Transcript U / ∂V

Faculty of Engineering
Cairo University
Physical Chemistry
Second year Metallurgy
CHAPTER TWO
The First Law of Thermodynamics
Prof. Dr. Fawzy
Prepared by :
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Ahmed Ibrahim Ali.
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Amina Abd El-Hai.
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Ibrahim Mahmoud.

Mohamed Ayman.

Mohamed Hafez.
Contents:
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2.1 Introduction
2.2 The Relationship Between Heat and Work
2.3 Internal Energy and The First Law of
Thermodynamics
2.4 Constant – Volume Processes
2.5 Constant – Pressure Processes & Definition of
Enthalpy (H)
2.6 Heat Capacity
2.7 Reversible adiabatic processes for a real gas
2.8 reversible isothermal processes
2.1 Introduction
In a frictionless kinetic system of interacting elastic bodies, kinetic energy is
conserved; i.e., a collision between two of the bodies of this system results in a
transfer of kinetic energy from one to the other; thus, the work done by one equals
the work done by the other, and the total kinetic energy of the system is unchanged
as a result of the collision
If the kinetic system is in the influence of the gravitational filed , thus the
sum of the kinetic and potential energies of the bodies is constant ;
however, as the result of possible interactions kinetic energy may be
converted to potential energy and vice versa but the sum of the two
dynamitic energies (kinetic and potential) remains constant
If, however, frictional energies are operative in the system then with continues collision
and interaction among the bodies, the total dynamic energy of the system decreases
and heat is produced. It is thus reasonable to expect that there is a relationship
between the dynamic energy lost and heat produced
The establishment of this relationship laid the foundations for the development of
the thermodynamic subject
2.2 The Relationship Between Heat and Work
In 1798, Count Rumford suggested the first relation between heat and work; he
noticed that during the boring of cannon at the Munich arsenal, the heat produced
was roughly proportional to the work performed during the boring process. He
suggested that heat is an invisible fluid (gas), called caloric, which resided between
the particles of the substance. In the caloric theory, it had been assumed that the
temperature of a substance is determined by the quantity of the caloric gas which it
contains; it was also assumed that the amount of caloric per unit mass is less for
smaller particles than larger particles. These two assumptions explain the flow of
the heat from the bodies with higher temperature to cold bodies; and the sensible
heat produced during poring larger pieces to form smaller pieces, metal turnings
In 1799, the caloric theory was discredited when Humphrey Davy melted two blocks
of ice by rubbing them together. Based on this experiment, it had been proven the laten
heat necessary to melt the ice was provided the mechanical work performed rubbing th
blocks together
In 1840, Joule conducted experiments in which work was performed in a certain
quantity adiabatically, i.e. the bath is defined contained water and measured the
resultant temperature rise of the water. He observed that a direct proportionality
exists between the work done and the temperature rise. He observed further that the
same proportionality exist, no matter what means were employed in the work
production.
▪ Methods for work production used are:
1. Rotating paddle wheel immersed in water.
2. Electric motor driving current through coil immersed in water.
3. Compressing a cylinder of gas immersed in water.
4. Rubbing together two metal blocks immersed in water.
These experiments placed on a firm quantitative basis for the thermodynamics
object. By defining the calorie, as a heat unit, as the amount of heat required to raise
the temperature of one gram of water from 14.5° c to 15.5° c it was found that:
1calorie = 4.184 joule
This constant, 4.184, is known as the mechanical equivalent of heat
2.3 Internal Energy and The First Law of Thermodynamics
The development of thermodynamics from its early beginnings to its present state was
achieved as a result of the definition of convenient thermodynamic functions of state. In
these section the first of these functions, the internal energy U, is introduced.
