Transcript Open system Closed System cork insulation Isolated System

Work, Heat and Internal
Energy: The First Law


System – the specific part of the
universe of interest to us
Surroundings – the part of the universe
not contained in the system

3 types of Systems
• open system – exchanges mass and energy
• closed system – exchanges energy but no
•
mass
isolated system – no exchange of either
mass or energy
cork
Open system
Closed System
insulation
Isolated System

State of a system
• the system is in a definite state when each of its
properties has a definite value.


Change in state
• initial state
• final state
Path
• initial and final states
• intermediate states

Process

Cyclic transformation
• reversible or irreversible transformation
• begins and ends at the same state variables.

Isothermal

Isochoric

Isobaric
• dT = 0
• dV = 0
• dP = 0

Work (w)
• any quantity that flows across the system’s
boundary and is completely convertible into
the lifting of a mass in the surroundings.

How much work was done?
dw  Fz dz
Unit of work = J = 1 kg m/s2
120
Force / N
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40
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0
0
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30
Distance / m
40
50
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A single-stage expansion process
mass (m)
h1
Piston
(T, P1, V1)
State 1
Direction of
piston

mass (m)
h2
Piston
(T, P2, V2)
State 2

The work done in the surroundings

The work done by the system

For an infinitesimal volume change
• wsurr= Pext DV
• wsys = - wsurr = - Pext DV
• dwsys = - Pext
dV

If the system is in equilibrium

For a simple system
• Fsys = -Fext
• P = Pext
• d wrev = - P dV

Ideal gas as the working fluid.
w rev   dw rev    PdV
c
nRT
P 
V
c

For an isothermal process (ideal gas as
working fluid)
w rev   dw rev
c
V 2 
 nRT ln 
V1 


dwirr = -Pext dV
for a constant external pressure
w irr 
c
2
dw irr    Pext dV
 Pext V 2 V1 
1

Heat - the quantity that flows across the
boundary of the system during a
change in state
• due to temperature difference between
•
system and surroundings
HOT to COLD (never the other way
around)!!!

Measured by determining the
temperature change of some known
object
dq  CdT
C - the heat capacity of the
system.

Integrate the infinitesimal heat flow
T2
q   dq   CdT   CdT
c
c
T1
q  CDT
heat
system


surroundings
Exothermic - system to surroundings
Endothermic – surroundings to system

Heat flows during phase changes latent heats
• Latent heat of vapourisation
• Latent heat of fusion

Subject our system to a cyclic
transformation
q cy cle   CdT   CdT
cy cle
w cy cle 
dw

cy cle

 dw

The following would be true for an exact
differential
df

0
if
df
is
exact


The infinitesimal change in the
internal energy
dU  dq  dw
 For a general process
DU   dq  dw  q  w
c

In general, we write U as a function of T
and V
 U 
dU  
 dT
 T V
 U 

 dV
 V T

Examine the first partial derivative
 U 
dU  
 dT  0
 T V

Define the constant volume heat
capacity, CV
 dq   U 
CV      
 dT V  T V

For a system undergoing an isochoric
temperature change
dU  CV dT
 For a macroscopic system
T2
DU  qV   C V dT
T1

Examine the second partial derivative
 U 
dU  
 dV
 V T
O
C
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Thermal insulation
O
O
F
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0
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10
0
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Valve
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A
T1, Vm,1, P1
B
Stirrer

The partial derivative
 T 


 V U
is known as the Joule coefficient, J.

The change in the internal energy
under isothermal conditions is related to
the Joule Coefficient
 U 
 T   U 

  
 

 V T
 V U  T V
 U 

  C V  J
 V T


For an adiabatic process, q = 0!!
The first law becomes
DU  w
dU  dw  CV dT

V 1

V
 2




For an ideal gas
undergoing a reversible,
adiabatic process
R /Cv
,m
T2

T1
 P2 
 
 P1 
R / C v ,m R 
T2

T1


Defining the enthalpy of the system
Re-examine the piston with the weight
on top
mass (m)
Piston
(T, P, V)

The first law
dU  dqP  PdV

Integrating
V2
2
dU

dq

P
dV
P



1
c
V1

We define the enthalpy of the system,
H
H  U  PV

In general, we write H as a function of T
and P
 H 
dH  
 dT
 T P
 H 

 dP
 P T

Examine the first partial derivative
 H 
dH  
 dT  0
 T P

Define the constant pressure heat
capacity, CP
 dq 
 H 
CP  
 

 dT P  T P

For a system undergoing an isobaric
temperature change
dH  C P dT
For a macroscopic system
T2
DH  q P   C P dT
T1

For an ideal gas
C P  CV  nR

In general
C P  CV
 TV

T
2

Examine the second partial derivative
 H 
dH  0  
 dP
 P T
O
C
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20
10
O
F
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Thermal insulation
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O
C
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0
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F
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O
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T1, P1, Vm,1
Porous Plug
60
T2, P2, Vm,2

The partial derivative
 T 


 P H
is known as the Joule-Thomson coefficient,
JT.

