Discuss the Joule-Kelvin effect for a Van der Waal's gas?

Question

Any idea on what is joule-kelvin effect?

Answer 1

I think it is the Joule-Thomson effect
William Thomson was honored with the title Baron of Kelvin at Largs for his achievements in science.

Edit:
The Joule-Thomson effect is the temperature change of an real gas, which undergoes a process of expansion at constant enthalpy. It is described with the Joule-Thomson coefficient:
µ = (∂T/∂p)h (lower case h denotes at constant enthalpy)

It can be evaluated from the equation of state as follows.
Because the enthalpy is function of state of the system H = H(T,p) its total differential is:
dH = (∂H/∂T)p dT + (∂H/∂p)t dp = 0
Hence:
µ = - (∂H/∂p)t / (∂H/∂T)p
The denominator is the heat capacity at constant pressure
(∂H/∂T)p ≡ Cp
The nominator can be derived from:
dH = T dS + V dp:
Therefore
(∂H/∂p)t = T (∂S/∂p)t + V
with the Maxwell relation
(∂S/∂p)t = - (∂V/∂T)p
you get
µ =(T(∂V/∂T)p - V) / Cp

You can evaluate the partial derivative from the van der Waals equation of state:
(p + a/V²)·(V-b) = RT
<=>
V = RT/p - a/(pV) + b - ab/(pV²)
For an approximate solution neglect the small term ab/pV² and set pV ≈ RT
V ≈ RT/p - a/(RT)
Hence:
(∂V/∂T)p = R/p + a/(RT²)
and
µ = ( 2a/(RT) - b ) / Cp
Since the denominator is always positive, the Joule-Thomson coefficient changes its sign for
T = 2a/(Rb), which is called inversion temperature.

The exact solution to this i cant find (seems to be very messy ;-) )