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Thermodynamics
The First Law
U 2 - U1 = q - w
work
change in
internal energy
of an object
heat
reservoir
object
b
U = q1-w1
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U = q2-w2
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For example one gram of H2O at 25oC is evaporated and
condensed; the condensed gram of water at 25oC will
have the same internal energy as it did previously.
If only pV work is done and the pressure of the system is
constant
w rev =
 pdV
Internal energy, heat and work
What is the energy required to vaporize water at 100oC???
when one mole of water is vaporized at 100oC the work is
w = p V = RT = 1.987 cal K-1 mole-1 x 373.15K
w= 741.4 cal mole-1
Enthalpy
U = q - pV)
at constant pressure
q= (U2 + pV2) - (U1+ pV1)
We define U + pV as the enthalpy, H
q = H2-H1 = H
2
or the heat adsorbed in a process at constant
pressure
Heat Capacity
Heat Capacity, C = ratio of heat absorbed/mole to the
temperature change = q/T
At constant pressure
q = U+pV = H
Cp = dH/dT
i.e. the calories of heat
adsorbed/mole by a substance/oC
so
H= Cp(T2-T1)
At constant volume
U = q - pV
U = q
Cv = dU/dT
What is the relationship between Cp and Cv?
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The Second Law
eff = (TH-TL)/TH
= (qH + qL)/qH
rearranging
qH qL

0
TH TL
qH qL

0
TH TL


qi
0
Ti
dqrev
0
T
define
dS = dq/T
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and
S2  S1  S 

2
1
dqrev
T
at absolute zero the entropy is assumed to be zero
When spontaneous processes occur there
is an increase in entropy
When the net change in entropy is zero the system is at
equilibrium
If the calculated entropy is negative the process will go
spontaneously in reverse.
The concept of free energy comes from the need to
simultaneously deal with the enthalpy energy and
entropy of a system
G = H -TS
G = U+PV - TS
dG= dU + PdV + VdP -TdS -SdT
dH = dU +pdV
at const temp and pressure
G= H -TS
Equilibrium Constants
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Starting with
dG= dU +VdP + pdV -TdS -SdT
for a reversible process
TdS = dq
dU -dq+dw = 0
so dG= +VdP -SdT
at const temperature, we were able to show
GO= -RT lnKeq
Equilibrium Constants and
Temperature
lnKeq = -H/RT + const
Chemical potentials
Closed Systems
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From the first law
dU = dq - pdV
from the definition of entropy
dS = dq/T
dU = TdS - pdV
If we differentiate by parts, i.e. separately hold dV and
dS constant
(
U
)V  T ; and ( U )S   p
S
V
dU  (
U
)V dS  ( U)S dV
S
V
Environmental systems are often open
systems, i.e. material is being added or
removed, and/or material is reacted
If a homogeneous system contains a number of
different substances its internal energy may be
considered to be a function of the entropy, the volume
and the change in the # moles
dU  (
U
)V dS  ( U)S dV
S
V
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k
U
U
dU  ( )V,ni dS  ( )S,ni dV   ( nU )
dni
i
S
V
S, V, n
i 1
j
k
the term
 ( nU )S, V, n
n i
.
i
is call the Chemical Potential,
j
k
dU  TdS  pdV   ( nU )S , V , n dni
i 1
i
j
k
 i   ( nUi )S, V , n
n i
j
k
dU  TdS  pdV   i dni
i 1
we could show that
k
i   ( )
i 1
U
ni S, V, n j
k
 (
i 1
G
ni T, P, n j
)
k
  ( nH )S, P, n
i 1
i
j
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k
i   ( nGi )T, P, n
i1
j
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What does this say about systems that are not at
equilibrium?
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Chemical Potentials and Pressure
If we go back to the expression for potential energy
k
U  TS  pV   i ni
i 1
dU = SdT+ TdS -Vdp-pdV+
k
k
i 1
i 1
 idni   nidi
For a closed system which only does pressure volume work we
said that
k
dU  TdS  pdV   i dni
i 1
subtracting
k
0 = SdT -Vdp+
 nidi
i 1
At constant temperature, one obtains the Gibbs-Duhem
Equation for gases
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k
Vdpi =  ni d ig
i 1
and so for just compound i
Vdpi/ni = (dig)T
substituting for V = nRT/pi
and integrating from a partial pressure of a compound
defined as pi0 to pi
ig = RT ln pi/pi0
if our boundary conditions or limits start at
standard states

