Download 1st Law Of Thermodynamics Part 2

ST
1
LAW THERMODYNAMICS
2nd part
ROHAZITA BT BAHARI
Basic concept:
1. Work, heat and energy
2. The internal energy
3. Expansion work
4. Heat transaction
5. Adiabatic changes
Thermochemistry:
1. Standard enthalpy changes
2. Standard enthalpies of
formation
3. The temperature dependence
of reaction enthalpies.
First Law
Thermodynamic
State of function and exact differentials:
1. Exact and inexact differentials
2. Changes in internal energy
3. The Joule – Thomson effect
THERMOCHEMISRTY
THERMOCHEMISTRY: BASIC TERMS
• Thermochemistry is the study
of energy changes that occur
during chemical reactions.
• System: the part of the universe
being studied.
• Surroundings: the rest of the
universe.
TYPES OF SYSTEMS:
– open (exchange of mass and energy)
– closed (exchange of energy)
– isolated (no exchange)
Open system
Close System
Isolated system
TYPES OF SYSTEMS
• Open: energy and
matter can be
exchanged with the
surroundings.
• Closed: energy can
be exchanged with
the surroundings,
matter cannot.
• Isolated: neither
energy nor matter
can be exchanged
with the
surroundings.
A closed system;
energy (not matter)
can be exchanged.
After the lid of the jar
is unscrewed, which
kind of system is it?
ENTHALPY (H)

Enthalpy can be defined as:
H  U  PV
Internal Pressure
energy

Volume
Enthalpy is a state function (values depend on
current state of the system, not on HOW the
system acquired that state, which is
independent of path.)
ENTHALPY (H)
• The energy possessed by a system is called the
enthalpy or heat content of the system and is
given the symbol H.
The change of the heat content is given by ΔH,
ΔH = Hfinal - Hinitial
ΔH = Hproduct - Hreactant
HEAT (Q)
Heat is energy transfer resulting from thermal
differences between the system and surroundings
“flows” spontaneously
from higher T 
lower T
“flow” ceases at
thermal equilibrium
EOS
ENTHALPY(H)
VS
HEAT (Q)
• When reactions take place at constant
pressure;
heat change = enthalpy change
q = ΔH
The quantity of heat energy liberated or absorbed
during a reaction taking place at constant pressure is
the enthalpy change of the reaction.
EXOTHERMIC REACTION
• A chemical reaction that releases/gives off heat
is called an exothermic reaction
During an exothermic process, heat flows
out of the system and into the surrounding.
Thus the energy of the system is negative.
ΔH = -ve
Eg : C(s) + O2(g)
CO(g) ΔH = -393.5 kJ/mol
EXOTHERMIC REACTION
Surroundings are at 25 °C
25 °C
Typical situation:
some heat is released
to the surroundings,
some heat is absorbed
by the solution.
Hypothetical situation: all heat
is instantly released to the
surroundings. Heat = qrxn
32.2 °C
35.4 °C
In an isolated system, all heat is
absorbed by the solution.
Maximum temperature rise.
ENDOTHERMIC REACTION
• A chemical reaction that absorbs/takes in heat
is called an endothermic reaction.
During an endothermic process, heat flows
into the system from the surrounding.
Thus the energy of the system is positive.
ΔH = +ve
ENTHALPY (CONSTANT VOLUME)

Consider a constant-volume process:
U  q  w
V2
 qV  P  dV
U  qV

Constant
volume
V1
For an infinitesimal change,
dU  dqV
INTERNAL ENERGY CHANGE AT CONSTANT VOLUME
• For a system where the
reaction is carried out at
constant volume, V = 0
and U = qV.
• All the thermal energy
produced by conversion
from chemical energy is
released as heat; no PV work is done.
ENTHALPY (CONSTANT PRESSURE)

Energy transferred as heat at constant
pressure is equal to the change in enthalpy of
the system.
H  qP
U

