Chapter 2 : Slide 1

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Chapter 2 The First Law of Thermodynamics (I)
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SYSTEMS The SYSTEM is the part of the universe
we are interested in. The rest of the universe is
the SURROUNDINGS. Properties of the system may
be INTENSIVE Do not depend on the quantity of
matter e.g. density, temperature,
pressure. EXTENSIVE Do depend on the
quantity of matter e.g. mass, volume. The
state of a system is uniquely defined in terms of
a few variables, that may be linked by an
EQUATION OF STATE.
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OUTLINE SECTION 2.1 - Work, heat and
energy SECTION 2.2 - The First Law SECTION 2.3 -
Expansion work SECTION 2.4 - Heat
transactions SECTION 2.5 - Enthalpy SECTION 2.6 -
Adiabatic changes SECTION 2.7 - Standard enthalpy
changes SECTION 2.8 - Enthalpies of
formation SECTION 2.9 - Temperature dependence of
reaction enthalpies
HOMEWORK ASSIGNMENTS EXERCISES 2.4 - 2.44, Part
(a)s only PROBLEMS 2.2, 2.6, 2.10, 2.16, 2.26
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Classification of systems If matter can be
transferred between the system and the
surroundings then it is called OPEN. If matter
cannot be exchanged with the surroundings, but
energy can be, the system is called CLOSED. If
neither energy nor matter can be transferred the
system is ISOLATED.
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WORK Work is done during a process if that
process could (in principle) be used to change
the height of a weight in the surroundings e.g.
gas pressure moving a piston, stretching a spring
etc. If this (imaginary) weight is raised then
we say "work has been done by the system". If
this (imaginary) weight is lowered then we say
"work has been done on the system". ENERGY Energy
is the capacity of a system to do
work. Compressing a gas or a spring enables more
work to be done i.e. it increases the energy of
the system. While a system does work, its energy
is reduced.
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HEAT We can change the energy of a system other
than by doing work. If the energy of a system
changes because of a temperature difference
between it and the surroundings, we say "energy
has been transferred as heat". If the system is
enclosed in walls that are good insulators, so
heat cannot be transferred even if there is a
temperature difference from the surroundings,
then the system is called ADIABATIC e.g. a
thermos or Dewar flask. If the walls permit easy
passage of heat the system is DIATHERMIC. Any
process that releases heat is called EXOTHERMIC,
e.g. combustion. Any process that absorbs heat
is called ENDOTHERMIC, e.g. evaporation.
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Molecular picture Heating the system is the
transfer of energy that depends on the thermal
motion of atoms in the surroundings - their
random molecular motion. Doing work on the
system is the transfer of energy which depends on
organized motion of the atoms in the surroundings
e.g. if a spring is compressed all the atoms move
in an organized way.
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The FIRST LAW The total energy of a system is
called the INTERNAL ENERGY, U, and is measured in
joules, J. 1 J 1 kg m2 s-2 In thermodynamics
we shall often deal with changes in internal
energy but not the absolute value ?U Uf - Ui
where "f" and "i" are for final and initial. U
is a state property which depends only on the
current state of the system and not on how that
state was reached. Other examples are p, V, T
etc. Variables like these that specify the state
of a system are called STATE FUNCTIONS or
FUNCTIONS OF STATE. When the state of a system
changes, the change in a state function depends
only on the initial and final states, and not on
the path taken.
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EXAMPLE Height h. Dh is
independent of the route taken so h is a
state function. The path length ? varies, so
? is not a function of state.
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If we add heat q to a system U increases DU
q. If work w is done on the system DU w
which is minus the work done by the system. In
general, DU q w which is the FIRST LAW OF
THERMODYNAMICS.
