Ideal gases (leacture 3)

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LECTURE 3
Ideal Gases
2
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The topics we will be covering are:
• Ideal Gases
• Dalton’s Law of Partial Pressure
• Raoult’s Law
• Henry’s Law
• Antoine Equation
2

Ideal Gases
- Relates the molar quantity and volume of a gas to temperature and
pressure.
- Simplest and most widely used relationship is the ideal gas equation of state
which is applicable to many problems involving gases at low pressures.
- However, some gases deviate from ideal behavior at certain conditions,
notably at high pressure and low temperatures.
- In such cases it is necessary to use more complex equation of state.
2

The Ideal Gas Equation of State
- Derived from kinetic theory of gases.
- Assumptions are: gas molecules have a negligible volume, exert no forces
on one another and collide elastically with the walls of the container.
- Equation is given as:
𝑷𝑽 = 𝒏𝑹𝑻
or
where,
𝑽 𝑽
𝑷 = 𝒂𝒃𝒔𝒐𝒍𝒖𝒕𝒆 𝒑𝒓𝒆𝒔𝒔𝒖𝒓𝒆 𝒐𝒇 𝒂 𝒈𝒂𝒔
= 𝒗𝒐𝒍𝒖𝒎𝒆 𝒗𝒐𝒍𝒖𝒎𝒆𝒕𝒓𝒊𝒄 𝒇𝒍𝒐𝒘 𝒓𝒂𝒕𝒆 𝒐𝒇 𝒈𝒂𝒔
𝒏 𝒏 = 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒎𝒐𝒍𝒆𝒔 𝒎𝒐𝒍𝒂𝒓 𝒇𝒍𝒐𝒘 𝒓𝒂𝒕𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒈𝒂𝒔
𝑹 = 𝒕𝒉𝒆 𝒈𝒂𝒔 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕, 𝒘𝒉𝒐𝒔𝒆 𝒗𝒂𝒍𝒖𝒆𝒔 𝒅𝒆𝒑𝒆𝒏𝒅𝒔 𝒐𝒏 𝒕𝒉𝒆 𝒖𝒏𝒊𝒕𝒔 𝒐𝒇 𝑷, 𝑽, 𝒏 𝒂𝒏𝒅 𝑻
𝑻 = 𝒂𝒃𝒔𝒐𝒍𝒖𝒆 𝒕𝒆𝒎𝒑𝒆𝒓𝒂𝒕𝒖𝒓𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒈𝒂𝒔

• The equation maybe written as: 𝑷𝑽 = 𝑹𝑻
where,
𝑉 =
𝑉
𝑛
𝑜𝑟
𝑉
𝑛
It is the specific volume of the gas.
• A gas whose PVT behavior is well represented by above equations is said to
behave as an ideal gas or perfect gas.
• 1 mol of an ideal gas at 0℃ and 1 atm occupies 22.415 liters.
• The gas constant R has units of 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑥 𝑣𝑜𝑙.
𝑚𝑜𝑙𝑒 𝑥 𝑡𝑒𝑚𝑝.
2

For an ideal gas at an arbitrary temperature T and pressure P,
𝑃𝑉 = 𝑛𝑅𝑇
for the same ideal gas at a specified reference temperature 𝑇𝑠and pressure
𝑃𝑠(also known as STP), we may write the equation as:
𝑃𝑠𝑉 = 𝑅𝑇𝑠
The first equation divided by the second, yields,
𝑃𝑠𝑉
=
𝑃𝑉 𝑛𝑅𝑇
𝑅𝑇𝑠
Standard Temperature and Pressure

Standard Temperature and Pressure
System 𝑻𝒔 𝑷𝒔 𝑽𝒔 𝒏𝒔
SI 273K 1 atm 0.022415 𝒎𝟑 1 mol
CGS 273K 1 atm 22.415 L 1 mol
American
Engineering
492⁰R 1 atm 359.05 𝒇𝒕𝟑 1 lb-mol
Standard conditions for gases
2

NOTE: You will frequently encounter problems that involve gases
at two different states – for example, at the inlet and outlet of a
process unit. A convenient way to determine an unknown variable
(P, V, n, or T) of the gas at one of the states is to write the gas law
for both states and divide one equation by other. Any variables
that are the same at both points will cancel, leaving an equation
containing only the variable you wish to determine and known
quantities.
2

