# Thermodynamics: Maxwell's Relations

Derive Maxwell's Relations from First and Second Laws of Thermodynamics and Thermodynamic Functions like Internal Energy, Helmholtz's Function, Enthalpy and Gibbs Free Energy.

And also explain how they are satisfied by an ideal monatomic gas?

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#### Solution Preview

According to the first and second laws of thermodynamics, we have respectively

dQ=dU+p dV

dQ=T dS

Combining the above two equations we get

dU=T dS-p dV [1]

Let us consider two independent variables, which determine the thermodynamic state of the system of a given gas, as x and y. Now, all thermodynamic variables, like U,S and V, will be functions of x and y.

Partial Differentiation of these thermodynamic variables with respect to x and y gives,

dU=∂U/∂x dx+∂U/∂y dy [2]

dS=∂S/∂x dx+∂S/∂y dy [3]

dV=∂V/∂x dx+∂V/∂y dy [4]

Putting [2], [3] and [4] in [1]:

∂U/∂x dx+∂U/∂y dy=T(∂S/∂x dx+∂S/∂y dy)-p(∂V/∂x dx+∂V/∂y dy)

Equating the coefficients of dx and dy from both sides of the equation, we get

∂U/∂x=T ∂S/∂x-p ∂V/∂x [5]

∂U/∂y=T ∂S/∂y-p ∂V/∂y [6]

Differentiating [5] with respect to y and [6] with respect to x, we get

(∂^2 U)/∂y∂x=∂T/∂y ∂S/∂x+T (∂^2 S)/∂y∂x-∂p/∂y ∂V/∂x-p (∂^2 V)/∂y∂x [7]

(∂^2 ...

#### Solution Summary

This solution is comprised of detailed step-by-step derivations and analysis of the Maxwell's Relations and provides students with a clear perspective of the underlying concepts.