23-Chem-A1 Process Balances and Chemical Thermodynamics · December 2018
Question 4 of 6: Part B — Chemical Thermodynamics
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exam 16-Chem-A1, December 2018 — open-book, 3 hours. Two parts: Part A (Process Balances, Q1–Q3) and Part B (Chemical Thermodynamics, Q1–Q3). Candidates answer TWO from each part; each question is of equal value. All six questions are solved in full below.
Reference texts: Felder, Rousseau & Bullard, Elementary Principles of Chemical Processes (4th ed., Wiley) — degree-of-freedom analysis, separation trains, humidity and condensation energy balances, recycle systems; Himmelblau & Riggs, Basic Principles and Calculations in Chemical Engineering (8th ed.) — psychrometrics and recycle-ratio calculations; Smith, Van Ness, Abbott & Swihart, Introduction to Chemical Engineering Thermodynamics (8th ed., McGraw-Hill) — residual properties from a real-gas EOS, reaction equilibrium from ΔG°, and generalized/EOS fugacity coefficients; supporting critical-property data from Poling, Prausnitz & O’Connell, The Properties of Gases and Liquids (5th ed.).
Part A — Process Balances
Part B — Chemical Thermodynamics
Question B1: Entropy Change and Mean $c_p$ for a Non-Ideal Gas (Part B — equal value)
Given. EOS $P(v-B)=RT+AP^2/T$, i.e. $v=B+RT/P+AP/T$. State 1: $P_1=4$ atm, $T_1=300$ K. State 2: $P_2=12$ atm, $T_2=400$ K. $c_p^{\,1\text{atm}}=8$ cal/(mol·K) (molar basis — the only dimensionally consistent reading, no molar mass supplied); $A=1$ L·K/(atm·mol); $B=0.08$ L/mol. Conversion $1$ L·atm $=24.217$ cal.
Find. (a) $\Delta S$ between the two states; (b) the mean $c_p$ at 12 atm.
Approach. Use the departure-function route: build $\Delta S$ from an ideal-gas term plus a residual (departure) evaluated from $(\partial v/\partial T)_P$, and get the pressure dependence of $c_p$ from $(\partial c_p/\partial P)_T=-T(\partial^2 v/\partial T^2)_P$.
Volume-explicit form and its temperature derivative. The EOS rearranges to $v=B+RT/P+AP/T$, so
$$\left(\frac{\partial v}{\partial T}\right)_P=\frac{R}{P}-\frac{AP}{T^2}.$$
Entropy change (departure method). Along $T_1,P_1\to T_2,P_2$,
$$\Delta S=\int c_p\frac{dT}{T}-\int\left(\frac{\partial v}{\partial T}\right)_P dP
= c_p\ln\frac{T_2}{T_1}-R\ln\frac{P_2}{P_1}+\frac{A}{2}\!\left(\frac{P_2^2}{T_2^2}-\frac{P_1^2}{T_1^2}\right).$$
The last (residual) group is the departure from ideal behaviour; the $R/P$ part of $(\partial v/\partial T)_P$ reproduces the ideal $-R\ln(P_2/P_1)$ term.
Evaluate each term (in cal/mol·K). With $c_p=8$ and $R=1.987$:
$$c_p\ln\tfrac{400}{300}=2.302,\quad -R\ln\tfrac{12}{4}=-2.183,\quad \frac{A}{2}\!\left(\tfrac{12^2}{400^2}-\tfrac{4^2}{300^2}\right)\!(24.217)=0.0109-0.0022=0.0087.$$
Summing,
$$\Delta S=2.302-2.183+0.009=\boxed{0.13\ \text{cal}\,\text{mol}^{-1}\text{K}^{-1}.}$$
The near-cancellation reflects that the temperature rise almost offsets the three-fold pressure rise.
Pressure dependence of $c_p$ (part b). Since $\left(\partial^2 v/\partial T^2\right)_P=2AP/T^3$,
$$\left(\frac{\partial c_p}{\partial P}\right)_T=-T\left(\frac{\partial^2 v}{\partial T^2}\right)_P=-\frac{2AP}{T^2}
\;\Rightarrow\; c_p(P)=c_p^{\,1\text{atm}}-\frac{A\,(P^2-1)}{T^2}.$$
Mean $c_p$ at 12 atm. Evaluating the correction at the mean temperature $\bar T=350$ K:
$$\Delta c_p=-\frac{(1)(12^2-1)}{350^2}(24.217)=-0.028\ \text{cal}\,\text{mol}^{-1}\text{K}^{-1},$$
$$c_p^{\,12\text{atm}}=8-0.028=\boxed{7.97\ \text{cal}\,\text{mol}^{-1}\text{K}^{-1}.}$$
The gas is only mildly non-ideal, so $c_p$ barely shifts from its 1-atm value.