U

q

w
2
1
HOW?
V
2
 qP  P  dV
V1
 qP  PV2  V1 
ENTHALPY (CONSTANT PRESSURE)
U 2  U1  qP  PV2  V1 

Rearranging,
qP  U 2  PV2   U1  PV1 
qP  H 2  H1
qP  H

For a measurable change
For an infinitesimal change,
dH  dqP
INTERNAL ENERGY CHANGE AT CONSTANT PRESSURE
• For a system where the
reaction is carried out at
constant pressure,
U = qP – PV or
U + PV = qP
• Most of the thermal
energy is released as
heat.
• Some work is done to
expand the system
against the surroundings
(push back the
atmosphere).
ENTHALPY AND ENTHALPY CHANGE
Enthalpy is the sum of
the internal energy and
the pressure-volume
product of a system:
 H = U + PV

The evolved H2
pushes back the
atmosphere; work
is done at
constant
pressure.
Mg + 2 HCl  MgCl2 + H2
For a process carried
out at constant
pressure,
so
qP = H
 qP = U + PV
Most reactions occur at constant
pressure, so for most reactions, the
heat evolved equals the enthalpy
change.
ENTHALPY DIAGRAMS
• Values of H are measured experimentally.
• Negative values indicate exothermic reactions.
• Positive values indicate endothermic reactions.
A decrease in enthalpy
during the reaction; H
is negative.
An increase in enthalpy
during the reaction; H
is positive.
THERMOCHEMICAL EQUATIONS
• A themochemical equation is an equation that
represents a reaction and shows the overall enthalpy
changes during the reaction.
• The following are three important rules for writing
thermochemical equations:
1. The physical states (s, l, g) of all reactants
and products must be specified.
2H2O(l)
2H2O(g) ΔH= 88.0 kJ/mol
2. WHEN BOTH SIDES OF A THERMOCHEMICAL
EQUATION IS MULTIPLIED BY A FACTOR, THEN ΔH
MUST ALSO BE MULTIPLIED BY THE SAME
FACTOR.
H2O(s)
H2O(l) ΔH= 6.01 kJ/mol
2H2O(s)
2H2O(l) ΔH= 12.02 kJ/mol
3. When a chemical equation is reversed,
the value of ΔH is reversed in sign
REVERSING A REACTION
• H changes sign when a process is reversed.
Same magnitude; different signs.
HOW TO MEASURE ENTHALPY CHANGE

1. DIRECT METHOD ( through experiment)
 Calorimetry
Heat absorb or evolve is measured using a
 calorimeter
 Two types of calorimeter :



Coffee-cup calorimeter
Bomb calorimeter
CALORIMETRY
Calorimetry is a technique used to measure heat
exchange in chemical reactions
Calorimetry – measuring the heat
flow associated with a chemical
reaction by measuring the
temperature.
EOS
CALORIMETRY
• We measure heat flow using calorimetry.
• A calorimeter is a device used to make this
measurement.
• A “coffee cup” calorimeter may be used for
measuring heat involving solutions.
A “bomb” calorimeter is used to
find heat of combustion; the
“bomb” contains oxygen and a
sample of the material to be burned.
CALORIMETRY, HEAT CAPACITY, SPECIFIC
HEAT
• Heat evolved in a reaction is absorbed by the calorimeter
and its contents.
• In a calorimeter we measure the temperature change of
water or a solution to determine the heat absorbed or
evolved by a reaction.
• The heat capacity (C) of a system is the quantity of heat
required to change the temperature of the system by 1 °C.
C = q/T (units are J/°C)
• Molar heat capacity is the heat capacity of one mole of a
substance.
• The specific heat (s) is the heat capacity of one gram of a
pure substance (or homogeneous mixture).
s = C/m = q/(mT)
q = s m T
HEAT CAPACITY: A THOUGHT EXPERIMENT
• Place an empty iron pot weighing 5 lb on the
burner of a stove.
• Place an iron pot weighing 1 lb and containing 4 lb
water on a second identical burner (same total
mass).
• Turn on both burners. Wait five minutes.
• Which pot handle can you grab with your bare
hand?
• Iron has a lower specific heat than does water. It
takes less heat to “warm up” iron than it does
water.
SPECIFIC HEAT CAPACITY
• The specific heat capacity (or “specific heat”)
is the heat required to raise the temperature
of 1 gram of a substance by 1 oC.
• Specific heat capacity of water :
4.184 J /(g oC).
4.184 J energy is needed to raise the
temperature of 1 g of water b 1 oC
HOW DOES A CALORIMETER MEASURE
HEAT (Q)
• Heat evolved in a reaction is absorbed by the
calorimeter and its contents (water or
solution).
• In a calorimeter the temperature change of
water or a solution is measured to determine
the heat absorbed or evolved by a reaction.
TO CALCULATE Q
• q is calculated using
q = mass x specific heat capacity x T
q = m c T
• If T is positive (temperature increases), q is
positive and heat is gained by the system.
• If T is negative (temperature decreases), q is
negative and heat is lost by the system
CALCULATION OF Q - EXAMPLE
• Calculate the heat absorbed when the
temperature of 15.0 grams of water is raised
from 20.0 oC to 50.0 oC. (The specific heat of
water is 4.184 J/g.oC.)
using the equation q = mc T
T = (50oC – 20oC ) = 30oC, therefore
q = 4.184 x 15 x 30 = 188.28 J
= 1.88 kJ
CALCULATING ENTHALPY USING DIRECT
METHOD
• When 23.6 grams of calcium chloride, CaCl2,
was dissolved in 500 mL water in a calorimeter,
the temperature rose from 25.0 oC to 38.7 oC.
Given the heat capacity of water is 4.18 J/g oC,
and density of water is 1 g/mL, what is the
enthalpy change per mole of calcium chloride?
HEATS OF REACTION: CALORIMETRY
– First, calculate the heat absorbed by the
calorimeter.
q = mc T
q= 500 g x 4.18 J/g oC x (38.7- 25) oC
q= 28633 J = 28.633kJ
• Now we must calculate the heat per mole of
calcium chloride.
HEATS OF REACTION: CALORIMETRY
• Calcium chloride has a molecular mass of
111.1 g, so
(1 mol CaCl2 )
(23.6 g CaCl2 )
 0.212mol CaCl2
111.1 g
• Now we can calculate the heat per mole of
calcium chloride.
qrxn
 28.633 kJ
H 

