# Properties of a Pure Substance

This is a topic from Thermodynamics:

## Introduction

• A pure substance is one composed of only one type of molecule - it is invariable in chemical composition.
• It is important to consider the properties of pure substances because they do not behave like an ideal gas in the majority of cases due to their phase.
• A system has a particular phase when it is invariable in both chemical composition and physical structure.
• Examples of phase include solid, liquid, vapour and gas.
• If a pure substance system is not homogeneous throughout, for example it is part liquid part vapour, it is said to contain two phases.
• Two phases coexist during the changes in phase known as vaporisation, melting and sublimation.
• A saturated liquid is a a fluid for which any decrease in pressure at the given temperature will cause it to boil and become steam.
• A saturated vapour is one for which any increase in pressure at the given temperature will cause it to condense and become a liquid.

### Discerning the Difference Between Gas and Vapour

• Gas is one of the 4 states of matter (solid, liquid, gas and plasma), while vapour does not classify as such.
• The primary distinction is that at room temperature, gas remains gaseous, while vapour is a liquid.
• Consider water vapour, which is always liquid below a certain temperature regardless of its pressure, in comparison to helium gas which is only liquid when both the temperature is adequately low and the pressure is adequately high.
• Vapour does not behave like a gas and so the ideal gas equations do not apply to vapours in the majority of cases.
• Another trivial distinction is that gases cannot be seen and vapours can, and gases always fill the container they occupy while vapours often settle towards the ground under the effects fo gravity.

## P-v-T Surface for Pure Substances

• The state of a substance is dependent upon the amount of thermal energy contained by the substance (proportional to T) as well as the physical constraint on the amount of space the substance can take up (specific volume, v).
• Pressure is dependant upon both T and v, such that if you know any two values out of P, v and T you can calculate the third.
• This dependance of state upon these three variables is captured in the three-dimensional graph known as the P-v-T surface, pictured below:
• The temperature at which a phase change takes place at a given pressure is known as saturation temperature.
• Conversely, the pressure at which a phase change takes place at a given temperature is known as saturation pressure.
• The triple point of a substance refers to the temperature and pressure at which the three phases (vapour, liquid solid), exist in thermal equilibrium.

### P-v Diagrams for Pure Substances

The difference between pure substances and ideal gases can be seen by projecting the P-v-T surface onto the pressure-specific-volume-plane to create the following P-v diagram: Notice that the isotherms are not hyperbolic like they are for isotherms in an ideal gas, because of the phase changes occuring between liquid, liquid and vapour, and vapour.

### T-v Diagrams

Projecting the P-v-T surface onto a temperature-specific-volume plane results in the follow T-v diagram: ## Steam Tables and the Thermodynamic Phases

When performing calculations to do with pure substances, in particular water, we use steam tables. Certain rules arise for how to use the steam tables based on the properties of the system containing the pure substance.

### Compressed Liquid Region

For a fluid, at a given pressure, if T < TSAT (see T-v diagram), then we have a compressed liquid.

• If P ≤ 5000kPa, properties are taken to be the saturated liquid value at that temperature.
• If P > 5000kPa, use the compressed water tables.

### Two-phase Region

For a fluid, at a given pressure, if T = TSAT, then we have two-phase or wet steam.

• Within this region, steam is said to have a quality x defined as:
x = mvapour / mtotal
= mvapour / (mvapour + mliquid)
= mg / (mg + mf)
where the subscript f represents fluid (liquid),
the subscript g represents gas (vapour).
• Consider internal energy:
UTotal = Uf + Ug
muT = mfuf + mgug
(mf + mg)uT = mfuf + mgug
(mf + mg)uT = (mf + mg - mf)uf + mgug
(mf + mg)uT = (mf + mg - mg)uf + mgug
(mf + mg)uT = (mf + mg)uf - mguf+ mgug
Dividing both sides by (mf + mg):
uT = uf - (mg/(mf + mg))uf+ (mg/(mf + mg))ug
uT = uf - xuf+ xug
And so we have the following: (with the subscript T removed for simplicity)
u = uf + x(ug - uf)
h = hf + x(hg - hf)
v = vf + x(vg - vf)
s = sf + x(sg - sf)
This last term refers to be entropy, which will be defined later in the notes for this course.

### Superheated Vapour Region

For a fluid, at a given pressure, if T > TSAT, then we have a superheated vapour

• Use the superheated tables directly.

## Linear Interpolation

The method involves approximating the relationship between v and T, u and T, h and T, or s and T as linear.
Consider the following example:

• Given P = 5 bar, and T = 230°C, find h
• At P = 5 bar, TSAT = 151.8°C, and so T > TSAT
• Using the steam tables for a superheated vapour at P = 5 bar, we can read off that the following values:
At 5 bar, at 200°C, h = 2857 kJ/kg
At 5 bar, at 250°C, h = 2962 kJ/kg
• Using this information, and assuming the linear relationship we can deduce that the ratio between the differences in temperature will equal the ratio in the difference in specific enthalpy:
(T - 200)/(250 - 200) = (h - 2857)/(2962 - 2587)
• Substituting and solving for h:
(230 - 200)/(250 - 200) = (h - 2857)/(2962 - 2587)
0.6 = (h - 2857)/(105)
h = 0.6*105 + 2857
h = 2920 kJ/kg

## Example Problem

The best way to prepare for steam table questions is to practice by doing examples. There are many variations of question which can be asked, of which the following example is just one.
The general method of steam table problems however is:

• Assess the given information to determine which steam table to use
• Apply interpolation and/or steam quality equations to values taken from the steam table one or more times to find all unknowns

Problem

• There exists a closed system containing pure water
• Pressure, P = 5 bar. Specific internal energy, u = 2700 kJ/kg
• Find T(°C), v (m3/kg), h (kJ/kg)

Solution

• From steam tables (saturated water and steam, from pg 2):
uf = 639 kJ/kg
ug = 2562 kJ/kg
• The given u > ug, therefore the water is superheated steam.
• From steam tables (superheated steam, from pg 6):
At 5 bar, at 200°C, u = 2644 kJ/kg
at 250°C, u = 2725 kJ/kg
• Using interpolation with these values and the given value for u, T = 234.57°C
• From steam tables (superheated steam, from pg 6):
At 5 bar, at 200°C, v = 0.4252 kJ/kg and h = 2857 kJ/kg
at 250°C, v = 2644 kJ/kg and h = 2962 kJ/kg
• Using interpolation with both pairs of values and the calculated value for T, v = 0.4593 m3/kg and h = 2930 kJ/kg