Alternative versions are The energy of an
isolated system is constant The internal
energy of a system is constant unless it is
changed by doing work or by heating The work
required to change an adiabatic system from a
specific state to another is the same however the
work is done
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We may consider infinitesimal changes in U dU
dq dw From physics we know that to move an
object a distance dz against an opposing force F
requires work to be done by the system, so the
work done on the system is dw -F dz (the
negative sign shows U of the system
decreases) If the object moves from zi to zf,
the total work done is
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Examples 1. A system raises a mass m by a
distance h. What is w? F -mg so w
-F(zf-zi) -mgh. 2. The system contains a
spring which is compressed by a distance x from
the equilibrium position, zi 0. What is
w? The force law (Hooke's Law) is F
-kz (k is the FORCE CONSTANT) so dw kz dz
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Types of work E.G. Electrical, surface tension
(change in liquid surface area), osmotic (change
in concentration), mechanical (especially volume
changes) etc. We will mainly focus on volume
changes here, work of compression and
expansion. Consider an expansion
Force pexA dV Adz dw -pexAdz -pexdV
For a compression dV lt 0 and thus dw gt 0. FREE
EXPANSION pex0 and w0.
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Types of changes If the system is kept in
equilibrium all the time a change is carried out
then that process is called REVERSIBLE.
Example Compressing a gas (the system) in a
cylinder by pushing in the piston very slowly. If
the piston were pushed in fast the gas near the
piston would heat up and the system would no
longer be in equilibrium. This would be an
IRREVERSIBLE process. Truly reversible processes
are ideal they are carried out infinitely
slowly. For such a process, infinitesimal
changes of conditions in opposite directions
cause opposite changes in the state of the system.
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Example Consider the reversible, isothermal
expansion of 1 mol of ideal gas.
Reversible, so p pex RT/V. dw -pdV -RT/V
dV
This is the minus the area on the indicator
diagram. Irreversible bring pex down instantly
from pi to pf. The work is less. Max work
performed is -wrev.
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This type of expansion/compression work is called
"pV workand in this context the First Law
becomes dU dq dw dq - pdV Constant
volume conditions dU dqV (dqV dq at
constant V) and dV 0 so DU U2 - U1
qV. Thus the increase in U of any system in a
rigid container is qV. Even though the p may
increase no pV work is done "the system cannot
do work on itself".
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Constant pressure conditions dqp dU
pdV qp ?dU ?pdV (U2 - U1) p (V2 -
V1) (U2 pV2) - (U1 pV1). We shall
define a new function, called ENTHALPY, H H
U pV so qP H2 - H1 DH. DH is the heat
transfer at constant pressure. Because U, p and
V are state functions, so is H.
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THERMOCHEMISTRY A general chemical reaction can
be written as (-?1)S1 (-?2)S2 .... ?nSn
?n1Sn1 .... ?j is the STOICHIOMETRIC
COEFFICIENT for the species Sj. ?j is Positive
for Products, and negative for reactants. The
overall reaction can be written as ??jSj
0. STANDARD STATES These are denoted by a
superscript o or ? . Ho(T) is the value of H
at the standard p and the specified T. Unless
otherwised specified, T is usually 298 K. The
standard pressure is 105 Pa 1 bar. The
substance must be in its most thermodynamically
stable form e.g. carbon in its standard state
is Cgraphite, not Cdiamond. oxygen is O2(g),
ethanol is C2H5OH(l) etc.
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For solutions, commonly met standard states
are a) a molality of 1 mol/kg (amount of
substance/mass of solvent) b) a molarity of 1
mol/dm3 (amount of substance/volume of (
1 M) solution).
Measurements of DH are measurements of heat
changes at constant pressure (qP). In practice
we used a sealed CALORIMETER and measure heat
changes at constant volume (qV).
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H U pV DH DU D(pV) which for ideal
gases becomes DH DU D(nRT) DU RT
Dn Dn is the change in the amount of gas during
the reaction. What about liquids and
solids? The calorimeter directly yields DU (via
qV), from which we can calculate DH. Other
methods to measure DH include i) The
temperature dependence of equilibrium constants
(see later). ii) Spectroscopic measurements.
