**Properties of**** ****Pure Substances**

**Properties of****Pure Substances**In this chapter the relationships between pressure, speciﬁc volume, and temperature will be presented for a pure substance. A pure substance is homogeneous, but may exist in more than one phase, with each phase having the same chemical composition. Water is a pure substance; the various combinations of its three phases (vapor, liquid, ice) have the same chemical composition. Air in the gas phase is a pure substance, but liquid air has a different chemical composition. Air is not a pure substance if it exists in more than one phase. In addition, only a *simple compressible substance*, one that is essentially free of magnetic, electrical, or surface tension effects, will be considered.

**2.1 ****The P-****v****-T Surface**

It is well known that a substance can exist in three different phases: solid, liquid, and gas. Assume that a solid is contained in a piston-cylinder arrangement such that

the pressure is maintained at a constant value; heat is added to the cylinder, causing the substance to experience all three phases, as in Fig. 2.1. We will record the temperature *T *and speciﬁc volume *v *during the experiment. Start with the solid at some low temperature, as in Fig. 2.2*a*; then add heat until it is all liquid (*v *does not increase very much). After all the solid is melted, the temperature of the liquid again rises until vapor just begins to form; this state is called the *saturated liquid *state. During the phase change from liquid to vapor,1 called *vaporization*, the temperature remains constant as heat is added. Finally, all the liquid is vaporized and the state of *saturated vapor *exists, after which the temperature again rises with heat addition. Note, the speciﬁc volumes of the solid and liquid are much less than the speciﬁc volume of vapor at relatively low pressures.

If the experiment is repeated a number of times using different pressures, a *T-**v *diagram results, shown in Fig. 2.2*b**. *At pressures that exceed the pressure of the *critical point*, the liquid simply changes to a vapor without a constant-temperature vaporization process. Property values of the critical point for various substances are included in Table B.3.

The experiment could also be run by holding the temperature ﬁxed and decreasing the pressure, as in Fig. 2.3*a *(the solid is not displayed). The solid would change to a liquid, and the liquid to a vapor, as in the experiment that led to Fig. 2.2. The *T**-**v *diagram, with only the liquid and vapor phases shown, is displayed in Fig. 2.3*b*.

The process of melting, vaporization, and *sublimation *(the transformation of a solid directly to a vapor) are shown in Fig. 2.3*c*. Distortions are made in all three diagrams so that the various regions are displayed. The *triple point *is the point where all three phases exist in equilibrium together. A constant pressure line is

**F****i****g****u****re 2.2 **The *T*-*v *diagram.

shown on the *T-**v *diagram and a constant temperature line on the *P-**v *diagram; one of these two diagrams is often sketched in problems involving a phase change from a liquid to a vapor.

Primary practical interest is in situations involving the liquid, liquid-vapor, and vapor regions. A saturated vapor lies on the *saturated vapor line *and a saturated liquid on the *saturated liquid line*. The region to the right of the saturated vapor line is the *supe**rheated region*; the region to the left of the saturated liquid line is the *compressed liquid region *(also called the *subcooled liquid region*)*. *A *supercritical state *is encountered when the pressure and temperature are greater than the critical values.

**F****i****g****u****re 2.3 **The (*a*) *P-**v*, (*b*) *T-**v*, and (*c*) *P-T *diagrams.

__2.2 The Liquid-Vapor Region__

__2.2 The Liquid-Vapor Region__

At any state (*T*, *v*) between saturated points *f *(state 1) and *g *(state 2), shown in Fig. 2.4, liquid and vapor exist in equilibrium. Let *v**f *and *v**g *represent, respectively, the speciﬁc volumes of the saturated liquid and the saturated vapor. Let *m *be the total mass of a system, *m*the amount of mass in the liquid phase, and *m *the amount of mass in the vapor phase. Then, for a state of the system represented by any (*T*, *v*), such as state 3, the total volume of the mixture is the sum of the volume occupied by the liquid and that occupied by the vapor, or

*V *= *V** _{f }*+

*V*

*or*

_{g }*m*

*v*=

*m*

_{f}*v*

*+*

_{f}*m*

_{g}*v*

*(2.1)*

_{g}The ratio of the mass of saturated vapor to the total mass is called the *quality *of the mixture, designated by the symbol *x*; it is

*x *= *m** _{g}/m *(2.2)

We often refer to the region under the saturation lines as the *quality region*, or the *mixture region*, or the *wet region*; it is the only region where quality *x *has a meaning.

Recognizing that *m *= *m** _{f }*+

*m*

*we may write Eq. (2.1), using our deﬁnition of quality, as*

_{g}*v *= *v ** _{f} *+

*x*(

*v*

*−*

_{g}*v*

*) (2.3)*

_{f}**F****i****g****u****re 2.4 **A *T**-v *diagram showing the saturated liquid and saturated vapor points.

