Answer
$h_{}=80.17\text{ Btu/lbm}$
Work Step by Step
(a) From the ideal gas table of nitrogen (Table A-18E) we read $$
h=2777.0\ \mathrm{Btu} / \mathrm{bmol}=99.18\ \mathrm{Btu} / \mathrm{bm}\left(M_{\mathrm{N}_2}=28\ \mathrm{lbm} / \mathrm{bmol}\right)
$$ at the specified temperature. This value involves $44.2 \%$ error.
(b) The enthalpy departure of nitrogen at the specified state is determined from the generalized chart to be (Fig. A-29) $$
\left.\begin{array}{l}
T_R=\frac{T}{T_{\mathrm{cr}}}=\frac{400}{227.1}=1.761 \\
P_R=\frac{P}{P_{\text {er }}}=\frac{2000}{492}=4.065
\end{array}\right\} \longrightarrow Z_h=\frac{\left(\bar{h}_{\text {ideal }}-\bar{h}\right)_{T, P}}{R_u T_{c r}}=1.18
$$ Thus, $$
\bar{h}=\bar{h}_{\text {sdeal }}-Z_h R_{\mathrm{u}} T_{\text {cx }}=2777.0-[(1.18)(1.986)(227.1)]=2244.8\ \mathrm{Btu} / \mathrm{lbmol}
$$ or $$
h=\frac{\bar{h}}{M}=\frac{2244.8\ \mathrm{Btu} / \mathrm{lbmol}}{28\ \mathrm{lbm} / \mathrm{lbmol}}=80.17\ \mathrm{Btu} / \mathrm{lbm}
$$ $(54.9 \%$ error$)$
