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PRT 140 PHYSICAL CHEMISTRY
PROGRAMME INDUSTRIAL
CHEMICAL PROCESS
SEM 1 2013/2014
THE FIRST LAW OF THERMODYNAMIC
BY
PN ROZAINI ABDULLAH
SCHOOL OF BIOPROSES ENGINEERING
©RBA FTK RY 20 2013
LEARNING OUTCOME:
In this chapter you may be able to:
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IMPORTANT CONCEPT IN THERMODYNAMIC
HEAT
ENERGY
WORK
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ENERGY
WORK
Expansion
Contraction
HEAT
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INTERNTIONAL SYSTEM OF UNITS (SI)
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GAS CONSTANT
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WORK
Work (w) is defined as the force (F) that produces
the movement of an object through a distance (d):
Work = force × distance
w=Fxd
(2.1)
Work also has units of J, kJ, cal, kcal etc.
1 J = 1 Nm = 1 kg m2/s2
The two most important types of chemical work are:
– the electrical work done by moving charged particles.
– the expansion work done as a result of a volume change in a system,
particularly from an expanding or contracting gas. This is also known
as pressure-volume work, or PV work.
PV work occurs when the force is the result of a volume change against an
external pressure.
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The example of PV work – in the cylinder of an automobile engine
The combustion of the gasoline causes gases within the cylinder to
expand, pushing the piston outward and ultimately moving the wheel of
the car.
The relationship between a volume change (∆V) and work (w):
W= -P ∆V
(2.2)
Where P is external pressure
The units of PV work are
L·atm; 1 L·atm = 101.3 J
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ΔV
Sign
Gas expands
positive
negative sign
(work energy is leaving the
system)
Gas contracts
negative
positive sign
(work energy is entering the
system).
No change in volume zero
no work done
(This occurs in reactions in
which there is no change in the
number of moles of gas)
Please remember sign!!!!
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EXPANSION WORK
The most common way work is done on a
thermodynamic system is by a change in the system’s
volume.
Expansion
Volume increase
Volume decrease
Contraction
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REVERSIBLE P-V WORK
Therefore, the work done when the system expands by dV against
a pressure pex is :
closed system,
reversible process
dw > 0
In a contraction, the
work done on the
system is positive
Dw<0
In an expansion, the
work done the system is
negative.
(2.3)
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REVERSIBLE P-V WORK
To obtain the total work done when the volume changes from Vi to
Vf , we integrate this expression between the initial and final volume:
Closed syst., rev. proc.
(2.4)
Equation 2.3 & 2.4 Only can be apply to
reversible expansion & contraction.
These equation is apply to mechanically
reversible volume changes.
The expansion at constant pressure.
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EXPANSION AGAINST CONSTANT PRESSURE
Expansion against constant pressure – now suppose that
the external pressure is constant throughout expansion.
Therefore, if we write the
change in volume as
∆V=Vf -Vi
w = - Pex ∆V
The work done by a gas when it
expands against a constant external
pressure Pex is equal to the shaded
area in this example of an indicator
diagram.
(2.5)
Constant
pressure with
no phase
change
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ISOTHERMAL REVERSIBLE EXPANSION
Consider the isothermal, reversible expansion of a perfect gas. The
expansion is made isothermal by keeping the system in thermal
contact with its surroundings (which may be a constanttemperature bath).
pV = nRT ,
p = nRT/V , with V volume at the stage of the
expansion
The temperature, T is constant in an isothermal expansion so
(together with n and R) it may be taken outside the integral.
It follows that the work of reversible isothermal expansion of a
perfect gas from Vi to Vf at a temperature T is
(2.6)
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(2.7)
• When the volume is greater than the initial volume, as
in expansion, the logarithm in above equation is positive
and hence w<0.
•In this case the system has done work on the surroundings
and there is a corresponding reduction in its internal energy.
