1. **Problem Statement:** Given an isentropic process with inlet pressure $P_1$ and temperature $T_1$, and exit pressure $P_2 = 4$ MPa, find the exit temperature $T_2$. The process is isentropic, and constant specific heat relations must not be used. 2. **Key Concept:** For an isentropic process in real gases, the entropy remains constant: $s_2 = s_1$. We use property tables or equations of state to find $T_2$ such that $s(T_2,P_2) = s(T_1,P_1)$. 3. **Method:** - From inlet conditions $(P_1, T_1)$, find entropy $s_1$ using thermodynamic tables or software. - At exit pressure $P_2 = 4$ MPa, find temperature $T_2$ such that $s(T_2, P_2) = s_1$. 4. **Explanation:** Since the process is isentropic, entropy does not change. We cannot use simplified formulas with constant specific heats, so we rely on real gas data or property relations. 5. **Final answer:** The exit temperature $T_2$ is the temperature at $P_2=4$ MPa where entropy equals $s_1$. This requires interpolation or software using real gas properties. **Note:** Without numerical values for $P_1$, $T_1$, or entropy data, the exact numeric $T_2$ cannot be computed here.