Tuesday, 7 May 2019

thermodynamics - Joule-Thomson effect: why does a gas cool if it's below the inversion temperature?


The Joule-Thomson coefficient is given by μJT=(TP)H=VCP(βT1),

where β is the coefficient of thermal expansion. If the inversion temperature is defined by Tinv=1β, then why is μJT>0 if T<Tinv, as stated by Wikipedia Joule-Thomson effect? I really don't get this, this should be basic math ? This results in not comprehending why a gas cools if it's below inversion temperature and vice versa.



Answer




If you look up the coefficient of thermal expansion of air, you will find that it decreases with increasing absolute temperature. In fact, it decreases rapidly enough that even the product βT decreases with increasing temperature. For air at room temperature and 1 bar, for example, the product is about 1.01 while, at 200 C, the product is about 0.99.


In terms of the compressibility factor z, the product of β and T is given by: βT=1+(lnzlnT)P

So (βT1)=(lnzlnT)P


Get yourself a plot of z vs reduced temperature and reduced pressure and note how the right hand side of this equation decreases with increasing reduced temperature. It is positive at low reduced temperatures, and reaches a value of zero at a reduced temperature of about 4.


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