• From the equation also show that more work is done for a
given change of volume when temperature is increased:
at higher temperature the greater pressure of the confined
gas needs a higher opposing pressure to ensure
reversibility and the work done is correspondingly
greater.
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Calculate the work done when 50 g of iron
reacts with hydrochloric acid to produce
FeCl (aq) and hydrogen in :
(a) A closed vessel of fixed volume
(b) An open beaker at 25oC.
1) We need to judge the magnitude of the
volume and then decide how the process
occur.
2) If there is no change in volume, there is
no expansion work however the process
takes place.
3) A general features of a process in which a
condensed phase changes into a gas is
that the volume of the former may usually
be neglected relative to that of the gas it
forms.
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(a)The volume cannot change, so no expansion work is done and
w = 0.
(b) W = -pex ∆V ≈ -pex x (nRT/pex) = - nRT
Fe (s) + 2 HCl (aq)  FeCl2 (aq) + H2 (g)
W ≈ - (50 g / 55.85 g mol-1 ) x 8.3145 J K-1 mol-1 x 298 K
≈ -2.2 kJ
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Heat
Heat Transfer
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TYPES OF HEAT
TRANSFER
CONDUCTION
CONVECTION
Most
appearance in
solid
conduction
Most
appearance in
liquid and gas
RADIATION
Only form that can
transmit heat over a
vacuum
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HEAT
A
+
B
A
B
Heat is a exchange of thermal energy between a system and its
surroundings caused by temperature difference.
Heat (q)
System
∆T
q α ∆T
The constant of proportionality between q and ∆T is called the
heat capacity.
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HEAT CAPACITY
Heat capacity
of the system
at
that
temperature
Heat capacity is an extensive property
The internal energy of a system increase
when its temperature is raised.
Heat capacity at constant pressure can be
define as:
(2.8)
Internal Energy Vs. Temperature
Heat capacity at constant volume can be
define as:
Eqs. 2.8 & 2.9 apply only
to reversible process
(2.9)
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HEAT CAPACITY AT CONSTANT VOLUME
(2.10)
HEAT CAPACITY AT CONSTANT PRESSURE
(2.11)
For closed system in equilibrium, P-V work only
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MOLAR HEAT CAPACITY
(2.12)
Pure substance, constant pressure
(2.13)
Pure substance, constant volume
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SPECIFIC HEAT
• The specific heat (c, or specific heat capacity, Cs) of an object, is the
quantity of heat required to change the temperature of 1 gram of a
substance by 1 °C (or K):
- Specific heat has units of J / g °C, and is an intensive property, which is
independent of the sample size.
http://oceanservice.noaa.gov/education/pd/oceans_weather_climate/media/specifi
c_heat.swf
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ENERGY
Energy is define as the ability to do
work.
Work is done when a force is exerted
through a distance.
Force through distance; work is done.
Energy is measured in Joules (J) or Calories (cal).
1 J = 1 kg m2 s-2
Energy may be converted from one to another, but it is neither created nor destroyed
(conversion of energy).
In generally, system tend to move from situations of high potential energy (less
stable) to situations having lower energy (more stable).
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UNIT OF ENERGY
• A calorie (cal) is the amount of energy needed to raise the
temperature of 1 g of water by 1°C.
1 cal = 4.184 J
• The nutritional unit Calorie (Cal) is actually a kilocalorie (kcal):
1 Cal = 1000 cal = 1 kCal = 4184 J
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INTERNAL ENERGY
•The internal energy, U of a system is the sum of the kinetic and potential
energies of all the particles that compose the system or the total energy of a
system.
•Internal energy is the state function, which means that its value depends
only the state of the system, not the how the system arrive at the state.
•Some examples include energy (and many other thermodynamic terms),
pressure, volume, altitude, distance, etc.
• An energy change in a system can occur by many different combinations of
heat (q) and work (w), but no matter what the combination, ΔU is always the
same — the amount of the energy change does not depend on how the
change takes place.
