Question 2 of 7: Entropy of Reversible vs. Irreversible Isothermal Compression
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2017 — 10-Met-A1 Metallurgical Thermodynamics. Three-hour, closed-book exam using an approved (Casio or Sharp) calculator; candidates were told to state any interpretive assumptions. Any five of the seven questions constitute a complete paper — all seven are solved below for completeness. All questions are of equal value (20 marks each out of 100).
Reference texts: Gaskell, D. R., Introduction to the Thermodynamics of Materials (2nd–5th ed.) — source of the first-law/second-law relations used throughout and of the attached Ellingham diagram (Fig. 9-3) and its underlying 298 K formation data used in Question 7; supporting 298 K entropies for CO(g), CO₂(g), H₂(g) and H₂O(g) from the NIST–JANAF Thermochemical Tables.
Question 2: Entropy of Reversible vs. Irreversible Isothermal Compression (20 marks)
Given. $n=1$ mol ideal gas, $T=325$ K (constant), $V_1=50$ L, $V_2=10$ L. (a) reversible isothermal compression. (b) irreversible compression against a constant external pressure $P_{ext}=250{,}000$ Pa.
Find. $\Delta S_{sys}$, $\Delta S_{surr}$ and $\Delta S_{total}$ for each of the two compression paths between the same two states.
Approach. $\Delta S$ is a state function, so $\Delta S_{sys}=nR\ln(V_2/V_1)$ is identical for (a) and (b); the paths differ only in how much heat crosses into the surroundings, found from the first law ($\Delta U=0$ for an isothermal ideal gas) using the reversible work in (a) and the irreversible constant-$P_{ext}$ work in (b).
(a)(i)–(ii) Reversible path. For a reversible isothermal process, $q_{rev}=-w_{on}=nRT\ln(V_2/V_1)$, and $\Delta S_{sys}=q_{rev}/T=nR\ln(V_2/V_1)$:
$$\Delta S_{sys}=(8.314)\ln\!\left(\dfrac{10}{50}\right)=\boxed{-13.38\ \text{J}\,\text{K}^{-1}}.$$
A reversible process is by definition entropy-neutral for the universe, so $\Delta S_{surr}=-\Delta S_{sys}$:
$$\Delta S_{surr}=\boxed{+13.38\ \text{J}\,\text{K}^{-1}}.$$
(a)(iii) Total.
$$\Delta S_{total}=\Delta S_{sys}+\Delta S_{surr}=-13.38+13.38=\boxed{0},$$
consistent with the reversible path being an idealised, zero-dissipation limit.
(b)(i) Irreversible path — system entropy. $\Delta S$ depends only on the end states, not the path, so it is unchanged from (a):
$$\Delta S_{sys}=nR\ln\!\left(\dfrac{10}{50}\right)=\boxed{-13.38\ \text{J}\,\text{K}^{-1}}.$$
(b)(ii) Irreversible path — surroundings entropy. The work done ON the gas by the constant external pressure is
$$w_{on}=-P_{ext}\Delta V=-(250{,}000)(0.010-0.050)=\boxed{10{,}000\ \text{J}}.$$
Since $\Delta U=0$ (isothermal ideal gas), $q=\Delta U-w_{on}=0-10{,}000=\boxed{-10{,}000\ \text{J}}$ — the system rejects 10 kJ to the surroundings. The surroundings receive $q_{surr}=-q=+10{,}000$ J reversibly (they act as a large reservoir at constant $T$), so
$$\Delta S_{surr}=\dfrac{q_{surr}}{T}=\dfrac{10{,}000}{325}=\boxed{30.77\ \text{J}\,\text{K}^{-1}}.$$
(b)(iii) Total.
$$\Delta S_{total}=-13.38+30.77=\boxed{17.39\ \text{J}\,\text{K}^{-1}}>0,$$
confirming the constant-$P_{ext}$ compression is irreversible: it dissipates more work as heat than the minimum reversible amount, generating net entropy.