Because the difference in saturated vapor and saturated liquid values frequently appears in calculations, we often let the subscript “*fg*” denote this difference; that is,

*v *_{fg}* *= *v** _{g }*−

*v*

*(2.4)*

_{f }Thus, Eq. (2.3) can be written as

*v *= *v _{ f}*

*+*

*x*

*v*

*(2.5)*

_{fg }The percentage liquid by mass in a mixture is 100(1 – *x*), and the percentage vapor is 100*x.*

__2.3 The Steam Tables__

__2.3 The Steam Tables__

Tabulations have been made for many substances of the thermodynamic properties *P*, *v*, and *T *and additional properties to be identiﬁed in subsequent chapters. Values are presented in the appendix in both tabular and graphical form. Table C.1 gives the saturation properties of water as a function of saturation temperature; Table C.2 gives these properties as a function of saturation pressure. The information contained in the two tables is essentially the same, the choice of which to use being a matter of convenience. We should note, however, that in the mixture region pressure and temperature are dependent. Thus, to establish the state of a mixture, if we specify the pressure, we need to specify a property other than temperature. Conversely, if we specify temperature, we must specify a property other than pressure.

Table C.3 lists the properties of superheated steam. To establish the state in the superheated region, it is necessary to specify two properties. While any two may be used, the most common procedure is to use *P *and *T *since these are easily measured. Thus, properties such as *v *are given in terms of the set of independent properties *P *and *T*.

Table C.4 lists data pertaining to compressed liquid. At a given *T *the speciﬁc volume *v *of of 100°C in Table C.1, the speciﬁc volume *v**f *of liquid is 0.001044 m /kg at a pressure of l00 kPa. At a pressure of 10 MPa, the speciﬁc volume is 0.001038 m /kg,

less than 1 percent decrease in speciﬁc volume. Thus, it is common in calculations to assume that *v *(and other properties, as well) of a compressed liquid is equal to *v**f *at the same temperature. Note, however, that *v**f *increases signiﬁcantly with temperature, especially at higher temperatures.

Table C.5 gives the properties of a saturated solid and a saturated vapor for an equilibrium condition. Note that the value of the speciﬁc volume of ice is relatively

insensitive to temperature and pressure for the saturated-solid line. Also, it has a greater value (almost 10 percent greater) than the minimum value on the saturated- liquid line.

Appendix D provides the properties of refrigerant R134a, a refrigerant used often in air conditioners and commercial coolers.

__EXAMPLE 2.1__

Determine the volume change when 10 kg of saturated water is completely vaporized at a pressure of (a) 1 kPa, (b) 260 kPa, and (c) 10 000 kPa.

__Solution__

Table C.2 provides the necessary values. The quantity being sought is Δ*V *= *m**v** _{fg }*where

*v*

*=*

_{fg}*v*

*–*

_{g}*v*

*. Note that*

_{f}*P*is given in MPa.

(a) 1 kPa = 0.001 MPa. Thus, *v** _{fg} *= 129.2 – 0.001 = 129.2 m

^{3}/kg.

.^{.}. Δ*V *= 1292 m^{3}

(b) At 0.26 MPa we must interpolate2 if we use the tables. The tabulated values are used at 0.2 MPa and 0.3 MPa:

The value for *v** _{f} *is, to four decimal places, 0.0011 m

^{3}/kg at 0.2 MPa and at 0.3 MPa; hence, no need to interpolate for

*v*

*. We then have*

_{f}*v *_{fg}* *= 0.718 − 0.0011 = 0.717 m^{3} / kg. ∴ Δ*V *= 7.17 m^{3}

(c) At 10 MPa, *v** _{fg} *= 0.01803 − 0.00145 = 0.01658 m

^{3}/kg so that

Δ*V *= 0.1658 m^{3}

Notice the large value of *v** _{fg} *at low pressure compared with the small value of

*v*

*fg*as the critical point is approached. This underscores the distortion of the

^{ufg}

*T-*

*v*diagram in Fig. 2.4.

__EXAMPLE 2.2__

Four kilograms of water are placed in an enclosed volume of 1 m3. Heat is added until the temperature is 150°C. Find the (a) pressure, (b) mass of the vapor, and (c) volume of the vapor.

__Solution__

Table C.1 is used. The volume of 4 kg of saturated vapor at 150°C is 0.3928 × 4 = 1.5712 m^{3}. Since the given volume is less than this, we assume the state to be in the quality region.

(a) In the quality region, the pressure is given as *P *= 475.8 kPa, the value next to the temperature.