Example -5.2-2 (Felder)
Butane (𝐶4𝐻10) at 360℃ and 3 atm absolute flows into a reactor at a rate of
1100 𝑘𝑔/𝑕
. Calculate the volumetric flow rate of this stream using conversion
from standard conditions.
Solution:
𝑃𝑉
𝑃𝑠𝑉
= 𝑛
𝑇
𝑇𝑠
𝑛 =
𝑚
𝑀𝑊
1100𝑘𝑔
= ℎ
58.1 𝑘𝑔
𝑘 𝑚 𝑜
𝑙
= 19 𝑘𝑚𝑜𝑙/h
⇒ 𝑉 = 𝑛𝑉𝑠
𝑇 𝑃
𝑇 𝑃
𝑠 𝑠
⇒ 𝑉 = 19 𝑘𝑚𝑜𝑙
𝑕
22.4 𝑚3
𝑘𝑚𝑜𝑙
633 𝐾
273 𝐾 3 𝑎𝑡𝑚
1 𝑎𝑡𝑚 = 𝟑𝟐𝟗 𝒎𝟑/𝒉

Example 5.2-3 (Felder)
Ten cubic feet of air at 70℉ and 1.00 𝑎𝑡𝑚 is heated to 610℉ and compressed
to 2.50 𝑎𝑡𝑚. What volume does the gas occupy in its final state?
Solution:
Process
𝑉1 = 10𝑓𝑡3
𝑇1 = 70℉
𝑃1 = 1 𝑎𝑡𝑚
𝑉2 =? 𝑓𝑡3
𝑇2 = 610℉
𝑃2 = 2.5 𝑎𝑡𝑚
𝒏 (𝒍𝒃 − 𝒎𝒐𝒍 𝒂𝒊𝒓) 𝒏 (𝒍𝒃 − 𝒎𝒐𝒍 𝒂𝒊𝒓)
2

𝑃2𝑉2 = 𝑛𝑅𝑇2
𝑃1𝑉1 = 𝑛𝑅𝑇1
𝑃2𝑉2 𝑛𝑅𝑇2
⇒ =
𝑃1𝑉1 𝑛𝑅𝑇1
𝑉2= 𝑉1
𝑃1 𝑇2
𝑃2 𝑇1
= 10 𝑓𝑡3 1 𝑎𝑡𝑚
2.5 𝑎𝑡𝑚 70 + 460 °𝑅
610 + 460 °𝑅
= 𝟖. 𝟎𝟖 𝒇𝒕𝟑
2

Example 4.3 (Himmelblau)
Calculate the volume occupied by 88 lb of 𝐶𝑂2 at a pressure of 32.2 ft of water
and at 15℃.
SOLUTION
So here we have two situations, i.e.,
88 lb 𝒇𝒕𝟑 at S.C.
𝒇𝒕𝟑 at 32.2 ft 𝑯𝟐𝑶
and 15℃
At S.C. (state 1)
𝑝 = 33.91 𝑓𝑡 𝐻2𝑂
2

IDEAL GAS MIXTURES
- Suppose 𝑛𝐴 moles of substance A, 𝑛𝐵 moles of B, 𝑛𝑐 moles of C and so one,
are contained in a volume V at a temperature T and total pressure P
- The partial pressure 𝑝𝐴 and pure component volume 𝑣𝐴 of A in the
mixture are defined as follows:
𝒑𝑨: the pressure that would be exerted by 𝑛𝐴 moles of A alone in the
same total volume V at the same temperature T
𝒗𝑨: the volume that would be occupied by 𝑛𝐴 moles of A alone at the
total pressure P and temperature T of the mixture
2

- Suppose that each of the individual mixture components and mixture as a whole
behave in an ideal manner, then:
- From the definition of partial pressure:
𝒑𝑨𝑽 = 𝒏𝑨𝑹𝑻
- Dividing these equations, we get,
𝑨
𝒑𝑨
=
𝒏𝑨
= 𝒚
𝑷 𝒏
𝒑𝑨 = 𝒚𝑨𝑷
(𝑡𝑕
𝑒 𝑚𝑜𝑙𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝐴 𝑖𝑛 𝑡𝑕
𝑒 𝑔𝑎𝑠)
or
2

𝒑𝑨 = 𝒚𝑨𝑷
that is, the partial pressure of a component in an ideal gas mixture is
the mole fraction of that component times the total pressure
or
the partial pressures of the components of an ideal gas mixture add up
to the total pressure (Dalton’s law)
𝒑𝑨 + 𝒑𝑩 + ⋯ = (𝒚𝑨 + 𝒚𝑩 + ⋯ )𝑷 = 𝑷
2

Vapor-Liquid Equilibria for Multi Component System
• In a two phase mixture at equilibrium, a component in one phase is in
equilibrium with the same component in the other phase.
• The equilibrium relationship depends on the temperature, pressure and
composition of the mixture.
Vapor-pressure liquid
equilibrium for a binary
mixture. The dashed lines
show the equilibrium
compositions (a) when the
total pressure is constant and
(b) when the temperature is
constant over the composition
range.

