Question

The general relation among the properties $x$, $y$ and $z$ at any state point can be expressed as \[ \left( \frac{\partial x}{\partial y} \right)_z \left( \frac{\partial y}{\partial z} \right)_x \left( \frac{\partial z}{\partial x} \right)_y = -1 . \] If $p$, $T$ and $h$ are continuous functions and \[ c_p = \left( \frac{\partial h}{\partial T} \right)_p , \] and $\mu$ is the Joule–Thomson coefficient, then \[ \left( \frac{\partial h}{\partial p} \right)_T \] is:

• A. $- \mu c_p$
• B. $c_p T$
• C. $-\dfrac{c_p}{T}$
• D. $\mu c_p$

Hint:

Always apply the cyclic identity \[ (\partial x/\partial y)_z (\partial y/\partial z)_x (\partial z/\partial x)_y = -1 \] when dealing with thermodynamic property derivatives—it simplifies Joule–Thomson and Maxwell relation problems significantly.

Answer 1

The Correct Option is A

Step 1: Use enthalpy as a function of $T$ and $p$.
Since $h = h(T,p)$, we can write the total differential: \[ dh = \left( \frac{\partial h}{\partial T} \right)_p dT + \left( \frac{\partial h}{\partial p} \right)_T dp . \] We are asked to find the value of $\left( \frac{\partial h}{\partial p} \right)_T$. Step 2: Use definition of $c_p$.
\[ c_p = \left( \frac{\partial h}{\partial T} \right)_p. \] Step 3: Use Joule–Thomson coefficient relation.
The Joule–Thomson coefficient $\mu$ is defined as: \[ \mu = \left( \frac{\partial T}{\partial p} \right)_h. \] To connect this with our derivative, we apply the cyclic identity among variables $(h, T, p)$: \[ \left( \frac{\partial h}{\partial T} \right)_p \left( \frac{\partial T}{\partial p} \right)_h \left( \frac{\partial p}{\partial h} \right)_T = -1. \] Substitute the known quantities: \[ \left( \frac{\partial h}{\partial T} \right)_p = c_p, \quad \left( \frac{\partial T}{\partial p} \right)_h = \mu. \] Thus, \[ c_p \cdot \mu \cdot \left( \frac{\partial p}{\partial h} \right)_T = -1. \] Step 4: Solve for the required derivative.
Rearrange: \[ \left( \frac{\partial p}{\partial h} \right)_T = -\frac{1}{c_p \mu}. \] Invert both sides to obtain $\left( \frac{\partial h}{\partial p} \right)_T$: \[ \left( \frac{\partial h}{\partial p} \right)_T = - \mu c_p. \] Step 5: Conclusion.
Thus, the correct expression is \[ \boxed{ \left( \frac{\partial h}{\partial p} \right)_T = - \mu c_p }. \]