Topic: Thermodynamics
A simple compressible substance inside a cylinder undergoes a change of state quasistatically, isothermally and at constant pressure of 1.5 MPa. If the enthalpy change for the system is 4000 kJ at 150 °C, what is the corresponding change in its entropy?
- A. 9.45 kJ K^-1
- B. 26.67 kJ K^-1
- C. 14.17 kJ K^-1
- D. 0
Answer: A. 9.45 kJ K^-1
Explanation: Since it is a phase change (temperature is constant), we use dH = TdS + VdP. At constant pressure (dP = 0), dH = TdS. Entropy change dS = dH / T = 4000 / (150 + 273) = 9.45 kJ/K.