MODULE 7

• Property-Any measurable characteristics of a substance

such as Press., Vol., or Temp., or a characteristic that can
be calculated or deduced such as internal energy.
• State of a system is the condition of the system as
specified by its properties.
• Values of properties of compounds and mixtures can be
obtained in many formats, including
• Experimental data
• Tables
• Graphs
• Equations
• There are techniques of correlating and predicting
physical properties that are involved in design, operation,
and troubleshooting in all processes
• Collecting data and validating before storage of such
data are very crucial.
• At any temperature and pressure, a pure compound can
exist as a gas, liquid, or solid.
Phase
• Is defined as a completely homogeneous and uniform
state of matter.
1. IDEAL OR PERFECT GAS:
• A gas may be defined as a substance of which the vaporization
from the liquid state is complete. Such substances are oxygen,
nitrogen, hydrogen, air etc.
• An ideal or perfect gas is one which strictly obeys the gas laws
under all conditions of temperature and pressure.
• For engineering purposes, an ideal or perfect gas is a substance
which remains in the gaseous state during the whole
thermodynamic cycle in an engine.
• In fact, no real gas behaves exactly as an ideal or perfect gas,
but gases like hydrogen, oxygen, nitrogen and even air may be
regarded as ideal or perfect gases.
• 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑚𝑢𝑠𝑡 𝑏𝑒 𝑖𝑛 𝐾 𝑓𝑜𝑟 𝑖𝑑𝑒𝑎𝑙 𝑔𝑎𝑠.
2. LAWS OF PERFECT GAS:
• I. Boyle’s law: The volume of a given
mass of a perfect gas varies inversely as
the absolute pressure, when the
temperature is constant.
1
• 𝑉∝ or pv = Constant if T=Constant
𝑃
• 𝑝1𝑣1 = 𝑝2 𝑣2 = … = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑝 = absolute pressure of gas and
v = volume of gas occupied by 𝑝
II. Charle’s law: The volume of a perfect gas varies directly
as its absolute temperature when the absolute pressure
remains constant.
𝑣∝𝑇
𝑣1 𝑣2 𝑣3
= = = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑇1 𝑇2 𝑇3
III. Gay-Lussac’s law: The absolute pressure of a perfect
gas varies directly as its absolute temperature when the
volume remains constant.
𝑃∝𝑇
𝑃1 𝑃2 𝑃3
= = = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑇1 𝑇2 𝑇3
Combination of Boyle’s law and Charles’ law-general gas
equation:
𝑝 1 𝑣1 𝑝2𝑣2 𝑝3𝑣3
= = = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑇1 𝑇2 𝑇3
• IV. Joule’s law: the changes in internal energy (𝑑𝑈) of a
perfect gas is directly proportional to the change of
temperature (𝑑𝑇).
𝑈 ∝ 𝑑𝑇 𝑂𝑅 𝑑𝑈 = 𝑚𝑐 (𝑇2 – 𝑇1)
• V. Avogadro’s law: equal volume of all gases at the same
temperatures and pressures contains equal number of
molecules.
𝑣 ∝ 𝑛 (𝑠𝑎𝑚𝑒 𝑝&𝑇), 𝑣/ 𝑛 = 𝐶
• Example 1: A cylinder contains 0.28 m3 of oxygen at 3.5
bar. What will be its volume when expanded to 1.5 bar at
constant temperature?
• Example 2: A closed vessel contains gas at 1.5 N/mm2
and 20°C. The gas is compressed till it acquires a
temperature of 280°C. Determine the pressure in MPa at
the end of compression.
• Example 3: A gas at a temperature of 20°C and a pressure
of 1.5 bar occupies a volume of 0.1 m3. If the gas is
compressed to a pressure of 7.5 bar and a volume of 0.04
m3, what will be the final temperature of the gas?
 Ideal Gases
• An “ideal” gas exhibits certain theoretical properties.
Specifically, an ideal gas …
• Obeys all of the gas laws under all conditions.
• Does not condense into a liquid when cooled.
• Shows perfectly straight lines when its V and T & P and T
• relationships are plotted on a graph.
• In reality, there are no gases that fit this definition perfectly.
• We assume that gases are ideal to simplify our calculations.
Combining these
• V ∝1/P (Boyle’s law)
• V ∝T (Charles’ law)
• V∝ n (Avogadro’s law)
we get the ideal gas law
𝑛𝑇
𝑉 ∝
𝑃
𝑛𝑇
𝑉 = 𝑅
𝑃
𝑃𝑉 = 𝑛𝑅𝑇
R is constant.
 𝑃𝑉 = 𝑛𝑅𝑇
P = Pressure (in kPa) V = Volume (in L);
T = Temperature (in K) n = moles;
𝑘𝑃𝑎 • 𝐿
𝑅 = 8.31
𝐾 • 𝑚𝑜𝑙
• If we are given any three of P, V, n, or T, we can solve for the
unknown value.
• Sometimes the law may be written as
𝑃Ṽ = 𝑅𝑇
• Ṽ = specific molar volume (volume per mole) of the gas.
• When gas volumes are involved in a problem, Ṽ will be the
volume per mole and not volume per mass.
Example
• A vessel of capacity 3 m3 contains air at a pressure of 1.5
