Answer
$μ_{}=0.0599\text{ R/psia}$
$μ=0.00929\text{ R/psia}$
Work Step by Step
$(a)$ The enthalpy of nitrogen at 120 psia and $350 \mathrm{R}$ is, from EES, $h=84.88 \mathrm{Btu} / \mathrm{bm}$. Approximating differentials by differences about the specified state, the Joule-Thomson coefficient is expressed as $$
\mu=\left(\frac{\partial T}{\partial P}\right)_h \equiv\left(\frac{\Delta T}{\Delta P}\right)_{b=84838 \mathrm{Btu} / \mathrm{bm}}
$$ Considering a throttling process from $130$ psia to $110$ psia at $h=84.88\ \mathrm{Btu} / \mathrm{lbm}$, the Joule-Thomson coefficient is determined to be $$
\mu=\left(\frac{T_{110 \text { psia }}-T_{130 \text { pia }}}{(110-130) \text { psia }}\right)_{h=8488 \text { Bou llaw }}=\frac{(349.40-350.60) \mathrm{R}}{(110-130) \text { psia }}=0.0599\ \mathrm{R} / \mathrm{psia}
$$ (b) The enthalpy of nitrogen at 1200 psia and $700 \mathrm{R}$ is, from EES, $h=170.14 \mathrm{Btu} / \mathrm{bm}$. Approximating differentials by differences about the specified state, the Joule-Thomson coefficient is expressed as $$
\mu=\left(\frac{\partial T}{C P}\right)_h \equiv\left(\frac{\Delta T}{\Delta P}\right)_{b=170148 \mathrm{a} / \mathrm{lbut}}
$$ Considering a throttling process from 1210 psia to 1190 psia at $h=170.14\ \mathrm{Btu} / \mathrm{lbm}$, the Joule-Thomson coefficient is determined to be $$
\mu=\left(\frac{T_{1190 \text { paia }}-T_{1210 p s i a}}{(1190-1210) \text { psia }}\right)_{b=17014 \text { Bru/laws }}=\frac{(699.91-700.09) \mathrm{R}}{(1190-1210) \mathrm{psia}}=0.00929\ \mathrm{R} / \mathrm{psia}
$$
