Discussion
Joule-Thomson coefficient for a gas, \(\mu_{j}\) obeying the relation \(p(v - b) = RT\) is
Answer: C
The Joule-Thomson coefficient is \(\mu_{j} = -\dfrac{b}{c_{p}}\).
In general
\[ \begin{aligned} \mu_{j} &= \left(\frac{\partial T}{\partial p}\right)_{h} \\ &= \frac{1}{c_{p}}\left[T\left(\frac{\partial v}{\partial T}\right)_{p} - v\right] \end{aligned} \]
From \(v = RT/p + b\), \((\partial v/\partial T)_{p} = R/p\), so \(T(R/p) - v = (v - b) - v = -b\), and \(\mu_{j} = -b/c_{p}\).
The negative value means such a gas always warms on throttling. The positive \(b/c_{p}\) has the wrong sign, and \(\pm c_{p}/b\) inverts the expression and does not even have units of K/Pa.
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