 135.06 kJ / mol
mol CaCl2
0.212 mol
CALCULATING ΔHΘ BY
CALORIMETRY
N
• 25.0 cm3 of 1.00 M HCl at 21.5oC were placed in a
polystyrene cup and 25.0 cm3 of 1.00 M NaOH at
21.5oC was added. The mixture was stirred, and the
temperature rise to 28.2oC. The density of each
solution =
4.18J/(K g).
Calculatethe standard
 enthalpy of neutralization.
 1. calculate heat change using
q  mcT

molar
Q= MCT
= 50.0 G X 4.18 J/(K G) X 6.7 OC= 1400 J = 1.4
KJ
2. Calculate heat change per mole (ΔHnθ )
mole of H2O formed in the reaction=
25.0 x10-3 dm3 x 1.0 M = 2.5 x10-2 mol
2.5 x10-2 mol H2O formed = 1.4 kJ
1 mol H2O formed = 1
x 1.4
2.5 x10-2 mol
= 56 kJ/mol
2. CALCULATING ENTHALPY USING DATA
FROM
Standard Enthalpies of Formation
• The law of summation of heats of formation
states that the enthalpy of a reaction is equal
to the total formation energy of the products
minus that of the reactants.
Ho   nHof (products)  mHof (reactants)
  is the mathematical symbol meaning “the
sum of”, and m and n are the coefficients of the
substances in the chemical equation.
USING ΔHθF 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
θ
USING ΔHFθ TO CALCULATE ΔH
RXN
2Al(s) + Cr2O3(s)
Al2O3(s) + 2 Cr(s)
ΔHfθ = 0
ΔHfθ = -1128
ΔHfθ = -1669
ΔHfθ =0
θ
θ
ΔHrxnθ = Σ of ΔHof
f products - Σ of ΔH of
f reactants
= [2(0) + (-1128)] – [(-1669) + 2(0)]
= 541 kJ/mol
HESS’S LAW
Hess’s law states the enthalpy change of a reaction is constant
whether the reaction is carried out directly in one single step or
indirectly through a number of steps.
System A
ΔH1θ
ΔH4θ
System B
ΔH2θ
ΔH3θ
System A’
System A’’
EXAMPLE 1 : HESS’S LAW
• Hydrogen iodide can be prepared from hydrogen and
iodine using two separate routes.
• Route I:
Eqn 1: H2(g) + I2(s) →2HI(g)
∆H = +52.1 kJ
• Route II:
eqn 2: I2(s) → I2(g)
∆H= +61.3 kJ
eqn 3: H2(g) + I2(g) → 2HI(g)
∆H= -9.2 kJ
Eqn 1 can be obtained when eqn 2 is added to eqn
3; eqn 1 = eqn 2 + eqn 3
eqn 2: I2(s) → I2(g)
eqn 3: H2(g) + I2(g) → 2HI(g)
H2(g) + I2(s) → 2HI(g
∆H= +61.3 kJ
∆H= -9.2 kJ
Hence ∆H1 = ∆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.
HESS’S LAW
• Example 2; ΔH for formation of SO3 cannot be
obtained directly but the enthalpy of these reactions
are known:
S(s)  O (g)  SO (g); Ho  -297kJ
2
2
2SO (g)  2SO (g)  O (g); Ho  198kJ
3
2
2
The above data can be used to obtain the enthalpy
change for the formation of SO3 according to the
following reaction?
2S(s)  3O2 (g)  2SO3 (g); Ho  ?
• 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.
o
2S(s)  2O2 (g)  2SO2 (g); H  (-297kJ)(2)
o
2SO2 (g)  O2 (g)  2SO3 (g); H  (198kJ)(-1)
2S(s)  3O (g)  2SO (g); Ho  -792 kJ
2
3
EXERCISE
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)
iii) 3Fe(s) + 2O2(g)
• (Answer : -22.0 kJ)
2Fe2O3(s) ∆Ho = -1648.8 kJ
Fe3O4(s) ∆Ho = -1118.4 kJ
THERMOCHEMICAL CYCLE / ENERGY
DIAGRAM
Reactant ( A + B)
route 2
∆Ho
2
route 1
∆Ho
route 3
Product (C + D )
∆Ho
3
Enthalpy diagram illustrating
Hess’s law.
HESS’S LAW: AN ENTHALPY DIAGRAM
We can find H(a) by
subtracting H(b) from H(c)
BORN HARBER CYCLE
• Born Haber cycle is an energy cycle used to
calculate the lattice energy which cannot be
obtained by direct experimental method. They
only can be obtained by applying Hess’s law in
this cycle which involving breaking and
forming bonds.
• LATTICE ENERGY
THE ENTHALPY CHANGE WHEN ONE
MOLE OF IONIC COMPOUND
(CRYSTALLINE SUBSTANCE) IS FORMED
FROM ITS GASEOUS IONS.
Na+(g) + Cl-(g)
3+
2Al (g) + 3O2 (g)
NaCl(s)
Al2O3(s)
∆Hlat = -788 kJ/mol
∆Hlat = -1596 kJ/mol
II) ENTHALPY OF
ATOMIZATION
•
The enthalpy change when one mole of gaseous
