Because the gas laws
are based on numbers of molecules, it is often convenient to express concentrations
in terms of the relative numbers of molecules of the components. The unit
in this case is called the molecular fraction, shortened to **mole
fraction**, which has been introduced in Chapter 2. The mole fraction
of a component in a mixture is the proportion of the number of molecules
of the component present to the total number of the molecules of all the
components.
In a mixture which
contains *w*_{A} kg of component *A* of molecular weight
*M*_{A} and *w*_{B} kg of component *B*
of molecular weight *M*_{B}, the mole fraction:
*x*_{A} = __ number
of moles of __*A*
number
of moles of *A* +
number of moles of B
= * *__ w___{A}__ /__*M*_{A}__ __ (9.1)
* w*_{A} /*M*_{A} + *w*_{B} /*M*_{B}
*x*_{B} = __ __* w*_{B}__
/__*M*_{B}__ __ (9.2)
* w*_{A} /*M*_{A} + *w*_{B}/*M*_{B}
Notice that (*x*_{A}
+ *x*_{B}) = 1, and so, *x*_{B} = (1 - *x*_{A})
The definition of
the mole fraction can be extended to any number of components in a multicomponent
mixture. The mole fraction of any one component again expresses the relative
number of molecules of that component, to the total number of molecules
of all the components in the mixture. Exactly the same method is followed
if the weights of the components are expressed in grams. The mole fraction
is a ratio, and so has no dimensions.
** ****EXAMPLE 9.1. Mole fractions of ethanol in water**
A solution of ethanol in water contains 30% of ethanol by weight. Calculate
the mole fractions of ethanol and water in the solution.
Molecular weight
of ethanol, C_{2}H_{5}OH, is 46 and the molecular weight of water, H_{2}O, is 18.
Since, in 100 kg of the mixture there are 30 kg of ethanol and 70 kg of
water,
mole fraction
of ethanol = __(30/46) / [(30/46 + (70/18)]__
=
0.144
mole fraction of water = __ (70/18) /
[(30/46 + (70/18)]__
= 0.856
= (1 - 0.144)
Concentrations of
the components in gas mixtures can be expressed as weight fractions, mole
fractions, and so on. When expressed as mole fractions, they can be related
to the partial pressure of the components. The **partial
pressure **of a component is that pressure which the component would exert
if it alone occupied the whole volume of the mixture. Partial pressures
of the components are additive, and their sum is equal to the total pressure
of the mixture. The partial pressures and the mole fractions are proportional,
so that the total pressure is made up from the sum of all the partial
pressures, which are in the ratios of the mole fractions of the components.
If a gas mixture exists under a total pressure *P* and the mixture
comprises a mole fraction *x*_{A} of component *A*, a mole fraction
*x*_{B} of component *B*, a mole fraction *x*_{C} of component
*C* and so on, then
* P = Px*_{A}
+ *Px*_{B} + *Px*_{C} + …..
= *p*_{A} + p_{B}* + p*_{C} + ….. (9.3)
where *p*_{A},
*p*_{B}, *p*_{C}, are the partial pressures
of components *A, B,
C *...
In the case of gas
mixtures, it is also possible to relate weight and volume proportions,
as **Avogadro's Law** states that under equal conditions
of temperature and pressure, equal volumes of gases contain equal numbers
of molecules. This can be put in another way by saying that in a gas mixture,
volume fractions will be proportional to mole fractions.
**EXAMPLE 9.2. Mole fractions in air **
Air is reported to contain 79 parts of nitrogen to 21 parts of oxygen,
by volume. Calculate the mole fraction of oxygen and of nitrogen in the
mixture and also the weight fractions and the mean molecular weight.
Since mole fractions
are proportional to volume fractions,
mole fraction of nitrogen = __79 / [79 + 21]__
=
__0.79__
mole fraction of oxygen = __21 / [79 + 21]__
= __0.21__
The molecular weight
of nitrogen, N_{2}, is 28 and of oxygen, O_{2}, is 32.
The weight fraction of nitrogen is given by:
__ weight
of nitrogen __ =
__ 79 x 28 __
weight of nitrogen + weight of oxygen 79
x 28 + 21 x 32
=
__0.77__
Similarly the weight
fraction of oxygen = __(21 x 32) / [(79 x 28)
+ (21 x 32)]__
=
__0.23__.
As the sum of the
two weight fractions must add to 1, the weight fraction of the oxygen
could have been found by the subtraction of (1
- 0.77) = 0.23.
To find the mean
molecular weight, we must find the weight of one mole of the gas:
0.79 moles of
N_{2} weighing 0.79 x 28 kg = 22.1 kg
plus 0.21 moles of O_{2} weighing 0.21 x 32kg = 6.7 kg
make up 1 mole of air weighing 28.8 kg and so
__Mean molecular weight of air is 28.8, say 29__.
Contact-Equilibrium
Processes - THEORY > GAS-LIQUID EQUILIBRIA
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