Answer
$h_{}=121.6\text{ kJ/kg}$
Work Step by Step
(a) From the ideal gas table of nitrogen (Table A-18) we read $$
h=5083.8\ \mathrm{~kJ} / \mathrm{kmol}=181.48\ \mathrm{~kJ} / \mathbf{k g}\left(M_{\mathrm{N}_2}=28.013\ \mathrm{~kg} / \mathrm{kmol}\right)
$$ at the specified temperature. This value involves $44.4 \%$ error.
(b) The enthalpy departure of nitrogen at the specified state is determined from the generalized chart to be $$
\left.\begin{array}{l}
T_R=\frac{T}{T_{\text {ct }}}=\frac{175}{126.2}=1.387 \\
P_R=\frac{P}{P_{\text {ct }}}=\frac{8}{3.39}=2.360
\end{array}\right\} \longrightarrow Z_{\mathrm{h}}=\frac{\left(\bar{h}_{\text {sdcal }}-\bar{h}\right)_{T P}}{R_{\mathrm{u}} T_{\mathrm{cr}}}=1.6
$$ Thus, $$
\bar{h}=\bar{h}_{\text {ileal }}-Z_b R_{\mathrm{u}} T_{\text {ct }}=5083.8-[(1.6)(8.314)(126.2)]=3405.0\ \mathrm{~kJ} / \mathrm{kmol}
$$ or, $$
h=\frac{\bar{h}}{M}=\frac{3405.0 \mathrm{~kJ} / \mathrm{kmol}}{28.013 \mathrm{~kg} / \mathrm{kmol}}=121.6 \mathrm{~kJ} / \mathrm{kg} \quad(3.1 \%\ \mathrm{ error})
$$