atoms is formed from its elements under standard
conditions
•
•
Na(s)
½ O2(g)
Na(g)
O(g)
H = +108 kJ/mol
H = +247 kJ/mol
iii) Electron Affinity
The enthalpy change when one mole of gaseous atom gains
•
one mole of electron to form anion.
O(g) + eO-(g) + e-
O-(g)
O2-(g)
H = -141.0 kJ/mol
(1st electron affinity)
H = -744.0 kJ/mol
(2nd electron affinity)
IV) IONIZATION ENERGY
THE STANDARD IONIZATION ENERGY IS THE
ENTHALPY REQUIRED TO REMOVE ONE
MOLE OF ELECTRONS FROM ONE MOLE OF
GASEOUS METALLIC ATOMS TO FORM ONE
MOLE OF POSITIVELY CHARGED IONS. THE
PROCESS IS ENDOTHERMIC BECAUSE
ENERGY IS ABSORBED TO RELEASE
ELECTRONS
FROM
AN
ATOM
.
+
st
(1 ionization energy)
Mg(g) - e
Mg (g)
2
Mg
+
(2nd ionization energy)
Mg (g) - e
+(g)
• BORN HABER CYCLE
FOR LIF
CALCULATE THE LATTICE ENERGY FOR NACL(S)
USING THE FOLLOWING DATA.
a. standard enthalpy of formation of NaCl(s) ∆Hf = -411kJ/mol
b. standard enthalpy of atomization of Na(s) ∆Ha = +108 kJ/mol
c. first ionization energy of Na(s)
∆HIE
d. standard enthalpy of atomization of Cl2(s)
e. first electron affinity of Cl2(g)
∆HEA
f. Standard lattice enthalpy of NaCl(s)
∆Hlat
= -+494 kJ/mol
∆Hf
= +121 kJ/mol
= -364 kJ/mol
= x kJ/mol
MEASURING ENTHALPY – KIRCHHOFF’S
LAW
The standard enthalpy of many important reactions have been measure at
Different temperatures.
However, in the absence of this information standard reaction enthalpies at
different temperature may be calculated from heat capacities and reaction
enthalpy at some at some other temperature.
When temperature is increased, the
enthalpy of the products and reactants
both increase, but may do so to different
extent.
In each case, the change in enthalpy
depends on the heat capacities of the
substances.
The change in reaction enthalpy reflects
the difference in the changes of the
enthalpy.
WHEN A SUBSTANCE IS HEATED FROM T1 TO T2, ITS ENTHALPY
CHANGES FROM H(T1) TO
Eqn 1
Assumed that no phase transition take place in the
temperature range of interest) because this equation
applies to each substance in the reaction the
standard reaction, the standard reaction enthalpy
changes from ∆Hrɵ(T1) to
Kirchhoff ’s
Law
Eqn 2
Where the
is the difference of the molar heat capacities of product and
Reactants under standard conditions weighted by the stoichiometri coefficient that
Appear in chemical equation:
Eqn 3
Example:
The standard enthalpy of formation of H2O (g) at 298K is 241.82kJ mol-1.
Estimate its value at 100oC given the following values of the molar heat capacities
At constant pressure:
H2O (g) = 33.58JK-1mol-1
H2(g) = 28.82 JK-1mol-1
O2 = 29.36 JK-1 mol-1
Assume that the heat capacities are independent of temperature.
Method: When
is independent of temperature in the range T1 to T2, the
Integral in eqn 2 evaluate to (T2-T1)
. Therefore,
To proceed, write the chemical equation, identify the stoichiometri coefficients, and
Calculate
from the data.
It then follows that
ANSWER: THE REACTION IS H
2
SO
(G) + ½ O2 (G) → H2(G),
State Function
and
Exact Differentials
The internal energy and enthalpy are two example of state functions.
Physical quantities that do depend on the path between two state are called
path function.
Example of path functions are the work and the heating that are done when
preparing a state.
We do not speak of a system in a particular state as possessing work or heat.
In each case, the energy transferred as work or heat relates to the path being
taken between states, not the current state itself.
A part of the richness to thermodynamic is that it uses the mathematical properties
