Download CH1101 2014/2015

CH1101 2014/2015
Tutorial: Thermodynamics
Energetic transformations take place
within regions of space. A specified
collection of material particles enclosed in
a specified real or hypothetical
boundary is said to constitute a
thermodynamic system.
The surroundings or environment is labelled
as any region outside the system.
We do our measuring in the surroundings.
The system and surroundings are separated
by a boundary.
In an isolated system there is no transfer
of either energy or matter
between the system and the surroundings.
In a closed system there is transfer of
energy but not of matter across
the system/environment boundary.
In an open system both energy and matter
can be transferred between
the system and the surroundings.
First Law of Thermodynamics: Internal energy U or E.
Internal energy (U or E) : is the total energy of the system at any given time.
Comes from the total kinetic and potential energy of molecules which compose the system.
Change in internal energy (U or E) : energy change as system goes from an initial state with
energy Ui to a final state with energy Uf. Hence U = Uf – Ui.
We suppose that a closed system undergoes a process by which it passes from a state A
to a state B. Then the change in internal energy U = UB – UA is given by DU = q +W.
The first law may be cast in differential form corresponding to the situation here there is
an infinitesimal change in internal energy dU caused by the addition of an infinitesimal quantity
of heat d’q and the performance of an infinitesimal amount of work d’w on the system. Hence
We can write: dU = d’q + d’w. We note that d’q and d’w are inexact differentials – they cannot be
evaluated from a knowledge of the initial and final states alone.
q
 U
W
System
Heat transfer to system
Surroundings
We area led therefore to the important
conclusion that the energy of an isolated
system is constant.
In an isolated system there is no transfer
of heat or work and so q = 0 and W = 0.
Hence U = 0 hence U = constant.
U  q  W
Increase in internal
energy
Work done on system
This is a mathematical statement of the
First Law of Thermodynamics.
Heat Capacity
Increasing temperature increases the internal
energy of a system. The exact increase depends
upon the heating conditions.
Heat cannot be detected or measured directly.
There is no ‘heat meter’.
One way to determine the magnitude of a heat
transfer is to measure the work needed to bring
about the same change in the thermodynamic state
of a system as was produced by heat transfer.
Another approach is to deduce the magnitude of a
heat transfer from its effects: namely, a
temperature change. When a substance is heated
the temperature typically rises. For a specified
energy q transferred by heating, the magnitude of
the resulting temperature change T depends on
the heat capacity C of the substance.
We can therefore simply measure the heat
absorbed or released by a system :
We determine the temperature change and use the
appropriate value of the heat capacity of the
system.
Units of C are J K-1.
Heat capacity is an extensive property.
q
C
T
q  CT
Can also define specific heat capacity c
according to c = C/m where m = mass of
substance with units
J K-1g-1 or the molar heat capacity Cm as Cm =
C/n with units: J K-1mol-1.
The heat capacity depends on whether a sample
is maintained at constant volume (C = CV) or
constant pressure (C = CP). The respective
molar quantities are CV,m and CP,m.
Loss of energy into the surroundings can be
detected by noting whether the temperature
changes as the process proceeds.
This is the principle of a calorimeter.
Specific heat capacity c = quantity of heat
Required to change the temp of 1g of a substance
By 1 K.
c
q
mT
Unit : Jg‐1K‐1
q  cmT
Molar heat capacity Cm =quantity of heat required
to change the temp of 1 mol of substance by 1 K.
q
Cm 
nT
Unit : Jmol‐1K‐1
c(H2O(l)) = 4.184 J g‐1 K‐1
Cm(H2O(l)) = 75.4 J mol‐1 K‐1
Enthalpy H
The change in internal energy is not equal to the energy transferred as heat when the system
is free to change its volume. Under such circumstances some of the energy supplied as heat
to the system is returned to the surroundings as expansion work.
We shall show that under constant pressure conditions, the
energy supplied as heat is equal to the change in another
thermodynamic property of the system, called the enthalpy.
From the first law of thermodynamics for a finite change
at constant pressure
U  q  w  qP  PV
U  U F  U I
Now
V  VF  VI
Hence
U F  U I  qP  PVF  VI   qP  PVF  PVI
qP  U F  PVF   U I  PVI   H F  H I  H
Here we have introduced the enthalpy function H as
H  U  PV
Hence the heat absorbed at constant pressure equals
the increase in enthalpy of the system.
qP  H
For an exothermic reaction U < 0, H < 0,
whereas for an endothermic reaction, U > 0
and H > 0.
The term V is significant for gases.
Relationship between U and H.
Assume ideal gas behaviour. At constant
T and P we have
The enthalpy is given by
PV  nRT
H  U  PV
For a finite change at constant pressure
H  U  PV   U  PV 
 U  PV  VP
 U  PV
Since
H  q P
For reactants and products
PVRx  nRx RT
PVPr  nPr RT
P=0
Constant P
U  qV we note that
Hence
PVPr  VRx   nPr  nRx RT  nRT
where
 n  n Pr  n Rx
qP  qV  PV
Hence for gas phase species
Hence the heat absorbed at constant pressure
Exceeds that absorbed at constant volume by the
Amount PV.
For condensed phases (liquids, solids) V= Vproducts-Vreactants
Will be very small and so V ~0. Hence
H  U
Condensed
phases
Temperature variation
of enthalpy.
H  U  nRT
qP H

T T
H dH  H 
CV  Lim



T 0 T
dT  T  P
CP 
The specific heat capacity of liquid water is
4.18 JK-1g-1. Calculate the energy required to
heat 1.0 mol of water from 298 K to 363 K.
c
q
mT
q  cmT
q  1.0 mol 18 g mol 1    4.18 JK 1 g 1    (363  298) K   4891J  4.9kJ
The standard enthalpy change of vaporization of
ethanol C2H5OH is 43.5 kJmol-1. Calculate the
enthalpy change when 0.5 g of ethanol vaporizes at
its boiling point at 1 bar pressure.
Enthalpy change when 1 mol of ethanol C2H5OH vaporizes = 43.5 kJmol-1.
We calculate amount in mole corresponding to 0.5 g EtOH.
M C2 H5OH  46.07 gmol 1
1 mol C2 H 5OH  46.07 g
1
 2.17 102 mol
46.07
0.5 g C2 H 5OH  0.5  2.17 102  1.085 102 mol
1g C2 H 5OH 
H vap (C2 H 5OH )  (43.5 kJmol 1 )  (1.085 102 mol )  0.472 kJ
The change in internal energy for the combustion of
naphthalene at 298 K which proceeds according to
C10H8(s) + 12 O2(g)  10 CO2(g) + 4 H2O (l) was
measured in a bomb calorimeter as U0 = - 5142 kJ
mol-1. Calculate the enthalpy of combustion of
naphthalene H0 at 298 K.
H 0  U 0  nRT
C10H8(s) + 12 O2(g)  10 CO2(g) + 4 H2O (l)
n   nProducts   nReactants
 10  12  2
We only take gas phase reactants
& products into account.
U 0  5142 kJ mol 1
R  8.314 Jmol 1 K 1
T  298 K
H 0  5142  103 J mol 1  (2 mol )   8.314 Jmol 1 K 1    298K   5142 103 J mol 1  4.955 103 J mol 1
 5.147 103 J mol 1  5147 kJ mol 1
Annual 2012.