Answer
$μ_{}=0.317\text{ R/psia}$
Work Step by Step
The Joule-Thomson coefficient is defined as $$
\mu=\left(\frac{\partial T}{\partial P}\right)_h
$$ We use a finite difference approximation as $$
\mu \equiv \frac{T_2-T_1}{P_2-P_1} \text { (at constant enthalpy) }
$$ At the given state (we call it state $1$), the enthalpy of R-134a is $$
\left.\begin{array}{l}
R_1=40\ \mathrm{psia} \\
T_1=60^{\circ} \mathrm{F}
\end{array}\right\} h_1=113.80\ \mathrm{Btu} / \mathrm{lbm} \quad(\text { Table A }-13 \mathrm{E})
$$ The second state will be selected for a pressure of $30\ \mathrm{psia}$. At this pressure and the same enthalpy, we have $$
\left.\begin{array}{l}
P_2=30\ \mathrm{psia} \\
h_2=h_1=113.80\ \mathrm{~kJ} / \mathrm{kg}
\end{array}\right\} T_2=56.83^{\circ} \mathrm{F} \quad \text { (Table A - 13E) }
$$ Substituting, $$
\mu \equiv \frac{T_2-T_1}{P_2-P_1}=\frac{(56.83-60) \mathrm{R}}{(30-40)\ \mathrm{psia}}=0.317\ \mathrm{R} / \mathrm{psia}
$$
