23-Chem-A1 Process Balances and Chemical Thermodynamics · December 2017
Question 4 of 6: van der Waals Residual Properties of CO₂
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2017 — 16-Chem-A1 Process Balances and Chemical Thermodynamics. Three-hour, open-book exam; any non-communicating calculator permitted. Format: two parts — Part A (Q1–Q3) Process Balances, Part B (Q4–Q6) Chemical Thermodynamics; the candidate answers two questions from each part (four constitute a complete paper, equal value). All six questions are solved below for completeness. Property data not printed on the paper (molar volumes at STP, gas constant, Rankine conversion) are stated explicitly in each Given block as open-book look-ups.
Reference texts: Felder, Rousseau & Bullard, Elementary Principles of Chemical Processes (4th ed., Wiley) — recycle balances, humidity and adiabatic-flame (energy) balances; Himmelblau & Riggs, Basic Principles and Calculations in Chemical Engineering (8th ed.) — multi-unit separation balances; Smith, Van Ness, Abbott & Swihart, Introduction to Chemical Engineering Thermodynamics (8th ed., McGraw-Hill) — cubic equations of state, residual properties, reaction/phase equilibrium and gas-cycle work; supporting property data from Perry’s Chemical Engineers’ Handbook (9th ed.).
Question 4: van der Waals Residual Properties of CO₂ (Part B — equal value)
Given. CO₂ at $T=373.15$ K, $P=40.53$ bar; $T_c=304.2$ K, $P_c=73.8$ bar, $V_c=94$ cm³/mol; $R=83.14$ cm³·bar·mol⁻¹·K⁻¹. van der Waals EOS $P=\dfrac{RT}{V-b}-\dfrac{a}{V^{2}}$.
Parameter set
$a$ (bar·cm⁶/mol²)
$b$ (cm³/mol)
(a) from $T_c,V_c$: $a=\tfrac{9}{8}RT_cV_c,\ b=\tfrac{V_c}{3}$
2.675×10⁶
31.33
(b) from $T_c,P_c$: $a=\tfrac{27R^2T_c^2}{64P_c},\ b=\tfrac{RT_c}{8P_c}$
3.657×10⁶
42.84
Find. $Z$, $G^{R}$, $H^{R}$, $S^{R}$ for each parameter set.
Approach. Get $a,b$ from the two critical-property pairs, solve the vdW cubic for the vapour molar volume $V$, then apply the closed-form vdW residual functions $H^{R}=RT(Z-1)-a/V$, $S^{R}=R\ln\!\big[Z(V-b)/V\big]$, $G^{R}=H^{R}-TS^{R}$.
vdW residual functions. Integrating $\big[T(\partial P/\partial T)_V-P\big]$ and $\big[(\partial P/\partial T)_V-R/V\big]$ from $\infty$ to $V$ for the van der Waals EOS gives, once and for all:
$$H^{R}=RT(Z-1)-\frac{a}{V},\qquad S^{R}=R\ln\!\frac{Z(V-b)}{V},\qquad G^{R}=H^{R}-T S^{R}.$$
These are applied to each parameter set below (unit note: $1\ \text{bar}\cdot\text{cm}^3=0.1\ \text{J}$).
Case (a), parameters from $T_c$ and $V_c$. With $b=V_c/3=31.33$ and $a=\tfrac98RT_cV_c=2.675\times10^{6}$, the cubic $PV^{3}-(Pb+RT)V^{2}+aV-ab=0$ has vapour root $V=707.7$ cm³/mol, so
$$Z=\frac{PV}{RT}=\frac{40.53(707.7)}{83.14(373.15)}=\boxed{0.925}.$$
$$H^{R}=RT(Z-1)-a/V=-6122\ \text{bar}\cdot\text{cm}^3/\text{mol}=\boxed{-612\ \text{J/mol}},$$
$$S^{R}=R\ln\frac{Z(V-b)}{V}=\boxed{-1.03\ \text{J/mol}\cdot\text{K}},\qquad G^{R}=H^{R}-TS^{R}=\boxed{-228\ \text{J/mol}}.$$
Case (b), parameters from $T_c$ and $P_c$. With $b=RT_c/(8P_c)=42.84$ and $a=27R^2T_c^2/(64P_c)=3.657\times10^{6}$, the vapour root is $V=684.8$ cm³/mol:
$$Z=\frac{40.53(684.8)}{83.14(373.15)}=\boxed{0.895}.$$
$$H^{R}=\boxed{-861\ \text{J/mol}},\qquad S^{R}=\boxed{-1.46\ \text{J/mol}\cdot\text{K}},\qquad G^{R}=\boxed{-315\ \text{J/mol}}.$$
Compare the two parameterisations. The $(T_c,P_c)$ set forces the EOS through the measured critical pressure and predicts a larger $b$ and stronger non-ideality ($Z=0.895$), whereas the $(T_c,V_c)$ set anchors on the (less reliably measured) critical volume and gives $Z=0.925$. Both show CO₂ is modestly non-ideal ($Z<1$, attractive forces dominate) at these near-critical reduced conditions ($T_r=1.23$, $P_r=0.55$).
