Answer
$c_{p}=0.250\text{ Btu/lbm⋅R}$
$c_{v}=0.179\text{ Btu/lbm⋅R}$
Work Step by Step
The $c_p$ and $c_v$ of ideal gases depends on temperature only, and are expressed as $c_p(T)=d h(T) / d T$ and $c_v(T)=$ $d u(T) / \mathrm{d} T$. Approximating the differentials as differences about $600 \mathrm{R}$, the $c_p$ and $c_v$ values are determined to be $$
\begin{aligned}
& c_p(800 \mathrm{R})=\left(\frac{d h(T)}{d T}\right)_{T-800 \mathrm{R}} \cong\left(\frac{\Delta h(T)}{\Delta T}\right)_{T=800 \mathrm{R}} \\
& =\frac{h(820 \mathrm{R})-h(780 \mathrm{R})}{(820-780) \mathrm{R}} \\
& =\frac{(5704.7-5424.2) / 28.013 \mathrm{Btu} / \mathrm{lbm}}{(820-780) \mathrm{R}}=\mathbf{0 . 2 5 0} \mathrm{Btu} / \mathrm{lbm} \cdot \mathbf{R} \\
&
\end{aligned}
$$ (Compare: Table A-2Eb at $800 \mathrm{R}=340^{\circ} \mathrm{F} \rightarrow c_p=0.250 \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R}$ )
$$
\begin{aligned}
c_v(800 \mathrm{R}) & =\left(\frac{d u(T)}{d T}\right)_{T-800 \mathrm{R}} \cong\left(\frac{\Delta u(T)}{\Delta T}\right)_{T=800 \mathrm{R}} \\
& =\frac{u(820 \mathrm{R})-u(780 \mathrm{R})}{(820-780) \mathrm{R}} \\
& =\frac{(4076.3-3875.2) / 28.013 \mathrm{Btu} / \mathrm{lbm}}{(820-780) \mathrm{R}}\\&=\mathbf{0 . 1 7 9 ~ B t u / l b m} \cdot \mathbf{R}
\end{aligned}
$$