Joule’s experiments resulted in a statement that ” the change of a state of a body
inside an adiabatic enclosure from a given initial state to a given final state
involve the same amount of work by whatever means the process is carried
out”
This statement is a preliminary formulation of the
“First Law of Thermodynamics”
Similar to the following relation:
W=potential energy of body m at position 2 (mgh2)-potential energy of
body m at position 1 (mgh1)
For a body of mass m when lifted in a gravitational field, g, from height h, to
height H (where W is the work done on the body); and also when particle
with charge q, is moved in a electric field from point at potential Q1, to point a
potential Q2, the work done, W, on the charged particle is given as,
W=electric energy at point 2(qQ2) - electric energy at point 1(qQ1)
It is possible to write, for the joule adiabatic process in which W is done on
a body,as a result of which it’s state moved from state A to state B W=UB-UA
In describing the sign of work in this case, W will be positive by convention, if
the work is done on the system, UB > UA and thus W will be negative if the
work is done by the system UB <UA
In Joule’s experiment, the change in the state of the adiabatically contained water
was measured as a temperature of the water. The same temperature rise, and hence
the same change of state, could have been produced by placing the water in a
thermal contact with a source of heat and allowing heat q to flow into the water. In
describing the sign of q, it is by convention to assign a positive value of q if heat flows
into the body (endothermic process) and negative value for q if heat flows out of the
body (exothermic process); hence q = (UB –UA).
Now for a body which absorbs heat q and work W is performed on it, it is possible to
show that:
∆U = q + w → (1)
Let us first consider that a thermodynamic system changes its state A to state B` as a
result of heat flow q from a heat reservoir to the system, then:
Q = UB` - UA
If the body is adiabatically isolated, i.e. the path is defined, the work W done on the
system causes a change from state UB` to state UB, thus:
W = UB - UB`
Therefore,
Or
q + w = (UB` - UA) + (UB - UB`)
= UB – UA
∆U = q + w
For infinitesimal changes, we can write:
dU = δq + δw → (2)
Equations (1) and (2) are the mathematical statement of the First Law of
Thermodynamics.
From figure 1 (ref. figure 2.1 p. 20) it can be noticed that:
∆U = U2 – U1,
I.e. U is a state function,
w = ∫12PdV; thus dependant on the path;
i.e. W is not a state function because it is
path dependent; and accordingly, q is a
path dependent and accordingly it is not a
state function.
Fig. 2.1 Three process paths taken by a fixed quantity of
gas in moving from state 1 to state 2
Since U is a state function; thus,
U = U (P, V) = U (P, T)
= U (V, T)
Thus; the complete differentials of U in terms of its partial derivatives are
given by:
dU = (∂U/∂P)V dP + (∂U/∂V)P dV
= (∂U/∂P)T dP + (∂U/∂T)P dT
= (∂U/∂V)T dV + (∂U/∂T)V dT
Also we have:
∫12dU = U2 – U1, and
∫ dU = 0, i.e. the cyclic integral equals zero
2.4 Constant – Volume Processes
Since δw = Pdv, then δw = 0 for constant – volume process; and from the first law
of thermodynamics, we have :
dU =
δqv
And by integration we have:
ΔU = qv
2.5 Constant–Pressure Processes & Definition of Enthalpy (H)
If the pressure of the system maintained constant during a process which takes the
system from state 1, to state 2, and if the process is reversible ,i.e. Pexternal = Pinternal
, then the work done by the system is given as :
W = -∫v1v2 Pdv = -P (v2 –v1)
Then:
Then:
ΔU = qp + [-p(v2 – v1)]
qp = (u2 + pv2) – (u1 + pv1)
As the expression (u + pv) contains only state function, then the expression itself
is a state function; this expression is termed the enthalpy H, thus :
H = U + PV
Hence, for a constant pressure process ,
qp = ΔH
2.6 Heat Capacity
The heat capacity C, of a system is the ratio of the heat added or withdrawn from the
system to the resultant change in the temperature of the system, thus;
C = q / ΔT
Or, for infinitesimal change:
C = δq / ∂T
The concept of heat capacity is only used when the addition of heat to or withdrawal of
Heat from the system produces a temperature change; thus, the concept is not used
when the process is isothermal or when a phase change occurs.
The change of temperature of a system from T1 to T2 due to the admission of a certain
quantity of heat to the system is not complete thermodynamic statement since the state
of a system is defined by two independent variables; thus, this second independent
variable could be varied in a specified manner or cold be maintained constant during
the change.
Thus, we define the heat capacity term either at constant volume, Cv, or and constant
pressure, Cp as follows:
Cv = (δq/∂T)v
Cp = (δq/∂T)p
The heat capacity as defined is an extensive function; however, by defining the specific
heat capacity, heat capacity of the system per gram, or the molar heat capacity, heat
capacity per mole; thus, the heat capacities per unit quantity of the system are intensive
functions.