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Hess's Law A restatement of the idea that
enthalpy is a function of state. DH for a
reaction is the same however we get from
reactants to products, by whatever intermediate
steps (all chemical reactions must of course
balance). We can set up Hess cycles to calculate
DH for reactions which are difficult to measure
directly.
DH1 DH3 DH2
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We do not want to have to tabulate DHo for every
possible reaction. Instead, we can use a much
shorter list of STANDARD ENTHALPIES OF FORMATION,
DfHo. DfHo is the enthalpy for forming 1 mole of
a substance from its elements in their standard
states. Examples DfHo (CH4(g)) is DHo for C(gr)
2 H2(g) ? CH4(g) -75 kJ mol-1 DfHo
(H2O(l)) is DHo for H2(g) 1/2 O2(g) ? H2O(l)
-286 kJ mol-1 DfHo (CO2(g)) is DHo for C(gr)
O2(g) ? CO2(g) -394 kJ mol-1 DfHo for
any element in its most stable form is zero by
definition. Combustion example
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For any reaction DHo ?i ?i DfHoi
DfHo(products)-DfHo(reactants) Example 2 CO
O2 ? 2 CO2 DfHo/kJ mol-1 -111 0 -396 ?
-2 -1 2 so DHo/kJ mol-1 (-2 x -111)
(-1 x 0) (2 x -396) -570. The system has
lost enthalpy the reaction is EXOTHERMIC. ENDOTH
ERMIC reactions have DHo gt 0.
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BOND STRENGTHS Imagine the ATOMIZATION
process CH4(g) ? C(g) 4 H(g) i.e. dissociation
of all the bonds, to form gaseous atoms. DHo for
this process is the ENTHALPY OF ATOMIZATION,
DatHo, which for this example is 1664 kJ
mol-1. This is four times the MEAN BOND
DISSOCIATION ENTHALPY for the C-H bonds in
CH4. The MEAN BOND DISSOCIATION ENTHALPY is
1664 kJ mol-1 / 4 416 kJ mol-1. Note
individual bond strengths for successive C-H
bonds, for stepwise removal of H atoms, are
different. They total 1664 kJ mol-1.
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HEAT CAPACITY Note difference between molar and
specific quantities. The heat capacity at
constant volume is CV CV dqV/dT dU/dT at
const. vol. (?U/?T)V Here, dqV CVdT dU
which, provided CV is not a function of T, is
CV(T2T1). Similarly, the heat capacity at
constant pressure is Cp Cp dqp/dT dH/dT at
const. press. (?H/?T)p
For liquids and solids V hardly changes with T
(usually), so CV?Cp. For ideal gases we will
prove later that Cp CV nR (the ratio
Cp/CV is called ?).
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Two Examples We will be considering one mole of
an ideal gas. A. Reversible compression at
constant pressure. Initial state A p1, V1 and
T1. Final state B p1, V2 and T2. Remove heat
reversibly from the system (e.g. by infinitely
slow reduction in T of the surroundings).
p
V
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We have specified the path so w and q can be
calculated.
R(T1-T2)
Sign check V1gtV2 so wrevgt0 T1gtT2 so wrev still
gt0.
For an ideal gas Cp and CV are independent of T
(for proof see later) so qp Cp(T2 - T1).
The heat absorbed at constant p is equal to ?H,
thus ?H Cp(T2 - T1) which is negative.
?U q w always Cp(T2 - T1) R(T1 -
T2) (Cp - R)(T2 - T1) CV(T2 - T1) using
the result Cp CV R
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As a check, is the above consistent with "?H ?U
?(pV)" ? ?U p1 ?V (Cp - R)(T2 - T1)
p1(V2 - V1) (Cp - R)(T2 - T1) R(T2 -
T1) Cp (T2 - T1) ?H
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B. Reversible pressure drop at constant
volume. p1, V1, T1 ? p2, V1, T2
wrev 0 no pV work at constant volume qV
CV(T2 - T1) ?U here this is
negative.
