Question

Prove that, (dS / dP)T = - (dV/dT)p​

Answer 1

Answer:

Explanation:

Gibbs equation is

du = Tds - Pdv

The enthalpy h can be differentiated,

dh = du + pdv + vdP

Combining the two results in

�

dh = Tds + vdP

The coefficients T and v are partial derivative of h(s,P),

�

Since v > 0, an isentropic increase in pressure will result in an increase in enthalpy.

We introduce Helmholtz function

a = u � Ts

Combine Gibbs equation with the differential of a,

da = -Pdv � sdT

The coefficient �P and �s are the partial derivatives of f(v,T), so

Similarly, using the Gibbs function

g = h � Ts

dg = vdP � sdT

Consequently,

Answer 2

Using Gibbs equation and function we can prove the equation.

Explanation:

• To derive the relationships between the various thermodynamic
• variables,we first take s and V as independent,

• Gibbs equation is $du = Tds - Pdv$

The enthalpy h can be differentiated,$dh = du + pdv + vdP$

Combining the two results in

$dh = Tds + vdP$

$dF = d(u -T s) = Tds- P dV- T ds - sdT = -P dV - sdT$∂F

(∂T/∂F)v$=-s$  and

∂F/∂V)T = −P

• The coefficients T and v are partial derivative of h(s,P),

Since v > 0, an isentropic increase in pressure will result in an increase in enthalpy.

• We introduce Helmholtz function

$a = u+ Ts$

• Combine Gibbs equation with the differential of a,

$da = -Pdv + sdT$

The coefficient ΔP and Δs are the partial derivatives of f(v,T), so

Similarly, using the Gibbs function

$g = h + Ts$

$dg = vdP + sdT$

thus on

$(dS / dP)T = - (dV/dT)p$