of state functions to draw far-reaching conclusions about the relations between
physical properties and thereby establish connections that may be completely
unexpected.
The practical importance of this ability is that we can combine measurements of
different properties to obtain the value of a property we required.
EXACT AND INEXACT DIFFERENTIALS
Initial state
of the
system
Work is done by the
system as it expands
adiabatically to a state
f.
Initial and final state
are the same as
those in Path 1 but in
which expansion is
not adiabatic.
In these state the
system has an internal
energy Uf and the work
done on the system
along Path 1 from I to f
of is w.
U: property of the state, w=property of the path
However, in the second path an energy q’ enter the system as
heat and the work w’ is not the same as w. The work and the
heat are path functions.
Initial and final states
are the same as
before (because U is
the state funtion).
If a system is taken along a path (for example, by heating it) U changes from Ui
to Uf, and overall change is the sum (integral) of all the infinestimal changes along
the path:
Eqn 4
The value of ∆U depends on initial state and final state of the system but is
independent of the path between them.
An exact differential is an infinestimal quatity that when integrated gives a result
that is independent of the path between the initial and final state.
When the a system is heated the total energy transferred as heat is the sum of all
Individual contributions at each point of the path:
Eqn 5
Inexact differential is an infinitesimal quantity that, when integrated gives a
result that depends on the path between the initial and final state.
The differences between Eqn 4 and Eqn 5:
1. We do not write ∆q because q is not a state function and energy supplied as heat
cannot be expressed as qf-qi.
2. We must specified the path of integration because q depends on the path selected
(example: and adiabatic path has q=0, whereas on the non-adiabatic path between
the same two state would have q≠0).
In general, an inexact differential is an infinestimal quantity that, when integrated,
gives a result that depends on the path between the initial and final state.
The work done on a system to change it from one state to another depends on the
path taken between the two specified states.
example: in general the work is different if the change takes place adiabatically and
non-adiabatically. It folows that dw is an exact differential.
CHANGES IN INTERNAL ENERGY
In internal energy U can be regarded as a function of V,T, and p, but because
there is an equation of state, stating the values of two of the variables fixes the
value of the third.
Therefore, it is possible to write U in terms of just two independent variables:
V and T, p and T, or p and V.
Expressing U as a function of volume and temperature fits the purpose of our
discussion.
GENERAL EXPRESSION FOR A CHANGE IN U WITH T AND V
Eqn 6
Definition of internal pressure
Eqn 7
In terms of the notification Cv and
,eqn 5 can now write as
Eqn 8
THE
JOULE
EXPERIMENT
James Joule thought that he could measure
by observing the chenge in
Temperature of a gas when it is allowed to expand into vacuum.
When the stopcock opened and
the air expanded into a vacuum.
Filled with air
at about 22
atm
No change in
temperature
A schematic diagram of the apparatus used by
Joule in an attempt to measure the change in
internal energy when a gas expands isothermally.
The heat absorbed by the gas is proportional to the
