Thermodynamics

The First Law

U2 - U1 = q - w

work

change in

internal energy heat

of an object

reservoir

object

b

U = q1-w1

U = q2-w2

a

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

wrev =  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

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?
The Second Law

eff = (TH-TL)/TH= (qH + qL)/qH

rearranging

define

dS = dq/T

and

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

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,

so dG= +VdP and dG = RT dP/P

For a reaction of AB where Keq = [B]/[A]

For A GA= GAo +ln {PA/PAo}

For BGB= GBo +ln {PB/PBo}

The Free energy for the reaction is GB-GA

If we take GBo , GAo as being at standard temperature (273.15K) and pressure (1 atmosphere)

GA-GB = GBo-GAo+ ln {PB/PBo } - ln {PA/PAo }

we were able to show using PV=RT for one mole that

GO= -RT ln[B]/[A] =- RT lnKeq

Equilibrium Constants and Temperature

lnKeq = -H/RT + const

Chemical potentials

From the first law

dU = dq - pdV

from the definition of entropy

dS = dq/T

dU = TdS - pdV

If the number of moles are changing in a reaction

Say A B+C

The term is call the Chemical Potential,

We can also show that

So chemical potential is really the change in free energy per mole

And before we showed that:

dG = RT dP/P

So on a molar basis

d = RT dp/P

integrating

ig = oig + RT ln pi/poi

What if the system is not ideal?

Van der Waal’s equation

inter molecularoccupied molecular

attractionvolume

For a non-ideal system we could attempt to substitute for V in the chemical potential relationship

Vdpi/ni = (di)T

ideal:ig = oig + RT ln pi/poi

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

i is a fugacity coef.

------

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

fi pure liquid = i pure liquid piL*

Where i is called an activity coefficient

if we discuss compound iin a liquid mixture

fiL = i Xipi*L (pure liquid)

the fugacity of compound iwith 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

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, Xior 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.

at equilibrium

iH2O = 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 lnfi hx /fiL*(pure liquid) = RT lnfi H2O/fiL*(pure liquid)

fihx = fi H2O

Hint For your homework: You will need to calculate the activity coefficient for toluene, so let’s do if 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 hxXi hx= RT lniH2OXi H2O

What is the activity coefficient and mol fraction of hexane in hexane?

This gives the important result: iH2O=1/Xi H2O

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 =  Xi Vi ;

typically organics have a Vi of ~0.1 L/mol

Vmix0.1 Xi + 0.0182 XH2O

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