Question

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

Answer 1

For what kind of gas?

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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 `mu_(JT)` of zero. Note that for ideal gases:

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

And thus, `alphaT = 1` and `mu_(JT) = 0` for an ideal gas, regardless of temperature (as long as `T ne "0 K"`).

For real gases, when `mu_(JT)` is zero, it is known to be at the inversion temperature `T_"inv"`:

`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 `T = "0 K"` for the sign change to occur, the heat capacity does not reach zero:

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

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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 `T` and `P`, so consider the ubiquitous relation in a constant-temperature system:

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

By taking the partial derivative with respect to `P` at constant `T` of `(4)`, we get:

`((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 `(7)` and `(8)`,

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

Next, from the Maxwell Relation `(6)`,

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

As a result, we so far have, by combining `(9)` and `(10)` into `(5)`:

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

Therefore:

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

Lastly, plug this back into the cyclic rule `(1)` to find:

`((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)" ")|)`