and temperatures commonly experienced in

engineering. At low pressures and high temperatures the approximation is not so good. Another name for a perfect gas is an ideal gas.

The perfect gas equation is derived by considering Boyle’s law and Charles’s law. Boyle’s law states that, when temperature is constant, the volume occupied by a gas is inversely

A sealed container is completely filled with a gas at an absolute pressure of 3370 mbar. Air pressure outside the container is 1.1 bar. What is the gauge pressure of the gas? The container is then taken to the top of a mountain where air pressure is 0.9 bar. If the container does not deform and the ambient temperature does not change, what is the gauge pressure of the gas now? Why would the answer be different if the ambient temperature changed?

First, express both pressures in the same units:

3370 mbar is 3.37 bar Gauge pressure absolute pressure  atmospheric pressure:

Pgauge, ground Pabs, ground Pat, ground

Pgauge, ground 3.37  1.1  2.27 bar

At ground level the gauge pressure of the gas is 2.27 bar. This means that it has a pressure of 2.27 bar more than atmospheric.

When the container is taken to the top of the mountain, the pressure of the gas inside it remains the same. This is because the container is sealed and so there is the same mass of gas inside. The container does not deform and so has the same volume. If the temperature does not change either, then the gas inside the container will be in exactly the same state as it was at the ground, and its absolute pressure is therefore still 3.37 bar.

Atmospheric pressure has changed, however, and so there is a new gauge pressure:

Pgauge, mountain Pabs, mountain Pat, mountain

Pgauge, mountain 3.37  0.9  2.47 bar

If the temperature of the gas changes, then its pressure will change too. The change in pressure can be calculated using the perfect gas law (see later in this section).

proportional to its pressure, and Charles’s law states that, when pressure is constant, the volume occupied by a gas is proportional to its absolute temperature (in kelvin). Together, these give

p T

V  constant ^(3.7)

The constant depends on the quantity of gas, and Avogadro’s hypothesis states that equal volumes of all gases at a given temperature and pressure contain the same number of molecules.

Therefore, at a particular temperature and pressure, a certain number of atoms or molecules of any gas will occupy the same volume. For engineering calculations the quantity of atoms or molecules usually referred to is the kilomole, usually written as kmol. A kilomole of molecules always occupies the same volume at a particular temperature and pressure, and this volume is 22.41 m³at the reference temperature and pressure (0 °C and 1.013 25 bar; these conditions are known as standard temperature and pressure, or STP). Thus, for 1 kmol of gas, the constant in equation (3.7) is 8314 J kmol^1K^1, and it is known as the molar or universal gas constant and given the symbol R⬃.

Thus, in general, for n kmol of gas, equation (3.7) becomes

pV nR^⬃T (3.8)

The mass of 1 kmol of a substance depends on the molar mass, and this is determined from the atomic weights of the elements in the molecules. For example, the molar mass of hydrogen gas (H₂), in which there are two atoms per molecule, is 2.0 kg kmol^1. For oxygen gas (O₂) the molar mass is 32 kg kmol^1. The number of moles of a quantity of a substance, n, is its mass divided by the molar mass. So

n Substituting this relationship into equation (3.8) gives:

pV mRT ^(3.9)

where

R

This is known as the specific gas constant and is dependent on the particular gas under

consideration. For air, which is approximately 79 per cent nitrogen and 21 per cent oxygen, the average molar mass is 29 kg kmol^1, and so the specific gas constant is 287.1 J kg^1K^1. (It is worth noting that many people think that use of the word specific in the term specific gas constantmeans that the constant is specific to a particular gas, which it is. However, it is really used in its other meaning, where specific quantities are quantities per unit mass; and the specific gas constant has units J kg^1K^1and so refers to a kilogram of a particular gas.)

Dividing equation (3.9) by the mass of gas gives

pv RT ^(3.10)

where v (lower case) is the specific volume, or the volume of 1 kg of gas. It has the units m³kg^1. It is the reciprocal of density  (which has the units kg m^3). Therefore, an alternative form of the perfect gas equation is

p RT ^(3.11)

Therefore, the density of a gas can conveniently be determined from pressure and temperature as

  ^(3.12)

In using the various forms of the perfect gas equation, or the equation of state for a gas, as it is otherwise termed, it is important to remember to use the correct units: temperature must be expressed as absolute temperature in kelvin, and pressure must be expressed as absolute pressure in N m^2or Pa.

p RT R⬃

⬃m

m m⬃

The perfect gas equation only applies to gases at reasonable pressures and temperatures found in engineering. It does not apply to liquids, which for most cases can be assumed to be

incompressible, so the density does not vary significantly with pressure or temperature.

Compressibility

In general, a pressure increase in a substance causes a reduction in the volume occupied by a given mass; thus the density increases. For solids and liquids the reduction in volume is very small but it is important in some cases. For gases, even a small change in pressure can cause a significant change in volume and therefore density.

The equation that relates the change in volume to the change in pressure is given below, where K is the bulk modulus of elasticity:

p  K

V

V ^(3.13)

Remember temperatures must be in kelvin in the perfect gas equation.

To convert from °C to K, add 273.15.

What is the specific gas constant for oxygen gas?

Oxygen is normally present in the atmosphere as molecules of two atoms (O₂). The atomic mass of oxygen is 16 kg kmol^1, so the molar mass of molecular gaseous oxygen is 32 kg kmol^1.

R_oxygen  83 3 1 2

  259.8 J kg4 ^1K^1 R⬃

m⬃

oxygen

A helium-filled weather balloon is to expand to a sphere of 20 m diameter at a height of 30 km where the absolute pressure is 1200 Pa and the temperature is

47 °C. If there is to be no stress in the fabric of the balloon, what volume of helium must be added at ground level where the absolute pressure is 101.3 kPa and the temperature is 15 °C?

There are two key points to note here: first, if there is no stress in the fabric of the balloon, then the pressure inside the balloon is equal to the pressure outside. Secondly, there is the same quantity (or mass) of hydrogen in the balloon at ground level as at 30 km.

Apply

pV mRT The parameters m and R are constant, so

冢^pTV冣^ground冢^pTV冣^{30 km}

and

V_ground冢^ppg 3

r 0 ou

k n m

d

冣冢^TTg3 r 0

ou k n m

d冣^{V30 km}

冢^1011203000冣冢^{227733..11551457}冣冢^2603冣

 63.2 m³

K does vary slightly with pressure, but, over ranges where it can be regarded as constant, we can write

p  K V

V

For solids, K is of the order of 10¹¹Pa, and for liquids it is of the order of 10⁹Pa (increasing with pressure). For example, K for water is 2 10⁹Pa, and therefore a pressure increase of 100 bar will only cause a volume reduction of 0.5 per cent:

 V

V   K

  p 10 2

0

 1

1 0 0

9

  0.0055

It is for this reason that liquids are generally regarded as incompressible and will be treated as such in this chapter.

For gases, K is approximately equal to the pressure of the gas, provided that the pressure is not excessively high. Thus, for a gas at 1 bar, K is of the order of 1 bar, and a 0.1 per cent

volumetric strain only requires a pressure change of approximately 0.1 per cent of 1 bar (100 Pa). Gases can only really be treated as incompressible if the pressure changes involved are very small. However, in this chapter only fluids where the behaviour can be considered to be incompressible are dealt with. In the chapter of this book describing thermodynamics, the behaviour of gases subject to large pressure changes is considered.