Answer
For air: $s_{2}-s_{1}= 0.223\text{ kJ/kg}\cdot\text{K}$
For helium: $s_{2}-s_{1}= 1.27\text{ kJ/kg}\cdot\text{K}$
Work Step by Step
For air, the entropy change across the shock is determined to be $$
s_2-s_1=c_p \ln \frac{T_2}{T_1}-R \ln \frac{P_2}{P_1}=(1.005 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}) \ln (2.2383)-(0.287 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}) \ln (7.7200)=\mathbf{0 . 2 2 3 k J} / \mathbf{k g} \cdot \mathbf{K}
$$ For helium, the entropy change across the shock is determined to be
$$ s_2-s_1=c_p \ln \frac{T_2}{T_1}-R \ln \frac{P_2}{P_1}=(5.1926 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}) \ln (2.9606)-(2.0769 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}) \ln (8.2009)=\mathbf{1 . 2 7} \mathbf{~k J} / \mathrm{kg} \cdot \mathbf{K}
$$