bar and a temperature of 25°C. Additional air is now
pumped into the system until the pressure rises to 30 bar
and temperature rises to 60°C. Determine the mass of air
pumped in and express the quantity as a volume at a
pressure of 1.02 bar and a temperature of 20°C. If the
vessel is allowed to cool until the temperature is again
25°C, calculate the pressure in the vessel.
• Several equivalent standard states known as standard
conditions (S.C., or S.T.P., an acronym for “standard
temperature and pressure”) of temperature and pressure
have been specified for gases by custom. The standard
conditions for the SI, Universal Scientific, and American
Engineering systems are exactly the same conditions, but
in different units. On the other hand, the natural gas
industry uses a different reference temperature (15°F) but
the same reference pressure (1 atm).
What is R for 1 g mol of ideal gas with a volume in cubic
centimeters, pressure in atmospheres, and temperature in
kelvin?
Examples
• How many moles of H2 molecules are in a 3.1 L sample of H2
measured at 300 kPa and 20°C? What is the mass of that
sample?
• At 150 C and 100 kPa, 1.00 L of a compound has a mass of
2.506 g. Calculate its molar mass.
• How many moles of CO2(g) are in a 5.6 L sample of CO2
measured at STP?
• a) Calculate the volume of 4.50 mol of SO2(g) measured at STP
b) What volume would this occupy at 25 C and 150 kPa?
• How many grams of Cl2(g) can be stored in a 10.0 L container at
1000 kPa and 30 C?
• 98 mL of an unknown gas weighs 0.087 g at STP. Calculate the
molar mass of the gas. Can you determine the identity of this
unknown gas?
Density of a gas = mass per unit volume (kg/m3, Ib/ft3, g/l,
etc.). Since mass of gas varies with temperature and
pressure, the two conditions must be specified in
calculating density
Specific gravity of a gas is ratio of density of the gas at a
desired temperature and pressure to that of air (or any
other specified reference gas) at a certain temperature and
pressure. For example, for methane;
• Sp. Gr. = density of methane at S.C./density of air at S.C.
• Example: What is the specific gravity of N2 at 89 oF and
75 mmHg compared to air at 80 oF and 75 mmHg
Examples
• Calculate the volume in ft3 of 10 Ib mol of an ideal gas at
68 oF and 30 psia.
• A steel cylinder of volume 2 m3 contains methane gas
CH4 at 50 oC and 250 kPa absolute. How many kilograms
of methane are in the cylinder?
• 22 kg/h of CH4 are flowing in a gas pipeline at 30 oC and
920 mmHg. What is the volumetric flow rate of the CH4 in
m3 per hour.
• What is the specific gravity of CH4 at 70 oF and 2 atm
compared to air at S.C.
Ideal Gas Mixtures and Partial Pressure
• The ideal gas law can be applied to mixtures of gases by
interpreting
• p as the absolute pressure of the mixture;
• v as the volume occupied by the mixture,
• n as the total number of moles of all components in the
mixture,
• T is the absolute temperature of the mixture
• Air (made of gases like N2, O2, Ar, CO2, Ne, He, and other
trace gases) can be treated as a single compound in
applying ideal gas law
• The partial pressure of Dalton, Pi, is the pressure that
would be exerted by a single component in a gaseous
mixture if it existed alone in the same volume as that
occupied by the mixture and at the same temperature of
the mixture
𝑃𝑖𝑉𝑡𝑜𝑡𝑎𝑙 = 𝑛𝑖𝑅𝑇𝑡𝑜𝑡𝑎𝑙
• Pi is the partial pressure of component i in the mixture.
𝑃𝑖 = 𝑃𝑡𝑜𝑡𝑎𝑙 × 𝑛𝑖/𝑛𝑡𝑜𝑡𝑎𝑙 = 𝑃𝑡𝑜𝑡𝑎𝑙 𝑦𝑖
yi = mole fraction of component i.
• For example, in air, the percent oxygen is 20.95, therefore,
at S.C. of 1 atm, the partial pressure of oxygen is
𝑃𝑂2 = 0.2095 × 1 𝑎𝑡𝑚 = 0.2095 𝑎𝑡𝑚
 𝑃1 + 𝑃2 + ⋯ + 𝑃𝑛 = 𝑃𝑡𝑜𝑡𝑎𝑙
• The partial pressure of a gaseous component cannot be
measured directly with an instrument but can be
calculated from equations.
• The partial pressures of gases are hypothetical pressures
that the individual gases would exert if they were each put
into separate but identical volumes at the same
temperature.
Partial Pressure in gases
• Dalton’s law states that the pressure exerted by a mixture
of gases is the sum of the partial pressures of the
individual gases.
𝑃𝑡𝑜𝑡𝑎𝑙 = 𝑃𝐴 + 𝑃𝐵 + 𝑃𝐶 + … . .
Vapor Pressure is the pressure exerted by a substance’s
vapor over the substance’s liquid at equilibrium.
Example
1. A gas has the following composition at 120 oF and 13.8 psia