Since:
Cv = (δq/∂T)v
Thus:
Cv = (∂U/∂T)v
Or
dU = Cv dT
Also, as:
Cp = (δq/∂T)p
Thus:
Cp = (∂H/∂T)p
Or
dH = Cp dT
It is expected that, for any substance, Cp will be of greater magnitude than Cv since
the heat required the increase the body temperature by one degree involved the
amount of heat required to cause expansion of the body against constant pressure
per degree of temperature increase.
The relation between Cp and Cv (Cp – Cv) can be calculated as follows:
Cp – Cv = (∂H/∂T)p – (∂U/∂T)v
= (∂U/∂T)p + P (∂V/∂T)p - (∂U/∂T)v
As:
Then:
U= f (V, T)
dU= (∂U / ∂v)T.dV + (∂U / ∂T)V.dT
Thus:
(∂U/ ∂T)p= (∂U/ ∂V)T * (∂V / ∂T)p + (∂U / ∂T)v
So:
Cp – Cv = (∂U/ ∂ V)T * (∂V/ ∂T)p +(∂U/ ∂T)v
+P *(∂V/ ∂T)p – (∂U/ ∂T)v
Cp –Cv = (∂V/ ∂T)p *(P + (∂U/ ∂V)T)
Joule's proved through experimentation that (∂U/ ∂V)T = 0 for ideal gas; thus
1. Cp – Cv =P * (∂V/ ∂T)p =P * ∂/ ∂T (RT/P)p =R
2. For ideal gases : U=F (t) since (∂U/ ∂V)T = 0
Joule's experiment involved two copper vessels; one is filled with a gas at same
pressure connected to a similar evacuated vessel via a stopcock.
The two-vessel system was immersed in a quantity of adiabatically contained water;
the stopcock was then opened, thus allowing free expansion of the gas into the
evacuated vessel.
After this expansion, Joule could not detect any change in the temperature of the
system. As the system was adiabatically contained and no work was performed;
Then, from the first law:
∆U = 0
and hence:
dU=(∂U /∂V)T .dV + (∂U/∂T)V.dT = 0
Since ∂T = 0, and ∂V≠0, thus (∂U /∂v)T = 0,
i.e. The Internal energy of a perfect gas is not function of volume (and hence pressure);
it is function only of temperature.
The reason for Joule's not observing a temperature rise in the original experiment was
that the heat capacity of the copper vessels and the water was considerably greater
then the heat capacity of the gas; thus, the small heat changes which actually occurred
in the gas were absorbed in the copper vessels and the water.
The Cp/Cv relation is given by :
Cp- Cv = P * (dV/dT)p +(dU/dV)T *(dV/dT)p
The term P * (dV/ dT)p represents the work done by the system per degree rise in
temperature due to expending against the constant pressure P acting on the system.
The term (∂U/ ∂V)T *(∂V/ ∂T)p represents the work done per degree rise in temperature
in expanding against the internal cohesive forces acting between the particles of the
substance.
Thus, for perfect gases the value of (∂U/ ∂V)T *(∂V/ ∂T)p = 0,for real gases the
magnitude of P *(∂V/ ∂T)p is much greater then the value of (∂U / ∂V)T * (∂V/ ∂T)p which
approaches the zero value, and for condensed phase, the magnitude of
(∂U/ ∂V)T *(∂V/ ∂T)p is much greater than the value of P(∂V/ ∂T)p which approaches
the zero value.
2.7 Reversible adiabatic processes for a real gas
In a reversible adiabatic process for a real gas the value of q=0 and the external
pressure is almost equal the gas internal pressure. Thus from the first law of
thermodynamics and by using the equation of state of a real gas, we can show that:
PVγ = constant
TVγ-1 =constant
TP(1-γ)/γ =constant
2.8 reversible isothermal processes
As dT= 0, then dU= 0; accordingly:
δq= -δw = p dV
δq= -δw = p dV
Thus:
q= -w = RT ln( V2/V1)= RT ln(p1/p2)
A reversible isothermal process and a
reversible adiabatic process are shown
on the
P-V diagram of figure 2 in which it
seen that, for a given pressure
decrease; the work
done by the reversible isothermal
process exceeds that done by the
reversible
adiabatic process.