∆U = q + w
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A system can exchange energy with its surroundings through heat and work:
Heat (q)
System
Work (w)
Surroundings
According to the first law thermodynamic, the change in the internal energy of the
system (∆U) must be the sum of the heat transferred (q) and the work done (w):
∆U = q +w
(2.14)
Sign of conventions for q, w, and ∆U
q (heat)
w (work)
∆U
(change in
internal
energy)
(+) system gain thermal
energy
(-) System loses thermal
energy
(+) work done on the system (-) Work done by the system
(+) energy flows into the
system
(-) Energy flows out of the
system
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• For an isolated system, with no energy flowing in or out of the system, the
internal energy is a constant.
– First Law of Thermodynamics (restated): The total internal energy of an
isolated system is constant.
• It is impossible to completely isolate a reaction from its surroundings, but it is
possible to measure the change in the internal energy of the system, ΔU, as
energy flows into the system from the surroundings or flows from the system
into the surroundings.
∆U = Uf - Ui
(2.15)
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ENTHALPY
•Most reactions are not done in sealed containers: they are carried
out in open vessels at constant pressure, with the volume capable of
changing freely, especially if the reactants or products of the reaction
involve gases.
• In these cases, ΔV ≠ 0, and the energy change may be due to both
heat transfer and P-V work.
• In order to eliminate the contribution from P-V work, a quantity
called enthalpy, H is defined as internal energy (U) plus the
product of pressure and volume:
H = U + PV
(2.16)
P= pressure, V=volume
Because p,V,U are all state function, the enthalpy is a state function
too.
• The change in enthalpy (ΔH) is:
∆H = qP
(2.17)
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THE MEASUREMENT OF ENTHALPY CHANGE
1) Calorimeter
Monitoring the temperature
change that accompanies a
physical
or
chemical
change
occurring
at
constant pressure.
Example:
thermally
insulated vessel open to
the atmosphere: the heat
released in the reaction is
monitor by measuring the
change in temperature of
the content.
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THE MEASUREMENT OF ENTHALPY CHANGE
2) Bomb Calorimeter
Measuring the internal energy change.
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THE MEASUREMENT OF ENTHALPY CHANGE
3) Differential Scanning Calorimeter
The most sophisticated way to measure enthalpy changes.
Reference pan
Sample pan
pan
Sample
Reference
material
Persistence
thermometer
Heaters
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Example 2:
When 2.0 mol CO2 is heated at the constant
pressure of 1.25 atm, it’s temperature increase
from 250 K to 277K. Given the molar heat
capacity of CO2 at constant pressure is
37.11Jmol-1K-1, calculate q, ∆H and ∆U.
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Solution:
1)
∆ H = qp = (2.0 mol ) x (37.11 Jmol-1K-1) x (277-250) K
= 2.0 x 103 J
2)
∆ U = 2.0 x 103 J – (2.0 mol x 8.3145 J mol-1 K-1) x (277 – 250) K
= 1.983 X 103 J
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ADIABATIC CHANGES
For the reversible adiabatic expansion of a perfect gas, pressure and volume
are related by an expansion that depend on the ratio of heat capacities.
Step 1:
The
volume
changes
and
the
temperature is constant at its initial
value. The system expands at the
constant temperature; there is no
change in internal energy if the
system consist of a perfect gas
Step 2:
The temperature of the system is
reduced at constant volume.
(2.18)
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ADIABATIC CHANGES
q=0
∆U = q + w
= 0 +w
∆U=wad
Therefore, by equating the
two expansion we have
obtained for ∆U we obtained
Wad = Cv ∆T
(2.19)
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Work done during adiabatic expansion of a perfect gas is
proportional to the temperature difference between the initial & final
state.
So a change in internal energy arising from temperature alone is
also expected to be proportional to ∆T.
