1.4: Joule-Thompson Coefficient
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A useful derivative for understanding cooling or heating of gases when they expand is the Joule-Thomson coefficient (\(\mu_{JT}\)). In 1852, working with William Thomson (who would later become Lord Kelvin), Joule conducted an experiment in which they pumped gas at a steady rate through a lead pipe that was cinched to create a constriction. On the upstream side of the constriction, the gas was at a higher pressure than on the downstream side of the constriction. Also, the temperature of the gas was carefully monitored on either side of the constriction. The cooling that they observed as the gas expanded from a high pressure region to a lower pressure region was extremely important and lead to a common design of modern refrigerators.
Definition
Not all gases undergo a cooling effect upon expansion. It depends on the the pressure, temperature and the identity of the gas. Some gases, such as hydrogen and helium, will experience a warming effect upon expansion under conditions near room temperature and pressure. The direction of temperature change can be determined by measuring the Joule-Thomson coefficient, \(\mu_{JT}\). This coefficient has the definition
\[ \mu_{JT} \equiv \left( \dfrac{\partial T}{\partial P} \right)_H \nonumber \]
This is a useful derivative because it appears in the expression for the infinitesimal change of H with respect to pressure and temperature:
\[dH= \left( \dfrac{\partial H}{\partial P} \right)_T dP+ \left( \dfrac{\partial H}{\partial T} \right)_P dT \label{totalH} \]
You should recognize the second derivative as CP. Using the cyclic derivative relationship the first derivative can be rewritten to depend on CP and \(\mu_{JT}\).
\[\left( \dfrac{\partial H}{\partial P} \right)_T = -\frac{1}{\left(\frac{\partial P}{\partial T}\right)_H\left(\frac{\partial T}{\partial H}\right)_P}\nonumber\]
Recognizing that the differentials on the bottom are the inverse of CP and \(\mu_{JT}\) leads to:
\[\left( \dfrac{\partial H}{\partial P} \right)_T = -\mu_{JT}C_P.\label{isothermJT}\]
This allows dH to be written in terms of measurable quantities:
\[dH= -\mu_{JT}C_P dP+ C_P dT \label{totalH2} \]
Measurement of \(\mu_{JT}\)
The Joule-Thomson coefficient can be measured by recording the adiabatic temperature drop or increase a gas undergoes for a given pressure drop. This can be visualized as two pistons moving as in figure \(\PageIndex{1}\)) to maintain the pressures on either side as constants. The apparatus is insulated to make it adiabatic. This is clearly an adiabatic process, but is isenthalpic, as indicated in the definition?
This process is isenthalpic as shown below.1
Since the gas expands adiabatically, q = 0. Thus, \(\Delta U = w\). Imagine starting with all the gas on the high pressure (up stream) side. The initial conditions are \(P_1, V_1\) and \(T_1\). After all the gas has passed through it will be at \(P_2, V_2\) and \(T_2\). The work on the upstream side was:
\[w_1 = -P_1\Delta V_1 = -P_1(V_{1f} - V_{1i}) = -P_1(0 - V_1) = P_1 V_1\nonumber\]
On the downstream side the work is:
\[w_2 = -P_2\Delta V_2 = -P_2(V_{2f} - V_{2i}) = -P_2(V_2 - 0) = -P_2 V_2.\nonumber\]
Summing these together gives us the total work and \(\Delta U\):
\[\Delta U = U_2 - U_1 = w_1 + w_2 = P_1 V_1 - P_2 V_2. \nonumber\]
Collecting all the upstream quantities on one side and the downstream on the other yields:
\[U_2 + P_2 V_2 = U_1 + P_1 V_1.\label{isenthalpic}\]
Since H = U + PV equation \(\ref{isenthalpic}\) means that the enthalpy on the two sides is the same, making the process isenthalpic.
In an actual laboratory setting constant flow of gas at a fixed pressure and temperature into the high pressure side and removal of gas from the low pressure side is used. An alternative to measuring the temperature change is to measure the isothermal Joule_Thomson coefficient \(\left(\frac{\partial H}{\partial P}\right)_T\) (equation \(\ref{isothermJT}\)) directly by determining how much heat is needed to keep the temperatures on the two sides equal. Equation \(\ref{isothermJT}\) can then be used to calculate \(\mu_{JT}\).
References
1. P. Atkins and J. De Paula, Physical Chemistry, 9th Ed. (W.H. Freeman and Co., New York, 2010), Pp. 79 - 83.
Real gas behavior
The typical behavior of the Joule-Thomson coefficient can be summarized in Figure \(\PageIndex{2}\). At the combinations of \(T\) and \(P\) for which \(\mu_{JT} > 0\) (inside the shaded region), the sample will cool upon expansion. At those \(P\) and \(T\) conditions outside of the shaded region, where \(\mu_{JT} < 0\), the gas will undergo a temperature increase upon expansion. And along the boundary, a gas will undergo neither a temperature increase not decrease upon expansion. For a given pressure, there are typically two temperatures at which \(\mu_{JT}\) changes sign. These are the upper and lower inversion temperatures.
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Joule-Thompson cooling is used in the Linde (you may have heard of a company by that name) refrigeration technique to liquefy atmospheric gases. The gas to be cooled is pumped through a cooling coil inside an insulated container. The end of the coil is open to the box. The gas that expands through a throttle out of the end of the coil cools as long as the temperature and pressure are within the cooling region (\(\mu_{JT} \gt 0\)). This cooler gas cools the gas within the coil and is then pumped out of the insulated container. The gas pumped out of the container is compressed to raise its temperature above ambient and run through a heat exchanger to cool it back down to close to ambient. This is then run through the coil inside again. Eventually, the gas coming out the throttle is so cool that it condenses and drips to the bottom of the insulated container, where it can be collected.
Relation of \(\mu_{JT}\) to other measurable quantities
Recall that equation \(\ref{isothermJT}\) gives us a relation between \(\mu_{JT}\) and a derivative of the enthalpy:
\[\left( \dfrac{\partial H}{\partial P} \right)_T = -\mu_{JT}C_P.\nonumber\]
From the fundamental differential dH = TdS + VdP we can divide by dP while holding T constant to get:
\[\left( \dfrac{\partial H}{\partial P} \right)_T= T \left( \dfrac{\partial S}{\partial P} \right)_T + V, \nonumber \]
where we have used the general result that du/du = 1. Looking at the Maxwell relations you will recognize that \(\left( \frac{\partial S}{\partial P} \right)_T = - \left( \frac{\partial V}{\partial T} \right)_P\). So:
\[\left( \dfrac{\partial H}{\partial P} \right)_T = - T\left( \frac{\partial V}{\partial T} \right)_P + V = -TV\alpha + V = V(1-T\alpha ). \nonumber \]
Setting this equal to \(-\mu_{JT}C_P\):
\[ V( 1-T\alpha) = -\mu_{JT}C_P \implies \mu_{JT} = \frac{V}{C_P}(T\alpha -1 ) \nonumber \]
For an ideal gas, \(\alpha = 1/T\), so \(\mu_{JT} = 0\). Thus, gases will only show non-zero values for \(\mu_{JT}\) when they deviate from ideal behavior.
Contributors
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Jonathan Gutow, UW Oshkosh
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Patrick Fleming, Cal State Eastbay