Download Thermodynamics: Lecture 2

Thermodynamics: Lecture 2
Review of Lecture 1
o Concept of system and Surroundings
o Isolated, closed and open
o Concept of work
o Sign convention
o Mechanical work in stretching DNA
Today
o Concept of heat
o Heat capacity of an object.
o 1 st law of thermodynamics.
o Characterizing state of System.
o State variables.
 Intensive.
 Extensive.
 Enthalpy and Internal energy
o Equation of state (i.e. relation between state
variables)
 Solids/liquid
 Gases
o Path connecting different states
 Reversible vs. Irreversible
 Examples.
 Isothermal, isobaric, adiabatic.
 Examples.
 Phase change
Heat
When we lift weights or exercise a common sensation
is that we feel warm. This sensation of warmth or heat is
measured by temperature.
Hot coffee becomes lukewarm if we do not drink it
quickly. More succinctly, we can say that when two bodies
at different temperature when brought in contact will
develop identical temperature through exchange of heat.
How do we quantify heat? A common observation is that it
takes lot more heat to warm water than corresponding
volume of air (Think about heating water versus air with a
hair dryer). Thus, we measure heat by defining a special
quantity, heat capacity. Heat capacity of an object is the
amount of heat needed to raise it’s temperature by 1 degree.
dq q
C

dT T
Where C is the heat capacity, and q, T are heat and
temperature. Since we can only measure temperature we
can rewrite:
q   C.dT  C (T2  T1 )
Generally C does not vary much with temperature except
near phase transition. T2 and T1 are the final and initial
temperatures. So if system cools then the sign of q is
negative since it looses heat to surrounding and vice versa.
Units of C are J/K. Similarly we can define specific molar
heat capacities depending on whether it is measured under
conditions of constant pressure or volume.
Relation between Work and Heat
The first law of thermodynamics.
In 1840, Joule performed pioneering set of experiments
to show:
W  Jq
Where a direct proportionality between work and heat
was established. The conversion factor J, is
4.18Joules/calorie in honor of Joule.
Some of his experiments were very simple. He let
fixed weight to rotate paddle in water and measured the
rise in temperature. Similarly he passed current
through resistor and measured the rise in temperature.
Now we know that in a closed system, energy can
flow in either as work done on the system or as heat
flow. If the energy flows into the system it must be
provided by surroundings. Otherwise we will have
invented a perpetual motion machine. Although
different forms of energy may be inter-converted, the
first law states that the total energy of system and
surroundings is conserved. For a closed system we may
therefore write:
E  q  W
H. Von Helmhotz stated this in 1847. For isolated
system ΔE=0. What about an open system?
Characterizing the State of System: State variables
So far we have seen how a system in thermodynamics
is identified and how different processes such as work or
heat can influence it. However to be more specific, we
have to identify certain key properties of system. We do
that using “state variables”.
Easiest way to think of state variables of system is to
think of driver’s license. It gives very pertinent
information, like date of birth, height, weight and picture
etc. Thus, it is useful to decide on what type of information
we need to specify about the system.
As a first try, we may specify P, V, T, and
composition of system to define it specifically. However,
we set an important criterion for the selection variables of
state: State variables are those variables that only depend
on state, and not on how the system arrived at that state.
This excludes W the work as a state variable and also q!
However E the energy is a state variable! We can also
define new state variable by using already known state
variables. For example, Enthalpy
H  E  PV
A further subdivision of state variables involves on whether
they depend on the size of system. Those that depend on
the size are called extensive variable, e.g., V,H, E; and
those which do not, such as T and P, are called as intensive
variables. An extensive variable can be converted to an
intensive variable by appropriate division, e.g. density.
Equation of State
As you can see, we may have too many state variables. One
of the ways we can eliminate the redundant variables is
through equation of state. Simply stated equation of state
represents a relationship between state variables.
For example, if our system is made up of ideal gas then we
may use PV= nRT as way to eliminate one of the 4 state
variables. So to specify the state of an ideal gas we only
need 3 state variables. But this process of identifying
equation of state is complex and involves approximations,
which may not be justified. Therefore, it is essential to
experimentally establish an empirical the equation of state
as shown in figure 2.3.
Once one has an equation of state, we can predict how the
state will change as the state variables are changed.
However, in practice there are many ways to change the
state of a system, some reversible and some irreversible. In
thermodynamics, the reversible process are useful in that
they allow us to calculate precisely the work, or internal
energy changes involved, with aid of the equation of state.
