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

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

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Answer 1 · 2017-09-09T15:38:04

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:

![http://img.tfd.com/]( useruploads.socratic.org )

Any gas is then described by the Joule-Thomson coefficient

$μ_{J T} = {(\frac{\partial T}{\partial P})}_{H}$ $mu_(JT) = ((delT)/(delP))_H$ ,

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:

$μ_{J T} = {(\frac{\partial T}{\partial P})}_{H} = \frac{V}{C_{P}} (α T - 1)$ $mu_(JT) = ((delT)/(delP))_H = V/C_P(alphaT - 1)$

An ideal gas always has a Joule-Thomson coefficient $μ_{J T}$ $mu_(JT)$ of zero. Note that for ideal gases:

$α = \frac{1}{V} {(\frac{\partial V}{\partial T})}_{P} = \frac{n R}{P V} = \frac{1}{T}$ $alpha = 1/V ((delV)/(delT))_P = (nR)/(PV) = 1/T$

And thus, $α T = 1$ $alphaT = 1$ and $μ_{J T} = 0$ $mu_(JT) = 0$ for an ideal gas, regardless of temperature (as long as $T \neq 0 K$ $T ne "0 K"$ ).

For real gases, when $μ_{J T}$ $mu_(JT)$ is zero, it is known to be at the inversion temperature $T_{inv}$ $T_"inv"$ :

$0 = \frac{V}{C_{P}} (α T_{inv} - 1)$ $0 = V/C_P(alphaT_"inv" - 1)$

Since the volume of any gas is never zero and since assuming we don't need to reach $T = 0 K$ $T = "0 K"$ for the sign change to occur, the heat capacity does not reach zero:

$T_{inv} = \frac{1}{α}$ $color(blue)(T_"inv" = 1/alpha)$

DISCLAIMER: LONG DERIVATION BELOW!

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

${(\frac{\partial T}{\partial P})}_{H} {(\frac{\partial H}{\partial T})}_{P} {(\frac{\partial P}{\partial H})}_{T} = - 1$ $((delT)/(delP))_H((delH)/(delT))_P((delP)/(delH))_T = -1$ $(1)$ $" "" "bb((1))$

By definition, the constant-pressure heat capacity is:

$C_{P} = {(\frac{\partial H}{\partial T})}_{P}$ $C_P = ((delH)/(delT))_P$ , $(2)$ $" "" "bb((2))$

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

${(\frac{\partial P}{\partial H})}_{T} = \frac{1}{{(\frac{\partial H}{\partial P})}_{T}}$ $((delP)/(delH))_T = 1/((delH)/(delP))_T$ $(3)$ $" "" "bb((3))$

The Gibbs' free energy is a function of $T$ $T$ and $P$ $P$ , so consider the ubiquitous relation in a constant-temperature system:

$d G = d H - T d S$ $dG = dH - TdS$ $(4)$ $" "" "bb((4))$

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

${(\frac{\partial G}{\partial P})}_{T} = {(\frac{\partial H}{\partial P})}_{T} - T {(\frac{\partial S}{\partial P})}_{T}$ $((delG)/(delP))_T = ((delH)/(delP))_T - T((delS)/(delP))_T$ $(5)$ $" "" "bb((5))$

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

$d G = - S d T + V d P$ $dG = -SdT + VdP$ $(6)$ $" "" "bb((6))$

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

${(\frac{\partial S}{\partial P})}_{T} = - {(\frac{\partial V}{\partial T})}_{P}$ $((delS)/(delP))_T = -((delV)/(delT))_P$ $(7)$ $" "" "bb((7))$

By definition, the coefficient of thermal expansion is:

$α = \frac{1}{V} {(\frac{\partial V}{\partial T})}_{P}$ $alpha = 1/V ((delV)/(delT))_P$ $(8)$ $" "" "bb((8))$

which is experimentally-determined. So, using $(7)$ $(7)$ and $(8)$ $(8)$ ,

${(\frac{\partial S}{\partial P})}_{T} = - V α$ $((delS)/(delP))_T = -Valpha$ . $(9)$ $" "" "bb((9))$

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

${(\frac{\partial G}{\partial P})}_{T} = V$ $((delG)/(delP))_T = V$ . $(10)$ $" "" "bb((10))$

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

$V = {(\frac{\partial H}{\partial P})}_{T} + T \cdot V α$ $V = ((delH)/(delP))_T + TcdotValpha$

Therefore:

${(\frac{\partial H}{\partial P})}_{T} = V - V α T = - V (α T - 1)$ $((delH)/(delP))_T = V - ValphaT = -V(alphaT - 1)$

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

${(\frac{\partial T}{\partial P})}_{H} \frac{C_{P}}{- V (α T - 1)} = - 1$ $((delT)/(delP))_HC_P/(-V(alphaT - 1)) = -1$

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

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

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

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