Deriving thermodynamic relations

[mclame22]

1. The problem statement:

Show that

a) (∂H/∂T)_V = C_V(1 - βμ/κ)

b) (∂H/∂V)_T = μC_P/Vκ

c) (∂T/∂V)_H = μ/(V(μβ - κ))2. Homework Equations :

i) β = (1/V)(∂V/∂T)_P

ii) κ = -(1/V)(∂V/∂P)_T

iii) β/κ = (∂P/∂T)_V

iv) C_V = (∂U/∂T)_V

v) C_P = (∂H/∂T)_P

vi) C_P - C_V = TVβ²/κ

vii) η = (∂T/∂V)_U = (1/C_V)(P - Tβ/κ)

viii) μ = (∂T/∂P)_H = (V/C_P)(βT - 1)3. The Attempt at a Solution :

a) H = U + PV
(∂H/∂T)_V = (∂U/∂T)_V + V(∂P/∂T)_V

Using (iv) and (iii):
(∂H/∂T)_V = C_V + Vβ/κ --> stuckb) H = U + PV
(∂H/∂V)_T = (∂U/∂V)_T + P + V(∂P/∂V)_T

The change in internal energy with respect to volume at constant temperature for an ideal gas is 0, and using (ii):
(∂H/∂V)_T = P - 1/κ --> stuckc) I have no idea how to get this one.

 

[mark726]

To solve for (∂T/∂V)H, we can use the relationship between specific heat at constant pressure (CP) and specific heat at constant volume (CV), which is given by CP - CV = TVβ^2/κ (from equation vi). Rearranging this equation, we get:

TVβ^2 = (CP - CV)κ

Substituting this into the definition of μ (from equation viii), we get:

μ = (V/CP)(βT - 1)
βT = (μCP/V) + 1

Now, using the definition of β (from equation i), we get:

β = (1/V)(∂V/∂T)P = (1/V)(∂T/∂V)P^-1

Substituting this into our previous equation, we get:

(∂T/∂V)P^-1 = (μCP/V) + 1

Rearranging, we get:

(∂T/∂V)P = (μCP/V)^-1 + 1

Finally, using the definition of κ (from equation ii), we get:

(∂T/∂V)H = (μ/(V(μβ - κ)))