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*Combined First and Second Laws of Thermodynamics:*

First law of Thermodynamics is

dQ = dU + dW

And second law of thermodynamics is:

dQ = TdS

But differential work

dW = PdV

Combination of the three equations in any infinitesimal reversible process, for PVT system provides

TdS = dU + PdV

It is the combined form of first and second laws of thermodynamics for PVT system. Few useful thermodynamics relations can be derived from by selecting T and V, T and P, P and V as independent variables. It is significant to note that state of the pure substance can be stated or specified by any two of its properties.

*T and V Independent:*

Using equation TdS = dU + PdV we can write:

dS = 1/T (dU + PdV)

Consider internal energy U as the function of T and V, i.e. U(T,V), then derivative of U is

dU = (∂U/∂T) VdT + (∂U/∂V) TdV

Further we get:

1/T(∂U/∂T)V = (∂S/∂T)V

1/T[(∂U/∂V)T + P](∂S/∂V)T

If differentiate equation partially with respect to V at constant T and differentiate partially with respect to T and constant V, equating mixed second-order partial derivatives of S provides

(∂U/∂V)T = T(∂P/∂T)V - P

It can be shown that (∂P/∂T)V = β/k

Thus equation becomes:

(∂U/∂V)T = Tβ/k - P

Equations show dependence of internal energy of the system on volume, at constant temperature and this can be computed from equation of state, or from values of β, κ, T and P.

By making use of other equations

C_{P} - C_{V} = T(∂P/∂T)V(∂V/∂T)P = β^{2}TV/k

Therefore difference (C_{P} - C_{V}) can be computed for any substance, from equation of state or from values of T, V, β and κ.

*T and P Independent:*

In terms of enthalpy H = U + PV, equation can be rewritten as:

dS = 1/T(dH - VdP)

Consider enthalpy H as the function of T and P, i.e. H(T, P), then derivative of H is:

dH = (∂H/∂T)PdT + (∂H/∂P)TdT

Also consider entropy S as function of two independent variables T and P, i.e. S(T, P), then derivative of S is:

dS = (∂S/∂T)pdT + (∂S/∂P)TdP

By comparing equation, we get:

(∂S/∂T)p = 1/T(∂H/∂T)p and (∂S/∂P)T = 1/T[(∂H/∂P)T - V]

By further solving

(∂H/∂P)T = -βVT + V

This equation illustrates dependence of enthalpy on pressure, at constant temperature and this can be computed from equation of state or from values of β, T and V.

Also as we know that (∂H/∂T)p = CP,

Thus we get:

dh = CPdT - [T(∂V/∂T)P - V]dP

*P and V Independent:*

Consider entropy S as the function of P and V, i.e. S(P, V), derivative of S is

dS = (∂S/∂P)VdP + (∂S/∂V)PdV

Also consider internal energy U as the function of P and V, i.e. U(P, V), derivative of U is:

dU = (∂U/∂P)VdP + (∂U/∂V)PdV

Further solving we get:

(∂S/∂P)V = 1/T (∂U/∂P) V and (∂S/∂V) P = 1/T (∂U/∂V) P + P/T

(∂U/∂P)V can be written as (∂U/∂T) V (∂T/∂P) V then equation becomes:

(∂S/∂P)V = 1/T (∂U/∂T) V (∂T/∂P) V

But (∂U/∂T) V = CV, and (∂T/∂P) V = k/β

Solving equation we get:

(∂S/∂P)V = (CV/T) (k/β)

Equation provides change in entropy with respect to pressure at constant volume (∂S/∂P) V in terms of measurable quantities CV, k, β, and T.

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