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An ideal mixture of gases is one in which there is no preference for A to be near B compared to A
near A and B near B. In an ideal mixture of ideal gases, there are no energetic interactions whatsoever
among molecules. The intermolecular potential energy is zero.
Dalton mixing process: Take nA moles of Gas A at (PA,V,T) and nB moles of Gas B at (PB,V,T) and
do (positive or negative) work to make a mixture at (P,V,T), as shown in the figure below.
Gas A: PA = f(nA,V,T) = nA R T/V if ideal gas.
(1)
Gas B: PB = g(nB,V,T) = nB R T/V if ideal gas.
(2)
Mix them together isothermally, i.e. let the temperature constantly equilibrate with an external
thermal reservoir at temperature T. You can do this reversibly, for at least some species A and B, by
creating semi-permeable pistons: the dark ones in the figure are permeable only to B and the light ones
are permeable only to A. (You might imagine that you could do this practically if, say, A and B were
right-handed and left-handed chiral molecules, or if A were water-soluble and B were oil-soluble.) Pairs
of pistons are connected by handles so that the volume between the dark pistons remains constant and
the volume between the light pistons remains constant.
What is the final P? The answer is that for an "ideal mixture" of gases, i.e. gases that neither attract nor
repel each other significantly,
P=PA + PB.
(3)
Each gas completely ignores the other; the total force on the walls is the sum of the forces that the
individual gases exerted; the force from each individual gas is unchanged by the presence or absence of
the other. Additionally, if they are both ideal gases, then we can divide (1)and (2) to obtain
PA/ PB = yA/yB
(4)
where yi is mole fraction (ni/ntot) (for example, N2 and O2 in air at S.T.P., yN2≈ 0.8; yO2≈0.2)
PA and PB are called the Partial Pressures of the gases within the mixture. We'll use this concept a
lot.
In general PA, PB, and P are not necessarily related by (3). If A and B have significant attractive
forces toward each other, the final P will be less than PA+ PB. When A molecules go to hit the left-hand
wall, they find all the B molecules on their right are pulling them to the right so they slow down as they
approach the wall. And vice-versa.
An extremely non-ideal case would be 2 moles of water vapor (at low P, high T, high v so it acts as
an ideal gas) formed by "mixing" (i.e. reacting) 2 moles H2 + 1 mole O2. If the O2 and the H2 are also in
the ideal gas regime, each of the three gases obeys the ideal gas EOS:
PV = n R T
Since there are the same number of moles of H2O after as there were of H2 before, P = PA = 2PB.
Note the double meaning of the term "ideal" if there's a molecular reaction. Each molecule can act
as an ideal gas on its own, but the reaction product H2O cannot be considered an ideal mixture of H2
and O2! If you're careful, you can mix H2 and O2 and not get any reaction (don't light a match though!)
Then, if all the gases are low enough density, they will behave as an ideal gas mixture, P(after) =
PA+PB(before).
Note that if the Dalton mixing process is quasistatic and frictionless and if the gases are an ideal
mixture, no work is done in 1Æ2. (Why? Because A does positive work on the right dark piston and an
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equal amount of negative work on the left dark piston and, in the absence of friction, A does no net work
on the left light piston. You should construct a similar argument for B to make sure you understand
this.) We know that there is no change in U and H because for ideal gases, u and h are functions solely
of T. So by the First Law ∆U=Q-W, no heat must have been transferred to or from the surroundings.
Finally, ∆S12=(Q/T)rev, and we've already found a reversible path, and decided Q=0 on it, so there is
no entropy change on mixing at constant V (i.e. in the Dalton mixing process, figure on previous page).
Therefore U, H, and S in the mixture are the sum of the U, H, and S of each separate gas in the premixed condition at the same V and T, which we know how to calculate or measure because we
understand everything about pure ideal gases. That's all that SBVW are trying to say in the paragraphs
containing Eqs. (12.14)-(12.16). But they jump through hoops to use mass-denominated units - ugh!
Use moles! That's why they were invented!
Amagat mixing process: This isn't in the text but it's very instructive to contrast it with the Dalton
mixing process. Take nA moles of Gas A at (P,VA,T) and nB moles of Gas B at (P,VB,T), as shown in
the figure below, and do (positive or negative) work isothermally to make a mixture at (P,V,T). Note the
difference between this and the Dalton mixing process.
For P to be the same after as before mixing (enforced by the totally impermeable piston on the left
with force F on it), how must the volumes of the separated gases be related to that of the mixture?
Gas A: VA = f(nA,P,T) (=, if ideal gas, nA R T/P).
(5)
Gas B: VB = g(nB,P,T) (=, if ideal gas, nB R T/P).