a) What is the partial pressure of each component

b) What is the volume fraction of each component
2. a) If the C2H6 were removed from the gas in example 1,
what would be the subsequent pressure in the vessel
b) What would be the subsequent partial pressure of the N2
Real Gases
• Real gases are non-hypothetical gases (something that is
real) whose molecules occupy space and have
interactions; consequently, they adhere to gas laws
• Real gases behave ideally at ordinary temperatures and
pressures.
• At low temperatures and high pressures real gases do not
behave ideally. The reasons for the deviations from ideality
are:
• 1. The molecules are very close to one another, thus their
volume is important.
• 2. The molecular interactions also become important.
• Real gases assume that a gas sample is compressed
using a piston.
• As the gas undergoes compression the individual
molecules are brought closer together : the finite volume
of the individual molecules will become important and
these molecules will interact with one another.
• Hence finite molecular size and intermolecular
interactions will be important in the description of real
gases.
• Deviations from ideal gas behaviour will therefore be
observed as the gas becomes more dense.
• Molecular interactions in dilute gases may be
neglected. Volume of individual gas molecules <<
overall gas volume.
• Significant molecular interactions present in
dense gas. Volume of individual gas molecules
cannot be neglected compared with overall
volume of gas.
• Intermolecular forces may be either attractive or
repulsive.
• Repulsive forces assist expansion of gas:
significant when molecules are close together
(within single molecular diameter) and operative
at high pressure.
• Attractive forces assist compression of gases: can
have influence over a long distance (close but not
touching). Operative at moderate pressures.
 Van der Waals Equation of State
• To model real gases, the volume occupied by the gas particles
themselves must be taken into account:
𝑛𝑅𝑇
𝑃=
𝑉 − 𝑛𝑏
where the constant b is related to the molar volume of the gas
particles
• Must also account for the interparticle attractions among the gas
particles, which effectively act to decrease the observed pressure:
𝑛 2
𝑃𝑜𝑏𝑠 = 𝑃 − 𝑎
𝑉
where the constant a is a quantitative measure of the potential
energy of gas particle interactions
• Thus, applying both the “volume effect” and the
“interactive effect” to the ideal gas law gives:
𝑛𝑅𝑇 𝑛 2
𝑃𝑜𝑏𝑠 = −𝑎
𝑉 − 𝑛𝑏 𝑉
• Which can be rearranged to the usual form of the van der
Waals Equation of State:
𝑛 2 a
𝑃+𝑎 V − nb = nRT; OR P + 2 𝑉𝑚 − 𝑏 = 𝑅𝑇
𝑉 𝑉𝑚
where
Corrected Corrected 𝑉
Pressure 𝑉𝑚 =
Volume 𝑛
Van der Waal and critical phenomenon
• Critical temperature, Tc: Critical temperature of a gas is that
temperature above which it cannot be liquefied no matter
how great the pressure the pressure applied. It is given by the
8𝑎
expression: 𝑇𝑐 =
27 𝑅𝑏
• Critical pressure, Pc: is the minimum pressure required to
liquefy a gas at its critical temperature. It is given as
𝑎
𝑃𝑐 =
27 𝑏2
• Critical volume, Vc: it is the volume occupied by 1 mole of it at
its critical temperature and pressure. It is expressed as:
𝑉𝑐 = 3𝑏
• The van der Waals constant a and b can be expressed in
terms of critical constants by means of the following
