lecture 5-6 (statistical physics )
THERMAL EQUILLIBRIUM :
we have to find condition for a system which is in thermal equillibrium .
Consider an isolated system of perfect gas which is subdivided into two systems by a conducting wall as shown in given figure .
Let S1 and S2 be the entropies of two components then total entropy is S=S1 +S2
When system is at equillibrium then ih has maximum entropy ( and change in entropy will be zero )
So in this case the system entropy changed due to the transfer of heat from one component to other . So we can take entropy as a function of internal energy .
S=S(U)
let us define a quantity
dS/dU =1/T
It means change of entropy with respect to cahnge in internal energy must be a function of temperature .also in the present case S=S(U1 +U2 )
as S =S1+S2
dS = dS/dU1 X dU1 + dS/dU2 X dU2 [eq 1 ]
for isolated system :
U=U1+U2
dU1 +dU2 =0
dU1 = – dU2
dU2 = -dU1
from eq 1
dS = dS/dU1 X dU1 + dS /dU2 X -dU1
dS = dS/dU1 X dU1 – dS /dU2 X dU1
As dS= 0
0= dS/dU1 X dU1 – dS /dU2 X dU1
0= (dS/dU1 -dS /dU2 ) dU1
divide dU1 on both sides
dS / dU1 – dS/ dU2 = 0
dS / dU1 = dS /dU2 [eq 2 ]
AS dS/dU1 =1/T
so putting in eq 2 we get
1/T1 =1/T2
T1=T2
This is required condition of equillibrium.
MECHANICAL EQUILLIBRIUM :
Consider a system of a perfect gas which is subdivided into two system by a movable wall then due to colonials between molecules pressure of one side will increase the other one and all will be shifted to word low pressure but when pressure of both side become equal we say that the system is in mechanical equilibrium.
S=S1+S2
and
V=V1+V2
In this case we have
S= S(V)
Let us define a quantity
dS/dV = P /T
For isolated sysytem we know that
V=V1+V2
dV=dV1+dV2
dS = dS /dV1 X dV1 +dS/dV2 X ( – dV2)
dS = dS /dV1 X dV1 – dS/dV2 X ( dV1 )
As dS =0
0=(dS /dV1-dS/dV2 )dV1
Divide dV1 on both sides
dS /dV1-dS/dV2
dS /dV1= dS/dV2
As dS /dV = P/T
whereP 1/T1 =P 2/T2
T1 = T2
P1 = P2
This is required condition
CHEMICAL EQUILLIBRIUM ”OR ” PARTICLE EQUILLIBRIUM :
To find the condition for a system which is in particle equillibrium or chemical equllibrium .
Consider an isolated system of a perfect gas which is subdivided into two system by a conducting wall having small pores in it
S=S1+S2
and total no.of particles
N=N1+N2
N1 is no .of particles in 1st compartment
N2 is no .of particles in 2nd compartment
As S(N) ,means entropy in a function of Numbet of particles .
Let us define a quantity
dS/dN = – μ /T
where μ is chemical potential .
as S=S( μ 1 , μ2 )
N=N1+N2
dN=dN1+dN2
dS = dS /dN1 X dN1 +dS/dN2 X ( – dN2)
dS = dS /dN1 X dN1 – dS/dN2 X ( dN1 )
As dS =0
0=(dS /dN1-dS/dN2 )dV1
Divide dN1 on both sides
dS /dN1-dS/dN2
dS /dN1= dS/dN2
As dS /dN = –μ/T
where μ1/T1 =μ2/T2
T1 = T2
μ1 = μ2
This is required condition
THERMODYNAMICAL EQULLIBRIUM:
Thermodynamic equilibrium when three conditions of equilibrium thermal mechanical and chemical are satisfied the one say that the system is in thermodynamic equilibrium
thermodynamic equilibrium = thermal equllibrium + mechanical equllibrium+ chemical equllibrium .
thus
S=S(U,V,N)
dS = dS/dU X dU +dS/dV X dV +dS/dN X dN
dS = 1/T X dU +P/T X dV – μ /T dN
multipling above eq by ”T” we get
TdS = dU + PdV – μ dN .
(a) Internal Energy .
(b) Helmnhotz Free Energy
(c) Gibb’s Free Energy
(d)Enthalpy
CHEMICAL POTENTIAL :
Chemical potenial in case of internal energy .
μ = (dU/dN )S,P
In case of Helmhotz free energy .
μ = (dF/dN )V,T
In case of Gibbs free energy .
μ = (dG/dN )P,T
In case of enthalpy
μ = (dH/dN )S,P
Therefore
μ = (dU/dN )S,P = (dF/dN )V,T = (dG/dN )P,T μ = (dH/dN )S,P
U,F,G,H are free energies .
N is the number of particles .
thus the chemical potential is actually the change in free energy with respect to the number of particles .
Its unit is same as energy .
If M>0 it means density of the system increases means number of particles increases .
CHEMICAL POTENTIAL :
definition :-
simply it is the energy required to add a remove the particle to the system .
”or”
In other words chemical potential ( μ ) in the measure of how much free changes of a system changes (by dG ) if you add or remove a number dN then particles while the other parameters kept constant i.e temperature and pressure.
μ = (dG/dN )P,T