• We will consider two cases in which linear (ideal) equations exist to relate
the mole fraction of one component in the vapor phase to the mole fraction
of the same component in the liquid phase.
2

Henry’s Law
• Used primarily for a component whose mole fraction approaches zero, such
as dilute gas dissolved in as liquid:
𝒑𝒊 = 𝑯𝒊𝒙𝒊
𝐻𝑖 is the Henry’s law constant
Note that:
where,
𝑝𝑖 is the pressure in the gas phase of the dilute component at equilibrium at
some temperature, 𝑥𝑖 is the liquid phase mole fraction and
Note: 𝑯𝒊 values can be found in
several handbooks
𝑝𝑖
𝑝𝑡𝑜
𝑡
= 𝑦 𝐻𝑖𝑥𝑖
𝑖 ⇒
𝑝𝑡𝑜
𝑡
𝒊
⇒ 𝒚 =
𝒑𝒊
=
𝑯𝒊𝒙𝒊
𝒑𝒕𝒐𝒕 𝒑𝒕𝒐𝒕

Henry’s law
• Gas solubility increases as the partial pressure of a gas above the liquid
increases.
• Suppose a certain volume of water is in a closed container with the space
above it occupied by carbon dioxide gas at standard pressure.
• Some of the CO 2 molecules come into contact with the surface of the water
and dissolve into the liquid. Now suppose that more CO 2 is added to the
space above the container, causing a pressure increase.
• More CO 2 molecules are now in contact with the water and so more of them
dissolve. Thus the solubility increases as the pressure increases.
2

When carbonated beverages are packaged, they are done so under high
CO2 pressure so that a large amount of carbon dioxide dissolves in the liquid.
When the bottle is open, the equilibrium is disrupted because the
CO2 pressure above the liquid decreases.
Immediately, bubbles of CO2 rapidly exit the solution and escape out of the
top of the open bottle.
The amount of dissolve CO2 decreases.
If the bottle is left open for an extended period of time, the beverage becomes
“flat” as more and more CO2 comes out of the liquid.

Take for example 𝐶𝑂2 dissolved in water at 40℃ for which the value of H is
69,600 𝑎𝑡𝑚/𝑚𝑜𝑙 fraction. (The large value of H shows that 𝐶𝑂2(𝑔) is only
sparing soluble in water).
2
For example, if 𝑦𝐶𝑂 = 4.2 𝑥 10−6, the partial pressure of the 𝐶𝑂2 in the gas
phase is
𝑝𝐶𝑂2
= 69,600 4.2 𝑥 10−6 = 0.29 𝑎𝑡𝑚
2

Raoult’s Law
• Used primarily for component whose mole fraction approaches unit or for solutions
of components quite similar in chemical nature, such as straight chain
hydrocarbons.
• Raoult’s law states that a solvent’s partial vapor pressure in a solution (or mixture)
is equal or identical to the vapor pressure of the pure solvent multiplied by its mole
fraction in the solution.
• Mathematically, Raoult’s law equation is written as;
𝒑𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 = 𝒑𝒔𝒐𝒍𝒗𝒆𝒏𝒕
∗𝒙𝒔𝒐𝒍𝒗𝒆𝒏𝒕
• Where,
𝑝𝑠𝑜𝑙𝑢𝑡𝑖𝑜n = vapor pressure of the solution
𝑥𝑠𝑜𝑙𝑣𝑒𝑛𝑡 = mole fraction of the solvent
𝑝𝑠𝑜𝑙𝑣𝑒𝑛𝑡
∗= vapor pressure of the pure solvent

What is the Importance of Raoult’s law?
Assume that we have a closed container filled with a volatile liquid A. After
some time, due to evaporation, vapour particles of A will start to form. Then as
time passes, the vapour particles of A will be in dynamic equilibrium with the
liquid particles (on the surface). The pressure exerted by the vapour particles
of A at any particular temperature is called the vapour pressure of A at that
temperature.

