Introduction
Energies changes accompanies all chemical or physical transformations.
Examples:
•Drop of a stone
•The motion of billiard ball
•The burning of coal
•Reaction of complex mechanism
The most common form in which the energy appears and the form to which all others tend, is heat.
•Other type of energies:
1.Mechanical energy
2.Electrical energy
3.Radian energy
4.Chemical energy The energy stored in all substances.
All these different types of energies are related to one other and under certain conditions maybe transformed from one to other.
Thermodynamics means the study of interconversions of various forms of energy in a system.
Since thermodynamic laws deal with energy, they are applicable to all phenomena in nature.
Thermodynamic laws hold only on macroscopic systems rather than on the behavior of microscopic systems.
•Thermodynamics does not consider the time element it is interested in the initial and final state without any curiosity as to speed.
•Any system have two types of energy
1.Kinetic energy
2.Potential energy
The sum of both energies represent the total energy content of any system .
• equation E=mc2 calculate the absolute value of the total energy, this fact of little help in ordinary thermodynamics. Why?
•Therefore thermodynamics deals with the energy difference which accompany changes since these can be measured
The first law of thermodynamics
•It’s the law of conservation of energy its states that “energy can be neither created nor destroyed”
Mathematically it can express as:
q= ∆E + w ………. 1
∆E= E2 – E1 ………. 2
∆E depends only on the initial and final state
of a system and independent on the path .
However the value of q and w depends on the manner in which the process is conducted .
The quantities q, w, ∆E are measurable, but the absolute magnitude of E1 and E2 are not.
f= P x A
d w = f d l = p x A x d l
since A d l is the
element of volume d V
d w = f d l = P d V …(3)
by integration :
W= ∫ P d V …….(4)
∆E = q – W
= q - ∫ P d V ……..(5)
If d V = 0 ∆E = q ………6
if P is constant ∫ P d V = P (V2 – V1)
∆E = q – P (V2 - V1)
q = ∆E + P ∆V ………7
The enthalpy (heat content) of the system
•Thermal changes at constant pressure are expressed in terms of another function H called enthalpy of heat content where,
H = E + P V ……..8
this function is also like E completely independent of the manner in which the change was achieved it depends only on the initial and final state of the system thus
∆H = H2 - H1 ...........9
= (E2- P2V2) – (E1- P1V1)
= (E2-E1) + (P2V2 – P1V1)
= ∆E + (P2V2 – P1V1) ………..10
if P is constant
∆H = ∆E + P(V2-V1)
= ∆E + P ∆V ………..11
This mean that the change in enthalpy at constant pressure is equal to the increase in the internal energy plus any pressure – volume work done.
Heat capacity
•Heat capacity is defined as the amount of heat required to raise the temperature of the system 1 degree .
Thus C = dq /dT
Since ∆E = q - ∫ pdv
hence dq = dE - pdv
so C = ( dE + pdv ) / dT ……….12
If the volume is held constant ( dv = 0 )
Hence Cv = (dE / dT)v ……………..13
•So Cv is the rate of change of the internal energy with temperature at constant volume.
•When the heat absorbed at constant pressure equation 12 becomes:
Cp= (∂E/ ∂T)p + p ( ∂v / ∂T)p
since H = E + pv, by differentiation with respect to T at constant p we get:
(∂H/ ∂T)p= (∂E/ ∂T)p+ p( ∂v / ∂T)p
Consequently, Cp= (∂H/ ∂T)p
•So Cpis the rate of change of the enthalpy with temperature at constant pressure.
Cp-Cv= (∂H/ ∂T)p-- (∂E/ ∂T)v ………15
Which by theoretical treatment lead to the conclusion:
Cp-Cv= R (the general gas constant)……21
Cp/Cv= γ
The work of expansion of gases
•It was obviously illustrated that the work of expansion can be calculated from (4)
w = ∫v1v2 p dv …………..4
•P is not the pressure of the working gas but the pressure against which the gas is working.
•If dv = 0 or if p = 0 w = 0
•The later case involve free expansion (expansion into vacuum).
•Under these conditions ∆E = q
•Integration of equation (4)give:
w = p ( v2 – v1) ………..(22)
Isothermal and adiabatic expansion of ideal gases
•A) isothermal expansion in a system takes place at constant temperature.
Since T is a function of the internal energy, thus at constant temperature ΔE = 0
Appling the first law, then q = w………(23)
i. e the work performed occurs at the expense of the adsorbed heat .
q = w = ∫v1v2 p dv …………….(24)
If p is constant then,
q = w = p (v2- v1)…………(25)
B) adiabatic expansion takes place under condition that heat is neither absorbed nor evolved by the system.
i.e q = 0 then, w = - ∆E ……….(26)
This means that an adiabatic process is done at the expense of the internal energy, consequently the temperature drops.
Therefore pdv = - dE = - ncvdT ……..(27)
If the expansion is controlled that the external pressure differs from that of the internal only by an infinitesimal amount, then p may be substituted by p = nRT / v
Hence, - ncvdT = nRT dv /v
cv/R ∫T1T2 ( dT / T ) = - ∫ dv / v
by integration give:
Cv/R ln T2/T1 = - ln V2/V1
V1 T1cv/R = V2T2cv/R …………..(28)
other forms of equation (28) may be easily derived by eliminating dT or dv instead of p from (27)
a very common form is one involving p&v
P1V1y = P2V2y ...........(29)
Other form
T1/T2 = ( P1/P2)y-1/y
Post a Comment