Four important reversible paths for gases are (1) isothermal
(ΔT=0) (2) Isobaric (ΔP=0) (3) Adiabatic (q=0) (4)
Isochoric (ΔV=0) and (5) Cyclic (no change in the state of
system). Any irreversible path can be reconstructed using
combination of these reversible paths.
Example 1 Evaporation of water
Path 1:
Irreversible path
P=1atm
n=1mole(l)
T=25
P=1atm
n= 1 mole (vapor)
T=25C
Reversibly raise
Temperature
Reversibly
cool.
P=1atm
n=1 mole(v)
T=100C
P=1atm
n= 1 mole(l)
T= 100C
Isothermal evaporation
Path 2: Vacuum Evaporation
P=1atm
n= 1 mole(l)
T=25
Irreversible path
Isothermally
Reduce Pressure
P=0.001 atm
n=1 mole(l)
T= 25 C
Isothermal evaporation
P=1atm
n= 1mole (vapor)
T=25C
Isothermally
increase P
P=0.001atm
n=1mole(vapor)
T=25C
Note the final result of irreversible evaporation is the same.
Whether we boil the water as in example 1, or vacuum
evaporate the water as in example 2, the final state of the
system is indistinguishable.
So what? What is the big Deal?
The big deal is that we can calculate the work and heat, and
hence the energy changes that occur in any irreversible
process by breaking it down in terms of smaller reversible
steps. Consider again path 1.
Step 1: it is isobaric, and isochoric (volume change is
negligible). So the pV work done is 0.
ΔE1=q1=CLp(T2-T1), ΔH1=ΔE1.
Step 2: This requires us to provide latent heat of
evaporation and also it involves a massive change in
volume; so pV work is also involved.
ΔE2=qL +W= qL –P(VV-VL); ΔH=ΔE2+Δ(PV)= qL
Step 3: Involves isobaric cooling. Where only the volume
of gas decreases as the vapor is cooled.
ΔE3=q3 +W=CVp(T1-T2)-P(VT2-VT1) ; ΔH3=ΔE3+Δ(PV)=q3
This shows how an irreversible process can be broken
down in terms of series of reversible processes and hence
we can calculate associated changes in the state variables
such as energy and enthalpy.
More details, general comments
We have seen from the above examples how we can
calculate the new state of a system when the state variables
change. The specific paths we chose such as isothermal,
isobaric etc are particularly useful for the calculation of
heat and work.
From extensive body of thermodynamic work we know
(1) The p-V work is negligible if the transformation
involves solids or liquid unless we are studying
ultra-high pressure systems.
(2) In a given phase, e.g., liquid, gas, or solid, the
variation in specific heat (Cp or Cv) is essentially
negligible over wide ranges of temperature. The
only exceptions to this rule are phase transitions
such as melting or boiling, where specific heats
change dramatically.
(3) The p-V work is most significant for gaseous
systems. For a system of gas trapped in a cylinder
let us consider few specific paths
p-V work
1. Isobaric
W= -P(V2-V1) …(Expansion)
W= +P(V2-V1) …(Compression)
P
T1
V
2. Isochoric W=0 since there is no change in volume.
3. Isothermal reversible. In this case, to perform reversible
work we have to follow the equation of state
W    P.dV
P
Ideal Gas Law
PV  nRT V  P 
T1
V
W   nRT 
nRT
V
V 
dV
 nRT ln  2 
V
 V1 
For an isothermal
rev. process of an ideal gas the change in ΔE=0;q = -W
Phase Change
During phase transition (liquid-vapor) changes in both heat
and work occur. Normally, we study such transition at
constant pressure and temperature. (Think of melting point
measurement.) For liquid-liquid or solid-liquid or solidsolid phase transition the pV work is negligible, so ΔE=ΔH.
This energy goes to rearrange molecular packing in the
system. But if phase transition involves gas, (i.e., boiling)
there is considerable pV type work. Consider again path 1:
P=1atm
n=1mole(l)
T=25
P=1atm
n= 1 mole (vapor)
T=25C
P=1atm
n=1 mole(v)
T=100C
P=1 atm
n= 1 mole(l)
T= 100C
As we have shown before total energy and enthalpy
changes are:
ET  E1  E 2  E3
& H T  H 1  H 2  H 3
ET  (C PL  C PV )(T2  T1 )  q L  P(VTV2  VTL2 )
H T  (C PL  C PV )(T2  T1 )  q L
Note at constant pressure
q L  H (T2 )