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21-Mat-A1 Thermodynamics · May 2014

Question 3 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 2014 — 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.) — the exam's own Ellingham diagram (Fig. 9-3) is reproduced from this text, which also supplies the standard-state enthalpy/entropy data used below; supporting 298 K entropies for the elements, CO, H₂, H₂O(g) and CO₂ from the NIST-JANAF Thermochemical Tables.

Question 3: Entropy of Reversible vs. Irreversible Isothermal Compression (20 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. $n=1$ mol ideal gas, $T=300$ K throughout, $V_1=20$ L, $V_2=10$ L in both cases; case (b) is opposed by a constant external pressure $P_{ext}=200{,}000$ Pa.

Find. $\Delta S_{sys}$, $\Delta S_{surr}$, $\Delta S_{total}$ for (a) the reversible path and (b) the irreversible path between the same two states.

Approach. $S$ is a state function, so $\Delta S_{sys}=nR\ln(V_2/V_1)$ is identical in both cases; only $\Delta S_{surr}$ differs, because it depends on the actual heat exchanged, which is path-dependent. For the reversible case the process is internally reversible, so $\Delta S_{univ}=0$ pins $\Delta S_{surr}=-\Delta S_{sys}$ directly. For the irreversible case, compute the actual work from the constant $P_{ext}$, get $q$ from the first law, and divide the heat received by the (reservoir) surroundings by $T$.

  1. System entropy (both cases). $\Delta S_{sys}=nR\ln(V_2/V_1)=(1)(8.314)\ln(10/20)=\boxed{-5.76\ \text{J/K}}$ — a negative value, as expected for a compression.
  2. Part (a)(ii)–(iii) — reversible path. A reversible process generates no entropy, so $\Delta S_{univ}=0$ forces $\Delta S_{surr}=-\Delta S_{sys}=\boxed{+5.76\ \text{J/K}}$, and $\Delta S_{total}=\boxed{0}$.
  3. Part (b)(i) — system entropy again. Same start and end states as (a), so $\Delta S_{sys}=-5.76$ J/K unchanged (entropy is a state function; only the path to get there was different).
  4. Part (b) — actual work and heat. With $\Delta V=V_2-V_1=-10\ \text{L}=-0.01\ \text{m}^3$, $w=-P_{ext}\Delta V=-(2\times10^5)(-0.01)=+2000\ \text{J}$ (work done on the gas). Since $T$ is constant, $\Delta U=0$, so $q=\Delta U-w=\boxed{-2000\ \text{J}}$: the gas rejects 2000 J to the surroundings.
  5. Part (b)(ii) — surroundings entropy. The surroundings act as a reservoir at $T=300$ K and receive $q_{surr}=-q=+2000$ J reversibly, so $\Delta S_{surr}=q_{surr}/T=2000/300=\boxed{+6.67\ \text{J/K}}$.
  6. Part (b)(iii) — total entropy. $\Delta S_{total}=\Delta S_{sys}+\Delta S_{surr}=-5.76+6.67=\boxed{+0.90\ \text{J/K}}$, positive as required for an irreversible process (entropy is generated by the finite, non-quasi-static pressure difference).
CaseΔSsys (J/K)ΔSsurr (J/K)ΔStotal (J/K)
(a) reversible−5.76+5.760
(b) irreversible, Pext=200 kPa−5.76+6.67+0.90