ig = oig + RT ln pi/poi
New text books have elected to define one bar as the standard
state for pressure
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What if the system is not ideal?
Van der Waal’s equation
an 2
( p  2 )( V  nb )  nRT
V
inter molecular
attraction
occupied molecular
volume
For a non-ideal system we could attempt to substitute for
V in the chemical potential relationship
Vdpi/ni = (di)T
another way is to define a parameter related to pressure called
fugacity
where by analogy f
i = oi +RT ln fi/ foi
fi =i pi
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i is a fugacity coef.
--------------------------------------------------in a mixture of gas phase compounds
pi = x i pi *
the vapor pressure in bars
so what is vapor pressure???? which we
will call pi L*
one atmosphere =1.013 bars
one atmosphere supports a 76 cm column of Hg
one atmosphere = 760 mm Hg = 760 torr
one atmosphere =1.013x106 dynes/cm2
derived from the force of mercury on 1 cm2
1 bar = 105 pascals
133.3 pascals = 1torr
fig = xig ig piL*
where xi is the mole fraction
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xig 
nig
nj
j
Fugacitys of liquids
pi = Xi pi* (Raoult’s Law)
for two different liquids with the same
components
p1i
p2i
5%
10%
A in B A in B

2i = 1i +RT ln p2i/p1i
since p1i = x1 piL* and p2i = x2 piL*
2i = 1i +RT ln x2i/x1i (Ideal)
similarly
2i = 1i +RT ln f2i/f1i
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fi pure liquid = i pure liquid piL*
Where i is called an activity
coefficient
if we discuss compound i in a liquid mixture
fiL = i Xipi*L (pure liquid)
the fugacity of compound i with respect to the
fugacity of the pure liquid can also be written
as
fi = i Xifi*L (pure liquid) and
fi /fi*L= i Xi
for ideal behavior of similar compounds like benzene and
toluene in a mixture, i = 1
If we go back to the chemical potential
with respect to a pure liquid
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i = i pure liquid +RT ln fi/f*i pure liquid
fi = i Xif*i pure liquid
so
i = i pure liquid +RT ln iXi
where iXi is called the activity, a, of the
compound in a given state with respect to
some reference state
in iXi = ai the activity sometimes is
called the “apparent concentration” because it
is related to the to the mole fraction, Xi or the
“real” concentration via i
Phase Transfer Processes
Consider a compound, i ,which is dissolved in
two liquids which are immiscible like water and
hexane.
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at equilibrium
i H2O = i pure liquid +RT lniH2OXi H2O
i hx= i pure liquid +RT lni hx Xi hx
at equilibrium

i
H2O
=i hx
RT lni hx Xi hx = RT lniH2OXi H2O
substituting
i Xi = fi /fi*L (pure liquid)
RT ln fi hx /fiL*(pure liquid) = RT lnfi H2O /fiL*(pure liquid)
fi hx = fi H2O
Hint For your homework: You will need to
calculate the activity coefficient for toluene, so let’s do if
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for hexane, ie. what is i of hexane from its solubility in
water.
hexane has some low solubility in water in
grams/LH2O; 1st recall we derived
RT lni hx Xi hx = RT lniH2OXi H2O
What is the activity coefficient and mol fraction of hexane
in hexane?
This gives the important result:
i
H2O=1/ X H2O
i

to calculate the iH2O we need to know Xi H2O
to do this we need something called molar volume
1st the concept of molar volume
molar vol = liquid vol of one mole (L/mol)
/
Ci = sat. conc. = Xi molar volumemix (why???)
the VH2O = 0.0182 L/1 mol
Vmix =
XV ;
i
i
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typically organics have a Vi of ~0.1 L/mol


Vmix 0.1 Xi + 0.0182 XH2O
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