F = F ( x , y ) and if it is perfect differential then d F =M dx + Ndythen M and N will satisfy the condition Maxwell relations are relationship between two derivatives of thermodynamic variables, and energy due to the equivalence of potential second derivative under a change of operation order
Where F is thermodynamic potential and x and y are two of its natural independent variables.
Maxwell relations are extremely important for two reasons:
(i) First they show us that derivative of thermodynamic parameters are not all independent. This can serve as a consistency check in both experiments and in theoretical analysis.
(ii) Maxwell relations provide a method for expressing some derivative in other ways. This enables as to connect difficult to measure quantities to those which are readily accessible experimentally.
The measurement of entropy and chemical potential can not be directly measurable in lab but with the help of Maxwell relation there thermodynamic property can be determine theoretically.
For Maxwell relation,
Let us Legendre the independent variable as x , and y such that
U = U(x,y), S = S(x, y) V = V(x, y)
So From first law of thermodynamic
dU = TdS - PdV
Hence U,V, and S are perfect differential.
Then
Similarly
Equating equation (1) and (2)
Maxwell first relation:- put x = T , y = V
Maxwell Second Relation:- put x = T ,y = P
Maxwell Third Relation:- put x = s , y = v
Maxwell Fourth Relation:- put x = S, y = P
Thermodynamic potential is a scalar function used to represent the thermodynamic state of system. The concept of thermodynamic potentials was introduced by Pierre Duhem in 1886.
One main thermodynamic potential that has a physical interpretation is the internal energy. It is energy of configuration of a given system of conservative forces. Expression for all other thermodynamic energy potentials are drivable via Legendre transformation.
Thermodynamic potentials are different form of energy which can be used in different thermodynamic process. Thermodynamic potentials are path independent variables so they are perfect differential.
If F is unique thermodynamic potential defined by variables x and y as F = F ( x , y ) and if it is perfect differential then dF = Mdx + Ndy
then M and N will satisfy the condition
where is chemical potential, G is Gibbs energy and N is no of molecules
Example 1: Prove that internal energy U is given by
Solution: (a) F = U-TS =U = F + TSG = H-TS
Application of Maxwell Relation
First T - dS equation
Let T, and \/are independent variable S = S (T, V)
Second T - dS Equation
Let T and P are independent variable S = S(T, P).
From Maxwell relation
Third T-dS Equation
Let P , V are independent variable S = S (P ,V )
The First Energy Equation
Let T and V are independent variable U = U(T, V)
From first law of thermodynamics,
dU = TdS - PdV
Using Maxwell relation
Second Energy Equation
Using Maxwell relation,This is popularly known as second energy equation.
If U is function of independent variable of T and P.
Example 2: From relation dU = TdS - P d V
Derive Maxwell relation
Solution: dU = TdS - PdVHence U is exact differential
Example 3: A real gas which obey van der Waal's equation of state are kept in container which has temperature T0 and volume V0. if volume of container changes to V such that temperature of gas become T what is change in entropy?
Solution: Assume Cv is specific heat of constant volume
For van der Waal's gasFrom first T -dS equationwhere S0 is integration constant.
1. What are thermodynamic potentials? |
2. How are thermodynamic potentials related to thermodynamic variables? |
3. What is the significance of thermodynamic potentials in studying equilibrium states? |
4. Can you explain the concept of Legendre transformation in relation to thermodynamic potentials? |
5. How do thermodynamic potentials help in understanding phase transitions? |
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