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
CP - CV = T(∂P/∂T)V(∂V/∂T)P = β2TV/k
Therefore difference (CP - CV) 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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