(6)
The answer is that for an ideal mixture of
gases, i.e. gases that neither attract nor repel
each other significantly, we must have
V = VA + VB .
(7)
Additionally, if they are both ideal gases,
then we can divide (5) by (6) to obtain
VA/ VB = yA/yB.
(8)
However, if A and B really like each
other, V will be less than VA+VB. They
attract each other and pull together to spend
more time in the potential well.
Extremely non-ideal case: 2 moles H2O
= 2 moles H2 + 1 mole O2. PV = n R T, V =
VA = 2VB.
Unlike the Dalton mixing process,
however, the Amagat mixing process of an
ideal mixture of ideal gases causes a change
in S. That is , ∆S12 ≠0, where state 1 is the
two gases unmixed and state 2 is mixed at the
same P and T. This is a reversible process in
which the ideal gases do positive work and so
an equal amount of heat must be absorbed in
order to maintain constant temperature. So ∆S12=(Q/T)rev>0.
The following example shows exactly how much the entropy changes upon mixing. It's analogous
to the very important Example 12.2 of Sonntag but spelled out a bit more clearly here, I think. Let nA
moles of ideal gas A at a given pressure and temperature be mixed with nB moles of ideal gas B at the
same pressure and temperature in an isothermal, constant-pressure, reversible process (the Amagat
mixing process described above). Determine the change in entropy for this process.
Tds = du + Pdv (This is SBVW eq. (8.7))
(9)
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du P
+ dv
T T
The ideal gas energy relation and EOS imply, respectively, that
u (T , v ) = u (T ) only; du = Cv 0 dT
ds =
page 3
(10)
P R
=
T v
Substituting these last two equations into (10), we have
dT
dv
SBVW (8.21)
ds = Cv 0
+R
T
v
Or, on a per-mole basis,
dT
dv
ds = Cv 0
+R
T
v
Apply this to gas A as it undergoes expansion during the Amagat isothermal mixing process from
state 1 at volume VA to state 2 at volume V:
vA
V
1
(11)
s2A − s1A = R ln 2A = R ln
= R ln
= − R ln y A ,
v1
VA
yA
where we have used (7) and (8) to turn volume fraction into mole fraction. The total entropy change of
gas A during mixing is obtained by multiplying this last result by nA:
∆S A = S2A − S1A = nA ( − R ln y A )
(12)
The total entropy change on mixing at constant total pressure is obtained by adding to this equation the
equation obtained similarly for B:
∆S B = S2B − S1B = nB ( − R ln yB ) ;
∆Smixing = S2 − S1 = ∆S A + ∆S B = − R ( nA ln y A + nB ln yB )
⎛ nA
⎞
nB
= − R [ nA + nB ] ⎜
ln y A +
ln yB ⎟
nA + nB
⎝ nA + nB
⎠
∆S mixing = nR ( y A ln y A + yB ln yB )
or, dividing by the total number of moles mixed,
total
∆smixing
= − R ( y A ln y A + yB ln yB )
(13)
(14)
These last two equations are equivalent to Eq. (12.19) in §12.1 for the entropy of mixing of ideal
gases. This is a very important formula. Written
this way it depends only upon the mole fractions of
the species in the mixture. Note that since all y's
are < 1, ∆STOT > 0, as shown in the plot below. It
turns out that even though it was derived for ideal
gases, you get exactly the same formula for
isothermal, isobaric mixing of so-called ideal
liquid solutions and ideal solid solutions and we'll
do that later in the course. It's called the "ideal
entropy of mixing" and its use is very widespread,
because the gain per unit pain of going beyond
ideal behavior to predict "real" entropies of mixing
is small.
Moral: for ideal mixtures of ideal gases, u and
h of the separate components are the same as for
the pure components at the same T (independent of P or v). For the entropy you have to be more
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careful. The entropy of a component in a mixture at (T,V) is the same as that of the pure component at
(T,V). But in the tables you typically find entropy of pure components tabulated vs. T and P. The
entropy of a component in a mixture at (T,P) is greater than that of the pure component at (T,P) by eq.
(11), the net result being (13) or (14). That is to say, using component A as an example, analysis of the
Dalton mixing process showed that
sA (T ,V ) = spure A (T ,V ) ,
(15)
which we can rewrite with the additional (dependent) property, pressure, inside the parentheses for
clarity as
sA (T ,V , p = p A + pB ) = spure A (T ,V , p A ) .
(16)
On the other hand, analysis of the Amagat mixing process showed that
sA (T , p ) = spure A (T , p ) − R ln y A
(17)
which we can rewrite with the additional (dependent) property, volume, inside the parentheses for
clarity as
sA (T , pTot ,VTot = VA + VB ) = spure A (T , pTot ,VA ) − R ln y A .
(18)