(b) To ﬁnd the mass of the vapor we must determine the quality. It is found from Eq. (2.3), using *v *= 1/4 = 0.25 m^{3}/kg:

*v *= *v ** _{f }*+

*x*(

*v*

*−*

_{g }*v*

*)*

_{f}**0.25 = 0.00109 + x(0.3928 − 0.00109) ∴ x = 0.6354**

Using Eq. (2.2), the vapor mass is

*m** _{g} *=

*m*

*x*= 4 × 0

*.*6354 = 2

*.*542 kg

(c) Finally, the volume of the vapor is found from

*V** _{g} *=

*v*

_{g}*m*

*= 0.3928 × 2.542 = 0.998 m*

_{g}^{3}

Note: In mixtures where the quality is not close to zero, the vapor phase occupies most of the volume. In this example, with a quality of 63.5 percent, it occupies 99.8 percent of the volume.

__EXAMPLE 2.3__

Two kilograms of water are heated at a pressure of 220 kPa to produce a mixture with quality *x *= 0.8. Determine the ﬁnal volume occupied by the mixture.

*Solution*

Use Table C.2. To determine the appropriate values at 220 kPa, linearly interpolate between 0.2 and 0.25 MPa. This provides, at 220 kPa,

Note, no interpolation is necessary for *v**f *since, for both pressures, *v**f *is the same to four decimal places. Using Eq. (2.3), we now ﬁnd

The total volume occupied by 2 kg is

*V *= *m**v *= 2 kg × 0*.*662 m^{3 }/kg = 1.32 m^{3}

Two kilograms of water are contained in a constant-pressure cylinder held at 2.2 MPa. Heat is added until the temperature reaches 800°C. Determine the ﬁnal volume of the container.

__Solution__

Use Table C.3. Since 2.2 MPa lies between 2 MPa and 2.5 MPa, the speciﬁc volume is interpolated to be

*v *= 0.2467 + 0.4(0.1972 − 0.2467) = 0.227 m^{3} / kg

The ﬁnal volume is then

*V *= *m**v *= 2 × 0.227 = 0.454 m^{3}

The linear interpolation above results in a less accurate number than the numbers in the table. So, the ﬁnal number has fewer signiﬁcant digits.

__2.4 Equations of State__

__2.4 Equations of State__

When the vapor of a substance has relatively low density, the pressure, speciﬁc volume, and temperature are related by an *equation of state*,

*P**v *= *R**T *(2.6)

where, for a particular gas, the *gas constant *is *R. *A gas for which this equation is valid is called an *ideal gas *or sometimes a *perfect gas. *Note that when using the above equation of state the pressure and temperature must be expressed as absolute quantities.

The gas constant *R *is related to a *universal gas constant **R*, which has the same value for all gases, by the relationship

where *M *is the *molar mass*, values of which are tabulated in Tables B.2 and B.3. The *mole *is that quantity of a substance (i.e., that number of atoms or molecules) having a mass which, measured in grams, is numerically equal to the atomic or molecular weight of the substance. In the SI system it is convenient to use instead the kilomole (kmol), which amounts to *x *kilograms of a substance of molecular weight *x. *For instance, 1 kmol of carbon (with a molecular weight of 12) is a mass of 12 kg; 1 km__ol __of oxygen is 32 kg. Stated otherwise, *M *= 32 kg/kmol for O2.

The value *R *= 8.314 kJ/kmol·K. For air *M *is 28.97 kg/kmol, so that for air *R *is 1.287 kJ/ kg·K, or 287 J/kg·K, a value used extensively in calculations involving air.

Other forms of the ideal-gas equation are

*PV *= *mRT P *= *p**RT PV *= *n R T *(2.8)

where *n *is the number of moles.

Care must be taken in using this simple convenient equation of state. A low- density can result from either a low pressure or a high temperature. For air, the ideal- gas equation is surprisingly accurate for a wide range of temperatures and pressures; less than 1 percent error is encountered for pressures as high as 3000 kPa at room temperature, or for temperatures as low as −130°C at atmospheric pressure.

The *compressibility factor *Z helps us in determining whether or not the ideal-gas equation can be used. It is deﬁned as

and is displayed in Fig. 2.5 for nitrogen. This ﬁgure is acceptable for air also, since air is composed mainly of nitrogen. If *Z *= 1, or very close to 1, the ideal-gas equation can be used. If *Z *is not close to 1, then Eq. (2.9) may be used.

The compressibility factor can be determined for any gas by using a generalized compressibility chart presented in App. G. In the generalized chart the *reduced **pressure P *and *reduced temperature T *must be used. They are calculated from

**F****i****g****u****re 2.5 **The compressibility factor.

__EXAMPLE 2.5__

An automobile tire with a volume of 0.6 m3 is inﬂated to a gage pressure of 200 kPa. Calculate the mass of air in the tire if the temperature is 20°C using the ideal-gas equation of state.

__Solution__

Air is assumed to be an ideal gas at the conditions of this example. In the ideal-gas equation, *PV *= *mRT*, we use absolute pressure and absolute temperature. Thus, using *P*_{atm} = 100 kPa (to use a pressure of 101 kPa is unnecessary; the difference of 1 percent is not signiﬁcant in most engineering problems):

The mass is then calculated to be

Be careful to be consistent with the units in the above equation.