Question

Air at an initial temperature and pressure of 15 °C and 1 bar is heated in an irreversible process. The final temperature and pressure are 303 °C and 2 bar, respectively. Take gas constant $R = 287$ J/kg·K, $\gamma = 1.4$, and treat air as a calorically perfect gas. The change of entropy (in J/kg·K) in the process is \(\underline{\hspace{1cm}}\). [round off to nearest integer]

Hint:

For ideal gases with constant specific heats, use $\Delta s = c_p \ln(T_2/T_1) - R\ln(P_2/P_1)$ even for irreversible processes.

Answer 1

Correct Answer: 495 - 499

Initial temperature: \[ T_1 = 15^\circ\text{C} = 288\ \text{K} \] Final temperature: \[ T_2 = 303^\circ\text{C} = 576\ \text{K} \] Initial pressure: \[ P_1 = 1\ \text{bar} \] Final pressure: \[ P_2 = 2\ \text{bar} \] For a calorically perfect gas: \[ c_p = \frac{\gamma R}{\gamma - 1} = \frac{1.4 \times 287}{0.4} = 1004.5\ \text{J/kg} \centerdot \text{K} \] Entropy change: \[ \Delta s = c_p \ln\left(\frac{T_2}{T_1}\right) - R\ln\left(\frac{P_2}{P_1}\right) \] Compute terms: \[ \ln\left(\frac{576}{288}\right) = \ln(2) = 0.693 \] \[ c_p \ln(T_2/T_1)=1004.5 \times 0.693=696.1 \] \[ R\ln(P_2/P_1)=287 \ln(2)=287\times0.693=198.8 \] \[ \Delta s = 696.1 - 198.8 = 497.3\ \text{J/kg} \centerdot \text{K} \] Thus the entropy change lies between: \[ \boxed{495\text{ to }499\ \text{J/kg} \centerdot \text{K}} \]