The change in the enthalpy under
constant pressure conditions is related
to the Joule-Thomson Coefficient
 H 
 T   H 

  
 

 P T
 P H  T P
 H 

  C P  JT
 P T

The shorthand form for a chemical reaction
0   J J
J
J = chemical formula for substance J
J = stoichiometric coefficient for J

The enthalpy change for a chemical
reaction
Dr H   n J H m J 
J
Hm [J] = molar enthalpies of substance J
nJ = number of moles of J in the reaction

Reaction beginning and ending with
equilibrium or metastable states
D r H  H final  H initial

Dn J H m J 

J
Note – Initial and final states have the
same temperature and pressure!

We note that 1 mole of a reaction
occurs if
Dn J   J
Dr H 
 J H m J 

J


A reaction that begins and ends with all
substances in their standard states
The degree sign, either  or 
• P = 1.00 bar
• [aqueous species] = 1.00 mol/ kg
• T = temperature of interest (in data tables
25C or 298 K).
-

We note that for 1 mole of a reaction
under standard conditions
Dr H


 J H m J 

J



A "chemical thermodynamic reference
point."
For CO and CO2
C (s) + O2 (g)  CO2 (g)
C (s) + ½ O2 (g)  CO (g)

The formation reaction

Formation of 1.00 mole of Na2SO3(s)
2 Na(s) + S(s) + 3/2 O2 (g)  Na2SO3 (s)
‘Formation enthalpy of Na2SO3(s)’,
DfH°[Na2SO3 (s)]

• 1 mole of a compound
• constituent elements
• stable state of aggregation at that temperature.


DfH° is a measurable quantity!
Compare CO (g) with CO2 (g)
C (s) + 1/2 O2 (g)  CO (g)
DfH° [CO(g)] = -110.5 kJ/mole
C (s) + O2 (g)  CO2 (g)
DfH° [CO2(g)] = - 393.5 kJ/mole

Formation enthalpies - thermodynamic
reference point!
• Hom [J] = DfH [J]
• Hm [elements] = 0 kJ / mole.

Use the tabulated values of the
formation enthalpies

The enthalpy change for a given
reaction is calculated from the formation
enthalpies as
Dr H    J Df H J 


J

Notes


Reverse a reaction
Multiply a reaction by an integer


A calorimeter - device containing water
and/or another substance with a known
heat capacity
Calorimeters – either truly or
approximately adiabatic systems
DU = qv.
DH = qp


The enthalpy and the internal energy
both represent quantities of heat.
DU = qv.
DH = qp.
Relate the two state functions using the
following relationship
DU = DH - D PV




Enthalpy of solution
Enthalpy of dilution
Enthalpy of fusion
Enthalpy of vapourisation



DsolH - heat absorbed or released
when a quantity of solute is dissolved in
fixed amount of solvent
DsolH = Hm(sol’n) – Hm(component)
• H(component) =
Two definitions
• Standard
• Limiting
Hm(solid) + Hm(solvent)


For the process,
HCl (aq, 6 M)  HCl (aq, 1 M).
The Enthalpy of dilution of the acid.
DdilH = Hm(sol’n 2) – Hm(sol’n ,1)

Differentiate the reaction enthalpy with
temperature
Dr H 

 J H m J 

J
dDr H
d

dT
dT


J  J H m J 

Dr H T   Dr H 298 K   Dr C DT




p
DrCp - the heat capacity change for
the reaction
Dr C p 

 J C p J 

J



Examine a chemical reaction.
C (s) + O2 (g)  CO2 (g)
DU = U[CO2 (g)] – U[C(s)] –
U[O2(g)]
Note - DrH = -393.5 kJ/mole
D r U    J Df U J 


J
D r H   D r U   Dn g RT


Use tabulated values of formation
enthalpies to obtain DrH°.
May also estimate reaction enthalpies
using an indirect method.

Hess’s Law –
• the enthalpy change for
a given reaction is
the same whether the reaction occurs in a
single step or in many steps.

Examine the following reactions
H2 (g)  H (g) + H (g)
DU° = 433.9 kJ
Cl2 (g)  Cl (g) + Cl (g)
DU° = 239.5 kJ


Bond dissociation energies.
Enthalpy changes are designated D (HH) and D (Cl-Cl).
CO2 (g)  C (g) + 2 O (g)


DU = 740 kJ
DH of this reaction D(C=O)
What about dissociating methane into C
+ 4 H’s?
CH4(g)  C(g) + 4 H(g) DU° = 1640 kJ

4 C-H bonds in CH4 \ D (C-H)  410
kJ/mol

Note: all chemical reactions involve
the breaking and reforming of
chemical bonds
• Bonds break - we add energy.
• Bonds form - energy is released.

DrU°  S D(bonds broken) - S D(bonds
formed)



These are close but not quite exact.
Why?
The bond energies we use are
averaged bond energies !
This is a good approximation for
reactions involving diatomic species.
Can only use the above procedure for
GAS PHASE REACTIONS ONLY!!!