Joule-Thomson: adiabatic + isenthalpic expansion

2.13 Joule-Thomson: adiabatic + isenthalpic expansion

Performing an experiment as illustrated in Fig. 2.9, one typically observes a temperature change when pressing gas through a throttle.

Figure 2.9: Illustration of isenthalpic expansion: a) at beginning of experiment, b) after finishing the experiment.

Gas becomes cold / warm during expansion by counteracting attractive / repulsive forces, thus reducing / increasing \(U\).
Sign of \(\Delta T\) depends on the Joule-Thomson coefficient which depends on \(p\) and \(T\). Being adiabatic, we find

\begin{equation*} \Delta U = \Delta q + \Delta w = \Delta w \label{eq:DU_Joule-Thomson} \end{equation*} (2.38)

According to Fig. 2.9 a) we find for system 1 (left of diaphragm)

\begin{equation*} \Delta U_1 = 0 - U_1=-U_1=- p_1\, (0-V_1) = p_1\,V_1 \label{eq:DU1_Joule-Thomson} \end{equation*}

(2.39)

According to Fig. 2.9 b) we find for system 2 (right of diaphragm)

\begin{equation*} \Delta U_2 = U_2 - 0= U_2=- p_2\, (V_2-0) = - p_2\,V_2 \label{eq:DU2_Joule-Thomson} \end{equation*}

(2.40)

\begin{equation*} \begin{split} \mbox{Thus the total change is}\quad \Delta U_1 + \Delta U_2 & = - U_1 + U_2 = p_1\,V_1 - p_2 \, V_2\\ \Rightarrow & \quad \quad U_2 + p_2 \, V_2 = U_1 + p_1 \, V_1 \quad \Rightarrow \quad H_2 = H_1 \end{split} \label{eq:totalU_Joule-Thomson} \end{equation*}

(2.41)

So the \(\Delta\) in the above equation means state (after change) minus state (before change). Eq. (2.41) implies that the adiabatic expansion is also isenthalpic, and of course it is an irreversible process, because a tiny change in \(p_1\) will not change the gas flow direction.