Acetone Evaporated into Nitrogen and Compressed: 23-Chem-A1 Process Balances and Chemical Thermodynamics, December 2013, Q1

Question 1: Acetone Evaporated into Nitrogen and Compressed (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. Liquid acetone 400 L/min evaporated into N₂, then a second N₂ stream of 419 m³(STP)/min dilutes it; the mixture is compressed to 7.3 atm and 325 °C with acetone partial pressure 501 mm Hg. Property data (open-book): acetone density $\rho = 0.791$ g/mL and molar mass $M = 58.08$ g/mol; standard molar volume 22.414 L/mol at STP (0 °C, 1 atm).

Stream / datum	Value
Liquid acetone fed	400 L/min
Second N₂ (dilution)	419 m³(STP)/min
Final total pressure	7.3 atm = 5548 mm Hg
Acetone partial pressure (final)	501 mm Hg

Find. (a) the mole-fraction composition of the compressed stream; (b) the volumetric flow rate of the N₂ that enters the evaporator, evaluated at 27 °C and 1235 mm Hg.

Figure 1 — Acetone evaporates into nitrogen, is diluted by a second N₂ stream, then compressed. Acetone is conserved through the train.

Approach. Convert the liquid acetone to a molar flow, use the acetone partial-pressure ratio to fix the mole fractions and the total molar flow, then split the nitrogen between the two N₂ feeds and apply the ideal-gas law to the evaporator N₂ at the stated conditions.

Molar flow of acetone fed. The liquid rate times density over molar mass gives the moles evaporated (all of it ends up in the gas): $$\dot n_a = \frac{(400\ \text{L/min})(791\ \text{g/L})}{58.08\ \text{g/mol}} = \boxed{5447.7\ \text{mol/min}}.$$
Mole fraction of acetone in the product. By Dalton's law the acetone mole fraction equals its partial-pressure ratio at the compressor outlet ($P_{tot}=7.3\times760=5548$ mm Hg): $$y_a = \frac{p_a}{P_{tot}} = \frac{501}{5548} = 0.0903 \;\Rightarrow\; \boxed{9.03\%\ \text{acetone},\ 90.97\%\ \text{N}_2}.$$ That is part (a).
Total and nitrogen molar flows. Since acetone is 9.03 % of the total, the total molar flow and the total nitrogen follow: $$\dot n_{tot} = \frac{\dot n_a}{y_a} = \frac{5447.7}{0.0903} = 60{,}327\ \text{mol/min},\qquad \dot n_{\text{N}_2} = 60{,}327-5447.7 = 54{,}879\ \text{mol/min}.$$
Split the nitrogen between the two feeds. The dilution stream is 419 m³ at STP: $$\dot n_{\text{N}_2,\text{dil}} = \frac{419{,}000\ \text{L/min}}{22.414\ \text{L/mol}} = 18{,}694\ \text{mol/min},$$ so the nitrogen entering the evaporator is the difference, $$\dot n_{\text{N}_2,\text{evap}} = 54{,}879 - 18{,}694 = 36{,}185\ \text{mol/min}.$$
Volumetric flow of the evaporator nitrogen. Apply the ideal-gas law at $T=300.15$ K and $P=1235$ mm Hg $=1.625$ atm: $$\dot V = \frac{\dot n_{\text{N}_2,\text{evap}} R T}{P} = \frac{(36{,}185)(0.08206)(300.15)}{1.625} = 5.48\times10^{5}\ \text{L/min} = \boxed{548\ \text{m}^3/\text{min}}.$$ That is part (b).

Quantity	Result
(a) Compressor-outlet composition	9.03 mol% acetone, 90.97 mol% N₂
Total molar flow / total N₂	60,330 / 54,880 mol/min
(b) N₂ into evaporator (at 27 °C, 1235 mm Hg)	≈ 548 m³/min (36,185 mol/min)

Concept & Theory

Topic. Ideal-gas partial pressures and multi-stream mole balances on an evaporator/compressor train (no reaction).

Key relations. Dalton's law $y_i = p_i/P$; ideal-gas law $PV=nRT$; standard molar volume 22.414 L/mol at 0 °C and 1 atm; a species balance $\dot n_{\text{N}_2,\text{evap}} = \dot n_{\text{N}_2,\text{total}} - \dot n_{\text{N}_2,\text{dilution}}$.

Why this works. The single measured quantity that ties the whole train together is the acetone partial pressure at the end: because acetone is neither created nor destroyed, its molar flow is fixed by the liquid feed, and its mole fraction is fixed by $p_a/P_{tot}$. Those two numbers pin the total flow, and hence all the nitrogen. Everything downstream is bookkeeping.

Common pitfall. Forgetting that "m³(STP)" already implies 22.414 L/mol at 0 °C — students sometimes re-convert it with the 325 °C or 27 °C temperature. STP volumes are a molar surrogate and must be converted only at 0 °C, 1 atm.

Source. Felder, Rousseau & Bullard, Elementary Principles of Chemical Processes §5.1–5.2 (ideal-gas calculations, partial pressures and standard conditions).

Question 1 of 7: Acetone Evaporated into Nitrogen and Compressed

Question 1: Acetone Evaporated into Nitrogen and Compressed (equal value)