The initial and final temperature of a perfect gas that undergoes
reversible adiabatic expansion can be calculated :
ViTi = VfTf
The pressure of a perfect gas that undergoes reversible adiabatic
expansion from a volume Vi to a volume Vf is related to its initial
pressure by;
pfVf= piVi
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STANDARD ENTHALPY CHANGES
ENTHALPY OF
PHYSICAL
CHANGE
ENTHALPHY
OF
CHEMICAL
CHANGE
HESS LAW
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ENTHALPY OF CHEMICAL CHANGE
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (l)
∆rHɵ = -890kJmol-1
rH   nH (products )   mH (reactants )
o
o
f
o
f
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USING ΔHfΘ TO CALCULATE ENTHALPY OF REACTION
Calculate the enthalpy of reaction for the
following reaction:
2Al(s) + Cr2O3(s)
Al2O3(s) + 2 Cr(s)
given :
ΔHfθ (Cr2O3(s)) = -1669 kJ/mol
ΔHfθ (Al2O3(s)) = -1128 kJ/mol
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USING ΔHfΘ TO CALCULATE ΔHθrxn
2Al(s) +
Cr2O3(s)
ΔHfθ = 0
ΔHfθ = -1669
Al2O3(s) + 2 Cr(s)
ΔHfθ = -1128 ΔHfθ =0
ΔHrxnθ = Σ of ΔHfθ of products - Σ of ΔHfθ of reactants
= [2(0) + (-1128)] – [(-1669) + 2(0)]
= 541 kJ/mol
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HESS’S LAW
The standard enthalpy of an overall reaction is the
sum of the standard enthalpies of the individual
reactions into which reaction may be divided.
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EXAMPLE 1:
Hydrogen iodide can be prepared from hydrogen
and iodine using two separate routes.
Route I:
Eqn 1: H2(g) +
I2(s) → 2HI(g)
Route II:
eqn 2: I2(s) →
I2(g)
eqn 3: H2(g) + I2(g) → 2HI(g)
∆H = +52.1 kJ
∆H= +61.3 kJ
∆H= -9.2 kJ
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Eqn 1 can be obtained when eqn 2 is added to eqn
3; eqn 1 = eqn 2 + eqn 3
eqn 2:
eqn 3:
I2(s) →
I2(g)
H2(g) + I2(g) → 2HI(g)
H2(g) +
Hence ∆H1
∆H= +61.3 kJ
∆H= -9.2 kJ
I2(s) → 2HI(g)
= ∆H2
+
∆H3
= 61.3 – 9.2 = +52.1 kJ
The total enthalpy change for the route I is the
same as that for route II.
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EXAMPLE 2:
ΔH for formation of SO3 cannot be obtained
directly but the enthalpy of these reactions are
known:
S(s)  O 2 (g )  SO 2 (g ); H  -297 kJ
o
2SO 3 (g )  2SO 2 (g )  O 2 (g ); H  198 kJ
o
The above data can be used to obtain the enthalpy
change for the formation of SO3 according to the
following reaction?
2S(s )  3O 2 (g )  2SO 3 (g ); H  ?
o
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The third equation can be obtained by multiplying the first equation
by 2 and added to the reverse of the second equation, they will sum
together to become the third.
2S(s)  2O 2 (g )  2SO 2 (g ); H  (-297 kJ)  (2)
o
2SO 2 (g )  O 2 (g )  2SO 3 (g ); H o  (198 kJ)  (-1)
2S(s )  3O 2 (g )  2SO 3 (g ); H  -792 kJ
o
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EXERCISE 1:
Determine the heat of reaction;
Fe2O3(s) + FeO(s)
Fe3O4(s)
Using the information below:
i) 2Fe(s) + O2(g)
2FeO(s)
∆Ho = -554.0 kJ
ii) 4Fe(s) + 3O2(g)
2Fe2O3(s) ∆Ho = -1648.8 kJ
iii) 3Fe(s) + 2O2(g)
Fe3O4(s) ∆Ho = -1118.4 kJ
(Answer : -22.0 kJ)
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ENTHALPY DIAGRAM ILLUSTRATING HESS’S LAW
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THE JOULE & JOULE – THOMSON EXPERIMENT
What is Joule Thomson Experiment?
