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23-Chem-A1 Process Balances and Chemical Thermodynamics · May 2018

Question 6 of 6: Isothermal Expansion of Nitrogen — Ideal vs. van der Waals

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

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National Exams — May 2018 — 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 Mass and Energy 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 volume of an ideal gas, the gas constant, air molar mass, the psychrometric ratio 0.622, and the gas-phase heat-capacity polynomials in the attached Table C-4) 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) — combustion/excess-air, humidity and drying energy balances; Himmelblau & Riggs, Basic Principles and Calculations in Chemical Engineering (8th ed.) — psychrometric and dryer balances; Smith, Van Ness, Abbott & Swihart, Introduction to Chemical Engineering Thermodynamics (8th ed., McGraw-Hill) — generalized virial fugacity, reaction-equilibrium ΔG°(T) from heat-capacity data, and residual/real-gas property changes; supporting property data from the attached Perry’s / Poling “Properties of Gases and Liquids” Table C-4.

Question 6: Isothermal Expansion of Nitrogen — Ideal vs. van der Waals (Part B — equal value)

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 $N_2$, $T=400$ K (isothermal), $V_1=1$ L $\rightarrow V_2=2$ L. van der Waals $a=1.39$ L²·atm/mol², $b=0.0391$ L/mol. $R=0.08206$ L·atm/mol·K $=8.314$ J/mol·K; $1\ \text{L}\cdot\text{atm}=101.325$ J.

Find. $\Delta S$ and $\Delta G$ for the expansion, (a) ideal gas and (b) van der Waals gas.

Approach. For an isothermal ideal-gas expansion $\Delta H=0$ so $\Delta G=-T\Delta S$ with $\Delta S=R\ln(V_2/V_1)$. For the vdW gas, $\Delta S=R\ln\frac{V_2-b}{V_1-b}$ and $\Delta U=a(1/V_1-1/V_2)$; $\Delta H=\Delta U+\Delta(PV)$ with vdW pressures, then $\Delta G=\Delta H-T\Delta S$.

  1. Ideal gas (part a). Isothermal, so $$\Delta S=R\ln\frac{V_2}{V_1}=8.314\ln 2=\boxed{+5.76\ \text{J/K}},$$ and with $\Delta H=0$, $\Delta G=-T\Delta S=-400(5.763)=\boxed{-2305\ \text{J}}.$
  2. vdW entropy (part b). The configurational term replaces $V$ by the free volume $V-b$: $$\Delta S=R\ln\frac{V_2-b}{V_1-b}=8.314\ln\frac{1.9609}{0.9609}=\boxed{+5.93\ \text{J/K}}.$$
  3. vdW internal energy and $PV$ term. For a vdW gas $(\partial U/\partial V)_T=a/V^2$, so $\Delta U=a\!\left(\frac1{V_1}-\frac1{V_2}\right)=1.39(0.5)=0.695$ L·atm $=70.4$ J. The pressures are $$P_1=\frac{RT}{V_1-b}-\frac{a}{V_1^2}=32.77\ \text{atm},\quad P_2=\frac{RT}{V_2-b}-\frac{a}{V_2^2}=16.39\ \text{atm},$$ giving $\Delta(PV)=P_2V_2-P_1V_1=(32.78-32.77)\ \text{L}\cdot\text{atm}=1.4$ J.
  4. vdW enthalpy and Gibbs energy (part b). $\Delta H=\Delta U+\Delta(PV)=70.4+1.4=71.8$ J, so $$\Delta G=\Delta H-T\Delta S=71.8-400(5.930)=\boxed{-2300\ \text{J}.}$$ The result is barely different from the ideal gas: at 400 K nitrogen is only weakly non-ideal, so the attractive ($a$) and finite-size ($b$) corrections nearly cancel in $\Delta G$.
Quantity(a) Ideal gas(b) van der Waals
$\Delta S$+5.76 J/K+5.93 J/K
$\Delta H$0+71.8 J
$\Delta G$−2305 J−2300 J
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