equations
Compressibility (Compression) Factor
• We can express the extent of deviation from ideal behaviour as
a function of pressure (which is related to the density of the gas)
by introducing a quantity called the Compressibility or
Compression factor z.
• For an ideal gas z = 1, and real gases exhibit z values different
from unity. z values may be explained in terms of the operation
of intermolecular forces .
• At low pressures the molecules are far apart, and the
predominant intermolecular interaction is attraction . The molar
volume Vm is less than that expected for an ideal gas :
intermolecular forces tend to draw the molecules together and
so reduce the space which they occupy. Under such conditions
we expect that z < 1.
• As the pressure is increased the average
distance of separation between
molecules decreases and repulsive
interactions between molecules become
more important. Under such conditions
we expect that z > 1. When z > 1, the
molar volume is greater than that
exhibited by an ideal gas: repulsive forces
tend to drive the molecules apart.
• For the ideal gas equation of state, the compressibility factor, z,
is given by:
𝑃𝑉
𝑧= =1
𝑛𝑅𝑇
• For the Van der Waals’ equation of state, the compressibility
factor is:
𝑃𝑉 𝑉 𝑎 𝑛 1 𝑎 𝑛
𝑧= = − = 𝑛 −
𝑛𝑅𝑇 𝑉 − 𝑛𝑏 𝑅𝑇 𝑉 1 − 𝑏 𝑅𝑇 𝑉
𝑉
• At low P and high T, n/V becomes small and z approaches the
ideal value of unity.
• Repulsive forces, represented by the constant b, serve to
increase z.
• Attractive forces, represented by the constant a, serve to
decrease z.
Virial equation of state
• The observation of a Z factor different from unity can be used to
construct an empirical or observation-based equation of state, by
supposing that the ideal gas equation of state is only the first term
of a more complex expression which can be expressed in terms of
a mathematical power series. This is called the Virial equation of
state.
• The virial coefficients B, C, B’ and C’ are obtained by fitting
the experimental Z vs P data to the virial equation of state.
Their values depend on the identity of the gas and Reflect
the presence of intermolecular forces and interactions.
When the pressure P is small the molar volume Vm will be
very large and so the second and third terms in the virial
series will be very small and to a good approximation the
virial equation of state reduces to the ideal gas equation
of state.
• In a perfect gas dZ/dP = 0 (since Z = 1).
• In a real gas the result is different.

• At low T the initial dZ/dP < 0, B is negative.

• At high T the initial dZ/dP > 0 and B is positive.
• The temperature at which the initial slope is
zero is Termed the Boyle Temperature TB
where B = 0.
• At the Boyle temperature TB, dZ/dP = 0 and we
can show that TB= a/bR.
 - VdW equation of State
𝑃𝑉𝑚 𝑉𝑚 𝑎 𝑛 1 𝑎 𝑛
and 𝑧 = = − = 𝑛 −
𝑛𝑅𝑇 𝑉𝑚 −𝑛𝑏 𝑅𝑇 𝑉𝑚 1−𝑏 𝑅𝑇 𝑉𝑚
𝑉𝑚
However, therefore

Valid for moderate pressure where b/Vm << 1

Thus equal where
• Hence the second virial coefficient B incorporates terms
arising from repulsive and attractive forces. At high
temperature when bRT >> a, then repulsive forces
predominate, whereas at low temperature when bRT << a,
attractive forces predominate (gases condense to liquids
at low temperature due to attractive forces).
• When bRT = a, T = TB the Boyle temperature of the gas.