Answer
a) $c_{p}=1.045\text{ kJ/kg⋅K}$
b) $c_{p}=1.045\text{ kJ/kg⋅K}$
Work Step by Step
(a) We treat nitrogen as an ideal gas with $R=0.297 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}$ and $k=1.397$. Note that $P T^{\mathrm{k}(k-1)}=C=$ constant for the isentropic processes of ideal gases. The $c_p$, relation is given as $$
\begin{aligned}
c_p & =T\left(\frac{\partial P}{\partial T}\right)_x\left(\frac{\partial v}{\partial T}\right)_P \\
v & =\frac{R T}{P} \longrightarrow\left(\frac{\partial v}{\partial T}\right)_P=\frac{R}{P} \\
P & =C T^{k(k-1)} \longrightarrow\left(\frac{\partial P}{\partial T}\right)_x=\frac{k}{k-1} C T^{k(k-1)-1}=\frac{k}{k-1}\left(P T^{-k \cdot(k-1)}\right) T^{k(k-1)-1}=\frac{k P}{T(k-1)}
\end{aligned}
$$ Substituting, $$
c_p=T\left(\frac{k P}{T(k-1)}\right)\left(\frac{R}{P}\right)=\frac{k R}{k-1}=\frac{1.397(0.297 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})}{1.397-1}=1.045 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}
$$ (b) The $c_p$ is defined as $c_p=\left(\frac{\partial h}{\partial T}\right)_p$. Replacing the differentials by differences, $$
c_p \equiv\left(\frac{\Delta h}{\Delta T}\right)_{P=300 \mathrm{P}_2}=\frac{h(410 \mathrm{~K})-h(390 \mathrm{~K})}{(410-390) \mathrm{K}}=\frac{(11,932-11,347) / 28.0 \mathrm{~kJ} / \mathrm{kg}}{(410-390) \mathrm{K}}=\mathbf{1. 0 4 5}\ \mathbf{k J} / \mathbf{k g} \cdot \mathbf{K}
$$