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?H ?U ?(pV) ?U ?(RT) CV(T2 - T1)
R(T2 - T1) (CV R)(T2 - T1) Cp(T2 -
T1) These are the SAME ?U and ?H values we had
for case A. Thus U and H for an ideal gas are
functions of T only (not p nor V) and are
constant along any isotherm
Applies only to ideal gases.
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Adiabatic changes "ADIABATIC" means without any
heat transfer q 0. Consider the adiabatic
expansion of 1 mol of ideal gas.
1st law dU dq - pdV (here dq 0) and dU
CV dT therefore CV dT -p dV. Now for 1 mol
of ideal gas, p RT/V, thus CV dT/T -R/V
dV so
Thus CV ln(T2/T1) R ln(V1/V2) substitute R
Cp - CV ln(T2/T1) (Cp - CV)/CV ln(V1/V2)
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Let us define the ratio Cp/CV ? ln(T2/T1) (?
- 1) ln (V1/V2) so T2/T1 (V1/V2)? -1 but
T2/T1 p2V2 / (p1V1) so p2/p1 (V1/V2)? . In
other words, along an adiabat, p2V2? p1V1? or
pV? is constant. Because ? gt 1 (see later),
adiabats are steeper than isotherms. DH
Cp(T2-T1) and DU CV(T2-T1) q 0 so wrev
DU CVDT.
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The magnitude of ? For a monatomic ideal gas,
the total kinetic energy of the molecules
1.5 RT U so CV dU/dT 1.5 R. ? Cp/CV
(1.5 R R)/(1.5 R) 5/3. Molecular gases
have a larger CV, and thus a smaller ? (? always
gt 1).
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A practical example of adiabatic
expansion/contraction If a body of air in the
atmosphere changes height, the pressure and hence
volume change, leading to a temperature
change. Let pressure be p at a height h. Molar
mass M 0.029 kg mol-1, g acceleration due
to gravity 9.81 m s-1. In the atmosphere,
dp/dh -pMg/RT. What is dT/dh ? This
is called the ADIABATIC LAPSE RATE.
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TEMPERATURE DEPENDENCE OF DH AT CONSTANT P For a
reaction, DHo ??i DfHoi so d(DHo)/dT
d(DfH(products)-DfH(reactants))/dT
Cp(products)-Cp(reactants) which is defined as
DCp ??iCp,i . This yields Kirchoff's Law
Often, for a small range of T, it is a reasonable
approximation to take DCp as constant. For more
detailed work DCp is expressed as a function of
T. The form DCp a bT c/T2 often gives a
good fit to experimental data.
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Example What is DHo at 298 K and 2000 K for
the reaction 2 H2(g) O2(g) ? 2 H2O(g) ?
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Adiabatic flame temperatures Suppose we burn a
stoichiometric mixture of H2 and O2. How hot will
the flame be? Why is this an
overestimate? i) Real flames not truly
adiabatic. ii) Cp increases with T. iii) Reaction
does not go to completion. Compare to an H2/air
flame. The N2 is a diluent. It does not increase
DH, but still needs to be heated up. It increases
the heat capacity of the mixture, so the final
temperature is lower.
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• SUMMARY
• Systems may be open, closed or isolated.
• Definitions of work, internal energy (U) and
  work. First Law DU q w, or dU dq dw.
• U, and enthalpy, H, are state functions. H U
  pV.
• pV work and reversible/irreversible volume
  changes for an ideal gas at constant volume or
  pressure.
• Calorimetry and enthalpies of combustion.
• Thermochemistry Hess's Law, thermochemical
  cycles and standard enthalpies of formation.
• Relation to bond strengths and enthalpies of
  atomization.
• The heat capacities Cp and CV relation to
  temperature dependence of DH and DU for a process
  (Kirchoff's Law) and ? for adiabatic changes.