• Now imagine we are adding another liquid B (solute) to this container. This will
result in B particles occupying the space between A particles on the surface of the
solution.
• For any given liquid there are a fraction of molecules on the surface which will have
sufficient energy to escape to the vapour phase.
• Since now we have a lesser number of A particles on the surface, the number of
vapour particles of A in the vapour phase will be lesser. This will result in lower
vapour pressure of A.
• Now if we assume that B is volatile as well, we will have lesser number of B
particles in the vapour phase as compared to pure liquid B.

• This new pressure (partial pressure) of each (A and B) is given by Raoult’s
law and depends on the concentration of each component in the liquid
phase.
𝑝𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = 𝑝𝑠𝑜𝑙𝑣𝑒𝑛𝑡
∗𝑥𝑠𝑜𝑙𝑣𝑒𝑛𝑡
𝑝𝐴∞ 𝑥𝐴
𝒑𝑨 = 𝒑𝑨
∗𝒙𝑨
And similarly for B,
𝒑𝑩 = 𝒑𝑩
∗𝒙𝑩
2

Antoine Equation
The temperature dependence of vapor pressure maybe highly non-linear, the
logarithm of 𝑝∗ varies with T roughly the same way for a large number of
species.
A relatively simple empirical equation that correlates vapor pressure-
temperature data extremely well is the Antoine Equation
𝟏𝟎
𝒍𝒐𝒈 𝒑∗ = 𝑨 −
𝑩
𝑻 + 𝑪
values of A, B and C for several compounds are listed in different Tables
(given in books)
2

Antoine Equation
• The vapor pressure over a liquid is due to the thermodynamic equilibrium
between the gas and the liquid states of the component, which depends on
the cohesive forces linking the molecules. In a closed cell, the vapor
pressure of a pure component is a nonlinear relation of the temperature: the
more the component is volatile, the more the vapor pressure is high.
• For a given pure component in a closed cell, it calculates the saturated
vapor pressure P (mmHg) with the temperature T(⁰C),
𝟏𝟎
𝒍𝒐𝒈 𝒑∗ = 𝑨 −
𝑩
𝑻 + 𝑪
2

GAS-LIQUID SYSTEMS
- Systems containing several components, of which only one is capable of
existing as a liquid at the process conditions, are common in industrial
processes.
- Separation processes that involve such systems include evaporation, drying
and humidification – all of which involve transfer of the condensable species
from the gas to the liquid phase.
- Suppose liquid chamber is introduced into a chamber that initially contains
dry air and that the temperature and pressure in system are kept constant at
75℃ and 760 mmHg.
2

- Initially the gas phase contains no water (𝑝𝐻2𝑂 = 0) and water molecule
consequently begin to evaporate.
- The mole fraction of water in the gas phase, 𝑦𝐻2𝑂, increases, and hence so
does the partial pressure of water, 𝑝𝐻2𝑂 = 𝑦𝐻2𝑂𝑃.
- Eventually, however, the amount of water in the gas phase is such that the rate
at which water molecules enter the gas phase approaches zero, and thereafter
no change occurs in the amount or composition of either phase.
- The gas phase is then said to be saturated with water – it contains all the
water it can hold at the system temperature and pressure – and the water in
the gas phase is referred to as a saturated vapor.
2

RELATIVE VOLATILITY
• Vapor liquid equilibrium calculations can sometimes be simplified through
the use of the quantity called the relative volatility, which may be defined in
terms of the following depiction of vapor and liquid phases in equilibrium:
Vapor: 𝑦𝑖 , 𝑦𝑗 , 𝑦𝑘 , . . .
Liquid: 𝑥𝑖 , 𝑥𝑗 , 𝑥𝑘 , . . .
𝑦𝑖 = 𝑚𝑜𝑙𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛
𝑜𝑓 𝑠𝑝𝑒𝑐𝑖𝑒𝑠 𝑖 𝑖𝑛 𝑣𝑎𝑝𝑜𝑟
𝑥𝑖 = 𝑚𝑜𝑙𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛
𝑜𝑓 𝑠𝑝𝑒𝑐𝑖𝑒𝑠 𝑖 𝑖𝑛 𝑙𝑖𝑞𝑢𝑖𝑑
2

- The relative volatility of species i to species j is,
𝒊𝒋
𝒚𝒊/𝒙𝒊
𝜶 =
𝒚𝒋/𝒙𝒋
2

Ideal gases (leacture 3)

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