Entropy Change of an Ideal Gas Stirred at Constant Temperature: 04-BS-10, May 2013, Q8

Question 8: Entropy Change of an Ideal Gas Stirred at Constant Temperature (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. Closed, rigid tank of ideal gas at $T=40\ ^\circ\text{C}$ (constant throughout the process, by observation). Paddle-wheel work input $W_{in}=200$ kJ. Surroundings at $T_{surr}=25\ ^\circ\text{C}$.

Find. (a) $\Delta S_{gas}$; (b) $\Delta S_{surr}$.

Approach

No masses, moles, or volumes are given — and none are needed. For an ideal gas, entropy is a function of temperature and specific volume only, $s=s(T,v)$; since the tank is rigid ($v$ fixed) and the process is stated to be isothermal, both arguments are unchanged, so the gas's entropy change is exactly zero regardless of how much gas is present. The paddle-wheel work must then leave the system entirely as heat (first law, $\Delta U=0$ for constant-$T$ ideal gas), and that heat sets the surroundings' entropy change.

Entropy change of the gas (part a). For an ideal gas, $s_2-s_1=c_v\ln(T_2/T_1)+R\ln(v_2/v_1)$. Here $T_2=T_1$ (isothermal, as stated) and $v_2=v_1$ (rigid tank), so both logarithm terms vanish identically — independent of the gas's mass or identity: $$\boxed{\Delta S_{gas}=0\ \text{kJ/K}}.$$
Heat rejected to the surroundings. First law on the closed system, $\Delta U=Q-W$ (with $W$ the work done by the system; here work is done on the gas, so $W=-200$ kJ). Since the gas is ideal and $T$ is constant, $\Delta U=0$: $$0=Q-(-200)\quad\Rightarrow\quad Q=-200\ \text{kJ}\ (\text{i.e. 200 kJ leaves the system as heat}).$$
Entropy change of the surroundings (part b). The surroundings act as a large reservoir at constant $T_{surr}=298.15$ K, receiving the 200 kJ rejected by the gas: $$\Delta S_{surr}=\frac{Q_{surr}}{T_{surr}}=\frac{200}{298.15}=\boxed{0.6708\ \text{kJ/K}}.$$

Quantity	Result
(a) $\Delta S_{gas}$	0 kJ/K (exact)
(b) $\Delta S_{surr}$	0.6708 kJ/K

Concept & Theory

Topic. Entropy generation in an isothermal, constant-volume process driven by irreversible (paddle-wheel) work.

Key relations. Ideal-gas entropy change $\Delta s=c_v\ln(T_2/T_1)+R\ln(v_2/v_1)$; first law for a closed system $\Delta U=Q-W$; reservoir entropy change $\Delta S=Q/T_{res}$ (constant $T$).

Why this works. Entropy is a state property: because the initial and final states of the gas are identical in both $T$ and $v$ (the problem explicitly states $T$ is unchanged, and the tank is rigid), the gas's entropy change must be exactly zero no matter what irreversible process connects them — the paddle-wheel stirring is highly irreversible, but that irreversibility shows up entirely as entropy generated in the overall universe (via the heat rejected to the finite-temperature-difference surroundings), not as a state change of the gas itself. This is the entropy-generation-via-total-system check: $S_{gen}=\Delta S_{gas}+\Delta S_{surr}=0+0.671=0.671$ kJ/K $>0$, consistent with the second law for this clearly irreversible process.

Common pitfall. Assuming a process with entropy generation (irreversible stirring) must show up as a positive $\Delta S_{system}$ — entropy generation is a property of the process (the sum $\Delta S_{system}+\Delta S_{surr}$), not necessarily of the system alone; a system's own entropy can be unchanged (as here) or even decrease, provided the surroundings' entropy increases by at least as much.

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

Question 8 of 9: Entropy Change of an Ideal Gas Stirred at Constant Temperature

Question 8: Entropy Change of an Ideal Gas Stirred at Constant Temperature (15 marks)

Approach