derive the equation dE=[T(partialp/partialT)v-p] dV+CvdT?

Question

and from this show that (partialE/PartialV)t=a/V^2 for a van der walls gas

Answer 1

1. Proof
Take the specific internal energy of a pure substance as function of its molar volume and the temperature. Then the total derivative of the internal energy is
dE = (∂E/∂V)t dV + (∂E/∂T)v dT ((..)x stands for evaluation at constant X)
The second partial derivative in the expression is defined as the molar heat capacity at constant volume
Cv ≡ (∂E/∂T)v
The first part of the expression can be evaluated from first law of thermodynamics, which describes E as function of entropy and volume:
dE = (∂E/∂S)v dS + (∂E/∂V)s dV = TdS - pdV
Hence:
(∂E/∂V)t = (∂E/∂S)v (∂S/∂V)t + (∂E/∂V)s = T (∂S/∂V)t - p
and
dE = (∂E/∂V)t dV + (∂E/∂T)v dT
The derivative (∂S/∂V)s can be calculated from the Maxwell relations, which follow from the fact, that the order of differentiation in a second derivative of a thermodynamic potential function is irrelevant. For the Helmholtz free energy A follows:
dA = (∂A/∂T)t dT + (∂A/∂V)v dV = -SdT - pdV
The second derivatives are:
[∂(∂A/∂T)/∂V] = [∂(∂A/∂T)/∂V]
<=>
(∂S/∂V)t = (∂p/∂T)v
Therefore
(∂E/∂V)t = T (∂p/∂T)v - p
and
dE = [T(∂p/∂T)v - p] dV + Cv dT

2. Internal energy for a Van der Waals gas
The van der Waals equation of state is:
p = RT / (V-b) - a/V²
(∂E/∂T)v = T(∂p/∂T)v - p = RT/(V-b) - (RT/(V-b) - a/V² = a/V²
and
dE = a/V² dV + Cv dT