change in temperature of the bath.
The 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
His experiment, how ever is crude.
=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.
CHANGES IN INTERNAL ENERGY AT CONSTANT
PRESSURE
As an example, suppose we want to find out how the internal energy varies with
temperature when the pressure rather than volume of the system is kept
contant.
If we divided both sides of eqn 8 by dT and impose the condition of constant
pressure on the resulting differentials, so that du/dT on the left becomes
(∂U/ ∂T)p we obtain:
Eqn 9
It is usually in thermodynamic to inspect the output of a manipulation like this
to see if it contains any recognizable physical quantity. The partial derivative
on the right in this expression is the slope of the plot of volume against
temperature (at constant pressure). This property is normally tabulated as
the expansion coefficient, α, of a substance, which is defined as:
Definition of the
expansion
coefficient
Eqn 10
A large value of α means that the volume of the sample respond strongly to
changes in temperature.
List some experimental values of α
ISOTHERMAL COMPRESSIBILITY, KT( KAPPA)
WHICH DEFINE AS
Eqn 11
Isothermal compressibility is a measure of the fractional change in a volume
when the pressure is increase by a small amount; the negative sign in definition
ensure that the compressibility is positive quantity; because an increase of
pressure, implying a positive dp, bring about reduction of volume, a negative dV.
THE JOULE-THOMPSON EFFECT
The closed system of constant composition
Eqn 12
Where the Joule-Thompson coefficient, μ (mu) is define as
Eqn 13
This relation will prove useful for relating the heat capacities at constant volume
and for a discussion of liquefaction of gases.
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.
The apparatus used to for measuring
The Joule-Thompson effect. The gas
Expands through the porous barrier,
Which acts as a throttle, and the whole
apparatus is thermally insulated.
This arrangement corresponds to an
Isenthalpic expansion
(expansion at constant enthalpy).
Whether the expansion results in a
heating
or a cooling of the gas depends on the
conditions
DEFINITON OF THE
ISOTHERMAL JOULETHOMPSON
COEFFICIENT
The isothermal Joule –Thompson
coefficients is the slope of the enthalpy
with respect to changing pressure the
temperature being held constant.
By compressing Eqn 12 and above equation, we can see that the two coefficients
are related by
A schematic diagram of the apparatus used for measuring the
isothermal Joule-Thompson coefficient. The electrical heating required
to offset the cooling arising from expansion is interpreted as ∆H and
used to calculate
, which is then converted to μ.
Real gases have nonzero Joule-Thompson
coefficient. Depending on the identity of the
gas, pressure, the relative magnitudes of the
attractive and repulsive intermolecular forces,
and the temperature, the sign of the coefficient
may be either positive or negative.
+ve sign → dT is –ve when dp is –ve
(in which case the gas cool on expansion)
Heating effect (μ<0) at one temperature show at
cooling effect (μ>0) when the temperature is below
their upper inversion temperature,T1.
A gas typically has two inversion
temperatures, one at high
temperature and the other at low.
The principle of Linde refrigerator
The gas at high temperature is allowed to expand through a throttle, it cools
and is circulated past the incoming gas.
That gas is cooled, and its subsequent expansion cools it still further.