1. **Stating the problem:** Given initial pressure $P_1 = 9.75$ MPa, initial temperature $T_1 = 226$ K, and final pressure $P_2 = 4$ MPa, we want to find the final temperature $T_2$ assuming an ideal gas undergoing an isentropic process. 2. **Formula and rules:** For an isentropic process of an ideal gas, the relation between temperature and pressure is: $$\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{k-1}{k}}$$ where $k$ is the specific heat ratio (usually $k \approx 1.4$ for diatomic gases like air). 3. **Intermediate work:** Calculate the pressure ratio: $$\frac{P_2}{P_1} = \frac{4}{9.75} \approx 0.4103$$ Calculate the exponent: $$\frac{k-1}{k} = \frac{1.4 - 1}{1.4} = \frac{0.4}{1.4} \approx 0.2857$$ Calculate the temperature ratio: $$\left(0.4103\right)^{0.2857} \approx 0.774$$ Calculate final temperature: $$T_2 = T_1 \times 0.774 = 226 \times 0.774 \approx 175 K$$ 4. **Explanation:** The temperature decreases because the pressure decreases in an isentropic expansion. The exponent $\frac{k-1}{k}$ comes from thermodynamic relations for ideal gases undergoing reversible adiabatic processes. **Final answer:** $$T_2 \approx 175 \text{ K}$$