Part B — Chemical Thermodynamics: 23-Chem-A1 Process Balances and Chemical Thermodynamics, Undated paper, Q4

Question B1: Water Gas — Volume by Ideal-Gas and van der Waals Equations (25 points)

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. A six-component gas at $T=773.15$ K and $P=4$ bar, with the critical constants as printed. Masses are converted to moles with $n_i=m_i/M_i$, and $a_i=27R^2T_{c,i}^2/(64P_{c,i})$, $b_i=RT_{c,i}/(8P_{c,i})$ with $R=0.08314$ L·bar/(mol·K):

Species	Mass (g)	$M$ (g/mol)	$n_i$ (mol)	$y_i$	$a_i$ (L²bar/mol²)	$b_i$ (L/mol)
H₂	70.40	2.016	34.92	0.7872	0.2331	0.02580
CH₄	23.68	16.043	1.476	0.0333	2.303	0.04306
C₂H₄	35.84	28.054	1.278	0.0288	4.611	0.05821
CO₂	66.00	44.01	1.500	0.0338	3.657	0.04285
CO	94.92	28.01	3.389	0.0764	1.472	0.03948
N₂	50.40	28.014	1.799	0.0406	1.384	0.03886
Total			44.36 mol	78.7 mol% H₂

Find. the gas volume (a) as an ideal gas and (b) from the van der Waals equation.

Approach. Sum the moles and get the ideal volume from $PV=nRT$. Then combine the pure-species van der Waals constants with the one-fluid mixing rules and solve the cubic for the molar volume.

(a) Ideal-gas volume. $$V^{ig}=\frac{nRT}{P}=\frac{(44.36)(0.08314)(773.15)}{4}=\boxed{712.9\ \text{L}=0.713\ \text{m}^3}\qquad(V_m^{ig}=16.070\ \text{L/mol}).$$
Mixing rules. With the classical van der Waals one-fluid rules, $$a_{mix}=\Big(\sum_i y_i\sqrt{a_i}\Big)^2=0.4864\ \text{L}^2\text{bar/mol}^2,\qquad b_{mix}=\sum_i y_i b_i=0.02946\ \text{L/mol}.$$
(b) Solve the van der Waals cubic. Solve $\Big(P+\dfrac{a_{mix}}{V_m^2}\Big)(V_m-b_{mix})=RT$ with $RT=64.28$ L·bar/mol, starting from $V_m^{ig}=16.070$. Successive substitution in $V_m=b_{mix}+RT/(P+a_{mix}/V_m^2)$ converges in two passes to $$V_m=16.092\ \text{L/mol}\;\Rightarrow\;V^{vdW}=nV_m=(44.36)(16.092)=\boxed{713.9\ \text{L}=0.714\ \text{m}^3}.$$ Substituting back gives $(4+0.4864/16.092^2)(16.092-0.02946)=64.28$, so the root checks.
Size of the correction. To first order, $V_m\approx RT/P+b_{mix}-a_{mix}/RT=16.070+0.0295-0.0076=16.092$ L/mol. The volume is +0.14% above ideal: the co-volume $b$ outweighs the attraction $a$ because the gas is far above every component's critical temperature.

Quantity	Result
Total moles	44.36 mol
(a) Ideal-gas volume	712.9 L (0.713 m³)
$a_{mix}$ / $b_{mix}$	0.4864 L²bar/mol² / 0.02946 L/mol
(b) van der Waals volume	713.9 L (0.714 m³), +0.14% vs ideal

At 500 °C and 4 bar the reduced temperatures are all between 2.5 and 24 and the reduced pressures are below 0.31. At those conditions any real-gas equation should reproduce the ideal volume to within a fraction of a percent, and the van der Waals result does. A large correction here would signal an arithmetic error, not real-gas behaviour.

Concept & Theory

Topic. Real-gas volume of a mixture by the van der Waals cubic equation with one-fluid mixing rules.

Key relations. Pure-species constants: $a_i=27R^2T_c^2/64P_c$ and $b_i=RT_c/8P_c$. Mixing: $a_{mix}=(\sum y_i\sqrt{a_i})^2$ and $b_{mix}=\sum y_i b_i$. Equation of state: $\big(P+a_{mix}/V_m^2\big)(V_m-b_{mix})=RT$.

Why this works. The attraction term $a$ lowers the pressure a gas exerts, and the co-volume $b$ raises it. At a given $P$ and $T$, the volume shifts by roughly $b-a/RT$. At high temperature $a/RT$ becomes small, so $b$ wins and the volume is slightly above ideal (the Boyle-temperature argument).

Common pitfall. Mole-averaging $a$ directly instead of using the $\sqrt{a}$ combining rule, or forgetting to square the sum. Another is using °C in $RT$: at 500 °C that underestimates the volume by a factor of 1.55.

Source. Smith, Van Ness, Abbott & Swihart, Introduction to Chemical Engineering Thermodynamics Ch. 3 (cubic equations of state) and the mixing-rule treatment for cubic equations of state in the mixture chapters.

Question 4 of 6: Part B — Chemical Thermodynamics

Part A — Process Mass and Energy Balances

Part B — Chemical Thermodynamics

Question B1: Water Gas — Volume by Ideal-Gas and van der Waals Equations (25 points)