### Step-by-Step Solution To solve for the change in enthalpy (ΔH) for ethylene as it moves from state 1 (2 bar, 285 K) to state 2 (90 bar, 460 K), you generally need ethylene's heat capacity data and information about how enthalpy changes with both temperature and pressure. The change in enthalpy can be calculated using: \[ \Delta H = \int_{T_1}^{T_2} C_p dT + \left[ H^{\text{ig}}(T_2, P_2) - H^{\text{ig}}(T_2, P_1) \right] \] where \( C_p \) is the heat capacity at constant pressure. #### 1. Temperature Change (at constant pressure) First, calculate the enthalpy change due to temperature increase at constant pressure (from 285 K to 460 K at 2 bar): \[ \Delta H_T = \int_{T_1}^{T_2} C_p dT \] If you are not given \( C_p \) data in the question, you may use an average value for ethylene: \( C_p \approx 43.56 \) J/(mol·K) \[ \Delta H_T = C_p \cdot (T_2 - T_1) = 43.56 \, \text{J/(mol·K)} \cdot (460 - 285) \, \text{K} \] \[ \Delta H_T = 43.56 \cdot 175 = 7,623 \, \text{J/mol} \] #### 2. Pressure Change (at constant temperature) For an ideal gas, enthalpy does not depend on pressure. For a real gas, there is a pressure correction, but if not specified, we often neglect it or assume ideal gas behavior. \[ \Delta H_P \approx \] #### 3. Total Change in Enthalpy So, \[ \Delta H_{\text{total}} = \Delta H_T + \Delta H_P \approx 7,623 \, \text{J/mol} + = 7,623 \, \text{J/mol} \] #### 4. Convert to J/g Molar mass of ethylene (\( C_2H_4 \)) = 28.05 g/mol \[ \Delta H_{\text{total in J/g}} = \frac{7,623 \, \text{J/mol}}{28.05 \, \text{g/mol}} \approx 272 \, \text{J/g} \] ### Final Answer The change in enthalpy for ethylene from state 1 (2 bar, 285 K) to state 2 (90 bar, 460 K) is approximately **272 J/g**. If you are given more specific \( C_p \) data or real gas correction factors, use those for a more precise answer.