R-134a — Entropy Change and Feasibility of an Adiabatic Process: 04-BS-10, December 2014, Q6

Question 6: R-134a — Entropy Change and Feasibility of an Adiabatic Process (15 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. State 1: $P_1=700$ kPa, $x_1=0.50$. State 2: $T_2=16\ ^\circ$C, saturated liquid ($x_2=0$). $m=1$ kg.

Find. (a) $\Delta s$ [kJ/kg·K]; (b) feasibility of an adiabatic path.

Approach

Look up the entropy at each named state directly and take the difference; for part (b), recognize that a closed system undergoing an adiabatic process obeys $\Delta s=s_{gen}\ge0$, so any process with $\Delta s<0$ is impossible without heat removal.

State entropies. $s_1=1.42182$ kJ/kg·K (700 kPa, $x=0.50$). $s_2=1.07721$ kJ/kg·K (saturated liquid, 16°C).
Entropy change (part a). $$\Delta s=s_2-s_1=1.07721-1.42182=\boxed{-0.34461\ \text{kJ/kg}\cdot\text{K}}.$$
Adiabatic feasibility (part b). For a closed system with no heat transfer, $\Delta s=s_{gen}$, and the second law requires $s_{gen}\ge0$. Since $\Delta s=-0.34461$ kJ/kg·K $<0$, this process cannot be accomplished adiabatically — it requires heat to be removed from the refrigerant as it condenses and cools, which is exactly what would be expected physically (going from a warm two-phase mixture to a cooler saturated liquid).

Quantity	Result
(a) $\Delta s$	−0.34461 kJ/kg·K
(b) Adiabatic?	No — would require $s_{gen}<0$, violating the 2nd law

Concept & Theory

Topic. Using the entropy balance to test whether a specified closed-system state change can occur without heat transfer.

Key relations. Closed-system entropy balance: $\Delta S=\int\delta Q/T+S_{gen}$, with $S_{gen}\ge0$ always. Adiabatic case: $\Delta S=S_{gen}\ge0$.

Why this works. No calculation of the actual process path (irreversible or not) is needed — only the sign of the entropy change between the specified end states, compared against the second law's requirement that an adiabatic process cannot reduce entropy. A negative $\Delta s$ immediately rules out any adiabatic path between these two states, regardless of how the process might otherwise be imagined to proceed.

Common pitfall. Assuming feasibility must be checked via an energy balance (first law) instead of an entropy balance — the first law alone permits this state change (heat is simply removed to conserve energy), it's the second law's one-directional entropy requirement that rules out the adiabatic case specifically.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 7 (Entropy).

Question 6 of 9: R-134a — Entropy Change and Feasibility of an Adiabatic Process

Question 6: R-134a — Entropy Change and Feasibility of an Adiabatic Process (15 marks)

Approach