The objective of this experiment is to quantitatively measure the nonideality of gases using the Joule-Thomson coefficient and relating it to
the coefficients of equations for non-ideality and the Lennard-Jones
potential.
For an ideal gas, the internal energy is only a function of the absolute
temperature so in an isothermal process ΔE = 0. The same is true for
the enthalpy for such a process: ΔH = 0.
Thus:
 H   H 

 
 0
 V T  P T
These are non-zero for a non-ideal gas.
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For the perfect gas, the internal energy is independent of the volume (at
constant temperature).
If repulsion is dominant, the internal
energy decrease as the gas expand.
If attraction is dominant in a real gas, the
internal energy increase with volume
because the molecules become farther
apart on average.
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The internal pressure, πT is the slope
of U with respect to V with the
temperature T held constant.
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He thought that he can
measure
the
πT
by
observing the change in
temperature of a gas when
it is allowed to expand in a
vacuum.
A schematic diagram of the
apparatus used by Joule in an
attempt to measure the change in
internal energy when a gas
expand isothermally.
The heat absorbed by the gas is
proportional to the change in
temperature of the bath.
Water bath
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The thermodynamic implication as follow:
1.No work was done in the expansion into a vacuum, so w=0.
2.No energy entered of left the system (the gas) as heat because the
temperature of the bath did not change, so q = 0. Consequently, within
the accuracy of the experiment, ∆U=0.
Joule concluded that:
U does not change when a gas expand isothermally and therefore that
=0.
The heat capacity of the apparatus was so large that the temperature change
that gases do in fact cause was too small to measure.
Nevertheless, from his experiment Joule had extracted an essential limiting
property of a gas, a property of a perfect gas, without a small deviations
characteristic of a real gas.
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OBSERVATION OF THE JOULE-THOMPSON EFFECT
The analysis of the Joule-Thompson coefficient is central to the
technological problems associate with the liquefaction of gases. We
need to be able to interpret its physical and measured it.
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The pistons represent the upstream
and
downstream
gases,
which
maintain constant pressure either side
of throttle.
The transition from the top to bottom
diagram which represents the passage of a
given amount of gas through the throttle,
occurs without change of enthalpy.
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The gas on the left is compressed isothermally
by the upstream gas acting as piston.
The work done on the gas is:
The gas expands isothermally on the right
barrier against the pressure pf provided the
downstream gas acting like piston to driven
out. The volume changes from 0 to Vf, so the
work done on the gas at this stage:
The total work done on this stage:
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It follows the change of internal energy of the gas as it moves adiabatically
from one side of the barrier to the other is
Reorganization of this expression gives:
or
Therefore, the expansion occurs without change of enthalpy
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
The partial derivative
 T 


 P H
is known as the Joule-Thomson coefficient, JT.
©RBA FTK RY 20 2013
+ve sign → dT is –ve when
dp is –ve
(in which case the gas cool
on expansion)
A gas typically has two
inversion
temperatures,
one at high temperature
and the other at low.
©RBA FTK RY 20 2013
Renungan:
“Termodinamik merupakan satu subjek yang aneh. Kali
pertama anda mempelajarinya, anda tidak faham langsung.
Kali kedua anda mempelajarinya, anda fikir anda
memahaminya kecuali satu dua perkara. Kali ketiga anda
mempelajarinya,
anda
tahu
bahawa
anda
tidak
memahaminya, akan tetapi anda sudah terbiasa dengan
subjek itu maka subjek itu tidak akan menjadi masalah lagi
kepada anda”
Arnorld Sommerfeld (1868-1951)
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