# For what temperature is the Joule-Thomson coefficient for a gas zero?

##### 1 Answer

For what kind of gas?

The **Joule-Thomson experiment** essentially pushes a gas through a porous plug at constant enthalpy (insulated so that no heat is exchanged with the environment). This is done in such a way that all of the expansion work goes into changing the internal energy:

Any gas is then described by the **Joule-Thomson coefficient**

#mu_(JT) = ((delT)/(delP))_H# ,

that is, it is the change in temperature brought about by a change in pressure at constant enthalpy.

(This may be weird... but if you can perform the expansion slowly enough, the pressure can re-equilibrate and constant enthalpy is then possible with perfect insulation.)

A Joule-Thomson coefficient for any gas is given by:

#mu_(JT) = ((delT)/(delP))_H = V/C_P(alphaT - 1)#

An ideal gas ** always** has a Joule-Thomson coefficient

#alpha = 1/V ((delV)/(delT))_P = (nR)/(PV) = 1/T#

And thus,

For real gases, when *inversion temperature*

#0 = V/C_P(alphaT_"inv" - 1)#

Since the volume of any gas is never zero and since assuming we don't need to reach

#color(blue)(T_"inv" = 1/alpha)#

**DISCLAIMER:** *LONG DERIVATION BELOW!*

To derive this, begin at the **cyclic rule** of partial derivatives:

#((delT)/(delP))_H((delH)/(delT))_P((delP)/(delH))_T = -1# #" "" "bb((1))#

By definition, the constant-pressure heat capacity is:

#C_P = ((delH)/(delT))_P# ,#" "" "bb((2))#

which is an experimentally-determined quantity. Next, note that, as with all derivatives:

#((delP)/(delH))_T = 1/((delH)/(delP))_T# #" "" "bb((3))#

The Gibbs' free energy is a function of

#dG = dH - TdS# #" "" "bb((4))#

By taking the partial derivative with respect to

#((delG)/(delP))_T = ((delH)/(delP))_T - T((delS)/(delP))_T# #" "" "bb((5))#

For a similar reason, we then consider the Gibbs' free energy **Maxwell Relation**:

#dG = -SdT + VdP# #" "" "bb((6))#

Since for state functions, cross-derivatives are equal, we have:

#((delS)/(delP))_T = -((delV)/(delT))_P# #" "" "bb((7))#

By definition, the *coefficient of thermal expansion* is:

#alpha = 1/V ((delV)/(delT))_P# #" "" "bb((8))#

which is experimentally-determined. So, using

#((delS)/(delP))_T = -Valpha# .#" "" "bb((9))#

Next, from the Maxwell Relation

#((delG)/(delP))_T = V# .#" "" "bb((10))#

As a result, we so far have, by combining

#V = ((delH)/(delP))_T + TcdotValpha#

Therefore:

#((delH)/(delP))_T = V - ValphaT = -V(alphaT - 1)#

Lastly, plug this back into the cyclic rule

#((delT)/(delP))_HC_P/(-V(alphaT - 1)) = -1#

And so, the **Joule-Thomson coefficient** is found as:

#color(blue)(barul|stackrel(" ")(" "mu_(JT) = ((delT)/(delP))_H = V/C_P(alphaT - 1)" ")|)#