Discussion
The general relation among the properties \(x\), \(y\) and \(z\) at any state point can be expressed as
\[ \begin{aligned} &\left(\frac{\partial x}{\partial y}\right)_{z}\left(\frac{\partial y}{\partial z}\right)_{x} \\ &\times \left(\frac{\partial z}{\partial x}\right)_{y} = -1. \end{aligned} \]
If \(p\), \(T\) and \(h\) are continuous functions and \(c_{p} = \left(\frac{\partial h}{\partial T}\right)_{p}\), \(\mu\) is the Joule-Thomson coefficient, then \(\left(\frac{\partial h}{\partial p}\right)_{T}\) is
Answer: A
\(\left(\frac{\partial h}{\partial p}\right)_{T} = -\mu c_{p}\).
Apply the cyclic relation to h, p and T:
\[ \begin{aligned} &\left(\frac{\partial h}{\partial p}\right)_{T}\left(\frac{\partial p}{\partial T}\right)_{h} \\ &\times \left(\frac{\partial T}{\partial h}\right)_{p} = -1 \end{aligned} \]
Here \((\partial T/\partial h)_{p} = 1/c_{p}\) and \((\partial p/\partial T)_{h} = 1/\mu\), where \(\mu = (\partial T/\partial p)_{h}\) is the Joule-Thomson coefficient. Solving gives \((\partial h/\partial p)_{T} = -\mu c_{p}\).
\(\mu c_{p}\) loses the minus sign of the cyclic rule. \(c_{p}T\) and \(-c_{p}/T\) are dimensionally different from an enthalpy change per unit pressure. For an ideal gas \(\mu = 0\), so h does not vary with p at constant T, as expected.
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